# For AP Physics C Mech+EM

2017 AP Physics C Review Schedule:

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Ch. 31

OQ5. c, d

OQ6. b, d

OQ7. a

OQ9. i) b, ii) c, iii) a

OQ10. a, b, c, d

CQ6. Yes. When the aluminum enters and passes through the magnetic field, current (eddy current) is induced inside the aluminum. Because of the eddy current, the aluminum in the magnetic field experiences an upward magnetic resistance force.

4. 0.00982V Use Faraday’s law of induction.

9. 0.0618V Use Faraday’s law of induction.

36. a) N^{2}B^{2}w^{2}v/R to the left, b) 0, c) N^{2}B^{2}w^{2}v/R to the left The magnetic resistant force is N times Bwv only when there is magnetic flux change. When there is no magnetic flux change, there is no induced current and no magnetic force on the loop.

32. mgRsinθ/(B^{2}ℓ^{2}cos^{2}θ) At terminal v, F_{net} = 0. Look at forces along the incline only: the sliding down component of the gravity, mg sinθ, is balanced by the magnetic force's component along the incline. The magnetic resistant force equals IℓB and I = ɛ_{ind}/R = Bℓv_{^}/R where the v_{^} is the component of the v that is perpendicular to the B. Also, you need to be careful when you look for the component of the magnetic resistant force on the sliding bar along the incline.

68. Use ɛ_{ind} = Bℓv Without a conducting rail for the rod to slide on, there is still an emf induced in the rod that is ɛ_{ind} = Bℓv because of charge separation happening inside the rod (no loop for induced current to flow, but charges can move to end of rod and accumulate there).

39. a) 3.2×10^{-20}N First find the induced E at P_{1: }Pretend that you have a circular wire loop of radius r_{1} and use Faraday’s law of induction to find ɛ_{ind}. And use ɛ_{ind} = (the closed integral of **E·dL)**

b) downward and slightly to the right (perpendicular to r_{1}).

c) 1.33s The force is zero when E = 0 when the flux in the pretend circular wire loop is not changing. I.e. when B is not changing: i.e. dB/dt = 0.

41. a) E = 0.00987 cos (100πt) V/m, b) clockwise. a) Use ɛ_{ind} = - N dФ_{B}/dt= (the closed integral of **E·dL)**

13. a) μ_{0}IL ln((h+w)/h)/(2π). a) Use Ampere’s law to find the B by the long current and then chop the area into horizontal strips: use dA=Ldr for the integration of **B·dA** .

b) μ_{0}bL ln((h+w)/h)/(2π) = 4.8μV b) Use emf = - dФ_{B} /dt. After all constants are taken out of the derivative, we only have dI/dt left.

c) counterclockwise

63. a) R_{1}: 3.5A, R_{2}: 1.4A It’s like R_{1} and R_{2} are in parallel and they are in parallel with the ɛ_{ind} by the sliding rod.

b) 34.3W, c) 4.29 N Use the idea conservation of energy and that P = Fv.

72. a) πCKa^{2} Use Faraday’s law of induction.

b) upper plate, c) The changing magnetic flux causes an induced emf in the wire loop – which means there is induced E in the loop. The charges in E experience electric force to make them move in the wire loop.

Ch. 32

OQ3. b)

OQ6. a)

CQ2. a) size and shape of the coil, the number of coils, and what it has in its core (e.g. iron core). b) No.

4. a) 0.00197 H. b) 0.038A/s b) │ɛ_{L}│ =│L (dI/dt)│

9. a) 0.36V, b) – 0.18V, c) 3 s. a) and b) │ɛ_{L}│ =│L (dI/dt)│, c) when dI/dt = 0.

14. ɛ_{L}= –L (dI/dt) = – L (slope of the I as a function of t graph)

OQ4. V_{Inductor} > V_{1200ohms }> battery > V_{12ohms}

CQ4. i) a: Bulb turns on immediately and then gets dimmer and dimmer and then turns off. b: Bulb goes from dark to increasing brightness and then stays with the same brightness. c: Bulb goes from dark to increasing brightness and then stays with the same brightness. d: Bulb turns on immediately and then gets dimmer and dimmer and then turns off.

ii) a: Bulb turns off immediately. b: Bulb goes off (but not instantly). c: The bulb gets dimmer and dimmer to off. d: The bulb turns on suddenly and then gets dimmer and dimmer to off.

CQ7. a) I_{L} = 0, I_{C} = ɛ/R, I_{R} = ɛ/R, ɛ_{L} = ɛ, V_{C }= 0, V_{R }= ɛ

b) I_{L} = 0, I_{C} = 0, I_{R} = 0, ɛ_{L} = 0, V_{C }= ɛ , V_{R }= 0

31. a) 0.00566 s, b) 1.22 A, c) 0.0581s a) and b) The current is a 1 minus exponential decay equation. c) The current is an exponential decay equation.

80. a) 72V, b) b, c) I_{1}: Goes from 0.009A exponentially decays down towards zero. I_{2}: Goes from 0.003A before t = 0 suddenly jumps to 0.009A in the opposite direction (i.e. –0.009A) and then exponentially decays towards zero.

CQ8. When the charge on the C is zero the V_{C }= 0, the current in the circuit is a maximum (i.e. dI/dt = 0); therefore, the ɛ_{L} is also 0.

47. 0.281H Use energy conservation.

81. R 300 ohms Before S is opened, the I in the L is 0.267A. The moment after S is opened, the L still has 0.267A. Since we want the entire armature to be limited to 80V, we can find R. V_{armature} = V_{R}

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**Title**

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Saturday, March 4, 2017

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The Physics of Fantastic Worlds:

From *Star Wars* to *Harry Potter*

AP Physics C — Schedule for E&M Unit 4 (Ch. 29, 30)

Ch. 29

OQ1. c, e

OQ9. c

2. a) up, b) out of the page, c) no deflection, d) into the page

OQ7. i) b, ii) a

15. a) (√2)r_{p}, b) (√2)r_{p} Use ∆V to accelerate the charge from rest: U_{loss} = q∆V = K_{gain} = ½ mv^{2}. For the circular motion part, we have: qvB = m v^{2}/r.

22. a) 1.79×10^{-8} s, b) 35.1 eV a) For the circular motion part, we have: qvB = m v^{2}/r. The electron leaves after making half a circle. b) The maximum depth is the radius.

73. a) 1.04×10^{-4} m, b) 1.89×10^{-4} m a) The positron does constant velocity motion in the x direction, so ∆x = v_{x} t. The t is the time for the positron to go around one circle. I.e. t = the period of the circular motion. b) For the circular motion part, we have: qvB = m v^{2}/r.

OQ6. c

42. a) 2πrIB sinθ, b) upward The magnetic force on each segment of the current points upward and slanted towards the center. By symmetry, the net magnetic force is vertical, so we only have to add the force’s upward components.

61. 0.588T This one is just like the free response problem that was on your 2014 AP Physics B exam. The weight of the wire is canceled by the springs’ forces when the springs stretch 0.5 cm. The added magnetic force is canceled by the springs’ additional force. So: F_{magnetic }= ILB = 2k(∆x_{additional}).

24. a) 6.86 × 10^{6} Hz, b) 5.17 × 10^{7} m/s a) For the circular motion part, we have: qvB = m v^{2}/r. Period T = 2πr/v, so f = v/(2πr). b) The maximum speed corresponds to the radius 1.20 m.

29. 244000V/m Undeflected: so F_{magnetic} and F_{electric} must be equal and opposite so they can cancel. So: qvB =qE. The 750eV electron means the electron’s K = ½ mv^{2} = 750eV.

30. a) Yes, this tells us that the cathode rays must be in all those different cathode materials. b) Don't do part b. I don't think the question is a very correct one. ~~Yes, this means the cathode ray has a unique q/m ratio. See p.881 fig. 29.15 for diagram. When the deflection plates get a different voltage, the E = V/d in the velocity selector part is different. When you change E, because qvB = qE in the velocity selector, the v changes, so only particles with the new velocity can go through undeflected. When you turn off the E, the particles get deflected. ~~ c) The very small q/m told J.J. Thomson that cathode rays must be made of particles that have much smaller mass than any ions. Cathode rays are electrons. An electron is much lighter than even the lightest ion (with about 1836 times the mass of an electron). d) Accelerated by >100V means the electrons will have a KE that is >100eV. At 100eV, an electron would have a speed of > (5.93×10^{6})m/s. At this speed, the g = 9.8 m/s^{2} acceleration cannot produce visible deflection within the size of a cathode ray tube.

OQ13. A > C> B Same angle between the magnetic dipole moment and B. So we can just compare the magnitude of the magnetic dipole moment which is IA. They also have the same I, so we just need to compare the A, the areas of current loops.

CQ3. Yes, if B is in the same direction as or in the opposite direction to the normal vector of the area of the loop. I.e. if B is perpendicular to the plane of the loop.

51. a) 9.98 Nm, b) clockwise with ω in the –y direction (this is the direction you get when you use right-hand rule for ω). Find the direction of the magnetic force on each segment of the wire current first. Figure out the direction of the rotation, then specify the rotational axis, and then find the torque produced by each force.

Ch. 30

CQ5. Vertically upward. B lines come out of the north pole of a magnet. Note: the magnetic north pole is near the geographical South Pole.

CQ10. a) The magnetic repulsive force between the magnets cancels with the mg of the magnets above. b) To keep this equilibrium stable. Otherwise, with a little disturbance, the magnets can flip over and attract and stick to each other. c) For a magnet, if it has north pole on the top , its south pole would be at the bottom. d) Blue and yellow would stick together and they will still be levitated by the repulsive magnetic force from the red magnets.

OQ11. b) and c)

CQ4. Use Ampere’s law for this long straight current. If we make the circular Amperian loop inside, the I_{encl.} = 0, the Ampere’s law would lead to B = 0 inside. If we make the circular Amperian loop outside, the I_{encl.} = I in the wire, the Ampere’s law would lead to B 0 outside.

CQ6. Yes, Ampere’s law is valid for all closed paths, but most of the current situations and random choices of the made-up closed Amperian paths do not make Ampere’s law convenient to use for solving for B.

CQ9. 0 For both Figures 30.10 and 30.13, the B is always in the direction of ds, so there is no B lines going through those circular Amperian loop.

10. μ_{0}i/4πx Using Biot-Savart law, we know that the x direction current provides no B at P. For the rest of the current, it is half of a very long straight current (very long going up and down). Using Ampere’s law, we can find the B by the long straight current and then divide it by 2 to get the B by half of a long straight current.

13. 0.262 μT, into the page Using Biot-Savart law, we know that the straight parts of the current provide no B at P. For the arc part of the current, we can use Biot-Savart law again to find the B it produces at P. OQ5. a) and c).

OQ5. a) and c).

OQ9. c) and d).

15. b) 20 μT to the bottom of page, c) 0 There are 3 currents producing B at each location. Use superposition to add the 3 magnetic field vectors by the 3 currents together to get the net B.

OQ3. a) No, b) Yes, c) Yes, d) No It’s only possible when they all repel each other.

OQ10. b)

OQ12. a)

26. μ_{0 }I_{1}I_{2} a/[2πc(a+c)] to the left By symmetry, the magnetic force on the top segment and the magnetic force on the bottom segment of the I_{2}cancel with each other. The magnetic force on the left segment and the magnetic force on the right segment of the I_{2} are in opposite directions, but those 2 forces do not cancel.

OQ1. i) b, ii) d, iii) b, iv) c

31. a) 200 μT to top of page, b) 133 μT to bottom of page Use Ampere’s law.

34. By symmetry, the B on the right side of the sheet goes toward the top of the page and the B on the left side of the sheet goes toward the bottom of the page. Use Ampere’s law to find B. Use symmetry and choose a rectangular amperian loop that goes counterclockwise.

56. 2μ_{0}qω/(5(√5)πR) The moving charge makes a circular current loop. The current I = dq/dt = q/(period of the circular motion)

AP Physics C — Schedule for E&M Unit 3 (Ch.27, 28)

Ch. 27:

OQ1. d

2. qω/2π I = dq/dt or ∆q/∆t

3. 0.00013 m/s J = I/A = nqv_{d}

OQ4. i) a, ii) a

OQ5. c

OQ2. i) e, ii) d

OQ7. a

OQ3. c > a> b >d > e

OQ10. c

OQ12. i) a, ii) b

45. a) 0.66 kWh, b) $0.0726

63. a) Q/(4C), b) Q/4 on C and 3Q/4 on 3C, c) Q^{2}/(32C) on C and 3Q^{2}/(32C) on 3C, d) 3Q^{2}/(8C) After equilibrium is reached, I = 0, so V_{R} = IR = 0, so C and 3 C reach the same voltage.

Ch. 28:

OQ2. i) d, ii) b

1. a) 6.73 ohms, b) 1.97 ohms

OQ1. a

OQ4. b

OQ14. i) b, ii) a, iii) a, iv) b, v) a, vi) a

OQ15. . i) a, ii) d, iii) a, iv) a, v) d, vi) a

11. R_{1} = 1000 ohms, R_{2} = 2000 ohms, R_{3} = 3000 ohms

OQ12. i) a >d >b =c >e, ii) e> a= b> c =d

24. a) 0.909 A, b) *V _{b} *-

*V*= – 1.82 V Write 1 junction rule equation and 2 loop rule equations.

_{a}OQ7. d

37. a) 0.002 s, b) 0.00018 C, c) 0.000114 a) time constant = RC, c) use (1 – exponential decay) equation

39. a) 0.0616 A, b) 0.235 μC, c) 1.96 A a) Use the exponential decay equation, b) Use the exponential decay equation

63. a) 24.1 μC, b) 16.1 μC, c) 0.0161 A The 2 capacitors are in parallel, so we can treat the two as one capacitor that = the 2 in parallel.

65. a) 240 μC (1– e^{ – t/0.006} ), b) 360 μC (1– e^{ – t/0.006} ) Time constant = R_{eq} C_{eq}

41. a) 1.5 s, b) 1s, c) I_{1} + (I_{2} ) = 200μA + (100 μC (e^{ – t/1} )) a) Before S is closed, R_{1} and R_{2} are in series. B) After S is closed, the capacitor discharges through R_{2}. c) There is the I_{1} clockwise in left loop and the I_{2} counterclockwise in right loop.

43. a) 6 V, b) 8.29 μs The R in RC is the (4 + 2) ohms and (1 + 8) ohms in parallel.

75. a) Let 12000-ohm be the R_{1 }and 3000-ohm be the R_{3}. I_{R3} = 0, I_{R1} = I_{R2} = 0.000333A, b) 50 μC, c) (278 μA) *e*^{−t/ (0.180 s)} (for *t *> 0), d) 290 ms

AP Physics C — Schedule for E&M Unit 2 (Ch. 25, 26)

AP Physics C – For EM Unit 2 Homework

Ch. 25

OQ5. a>b=d>c Use: scalar U = kq_{1}q_{2} /r. Must plug in the signs for q.

OQ9. c>a>d>b Use: scalar U = kq_{1}q_{2} /r. Must plug in the signs for q. Add scalar U together for every pair of charges.

24. WK_{req} = ∆U (because no requirement on kinetic energy change)

OQ7. d>c>b>a Use: scalar V = kq/r. Must plug in the sign for q. Add scalar V together.

OQ10. b Same distance from all charges.

23. a) -4.83 m, b) at x= 0.667m and x= -2m V is a scalar.

26. Look at special locations and trend (how E changes with distance).

43. (1/√5 -1) kQ/R V is a scalar.

47. kλ(π+2 ln3) Have to integrate the scalar V.

OQ1. b

OQ4. d Uniform E, so we can use V = Ed and this d has to be parallel to the E. Also: when follow E line, V decreases.

36. E = -slope of the V as a function of x graph.

39. a) (-5 + 6xy) **i** + (3x^{2} – 2z^{2} )**j** – 4yz **k**, b) 7.07N/C** **a)** E _{x} = - ∂**V/

**∂**x, b) Plug in the coordinates to find

**E**’s 3 components, then use Pythagorean theorem to find the magnitude of E.

40. a) larger at A, b) 200N/C downward, c) E lines go outward and are always perpendicular to the equipotential surfaces.

For b), use V = Ed to estimate E.

OQ6. i) a, ii) c

21. a) 4(√2) kQ/a, b) 4(√2) kqQ/a b) WK_{req} = ∆U

OQ11. b

9. 0.3 m/s When q moves, the strings tension does not do any work (no displacement along the direction of tension.) So the K of q only comes from the electric field. Use U = qV and V = Ed (but the d has to be the displacement along the direction of E).

50. a) E = 0 inside conductor, V = V_{surface} = 1.67x10^{6} V, b) E = 5.84x10^{6} V/m radially outward, V = 1.17x10^{6} V, c) E = 1.19x10^{7} V/m radially outward, V = 1.67x10^{6} V

52. a) 450000V, b) 7.51 x10^{-6} C Use E = kq/r^{2} to find q first. Then use V = kq/r to find V.

60. a) –kq/(4a), b) when x>>a, V = -kq/(4.5a), so the difference is about 11%. Add the V scalar by the two charges together.

62. kQ^{2}/(2R) W = ∫Vdq = ∫(kq/R)dq integrate to build the charges from 0 to Q.

65. –λ ln(r_{2} /r_{1})/(2πє_{0}) E is not uniform, so we cannot use V = Ed. We have to integrate V = ∫Edr

72. 3kQ^{2}/(5R)

Ch. 26

2. a) 1μF, b) 100V

7. 4.43x10^{-6} m Have to combine a few equations: q = CV, q = σA, C = є_{0}A/d.

OQ6. b

13. a) 17μF, b) 9V, c) C_{1} has 45 μC, C_{2} has108 μC

14. a) 3.53μF, b) V_{1} = 6.35 V, V_{2} = 2.65 V, c) same q: 31.8μC

20. a) 2C, b) q_{1} > q_{3} > q_{2} , c) V_{1} > V_{2} = V_{3} , d) q_{1} and q_{3} increase, q_{2 }decreases.

24. a) 120μC, b) q_{1} = 80μC, q_{2} = 40μC b) the 2 capacitors share the charges on C_{1} and charges stop flowing between the two capacitors (when things reach static equilibrium), the two capacitors should have the same electric potential.

30. 4470 V

57. 0.00251 m^{3}, 2.51L.

OQ3. a

43. a) 13.3 nC, b) 272 nC

44. a) 3.4, b) nylon, c) between 25 and 85V.

OQ9. a

OQ11. b

36. a) C(∆V)^{2}, b) 4(∆V)/3, c) 4C(∆V)^{2}/3, d) Positive work must be done to pull the plates farther apart, because the 2 oppositely charged plates attract each other. The energy put into the system becomes extra stored U.

49. a) 40μJ, b) 500 V a) WK_{req} = ∆U

CQ6. An ideal voltmeter has infinite resistance. However, a real voltmeter has a finite large resistance. Therefore the capacitor slowly discharges through the voltmeter.

12. mgd tanθ/q Draw force diagram and write force equations for horizontal and vertical directions separately.

41. a) 400μC, b) 2500N/m V = Ed, and each plate’s charge contribute ½ of the E between the plates. To find the electric attractive force between the 2 charged plates, we can use one plate to produce E and place the other plate in this E, so F = qE.

48. 22.5V

59. 0.188 m^{2}

AP Physics C — Schedule for E&M Unit 1: Electric Force, Electric Field, and Gauss’s Law (Ch. 23, 24)

Ch. 23

13. a) 0.951 m, b) Yes, if the 3^{rd} bead is positively charged. a) The 2 electric forces on the 3^{rd} charge must be equal and opposite. b) Try displace the 3^{rd} charge a little bit in a certain way and see if the charge would experience a restoring force or not. Do this for every possible way the 3^{rd}charge can be displaced. If the 3^{rd} charge experiences a restoring force in all possible situations, the equilibrium is stable.

16. 0.299 m Draw a force diagram to show all of the forces acting on one of the 2 charges. The net force on the charge is zero in the x- and the y- directions.

19. kQ^{2} ((1/(2√2)**i** + (2 – 1/(2√2))**j**)/d^{2} We have to add the 2 electric force vectors together. Because one of the 2 forces is slanted, it can be more convenient for us to break that slanted force into x- and y- components. Then we can add the other force to it.

20. a) The net electric force can be written in the “–kx” format when a << d. b) (π/2) √(md^{3}/(kqQ)) , c) 4a√(kqQ/md^{3}) b) use k_{eq} to find ω. c) The midpoint is the equilibrium position. Use ω to find v_{eq} = v_{max}.

24. 2070 N/C, downward mg and electric force must be equal and opposite so they can cancel: mg = qE.

OQ12. a Use Coulomb’s law.

26. 3kq/r^{2} in the –y direction. Add the 3 electric fields produced by the 3 point charges.

29. 1.82 m to the left of the – 2.5μC The electric fields produced by each of the 2 charges at that location must be equal and opposite.

36. First find the E produced by each charge, then add the E together by adding vectors.

OQ13. c Electric field is a vector.

45. 2.16x10^{7} N/C to the left Have to chop the charges into point charges dq. Each dq provides dE. We have to integrate vector dE. Due to symmetry, we only have to integrate dE’s x-component.

82. First we have to find the E at a distance x from the center of the ring along the x-axis. Then we can find the electric force on the –q at that location: F = qE. Then we have to show that for x<<a, the F matches the –kx format.

49. a) -1/3, b) q_{1} is – and q_{2} is +. The number of E lines coming out of a + charge or going into a – charge is proportional to the amount of the net charge.

57. a) 111ns, b) 0.00567m, c) horizontal: 4.5x10^{5} m/s, vertical: 1.02x10^{5} m/s F = qE = ma

86. a) b) + z direction For example, at the center of the top face, the net E by the top 4 point charges is zero, so we only have to add the E by the lower 4 point charges. Because of symmetry, we only have to add the E_{z}.

Ch. 24

1. a) 1.98x10^{6} Nm^{2}/C, b) zero

4. a) -2340 Nm^{2}/C, b) 2340 Nm^{2}/C, c) 0 We can see that every E line entering the vertical surface comes out of the slanted the surface. That’s why parts a and b have the same amount of flux, but one is - and the other one is +.

OQ1. i) a, ii) b.

11. S_{1}: -Q/є _{0} , S_{2}: 0 , S_{3}: -2Q/є _{0} , S_{4} : 0.

CQ2. a) doubled, b) same, c) same, d) same, e) zero.

CQ3. Zero.

CQ6. +.

OQ4. i) c, ii) b. ii) If we move the charge to a point inside the box but extremely close to the center of one face, then half of the E lines coming out of this charge would go through that particular face.

20. (Q + 6q)/(6є _{0}) Because of symmetry, the flux through each of the 6 faces of the box is the same.

OQ6. i) e, ii) a, iii) c, iv) c, v) a

OQ8. C Inside conductor.

CQ7. a) No. b) The person will experience a shock when the charge on his body neutralizes the charges on the inner surface of the metallic sphere.

CQ8. The large net positive charge can induce charge separation on the other conducting sphere: with large net negative charges at the side near the other sphere and slightly more + charges at the far side, so the attractive force between the 2 spheres can be stronger than the repulsive force.

OQ2. c. The outer conducting shell is a Faraday’s cage. The external electric field induces charge separation on the outer surface of the conducting shell, so the electric field inside the shell is not affected.

OQ7. i) c, ii) e.

OQ9. a) A> B > D > C, b) B = D > A > C.

33. ρr/(2є_{0}), radially outward. Use Gauss’s law and choose a cylinder as Gaussian surface.

53. a) σ/(2є_{0}), b) 3σ/(2є_{0}), c) –σ/(2є_{0}) Use Gauss’s law and choose a cylinder (or a prism with any cross-sectional shape) as Gaussian surface.

56. a) 0, b) σ/є_{0} to the right, c) 0, d): a) 2σ/є_{0} to the left, b) 0, c) σ/є_{0} to the right. Use Gauss’s law and choose a cylinder (or a prism with any cross-sectional shape) as Gaussian surface.

59. σ/(2є_{0}), radially outward Superposition: E = E_{sphere }– E_{disk hole }by subtracting vectors. For finding E_{sphere} , we can use Gauss’s law or Coulomb’s law. For finding E_{disk hole} , we can use Gauss’s law and choose a cylinder (or a prism with any cross-sectional shape) as Gaussian surface. When we look for the E by a disk of charge at a location very very close to the disk, we can treat the disk as a very very large sheet of charge (large compared to the distance to the disk). Therefore, we can use Gauss’s law to find the E_{disk hole} .

60. a) 2kλ/r, radially outward, b) 2k(λ + ρπ(r^{2} – a^{2})) /r, radially outward, c) 2k(λ + ρπ(b^{2} – a^{2})) /r, radially outward. Use Gauss’s law and choose cylinders as Gaussian surfaces.

OQ3. e

AP Physics C — Schedule for Mechanics Unit 6 (Ch. 12, 15, 13)

AP Physics C – For Mechanics Unit 6 Homework

Ch. 12

OQ7. If use counterclockwise as + direction: D>C>E>B>A. If only compare the magnitude of torque: A>B=D>E>C=0.

11. Sam: 176N upward, Joe: 274N upward Use proportion or: Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

14. a) Horizontal: friction = (m_{1} g/2 + xm_{2} g/L)cotθ towards the wall, vertical: normal force = (m_{1} + m_{2})g upward Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

b) (m_{1}/2 + dm_{2}/L) cotθ /(m_{1}+m_{2}) μ = Friction/ normal force

15. a) 29.9 N, b) 22.2 N Draw force diagram with each force at the point of application and write force equations in x and y directions separately. For part a, you may look at the entire chain or half a chain. For part b) you will need to look at half of a chain.

39. 0.896m Use proportion or: Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

45. a) F_{g}(L+d)/((2L+d)sinθ), b) F_{x} = F_{g}(L+d)cotθ/(2L+d), F_{y} = F_{g}L/(2L+d) Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

49. a) 5080N, b) 4770N, c) 8260N Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

62. a) 120N, b) 0.3

c) 103N at an upper corner 31º above the horizontal, so the force is perpendicular to the diagonal of the rectangular cabinet. To minimize the force, we need to maximize the lever arm.

Ch. 15

5 a) 1.5Hz, b) 0.667s, c) 4m, d) π rad, e) 2.83m f = ω/(2π), etc. To find x at t=0.25s, make sure calculator is in radian mode.

8. a) 2.4s, b) 0.417Hz, c) 2.62 rad/s ω = 2πf

14. a) No mechanical energy is lost in elastic collisions, so the ball rebounds to 4m high each time. The motion repeats, so it’s periodic. b) 1.81s The period is the time it takes the ball to go down and then back up.

c) It’s not SHM because the net force on the ball is not in –kx format.

18. a) 1.26s, b) 0.15m/s, 0.75m/s^{2}, c) x(t) = 0.03cos(5t+π) = –0.03cos(5t), v(t) = –0.15sin(5t+π) = 0.15sin(5t) , a(t) = –0.75cos(5t+π) = 0.75cos(5t)

24. 0.153J, b) 0.784 m/s, c) 17.5m/s^{2} E = K + U = U_{max at end pt} = K_{max at equili.}, F = -kx = ma, so a_{max} happens at x_{max}.

66. μ_{s}g /(2πf)^{2} Write F_{net} = ma for the top box so you can find a_{max}. Then use a_{max} = ω^{2} x_{max}.

85. a) 0.5 m/s, b) 0.086m a) v at equilibrium is a max. To find v_{max}, we can use energy conservation or v_{max} = ω x_{max}.

b) It takes ¼ of a period for m_{1} to travel from equilibrium to the right end. During that time, m_{2} travels to the right at constant velocity.

OQ2 C, OQ3 A, OQ4 C, OQ5 D, OQ7 C, OQ8 B

OQ 16 i) B, ii) A, iii) C Use g_{apparent} instead of g.

38. I = mgd/(2πf)^{2}

39. 0.499

42. a) 2.09s I = I_{cm} + Mh^{2} where h is the distance between the c.m. of the meterstick and the pivot.

OQ13. D If the system is released from rest, the system’s c.m. stays at rest while the blocks oscillate. In this case, the 2 blocks are identical, so the c.m. is right in the middle. Because the c.m. does not move, we can treat the system as 2 oscillators one at each side of the c.m. Each oscillator is made of half of a spring and a block. Find the frequency of this oscillator - it is the same as the frequency of the original oscillator.

64. a) 2cm, b) 4s, c) π/2 rad/s, d) π cm/s, e) 4.93 cm/s^{2}, x(t) = 2 sin (πt/2) where x is in cm and t is in seconds.

68. a) Use 1/k_{eq} = 1/k_{1}+ 1/k_{2}.

b) The 2 springs are effectively in parallel – to compress one spring, you have to stretch the other spring by the same amount.

Ch. 13

11. 0.614m/s^{2} g = GM/r^{2}, where r is measured from the center of the earth, not from the surface.

12. 2/3 g = GM/r^{2}, where M = ρ•volume

OQ1. C There is one term between every pair of particles.

49. 2.52×10^{7} m It’s a maximum height problem. We can use conservation of energy. Because the projectile will go up very high (no longer considered “near the surface of the earth”), we need to use – GmM/r instead of mgy for gravitational PE.

38. For orbit speed, write F_{net} = ma. For escape speed, use energy conservation.

OQ7. i) E, ii) C, iii) A

OQ11. E Use Kepler’s law of periods: T^{2 }∝ a^{3} and the semi-major axis a = (max distance + min distance)/2.

CQ5. a) perihelion: when closest to the sun, b) aphelion: when farthest to the sun.

28. a) 6.02 × 10^{24} kg Circular orbit: start with F_{net} = ma.

b) Both the Moon and the Earth orbit around the center of mass of the Earth-Moon system. This center of mass is located close to the center of the Earth, but not at the center of the Earth. Therefore, the Moon’s orbit radius is a little less than the distance between the Moon and the Earth.

72. Circular orbit: start with F_{net} = ma.

CQ9. No. Air resistance causes the satellite to lose mechanical energy. Therefore, the satellite’s orbit gets closer and closer to the earth. With smaller orbit radius, the orbit speed increases.

42. GmM/(12R_{E}) WK_{nonconservative} = ∆E

43. a) 8.5× 10^{8} J WK_{nonconservative} = ∆E = ∆U, because the object does not have K up there.

b) 2.71× 10^{9} J WK_{nonconservative} = ∆E = ∆(K + U)

21. 1.26× 10^{32} kg Circular orbit: start with F_{net} = ma.

74. Circular orbit: start with F_{net} = ma.

AP Physics C — Schedule for Mechanics Unit 5 (Ch. 10, 11)

AP Physics C – For Mechanics Unit 5 Homework

Answers and hints for Ch. 10 problems: Try to do the problems without the hints first. It’s more fun that way!!

OQ2. b The centripetal acceleration is not zero.

OQ3. b = e > a = d > c = 0 a_{T} = rα and α is a constant. a_{r} = v^{2}/r = (rω)^{ 2}/r.

3. a) 5 rad, 10 rad/s, 4 rad/s^{2} , b) 53 rad, 22 rad/s, 4 rad/s^{2} ω = dθ/dt, α = dω/dt

17. a) 25 rad/s, b) 39.8 rad/s^{2}, c) 0.628 s ω = v/r. Need to convert rev. to radians.

70. a) ω(t) = ω + At + Bt^{2}/2, b) θ(t) = ωt + At^{2}/2 + Bt^{3}/6 ω = ∫αdt, θ = ∫ωdt .

18. a) 0.605 m/s ω = v/r, need to convert cm to m and rev/min to rad/s.

b) 17.3 rad/s The bike wheel has the same ω as the rear sprocket. The rear sprocket has the same v as the front sprocket because they are connected by a chain.

c) 5.82 m/s v = rω of the rear wheel.

d) 1.39 m/s The length of the pedal cranks being 17.5 cm long.

24. a (√(1 + π^{2} ))/g Friction is the only horizontal force to provide the horizontal acceleration. There are 2 horizontal acceleration components: a_{T} and a_{R}.

OQ6. i) a You want the mass to be close to the axis.

ii) c You want the mass to be far away from the axis.

OQ11. e I ∝ mr^{2}.

39. a) 5.8 kgm^{2} , b) the height of the door If you look at the top view of the door (looking down at the door from above), the door is just like a thin rod rotating about one of its ends.

43. 11 mL^{2}/12 Method 1: You can shift the axis to the center of mass of the 3-rod structure first and then shift the axis back and use the parallel-axis theorem to find the I. Method 2: You can look at one rod at a time. There are 3 rods. Look at one rod at a time: the one on x-axis is a thin rod rotating about one end, the one on y-axis is like a point mass, and the one on z-axis is a little more complicated. We can first find the I_{cm} of the z-axis rod and then shift the axis by L/2 and use the parallel-axis theorem to find the I.

27. clockwise 3.55 Nm

28. clockwise 168 Nm Lever arm is the distance between the line of force and the axis.

OQ7. a When the 50N weight accelerates downward, the tension in the string is less than the weight hung under.

29. 21.5 N Use τ_{net} = I α. The net torque is produced by the torque by T_{u} and T_{lower}. Those 2 tensions produce torques against each other.

46. 13 MR^{2}ω^{2}/24 K_{rot }= ½ I ω^{2} . You have to be careful when you find I. I = I_{cam} + I_{shaft}. I_{shaft} is just the I of a solid uniform cylinder rotating about its center axis. The I_{cam} is the I of a solid uniform cylinder with a cylindrical hole, so I_{cam} = I_{cylinder }– I_{hole}. For the I_{cylinder}, you can find I_{cm}first and then shift the axis from the c.m. to the actual axis using the parallel-axis theorem. Also note that the mass of the cam with hole is M, so the mass without hole is not M (it is 4M/3).

32. a) make sure that you draw each force from the point of application.

b) 0.309 m/s^{2}, c) Horizontal portion: T_{1} = 7.67 N, slanted portion: T_{2} = 9.22 N Write F_{net} = ma for each of the 2 boxes. Write τ_{net} = Iα for the pulley. Replace α with a/R. In order for you to be able to stack the 3 equations together and add them, you will need to divide by R on both sides of the τ_{net} = Iα equation, so you have force in newtons on the left side of the equation, so you can add it to the forces in the other 2 equations. When you add the 3 equations together, the internal forces T_{1} and T_{2} should cancel, so you can find acceleration conveniently.

37. 0.312 Use kinematics to find α and then write τ_{net} = Iα.

OQ9. i) a The basketball also has rotational KE.

ii) c p = mv

iii) a The basketball has more KE to turn into mgy.

OQ12. Less than Just look a small piece of mass (point mass) m attached to a light rod a distance R to the axis. When the rod is released from the horizontal position and swings down, we can use conservation of energy to find the ω at the bottom of the swing: mgR = ½ Iω^{2} = ½ (mR)^{2}ω^{2}, so we get ω^{ }∝ √(1/R), so ω is smaller the bigger the R. If the rod is not light, then we would consider all the mass on the rod that has smaller “R” than the “R” of the mass attached at the end. Therefore, the mass added at the end of the rod makes the ω at the bottom of the swing less than before.

56. Use conservation of energy, the lost mgy of the hanging mass is turned into KE of hanging mass and the KE_{rot} of the spool.

57. a) 2(√(Rg/3)) Lost mgy (y = R) is turned into KE_{rot }of the disk. Be careful that the I of the disk is not I_{cm}.

b) 4(√(Rg/3)) v = rω, and the lowest point is 2R from the axis.

c) √(Rg)

61. a) 2gsinθ/3 The disk does both translational motion and rotational motion, so we need to write both the F_{net} = ma and τ_{net} = Iα and for the disk.

b) hoop: gsinθ/2, c) tanθ/3

78. a) Mg/3, b) 2g/3 The disk does both translational motion and rotational motion, so we need to write both the F_{net} = ma and τ_{net} = Iα and for the disk.

c) (√(4gh/3)) The disk does constant acceleration motion, so you may use kinematics equations to find the final speed. Or, you may use conservation of energy to find it: lost mgy = KE_{translational} + KE_{rotational}.

81. (√(10(R-r)(1 – cosθ)/(7r^{2}))) Use conservation of energy. Be careful when you look for the height change. The radius of the circular motion of the center of mass of the sphere is (R-r).

82. a) 4F/3M to the right, b) F/3 to the right. The disk does both translational motion and rotational motion, so we need to write both the F_{net} = ma and τ_{net} = Iα and for the disk.

c) (√(8Fd/(3M)) Use kinematics equations.

83. a) 2.7 R First use forces to find the minimum speed for the sphere to make the loop at the top of the circle, then use energy conservation to find the starting height.

b) 20mg/7 to the left and 5mg/7 upward We have to look at both the a_{c} and a_{T}. The normal force provides the ma_{c} = mv^{2}/R. We can use energy conservation to find the speed of the sphere at P. (v_{P} = (√(20gR/7))) The a_{T }is provided by the upward friction and downward mg. So write the force equation for a_{T }as well as the τ_{net} = Iα for the sphere, use the 2 equations to find the a_{T}. m times a_{T }gives use the net force in the vertical direction.

73. a) (√(3g/L)) Use conservation of energy. Be careful when you look for the height change.

b) 3g/2L The rod is rotating about a fixed axis, so there is only rotational motion about that axis. There is no translational motion. So we only have to write τ_{net} = Iα. Be careful that the I is not the I_{cm}.

c) –(3g/2) **i** – (3g/4) **j** We have to look at both the a_{c} and a_{T}.

d) –(3Mg/2) **i** + (Mg/4) **j ** If we look at the entire rod, there are only 2 external forces acting on the rod: Mg and the force from the pivot. Together these 2 forces provide the acceleration we found in part c).

79. (√((2mgdsinθ + kd^{2})/(I + mR^{2}))) Use conservation of energy. Height change is dsinθ.

85. a) (√(3gh/4)) Use conservation of energy. The rod does not have a fix axis for rotation, so it’s easier if we treat the motion of the rod as: rotation about its center of mass and translational motion of the center of mass. So lost mgy = KE_{translational} + KE_{rotational}.

b) (√(3gh/4)) Use conservation of energy. This time the rod rotates about a fixed axis at one of its ends, so it’s easier if we treat the motion of the rod as: only rotation about its end and no translational motion. So lost mgy = KE_{rotational}. Be careful that the I in this case is not the I_{cm}.

92. a) Use conservation of energy. The lost E is friction times d.

b) 2Mg(sinθ – μcosθ)/(2M + m) Since it is constant acceleration motion, we can use kinematics equations to find a.

Ch. 11

10. a) No, b) **B** x **A** is perpendicular to both **B** and **A**. When 2 vectors are perpendicular to each other, their dot-product is zero. So we can find the dot product of (2, -3, 4) and (4, 3, -1). Since this dot-product is not zero, they are not perpendicular to each other.

12. – 22**k** kgm^{2}/s For a point mass, L = mvr_{┴}. And we can use right-hand-rule to find the direction for L. Or, we can use the formula in section 11.1.

13. (mxv_{y} – myv_{x})**k** For a point mass, L = mvr_{┴}. And we can use right-hand-rule to find the direction for L. Or, we can use the formula in section 11.1.

29. a) 1.57 x 10^{8} kgm^{2} We want the centripetal acceleration to equal to g. This will give us the final ω. Then we can use L = Iω.

b) 6260 s We can use τ_{ave} ∆t = ∆L.

OQ1. b L = constant

OQ8. d L = constant if τ_{net} = 0, p = constant if F_{net} = 0.

CQ3. To increase the moment of inertia of the walker, so it’s harder to change the walker’s state of rotational motion.

CQ6. The net torque (produced by gravitational force) on the c.m. of the “rider + motorcycle” system is zero, so the system’s angular momentum is constant. so if the wheels gain angular momentum in one direction, the rest of the system must gain angular momentum in the opposite direction to keep the total L in the system constant.

CQ8. It can still have mvr_{┴}, if r_{┴ }is not zero.

CQ9. a) I would increase, but only very slightly. b) Since L = constant, increase in I means decrease in ω. Therefore, the revolution period would increase, but only by a small fraction of a second.

30. a) I_{1}ω_{1}/ (I_{1 }+ I_{2}) It’s a collision problem with a fixed axle, so the linear momentum is not conserved; however, the angular momentum is conserved.

b) I_{1}/(I_{1 }+ I_{2})

31. 7.14 rev/min It’s a collision problem with a fixed axle, so the linear momentum is not conserved; however, the angular momentum is conserved, so: L_{i} = L_{f}. We can treat the child as a point mass.

33. a) No. Some calories (chemical PE) from the woman is turned into mechanical energy.

b) No. The turntable axle’s force and mg on the system do not cancel, so the net force on the system is not zero.

c) Yes. The net torque on the system is zero.

d) 0.36 rad/s counterclockwise. L_{i} = L_{f}.

e) 99.9 J It’s the KE gain.

39. a) 2mv_{i} d/((M + 2m)R^{2}) L_{i} = L_{f}

b) No, it’s a completely inelastic collision. Some KE is lost to heat in this case.

c) No. The fixed axle’s force does not cancel with mg. So the net force on the system is not zero.

49. a) 7md^{2}/3 Find I for each particle and then add all the I together.

b) mgd counterclockwise. There are three mg providing torque. Add the torque together to get net torque – remember that torque is a vector. Or: we can use the c.m. to represent the gravitational force of all 3 pieces of mass and find the torque produced by that gravitational force.

c) 3g/(7d) Use τ_{net} = Iα.

d) 2g/7 upward a_{T} = rα and a_{c} = v^{2}/r.

e) mgd That’s when the system has the lowest gravitational PE (when c.m. is the lowest), and the lost PE at that moment is the KE_{max}.

f) √(6g/(7d)) KE_{max} = KE_{rot} because the system is rotating about a fixed axis. There is no KE_{translational}.

53. a) r_{i}v_{i}/r L_{i} = L_{f}.

b) m(r_{i}^{2}v_{i}^{2})/r^{3} Tension provides the centripetal force for the puck.

c) ½ mv_{i}^{2}((r_{i}^{2}/r^{2}) – 1)) The work is turned into KE gain.

56. a) Mvd L_{total} = L_{1} + L_{2}

b) Mv^{2} K_{total} = K_{1} + K_{2}

c) Mvd L_{i} = L_{f}.

d) 2v L_{i} = L_{f}.

e) 4Mv^{2}, f) 3Mv^{2} The calories used is converted into the KE gain.

61. a) ω_{i }/3 That’s when v_{f} = Rω_{f}. Write force equation to find a and torque equation to find α. Because both a and α are constants, we can use kinematics equations to find the v_{f} and ω_{f}. Set v_{f} = Rω_{f} and find ω_{f}. (Note: When you write these equations, you can include μ. You will find that μ will cancel, so it does not show up in your answer for ω_{f }.)

b) – 2/3 fractional change = change/ initial value = (K_{f} – K_{i})/K_{i}

c) R ω_{i} /(9μg) The disk starts with v_{i} = 0, so ∆v = v_{f} - v_{i} = Rω_{f} . And F_{ave net} ∆t = ∆p = m ∆v, and F_{ave net} = F_{fr} = μmg.

d) R^{2} ω_{i}^{2} /(18μg) Can use kinematics equations.

63. 4(√(ga((√2) – 1)/3)) This problem involves 2 parts: a collision (L_{i} = L_{f}) and then the cube “swings up” about its lower right edge. So we have to do it as a 2-part problem: first the cube’s L_{i} = L_{f}, then we have conservation of energy for the swinging part. In order for the block to “swing” over the edge, the block’s c.m. has to get beyond the edge of the table – i.e. rising to a height that is half of the square’s diagonal – a height = a√2. Using formula on p. 304 and the parallel-axis-theorem, you should find that the I of the cube about one edge to be 8Ma^{2}/3. (If you have trouble finding the correct I, you may wish to just use I = 8Ma^{2}/3 to work on the problem.)

AP Physics C — Schedule for Mechanics Unit 4 (Ch. 9)

Answers and hints for Ch. 9 problems:

48. (11.7cm, 13.3cm) Because we can easily find the location of c.m. of a uniform rectangle, it can be convenient for us to divide this sheet metal into 3 rectangles. For each rectangle, we can use the c.m. of the rectangle as the location for this piece and use the area to represent the mass of this piece. Then we can find the center of mass of the 3 pieces combined using the center of mass equation. It’s a 2-d problem, so do the x and y separately.

50. 0.00673nm below the center of the oxygen Because the molecule is symmetric left and right to the dotted line in Fig. P9.50, we know that the c.m. has to be on the dotted line. Therefore, we can treat this problem as a 1-d problem with 2 pieces of mass: the oxygen (mass 16u) at the top and the combination of the 2 hydrogen (mass 2u) at the mid-point between the 2 H-atoms. Then we can just use proportion to find the location of the c.m. of the H_{2}O.

53. 0.7m If the friction between the boat and the water is negligible, the net external force on the system is zero, and the c.m. of the system should stay at rest the entire time. So, M∆**X**_{c.m.}= m_{1}∆**x**_{1} + m_{2}∆**x**_{2} = m_{1}∆**x**_{1 rel to 2} + (m_{1} + m_{2})_{ }∆**x**_{2} = 0. Use m_{1} for Juliet and m_{2} for (Romeo + boat). For “m_{1}∆**x**_{1 rel to 2} + (m_{1} + m_{2})_{ }∆**x**_{2}” part, because we have to consider that the boat carries Juliet with it, hence the (m_{1} + m_{2}).

OQ1. b>e>a=d>c The Frisbee “collides” with you on frictionless surface, so p_{i} = p_{f}, but that’s only true in the horizontal direction. So focus on the horizontal motion. The more horizontal velocity (and therefore, momentum) the Frisbee keeps in its original direction, the less momentum (and therefore, speed) you get in that direction. Note: For d, I guess the authors meant to say that the Frisbee ends up going vertically upward as viewed by an observer on the ground.

OQ2. a) No. It’s completely inelastic collision. b) No. When cart #1 rolled down, the net force on the cart is not zero. c) Yes because there is no friction. d) No. See part b. e) No. See part a. f) Yes. See part c.

CQ4. a) No. b) No. The ball cannot end with negative KE. c) Yes. Momentum is a vector, so the ball can end with negative p (e.g. if it turns around).

3. 7N **F**_{net} = d**p**/dt or Impulse by net force = Δ**p = F**_{ave net} Δt

5. 3263N toward homeplate and 3988N downward Look at the ball and use Impulse by net force on the ball: **J** = Δ**p = F**_{ave net} Δt. The force on the bat is the same amount as but in opposite direction to the force on the ball. In this case, we can ignore the mg of the ball because the force from the bat is so much stronger.

7. a) *v _{pi} *= −

*m*+

_{g}v_{gp}/(m_{g}*m*p = constant, so

_{p})*p*because we have to consider that the plank carries the girl with it, hence the

_{i}= 0 = p_{f}= m_{g}v_{gp}+ (m_{g}+ m_{p}) v_{pi}*(m*.

_{g}+ m_{p})

b) *v _{gp}+*

*v*=

_{pi}*v*−

_{gp}*m*+

_{g}v_{gp}/(m_{g}*m*The girl gets two velocities: one by herself relative to the plank, the other by the plank. Therefore: her net velocity relative to the ice is

_{p})*v*

_{gi}= v_{gp}+*v*.

_{pi}11. a) 6m/s to the left Use *p _{i} = p_{f}*.

b) 8.4 J Energy is conserved: energy stored in the spring is turned into the KE of the 2 blocks.

c) spring , d) The cord is burned before the blocks begin to move, therefore the cord’s forces do not have any displacement, so the cord’s forces do not do any work on the boxes. I.e. no energy is transferred from the cord to the boxes.

e) Yes. Because the net force on the whole system (consisting the 2 boxes) is zero, and that is the condition for momentum conservation. The spring’s forces on the two boxes are equal and opposite, so the net force by the spring on the 2-box system is zero.

15. a) x√(k/m) Use conservation of energy.

b) x√(km) Impulse by net force = Δ**p.**

c) No. Same work no matter what mass. Use conservation of energy.

17. a) 0.096 s Use kinematics equations for constant acceleration motion.

b) 365000N Impulse by net force = Δ**p = F**_{ave net} Δt

c) 265 m/s^{2} = 26.5g, d) No. The 11-kg toddler will seem to weigh 11×26.5 = 296 kg = 641 lb.

19. a) 12**i **Ns Impulse by net force = Δ**p = F**_{ave net} Δt = area of **F**_{net} as a function of time graph.

b) 4.8**i** m/s, c) 2.8**i** m/s Impulse by net force = Δ**p = **m Δ**v**.

d) 2.4**i** N Impulse by net force = Δ**p = F**_{ave net} Δt

20. a) 981 Ns, up Impulse by a force **= F** Δt = area of **F** as a function of time graph = ∫**F** dt

b) 3.43 m/s, down Use kinematics equations for constant acceleration or conservation of energy.

c) 3.83 m/s, up Impulse by net force **= **Impulse by (m**g** + **F**_{on her by platform}) = Δ**p**

d) 0.748 m Use kinematics equations for constant acceleration or conservation of energy.

21. 16.5 N The scale reads the weight of (the bucket + water inside) + the force related to the impulse of falling water. For the falling water part, consider **F**_{ave net} = Δ**p/**Δt = m Δ**v/**Δt = (m/Δt)Δ**v**.

61. 15 N The additional force is the force related to the impulse of the shooting water.

67. -260**i** N It’s 2-dim case, so we need to look at one dimension at a time. There is only momentum change in the x direction, so the force is only in the x-direction. Find components of **v _{i} **and

**v**

_{f}in the x-direction. Then do

**F**

_{ave net x}= Δ

**p**Δt.

_{x}/OQ3. i) c Newton’s 3^{rd} law of motion.

ii) a K = ½ mv^{2} = p^{2}/(2m) ∝ (1/m)

OQ7. a K = ½ mv^{2} = p^{2}/(2m) ∝ p^{2}

OQ8. d K = ½ mv^{2} = p^{2}/(2m) ∝ p^{2}

OQ11. b First use momentum conservation to find the speed of the block immediately after the impact. Then the KE is lost to friction.

OQ13. a Same F and same d, so the work done by the force are the same, so the ∆K are the same. Both particles start from rest, so K_{f} are the same.

OQ14. d Both the fall and bounce are interaction between the ball and the earth. The net external force on the ball + earth system is zero during the fall and the bounce.

CQ11. No **F**_{ave net} Δt = Δ**p**

CQ12. No WK_{Fnet} = ΔK

29. a) 4.85 m/s Kinematics or conservation of energy.

b) 8.41 m The basketball rebounds at the same speed and collides elastically with the tennis ball coming down toward it at the same speed. Find the final velocity of the tennis ball and then find the max. height.

30. (4M√(gL))/m Because the M is attached to a stiff rod (not a string), barely making the loop means the speed of the M at the top of the loop is zero. Then use conservation of energy to find the speed of the M at the bottom of the loop after the bullet has emerged. Then use conservation of momentum for the collision part to find the speed of the bullet before collision.

32. ((m+M)√(2μgd))/m For the collision part: p_{i} = p_{f}. For the sliding part: lost KE is taken away by friction.

33. 0.556 m For the sliding part: use conservation of energy to find the speed of m_{1} just before the collision. The blocks never touch à the collision is elastic. So use v_{approaching} = v_{separating} and p_{i} = p_{f} for the collision part. Find the velocity of m_{1} at the end of the collision and then use conservation of energy to find how high it rises.

34. a) 2.24 m/s to the right Use p_{i} = p_{f}. All 3 carts have the same velocity at the end.

b) No.

73. v_{m} = 2v_{i}, v_{3m} = 0 Use v_{approaching} = v_{separating} and p_{i} = p_{f} for the elastic collision.

35. a) v_{x} = 0.929 m/s, v_{y} = 0.53 m/s 2-dim case, so separate the x and y. Find the final v_{x} and v_{y} of the 0.2-kg puck. Then write p_{i} = p_{f} for x and y directions separately to find the final v_{x} and v_{y} of the 0.3-kg puck.

37. (3**i **– 1.2**j**) m/s Write p_{i} = p_{f} for x and y directions separately to find the final v_{x} and v_{y}.

57. a) The cart “suddenly” comes to a stop, so the top of the cart suddenly stops. However, the particle still has the v_{i} at that moment. So the particle swings up: use conservation of energy for this part.

b) at the lowest point The connection between the cart and the particle is the string. A string’s tension is along the direction of the string. So the tension has no horizontal component when the string is vertical.

69. a) 1.33 m/s to the right Use p_{i} = p_{f}.

b) 235 N to the left F_{fr} = μ_{k} F_{N}

c) 0.68 s Impulse by net force** = F**_{ave net} Δt = Δ**p = **m Δ**v**. In this case, the net force changing the person’s momentum is the friction.

d) Person: 160Ns to the left, cart: 160 Ns to the right.

e) 1.82 m For the person: F_{net} = F_{fr} = 235 = ma = 60a. We can find a (to the left), and we know v_{i} = 4 and v_{f} = 1.33, so we can find ∆x.

f) 0.454 m For the cart: F_{net} = F_{fr} = 235 = ma = 120a. We can find a (to the right), and we know v_{i} =0 and v_{f} = 1.33, so we can find ∆x.

g) -427 J, h) 107 J, i) (427 – 107) J is lost to friction.

72. a) There are 2 parts in this event: the collision and then the rise to maximum height. We can use conservation of momentum for the collision part and then conservation of energy for the rise to maximum height part.

b) ((m+M)√(2gh))/m

75. a) (m_{1}v_{1} + m_{2}v_{2})/( m_{1} + m_{2}) This problem shows a classic way to model elastic collision. During the collision, some KE is turned into the PE stored in the spring. At the end of the collision, the spring is relaxed again. All the energy stored in the spring during the collision is back in KE. So KE_{i} = KE_{f} à it’s an elastic collision. During this collision, when the spring has the maximum compression, the relative velocity between the 2 blocks is zero. Therefore at that moment the 2 blocks have the same velocity – kind of like the 2 blocks are stuck together at that moment.

80. a) 0.667 m/s to the left It’s like an explosion problem: one piece, the block and wedge combination) breaks into 2 pieces. So we can use p_{i} = p_{f}, although the momentum is only conserved in the horizontal direction.

b) 0.952 m The block and wedge has no KE at the beginning. The KE they gain at the end comes from the lost mgy of the block.

82. ((M+m)√(gd^{2}/(2h)))/m We know information at the end, so we can go backwards: Use projectile motion kinematics to find the speed of the (m+M) at the end of the collion. Then we can use p_{i} = p_{f} to find the initial speed of the bullet.

85. 0.403 They both swing down from the same height, so they meet at the lowest point with horizontal velocities of the same magnitude but in opposite directions. They collide and stick together, so we can use p_{i} = p_{f} to find their velocity immediately after the collision. Then we can use conservation of energy for their swing up part.

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AP Physics C — Schedule for Mechanics Unit 3 (Ch. 7, 8)

Hints for Ch. 7 Try to do the problems without the hints first. It’s more fun that way!!

1. a) 1590 J WK=Fd cos<

b) 1590 J Since friction stays the same, new F still has the same horizontal (parallel to d) component as before.

2. a) 0.0328 J WK=Fd cos<

b) -0.0328 J a = 0.

OQ 5. a>b=e>d>c cos<

CQ4. a) 90º ≥ θ > 0, b) 180º ≥ θ > 90º cos<

9. 16 For **A**-**B**, do each component separately. For scalar product, remember that cos 90º = 0

11. a) 16 J Work is a scalar product.

14. a) 24 J, b) -3 J, c) 21 J Area.

23. a) When one tray is removed, the force compressing the springs is reduces by the weight of one tray. If the springs are Hookes law springs, F = kx means ∆F = ∆(kx) = k ∆x. Same ∆F gives the same ∆x. We just have to make the ∆x for the weight of one tray equal to the thickness of the tray.

b) 316 N/m There are 4 springs in parallel supporting the trays.

c) 45.3cm by 35.6cm.

28. a) 9000J WK = ∫Fdx

OQ4. C WK = Fdcos<

CQ3. Sometimes. It’s true if the initial K is zero.

CQ8. a) Not necessarily. If the force does no work, no K change, ie. no speed change. (WK = 0 when F ┴ d, ie. F ┴ v, because v is always tangent to the path. So if F ┴ v, this F only provides centripetal (or radial) acceleration. No tangential acceleration means no speed change. )

b) Yes. F = ma. If F is not zero, a is not zero, so velocity has to change.

38. a) 23400N, opposite to the bullet’s v WK_{Fnet} = F_{net} d cos< = ∆K

b) 0.000191s Can use kinematics.

25. a) Draw force diagram and write force equation. Because the particle is being pulled at constant speed, there is no tangential acceleration. There is only centripetal acceleration.

b) mgR Use conservation of energy.

45. a) 125 J For the first segment of the path, only the x-component of the force does work. For the 2nd segment, only y-component of the force does work.

b), c) 125J

47. A/(r^{2}) F = -dU/dr

49. F_{x} = 7 – 9x^{2}y, F_{y} = - 3x^{3 }F_{x } = -∂U/∂x, F_{y } = -∂U/∂y

50. a) Ax^{2}/2 – Bx^{3}/3 U = - ∫Fdx

b) 2.5A – 6.33 B, c) - 2.5A + 6.33 B There is no non-conservative forces doing work, so E = constant, so ∆E = ∆K + ∆U = 0

52. a) A: 0, B: +, C: 0, D: -, E: 0 F = -dU/dx = - slope

b) A, E: unstable, C: stable, c)

OQ14. d U = ½ kx^{2} ∝ x^{2} , when x à2, U à4, so the extra U we need to provide is 4 – 1 = 3 times the old value.

CQ11. 2k When you pull on a spring with a force F, the entire spring has the same tension F. If a force F stretches the original spring by x, the same F would stretch half of the spring by ½ x.

59. 0.299 m/s Lost K turns into U in the spring. U also equals to the energy given to the spring by the car, i.e. the work done on the spring by the car. And WK = area of the F as a function of x graph. Or we can also find U by finding the energy stored in both springs.

60. The ball can only go up the incline by 0.881 m along the incline. That is only about half of the height of a professional basketball player. Need to convert k from N/cm to N/m and convert d from cm to m. E_{i }= ½ k x^{2} = mgy_{f} = mg (distance • sin θ)

64. E_{i }= ½ mv^{2} + mg (d + x) sin θ = E_{f} = ½ kx^{2}

Hints for Ch. 8

OQ1. a E_{i} = ½ kx^{2} = E_{f} = ½ mv^{2} , since you pull the slingshot for the same amount, E_{i }is the same for the pebble and the bean. This means mv^{2}à 1.

8. a) √(2(m_{1} – m_{2})gh/(m_{1} + m_{2})) Write E_{i} = E_{f}. Notice that one mass has mgy at the beginning and the other has mgy at the end. At the end, both boxes move at the same speed.

b) 2m_{1}h/(m_{1} + m_{2}) After m_{1} strikes the floor, m_{2} still has an upward velocity, so it keeps going up to max. h. before coming down.

11. √(8gh/15) The blocks have equal mass, and A is pulled up by 1T while B is pulled up by 2T. Therefore, block A will go down and B go up. Also, when A goes down by 1m, B goes up by ½ m causing their separation to be 1.5 meters. This means a separation of h means A goes down by 2h/3 and B goes up by h/3. Write E_{i} = E_{f}.

24. a) 0.381 m Write E_{i} = E_{f}. Note that the initial mgy has a y that is (1.2 + x) above the final position. You will get a quadratic equation from E_{i}= E_{f}. This means you will get 2 solutions for x, one +, one –. The answer is the + solution, of course. But the – solution also has a special meaning. What do you think it means?

13. v^{2}/(2μ_{k}g) ∆E = WK_{nonconservative force} or lost E = amount of energy take away by friction.

15. a) 0.791 m/s Write E_{i} = E_{f}.

b) 0.531 m/s ∆E = WK_{nonconservative force} or lost E = amount of energy take away by friction.

33. $145 (Power in kW) (time in h) = (energy used in kWh)

41. a) 10200 W P = WK/t = Fv, in this case, a = 0, so F = tension in cable = mg sinθ.

b) 10600 W In this case, T – mg sinθ = ma.

c) 5820000 J It’s easier to use conservation of energy.

45. h + d^{2}/(4h) Since the information at the end is given, we can work backwards to find the initial height H. First do the projectile part to find the speed at which the child goes off the slide. Then use conservation of energy to find H.

46. a) 2.49 m/s It’s convenient to use conservation of energy to find the speed of the blocks the moment m_{2} strikes the floor. After m_{2} strikes the floor, m_{1 }continues to move along the table at constant velocity until it goes off the table.

b) 5.45 m/s Can use conservation of energy to find landing speed. Look at only m_{1} and start from the moment it goes off the table to the moment before it lands.

c) 1.23 m It would be the ∆x of m_{1}’s projectile motion.

d) No, e) m_{2} loses its K when it lands.

47. a) K = 2 + 24t^{2} + 72t^{4} v = dx/dt

b) a = 12t, F_{net }= 48t a = dv/dt

c) P = 48t + 288t^{3} P = Fv

d) WK = 1250 J Can do WK = ∫Pdt, but it’s easier to use conservation of energy: WK = ∆K.

64. 1.24 m/s Write E_{i} = E_{f}. Note that block m_{2} will go down by h = 0.2 m, while m_{1} will have a vertical height increase of h sinθ. The 2 blocks have the same speed.

68. a) E_{i} = E_{f} because the peg’s force on the string does no work.

b) In order for the ball to complete the circle, the ball has to have at least a minimum speed that can be found if you write the force equation at the top of the circle. Then write E_{i} = E_{f}. v^{2} cannot be negative, so d has to be at least 3L/5.

84. a) For this problem, in order for 3m to be the correct answer, I think you will have to make the corner of the table frictionless – kind of like having a frictionless pulley at the corner for the chain to hang down from it, so the hanging part of the chain does not contribute to the normal force on the part of the table with friction. Then: Let the linear mass density of the chain be λ. I.e. every meter of the chain has a mass of λ. λ = Mass/Length. If the amount hanging over the edge has a length x, the mass of this hanging part would be x λ. OR: You may say the total mass of the chain is M and the hanging part has a mass that is a fraction x/L out of M.

Answers and hints for Ch. 9 problems:

48. (11.7cm, 13.3cm) Because we can easily find the location of c.m. of a uniform rectangle, it can be convenient for us to divide this sheet metal into 3 rectangles. For each rectangle, we can use the c.m. of the rectangle as the location for this piece and use the area to represent the mass of this piece. Then we can find the center of mass of the 3 pieces combined using the center of mass equation. It’s a 2-d problem, so do the x and y separately.

50. 0.00673nm below the center of the oxygen Because the molecule is symmetric left and right to the dotted line in Fig. P9.50, we know that the c.m. has to be on the dotted line. Therefore, we can treat this problem as a 1-d problem with 2 pieces of mass: the oxygen (mass 16u) at the top and the combination of the 2 hydrogen (mass 2u) at the mid-point between the 2 H-atoms. Then we can just use proportion to find the location of the c.m. of the H_{2}O.

53. 0.7m If the friction between the boat and the water is negligible, the net external force on the system is zero, and the c.m. of the system should stay at rest the entire time. So, M∆**X**_{c.m.}= m_{1}∆**x**_{1} + m_{2}∆**x**_{2} = m_{1}∆**x**_{1 rel to 2} + (m_{1} + m_{2})_{ }∆**x**_{2} = 0. Use m_{1} for Juliet and m_{2} for (Romeo + boat). For “m_{1}∆**x**_{1 rel to 2} + (m_{1} + m_{2})_{ }∆**x**_{2}” part, because we have to consider that the boat carries Juliet with it, hence the (m_{1} + m_{2}).

OQ1. b>e>a=d>c The Frisbee “collides” with you on frictionless surface, so p_{i} = p_{f}, but that’s only true in the horizontal direction. So focus on the horizontal motion. The more horizontal velocity (and therefore, momentum) the Frisbee keeps in its original direction, the less momentum (and therefore, speed) you get in that direction. Note: For d, I guess the authors meant to say that the Frisbee ends up going vertically upward as viewed by an observer on the ground.

OQ2. a) No. It’s completely inelastic collision. b) No. When cart #1 rolled down, the net force on the cart is not zero. c) Yes because there is no friction. d) No. See part b. e) No. See part a. f) Yes. See part c.

CQ4. a) No. b) No. The ball cannot end with negative KE. c) Yes. Momentum is a vector, so the ball can end with negative p (e.g. if it turns around).

3. 7N **F**_{net} = d**p**/dt or Impulse by net force = Δ**p = F**_{ave net} Δt

5. 3263N toward homeplate and 3988N downward Look at the ball and use Impulse by net force on the ball: **J** = Δ**p = F**_{ave net} Δt. The force on the bat is the same amount as but in opposite direction to the force on the ball. In this case, we can ignore the mg of the ball because the force from the bat is so much stronger.

7. a) *v _{pi} *= −

*m*+

_{g}v_{gp}/(m_{g}*m*p = constant, so

_{p})*p*because we have to consider that the plank carries the girl with it, hence the

_{i}= 0 = p_{f}= m_{g}v_{gp}+ (m_{g}+ m_{p}) v_{pi}*(m*.

_{g}+ m_{p})

b) *v _{gp}+*

*v*=

_{pi}*v*−

_{gp}*m*+

_{g}v_{gp}/(m_{g}*m*The girl gets two velocities: one by herself relative to the plank, the other by the plank. Therefore: her net velocity relative to the ice is

_{p})*v*

_{gi}= v_{gp}+*v*.

_{pi}11. a) 6m/s to the left Use *p _{i} = p_{f}*.

b) 8.4 J Energy is conserved: energy stored in the spring is turned into the KE of the 2 blocks.

AP Physics C — Schedule for Mechanics Unit 2 (Ch. 5, 6)

Answers and hints for Ch. 5:

CQ13. The scale reading oscillates up and down – more than (m_{weightlifter }g + m_{barbell} g) when his arms + barbell have an upward acceleration and less than (m_{weightlifter }g + m_{barbell} g) when his arms + barbell have a downward acceleration. When he throws the upward, the scale would read more than (m_{weightlifter }g + m_{barbell} g) when he accelerates the barbell upward. Once he lets go of the barbell, the scale would just read his weight.

CQ14. The sack moves up with the athlete whether the athlete is speeding up or slowing down. If you draw a force diagram for each of them, you will see that you have exactly the same force diagrams and same amount of forces acting on them. So they have the same a = F_{net}/m. They also both start from rest.

CQ22. a) Both students slide toward each other with B experiencing twice the force of A. b) Same as a). c) Same as a). d) Both students slide toward each other with B experiencing the force as A.

28. (If use g = 10 m/s^{2}) a) 50N, b) 50N, c) 100N, d) 25N

32. 112N Find accelerations x- and y- components. Then find the magnitude of the acceleration.

51. (If use g = 10 m/s^{2}) a) 720N, b) 828N, c) 720N, d) 662N

4. 8.71N x-components cancel.

26. a) 3 forces: one vertical (mg), one horizontal (magnetic force), one slanted (T) b) 1.03N, c) 0.805N to the right. Find components of the slanted force. Then write F_{net} = ma for x- and y-directions separately.

86. a) T = f/(2sinθ), b) 410N Draw forces for the midpoint of the cable. Take advantage of the symmetry: i.e. make one of your axis along f. Find components of the slanted force. Then write F_{net} = ma for x- and y-directions separately.

29. a) 7m/s^{2} to the right, b) 21N, c) 14N a) Can look all 3 boxes as one system. b) Can look at 1kg + 2kg combined. c) Can look at the 2kg box.

78. a) Note: In this problem the sagging of the rope is ignored. So even though there is no vertical force to cancel with the mg of the rope, the rope is drawn straight and horizontal. b) a = F/(m_{b} + m_{r}), c) T = F m_{b} /(m_{b} + m_{r}), d) For a massless string, the tension would be the same throughout.

42. b) 3.57 m/s^{2}, c) 26.7 N, d) 7.14 m/s. For m_{2}, only need to draw forces along the incline.

45. (If use g = 10 m/s^{2}) a) 20N, b) If F_{x} is a 80N pushing force to the left (or: F_{x} = - 80N), c) It’s a straight line with + slope for -80N 10) to 100N. (a = (F_{x} – 20)/10). For -100N to -80N, the line bents at -80N, the equation becomes: a = F_{x}/8.) b) The tension becomes zero when m_{1} is in free fall.

46. a) a_{2} = 2a_{1}, b) T_{1} = 2 m_{1}m_{2} g/(4m_{2} + m_{1}), T_{2} = m_{1}m_{2} g/(4m_{2} + m_{1}), and c) a_{1} = m_{1} g/(4 m_{2} + m_{1}), a_{2} = 2m_{1} g/(4 m_{2} + m_{1}) a) When P_{2} moves 1 m, m_{2}moves by 2 meters. b) Draw force diagram and write force equation for each box. Also T_{1} = 2T_{2}.

OQ1. d) Max friction also doubles.

OQ5. b) F_{net} = ma

OQ8. d) Can use F_{net} = ma and kinematics or use work-energy theorem.

OQ12. d) F_{net} = ma = 0

OQ13. a), c), d)

52. a) 14.7m, b) 10000kg and 20000kg

57. μ_{s} = 0.727, μ_{k} = 0.577 Look at forces along the incline.

58. a) 4.18, b) Applying any higher engine power would only make the car skid and therefore turning the friction to the smaller kinetic friction. If the driver keeps at that higher engine power, the wheels will spin faster (while skidding), lifting front end more and perhaps flip over.

60. a) The free-body diagram should include 4 forces. b) 55.2° c) 167N Find components for the slanted force and write one F_{net} = ma for each of the x and y directions.

66. a) between 31.7 N and 48.6 N. b) If P is too big, the block will slide up the wall. If P is too small, the block will slide down the wall. Draw force diagram – considering the 2 conditions described in answer for part b) separately. Find components for the slanted force and write one F_{net} = ma for each of the x and y directions.

69. 0.06 m Since we want to know the motion of the mug, let’s draw force and write force equation for the mug, so we can find the mug’s acceleration. Because the mug move during the pull of the tablecloth, the tablecloth has to move more than 30 cm relative to the table. However, we know how far the tablecloth has to move relative to the mug (30cm), we can use that to find out how much time it takes by looking at the relative motion between the mug and the tablecloth. We can then use this time to find the distance traveled by the mug within that time.

70. a) There are 4 forces acting on the 5kg and 6 forces on the 10kg. b) T = 9.8 N, a = 0.58 m/s^{2}. Since the 2 blocks have different accelerations, we should draw force and write force equation for the blocks separately.

81. a) b) 0.408 m/s^{2}, c) 83.3 N For b), look at Nick and the chair as one system. c) To find the force between Nick and the chair, we would have to look at Nick and the chair separately. Since there is only one unknown we want to find, we can choose to look at either Nick or the chair.

82. a) Since the child is lighter than Nick + chair, the child will acceleration upward. a = 0.426 m/s^{2} and T = 459 N. b) In this case, a = 0 and T = the weight of Nick + chair = 480N. This is more than 250N in 81 and the 459N in part a). a) This scenario is like the Atwood machine – a pulley with one mass hanging on each side.

85. a) b) T_{1} = T_{2} = T_{3} = Mg/2, T_{4} = 3Mg/2, and T_{5} = Mg, c) Mg/2 a = 0 for every object.

91. a) The cushion starts from rest and the acceleration is a slanted constant acceleration (downward g and a horizontal constant amount too), so the cushion will accelerate along the straight line that is slanted like the constant acceleration. b) No. See answer for a). c) 1.63 m from the base of the building. d) Parabola c) We can still look at the forces and kinematics one direction at a time – vertical and horizontal directions separately. d) The object starts with an initial velocity that is in a different direction than its constant acceleration (like a projectile), so its path is parabolic. The parabola is tilted.

92. a) a_{1} = 2a_{2} because when m_{2} goes down by 1 meter, the string has to increase by 1 m on each side of the pulley above m_{2}, so m_{1} has to move by 2 meters. b) a_{2} = 12.7N/(1.3kg + 4m_{1}) downward. c) a_{2} would be almost g downward: i.e. m_{2} would be almost like a falling object. d) a_{2} would approach 0, because the tension on m_{1} is a limited force. If m_{1} approaches infinity, the acceleration will have to approach to zero. e) 6.37N, f) Yes. See previous answers.

93. (M + m_{1} + m_{2}) m_{1} g/m_{2} m_{1} has no vertical acceleration. This is the information we can use to find the tension in the string. Knowing the tension, we will be able to find the acceleration of m_{2}. Then we can look at the whole 3-box system to find F.

Ch. 6

OQ1. a) A>C=D>B=E=0, b) A: north, B: west, C: south, c) A: west, B: nonexistent, C: east.

OQ4. a) Yes: A: to the right and downward: perpendicular to the string. b) Yes, A. See answer to part a) for acceleration. c) No. d) Yes, B: It’s to the right between (but not including) the direction along the string and the direction of the rightward tangent.

OQ5. b)

OQ6. a) No, because v = 0: turning point. b) Yes.

OQ7. i) c), ii) b): the angle has to do with acceleration (not velocity).

CQ4. a) The object moves in a circle at constant speed. b) The objet moves along a straight line at changing speed. a) Only radial acceleration and no tangential acceleration. b) Only tangential acceleration and no radial acceleration.

9. a) static friction, b) 0.085 Write force equation for the horizontal direction.

16. a) 24900N, b) 12.1 m/s b) almost lose contact, so normal force is zero.

42. Draw force diagram and write force equation: The acceleration is horizontal. It’s more convenient to choose an upright coordinate system. Find components for any forces that are slanted in your coordinate system.

54. a) m_{2}g, b) m_{2}g, c) √(m_{2}gR/m_{1}), d) m_{2} will go down and the radius of the puck’s circular motion will decrease. Because of angular momentum conservation, the puck will also speed up. e) m_{2} will go up and the radius of the puck’s circular motion will increase. Because of angular momentum conservation, the puck will also slow down.

63. 12.8 N Draw force diagram and write force equation: The acceleration is horizontal. In this case, either upright coordinate system or the slanted one will give us 2 slanted vectors in a coordinate system. So for either coordinate system you choose, you will have to find components for 2 vectors.

64. a) 2.63 m/s^{2}, b) 201 m, c) 17.7 m/s This problem is just like prob. 42. Draw force diagram and write force equation: The acceleration is horizontal. It’s more convenient to choose an upright coordinate system. Find components for any forces that are slanted in your coordinate system.

20. a) 3.6 m/s^{2} to the right, b) T = 0, c) Both observers in parts c and d will see the same mg and normal force. An observer in the car may think that in addition to the tension read by the spring scale there must also be a (fictitious) force of Ma to left acting on the m. d) An observer on the ground would think that the tension read by the spring scale is the only horizontal force acting on the m.

21. a) 17°, b) 5.12 N This problem is just like prob. 42 and 64. Draw force diagram and write force equation: The acceleration is horizontal. It’s more convenient to choose an upright coordinate system. Find components for any forces that are slanted in your coordinate system.

24. 2(vt – L)/[(g + a) t^{2}] Use kinematics and forces. Because of the acceleration of the elevator, the normal force on the backpack is not the same as its mg.

51. a = g(cosϕtanθ – sinϕ) Draw force diagram and write force equation. It can be convenient to choose your coordinate system tilted in the same way as the acceleration. You will have 2 slanted forces in this coordinate system. Be careful that the 2 forces slant at different angles.

**For 2016-2017 ERHS AP Physics C students: **

Those who took Physics S/T before senior year have summer homework! Your summer homework is: Watch and take notes on the following video lessons. Your notes are due on the first day of school.

First: AP Physics 1 & 2 Static Electricity Videos 5 to 41

Then: AP Physics 1 Rotation Videos 1 to 45

Those who took AP Physics 1 and AP Physics 2 before senior year have no summer homework!

# AP Physics C — Schedule for Mechanics Unit 1: Kinematics (Ch. 2, 4)

AP Physics C – For Mechanics Unit 1 Homework

Answers and hints for Ch. 2 problems: Try to do the problems without the hints first. It’s more fun that way!!

OQ11. b To obtain a(t), we need to first obtain v(t) from the x vs. t graph.

7. a) -2.4 m/s, b) about -3.8 m/s, c) about 4s a) ave. v = ∆x/∆t, c) v = slope of x vs. t graph, so v = 0 when the slope of the graph is zero.

11. a) 5m, b) 4875s Hint: Both hare and tortoise run at constant speed if you ignore the waiting time of the hare.

Solution: Total running time for hare = 1000m/(8m/s) = 125s, Total running time for tortoise = 1000m/(0.2m/s) = 5000s. So the hare has to wait 5000-125= 4875 s. At first the hare runs for 0.8km/(8m//s) = 100 s, so the hare must begin to run again 125-100 = 25s before the race ends. That means the tortoise must be (0.2m/s)(25s) = 5 m from finish line.

14. 13400m/s^{2} ave. a = ∆v/∆t. Pay attention to the direction of the velocities.

39. a) against (i.e. not equal), b) for (i.e. equal) Hint: The glider goes through constant acceleration motion, so its ave. v = (v_{0} + v) /2.

Solution: Since the glider goes through constant acceleration motion, its velocity changes at a constant rate, it reaches the (v_{0} + v) /2 half way through time. I.e. it reaches the ave. v half way through time. This is not the same as the v half way through space, because the glider speeds up, so it takes less time to travel the second half of the space than the time it takes to travel the first half of the space.

79. 1.6m/s^{2} From problem 39, we know that for constant acceleration motion, the average velocity is the velocity half way through time, and we have enough information to find the average v when each car goes by. So we can find the velocities of the cars when they pass Liz half way in time.

17. a) 1.3 m/s^{2}, b) at t = 3 s, 2 m/s^{2}, c) at t = 6s and t = about 10.3 s to 12 s, d) 8s, -1.5 m/s^{2} a) Use definition of ave. a. b) steepest part of the graph, c) where slope =0, d) steepest negative slope.

19. a) at 10s: 20m/s, at 20s: 5m/s, b) 263m a) use ∆v = area of graph, b) Can plot a v vs. t graph and then find area of graph: Since we are looking for distance, we have to add +∆x and the absolute value of -∆x together. Another way to do this is to look at the motion in 3 separate segments (0 to 10s, 10 to 15s, and 15 to 20s), and use kinematics equations to find the displacement traveled in each segment. In this case the object does not turn around, otherwise we have to be careful to pay attention to when the object turns around in order to find the correct distance traveled

21. a) 2m, b) -3m/s, c) -2 m/s^{2} a) Plug in t. b) Take time derivative and then plug in t. c) Take one more time derivative and then plug in t.

24. 160ft Use the v^{2} equation for constant acceleration motion. v_{0}^{2} is proportional to ∆x.

29. -16 cm/s^{2}

33. a) 35s, b) 15.7m/s Either plot a v vs. t graph or use kinematics equations and work through the 3 segments of the motion one at a time.

38. a) 2.56m, b) -3m/s a) What’s special about turning point? b) When x = 0 again.

59. a) –(1×10^{8})t + 3×10^{5} m/s^{2} , b) 3×10^{-3}s, c) 450 m/s a) a = dv/dt, b) make a = 0 and solve for t, c) plug the previous answer into the v(t) equation.

OQ14. b) The 2^{nd} ball’s motion is like the reverse of the 1^{st} ball’s. And they meet at the same time, so it must be half way in time.

OQ15. b) Can use kinematics or energy conservation.

OQ16. e)

OQ17. a) A=C=E>B=D, b) D>A=B=C=E

50. 7.96s (if use g = 9.8m/s^{2}) Find h(t=2) to find initial height. Take time derivative of h to find v(t) and plug in t = 2 to find initial velocity of the mailbag.

52. 0.6s When the two balls meet, they are at the same position at the same time. So if we use the same coordinate system for both balls and write one y(t) for each ball. The balls would meet when y_{1}(t) = y_{2}(t). Solve for t for time. (Of course, you can also write the upward displacement of 1^{st} ball as a function of t. Write distance fallen by 2^{nd} ball as a function of time. Then add these 2 together to equal to 15m. Solve for t.)

54. a) (h/t) + (gt/2), b) (h/t) –(gt/2)

55. (if use g = 9.8 m/s^{2}) b) 0.782s, a) 7.82 m b) Look at vertical motion.a) Look at horizontal motion.

56. a) v_{i} + gt, b) gt^{2}/2, c): a) The absolute value of (v_{i} – gt), b) gt^{2}/2 In this case the speed is the absolute value of the velocity. To find the distance between the helicopter and the package, we can find the difference in position of the two. So if we use the same coordinate system for both the helicopter and the package to write one y(t) for each. The absolute value of the difference in their y(t) would be the distance between the two.

85. (if use g = 9.8 m/s^{2}) a) 26.4 m, b) 28.2 m: 6.82% error a) Let t be the time it takes for rock to hit water, then the time it takes for sound to reach man is (2.4 –t). Use t to express the distance traveled by rock and the distance traveled by the sound. These 2 distances should be equal which gives us a quadratic equation. Solve the quadratic equation to find t. Then use t to find depth of the well. b) Treat 2.4 s as the time it takes for the rock to fall.

Ch. 4

OQ1. e) What is the definition of ave. a?

OQ4. d) Use the v^{2} equation.

OQ7. d) First find out by what factor the time changes. Then use ∆x = vt.

OQ11. a) T = 2πr/v

OQ14. d)

2. 2.5 m/s Which component of the velocity represents the velocity of the shadow?

3. a) (1**i** + 0.75**j**) m/s, b) v = (1**i** + 0.5**j**) m/s, speed = 1.12m/s, c) parabola a) b) Do the x- and y- components separately. c) because y(x) is a second degree equation.

CQ1. Parabola The spacecraft has no acceleration in one direction and has a constant acceleration in another direction.

CQ2. For the first loop, the velocity is always tangent to the circle and with increasing magnitude. The acceleration slants forward and towards the center of the circle. For the second loop, the velocity is always tangent to the circle and has a constant magnitude (no smaller than the largest velocity vector in the first loop). The acceleration always points toward the center of the circle and has constant magnitude. For acceleration, first look at tangential and radial accelerations separately.

CQ3. You can determine the average velocity, but not the instantaneous velocity.

CQ4. a) Follow a straight path. b) Follow a straight path (a = constant = 0), or follow a circular path (a = non-zero constant magnitude)

22. a) ∆x = 1.18 m, so the space is wide enough for a walkway. But a very tall person will have to walk carefully close to the wall to avoid getting his/her head wet. b) 0.491 m/s a) Projectile shot horizontally. b) ∆x = v_{x} t. When the height fallen changes by a factor of 1/12, the ∆y = 0 + ½ at^{2}changes by a factor of 1/12, so t^{2} changes by a factor of 1/12. Since we also want ∆x to change by a factor of 1/12, we can find the factor by which v_{x} has to change.

29. a) (0, 50m), b) (18m/s, 0), c) Constant acceleration motion with downward a = - g = - 9.8 m/s^{2}. d) Constant velocity motion with v = 18 m/s, e) v_{x} = v_{i }, v_{y} = -gt, f) x = v_{i}t_{ }, y = y_{0} - (1/2)gt^{2}, g) 3.19s, h) 36.1m/s, 60.1º below the horizontal.

15. 53.1º Start with an initial velocity v_{0} and initial angle θ. (Since v_{0 }does not matter, we can also say the initial velocity is 1m/s or 100m/s. It won’t matter.) Find max. height and ∆x in terms of v_{0}, θ, and g. Then make ∆x = (3 times the max. height). v_{0 }should cancel. Solve for the angle.

21. d(tanθ_{i} ) – (gd^{2} )/(2v_{i}^{2}cos^{2 }θ_{i}) Just like a normal projectile motion problem. Separate the horizontal and vertical sides. Use the horizontal side to find t and then use that t on the vertical side to find h.

23. a) The ball clears the crossbar by 0.89m. b) While falling. a) It’s just like problem 21. b) Using the initial velocity and the initial angle, we can find the time it takes the ball to reach maximum height. Compare this time to the time found in part a).

49. a) The acceleration has a downward component of 9.8 m/s^{2} and a southward component of 2.5 m/s^{2}. I.e. the acceleration is 10.1 m/s^{2}, 14.3º to the south from the vertical. b) a downward acceleration of 9.8 m/s^{2}. c) The bolt starts from rest and has a constant acceleration of 10.1 m/s^{2}, 14.3º to the south from the vertical. So its trajectory is a straight line that goes 14.3º to the south from the vertical. d) The bolt has an initial horizontal velocity that equals to the train’s velocity at the moment the bolt leaves the ceiling. So the bolt is like a projectile shot horizontally. The trajectory is parabolic.

53. 15.3m Since the observer standing on the ground observes the ball shot straight up, the ball must have been thrown with a horizontal velocity that equals to 10m/s backwards. Using trig, we can find the vertical component of the ball’s velocity which can be used to calculate for the ball’s max. height.

55. 54.4 m/s^{2} Radial acceleration is the same as the centripetal acceleration.

2016 AP Physics C Review Schedule:

# Assessing the Impacts of Urbanization on the US Surface Climate: Goddard Space Flight Center (GSFC) - Internship:

# https://intern.nasa.gov/ossi/mobile/opp.cfm?id=14331

Ch. 31

OQ5. c, d

OQ6. b, d

OQ7. a

OQ9. i) b, ii) c, iii) a

OQ10. a, b, c, d

CQ6. Yes. When the aluminum enters and passes through the magnetic field, current (eddy current) is induced inside the aluminum. Because of the eddy current, the aluminum in the magnetic field experiences an upward magnetic resistance force.

4. 0.00982V Use Faraday’s law of induction.

9. 0.0618V Use Faraday’s law of induction.

36. a) N^{2}B^{2}w^{2}v/R to the left, b) 0, c) N^{2}B^{2}w^{2}v/R to the left The magnetic resistant force is N times Bwv only when there is magnetic flux change. When there is no magnetic flux change, there is no induced current and no magnetic force on the loop.

32. mgRsinθ/(B^{2}ℓ^{2}cos^{2}θ) At terminal v, F_{net} = 0. Look at forces along the incline only: the sliding down component of the gravity, mg sinθ, is balanced by the magnetic force's component along the incline. The magnetic resistant force equals IℓB and I = ɛ_{ind}/R = Bℓv_{^}/R where the v_{^} is the component of the v that is perpendicular to the B. Also, you need to be careful when you look for the component of the magnetic resistant force on the sliding bar along the incline.

68. Use ɛ_{ind} = Bℓv

39. a) 3.2×10^{-20}N First find the induced E at P_{1: }Pretend that you have a circular wire loop of radius r_{1} and use Faraday’s law of induction to find ɛ_{ind}. And use ɛ_{ind} = to find E. Then we can find the force on the charge as F = qE.

b) downward and slightly to the right (perpendicular to r_{1}).

c) 1.33s The force is zero when E = 0 when the flux in the pretend circular wire loop is not changing. I.e. when B is not changing: i.e. dB/dt = 0.

41. a) E = 0.00987 cos (100πt) V/m, b) clockwise. a) Use ɛ_{ind} = - N dФ_{B}/dt= .

13. a) μ_{0}IL ln((h+w)/h)/(2π). a) Use Ampere’s law to find the B by the long current and then chop the area into horizontal strips: use dA=Ldr for the integration of **B·dA** .

b) μ_{0}bL ln((h+w)/h)/(2π) = 4.8μV b) Use emf = - dФ_{B} /dt. After all constants are taken out of the derivative, we only have dI/dt left.

c) counterclockwise

63. a) R_{1}: 3.5A, R_{2}: 1.4A It’s like R_{1} and R_{2} are in parallel and they are in parallel with the ɛ_{ind} by the sliding rod.

b) 34.3W, c) 4.29 N Use the idea conservation of energy and that P = Fv.

72. a) πCKa^{2} Use Faraday’s law of induction.

b) upper plate, c) The changing magnetic flux causes an induced emf in the wire loop – which means there is induced E in the loop. The charges in E experience electric force to make them move in the wire loop.

Ch. 32

OQ3. b)

OQ6. a)

CQ2. a) size and shape of the coil, the number of coils, and what it has in its core (e.g. iron core). b) No.

4. a) 0.00197 H. b) 0.038A/s b) │ɛ_{L}│ =│L (dI/dt)│

9. a) 0.36V, b) – 0.18V, c) 3 s. a) and b) │ɛ_{L}│ =│L (dI/dt)│, c) when dI/dt = 0.

14. ɛ_{L}= –L (dI/dt) = – L (slope of the I as a function of t graph)

OQ4. V_{Inductor} > V_{1200ohms }> battery > V_{12ohms}

CQ4. i) a: Bulb turns on immediately and then gets dimmer and dimmer and then turns off. b: Bulb goes from dark to increasing brightness and then stays with the same brightness. c: Bulb goes from dark to increasing brightness and then stays with the same brightness. d: Bulb turns on immediately and then gets dimmer and dimmer and then turns off.

ii) a: Bulb turns off immediately. b: Bulb goes off (but not instantly). c: The bulb gets dimmer and dimmer to off. d: The bulb turns on suddenly and then gets dimmer and dimmer to off.

CQ7. a) I_{L} = 0, I_{C} = ɛ/R, I_{R} = ɛ/R, ɛ_{L} = ɛ, V_{C }= 0, V_{R }= ɛ

b) I_{L} = 0, I_{C} = 0, I_{R} = 0, ɛ_{L} = 0, V_{C }= ɛ , V_{R }= 0

31. a) 0.00566 s, b) 1.22 A, c) 0.0581s a) and b) The current is a 1 minus exponential decay equation. c) The current is an exponential decay equation.

80. a) 72V, b) b, c) I_{1}: Goes from 0.009A exponentially decays down towards zero. I_{2}: Goes from 0.003A before t = 0 suddenly jumps to 0.009A in the opposite direction (i.e. –0.009A) and then exponentially decays towards zero.

CQ8. When the charge on the C is zero the V_{C }= 0, the current in the circuit is a maximum (i.e. dI/dt = 0); therefore, the ɛ_{L} is also 0.

47. 0.281H Use energy conservation.

81. R equal to or less than 300 ohms Before S is opened, the I in the L is 0.267A. The moment after S is opened, the L still has 0.267A. Since we want the entire armature to be limited to 80V, we can find R. V_{armature} = V_{R}

**Women in Engineering (WIE) DREAM Conference: UMCP: ****Saturday, February 27, 2016 from 9:00 AM - 2:30 PM**

**WHERE: Adele H. Stamp Student Union, University of Maryland, College Park**

The conference is **FREE** and includes lunch for participants. Plus, students that attend the event and subsequently the Clark School at the University of Maryland will be eligible to apply for DREAM scholarships valued at up to $2,500!

**For more information and to REGISTER, please visit http://www.wie.umd.edu/k12/dream-conference.**

**The Esperanza scholarship: **This year we are again awarding scholarships of at least $5000 and up to $20,000 to immigrant students and students with two immigrant parents in the District of Columbia, Maryland, and Virginia to attend public colleges and universities. Please visithttp://apps.esperanzafund.org/2016/.

We will also hold two application tip sessions. The first will be offered on February 28, at 3:15 pm at Central Library, 1015 N. Quincy Street, Arlington, VA 22201. I know this is short notice for the first information session, but students can attend even if they have not started an application. The second will be on March 12 at 10:15 am at the DC Tenley-Friendship Neighborhood Library, 4450 Wisconsin Avenue NW, Washington DC 20016.

**Physics is Phun: Friday, March 4, 2016 and Saturday, March 5, 2016. Shows begin at 7:00pm and end at 8:30pm: Program Flyer. Read about it. Turn in your notes by March 8 (for Dr. Rigal's classes) and on March 7th or 10th (for Ms. Twu's classes) for bonus points. **

AP Physics C — Schedule for E&M Unit 4 (Ch. 29, 30)

Ch. 29

OQ1. c, e

OQ9. c

2. a) up, b) out of the page, c) no deflection, d) into the page

OQ7. i) b, ii) a

15. a) (√2)r_{p}, b) (√2)r_{p} Use ∆V to accelerate the charge from rest: U_{loss} = q∆V = K_{gain} = ½ mv^{2}. For the circular motion part, we have: qvB = m v^{2}/r.

22. a) 1.79×10^{-8} s, b) 35.1 eV a) For the circular motion part, we have: qvB = m v^{2}/r. The electron leaves after making half a circle. b) The maximum depth is the radius.

73. a) 1.04×10^{-4} m, b) 1.89×10^{-4} m a) The positron does constant velocity motion in the x direction, so ∆x = v_{x} t. The t is the time for the positron to go around one circle. I.e. t = the period of the circular motion. b) For the circular motion part, we have: qvB = m v^{2}/r.

OQ6. c

42. a) 2πrIB sinθ, b) upward The magnetic force on each segment of the current points upward and slanted towards the center. By symmetry, the net magnetic force is vertical, so we only have to add the force’s upward components.

61. 0.588T This one is just like the free response problem that was on your 2014 AP Physics B exam. The weight of the wire is canceled by the springs’ forces when the springs stretch 0.5 cm. The added magnetic force is canceled by the springs’ additional force. So: F_{magnetic }= ILB = 2k(∆x_{additional}).

24. a) 6.86 × 10^{6} Hz, b) 5.17 × 10^{7} m/s a) For the circular motion part, we have: qvB = m v^{2}/r. Period T = 2πr/v, so f = v/(2πr). b) The maximum speed corresponds to the radius 1.20 m.

29. 244000V/m Undeflected: so F_{magnetic} and F_{electric} must be equal and opposite so they can cancel. So: qvB =qE. The 750eV electron means the electron’s K = ½ mv^{2} = 750eV.

30. a) Yes, this tells us that the cathode rays must be in all those different cathode materials. b) Don't do part b. I don't think the question is a very correct one. ~~Yes, this means the cathode ray has a unique q/m ratio. See p.881 fig. 29.15 for diagram. When the deflection plates get a different voltage, the E = V/d in the velocity selector part is different. When you change E, because qvB = qE in the velocity selector, the v changes, so only particles with the new velocity can go through undeflected. When you turn off the E, the particles get deflected. ~~ c) The very small q/m told J.J. Thomson that cathode rays must be made of particles that have much smaller mass than any ions. Cathode rays are electrons. An electron is much lighter than even the lightest ion (with about 1836 times the mass of an electron). d) Accelerated by >100V means the electrons will have a KE that is >100eV. At 100eV, an electron would have a speed of > (5.93×10^{6})m/s. At this speed, the g = 9.8 m/s^{2} acceleration cannot produce visible deflection within the size of a cathode ray tube.

OQ13. A > C> B Same angle between the magnetic dipole moment and B. So we can just compare the magnitude of the magnetic dipole moment which is IA. They also have the same I, so we just need to compare the A, the areas of current loops.

CQ3. Yes, if B is in the same direction as or in the opposite direction to the normal vector of the area of the loop. I.e. if B is perpendicular to the plane of the loop.

51. a) 9.98 Nm, b) clockwise with ω in the –y direction (this is the direction you get when you use right-hand rule for ω). Find the direction of the magnetic force on each segment of the wire current first. Figure out the direction of the rotation, then specify the rotational axis, and then find the torque produced by each force.

Ch. 30

CQ5. Vertically upward. B lines come out of the north pole of a magnet. Note: the magnetic north pole is near the geographical South Pole.

CQ10. a) The magnetic repulsive force between the magnets cancels with the mg of the magnets above. b) To keep this equilibrium stable. Otherwise, with a little disturbance, the magnets can flip over and attract and stick to each other. c) For a magnet, if it has north pole on the top , its south pole would be at the bottom. d) Blue and yellow would stick together and they will still be levitated by the repulsive magnetic force from the red magnets.

OQ11. b) and c)

CQ4. Use Ampere’s law for this long straight current. If we make the circular Amperian loop inside, the I_{encl.} = 0, the Ampere’s law would lead to B = 0 inside. If we make the circular Amperian loop outside, the I_{encl.} = I in the wire, the Ampere’s law would lead to B 0 outside.

CQ6. Yes, Ampere’s law is valid for all closed paths, but most of the current situations and random choices of the made-up closed Amperian paths do not make Ampere’s law convenient to use for solving for B.

CQ9. 0 For both Figures 30.10 and 30.13, the B is always in the direction of ds, so there is no B lines going through those circular Amperian loop.

10. μ_{0}i/4πx Using Biot-Savart law, we know that the x direction current provides no B at P. For the rest of the current, it is half of a very long straight current (very long going up and down). Using Ampere’s law, we can find the B by the long straight current and then divide it by 2 to get the B by half of a long straight current.

13. 0.262 μT, into the page Using Biot-Savart law, we know that the straight parts of the current provide no B at P. For the arc part of the current, we can use Biot-Savart law again to find the B it produces at P. OQ5. a) and c).

OQ5. a) and c).

OQ9. c) and d).

15. b) 20 μT to the bottom of page, c) 0 There are 3 currents producing B at each location. Use superposition to add the 3 magnetic field vectors by the 3 currents together to get the net B.

OQ3. a) No, b) Yes, c) Yes, d) No It’s only possible when they all repel each other.

OQ10. b)

OQ12. a)

26. μ_{0 }I_{1}I_{2} a/[2πc(a+c)] to the left By symmetry, the magnetic force on the top segment and the magnetic force on the bottom segment of the I_{2}cancel with each other. The magnetic force on the left segment and the magnetic force on the right segment of the I_{2} are in opposite directions, but those 2 forces do not cancel.

OQ1. i) b, ii) d, iii) b, iv) c

31. a) 200 μT to top of page, b) 133 μT to bottom of page Use Ampere’s law.

34. By symmetry, the B on the right side of the sheet goes toward the top of the page and the B on the left side of the sheet goes toward the bottom of the page. Use Ampere’s law to find B. Use symmetry and choose a rectangular amperian loop that goes counterclockwise.

56. 2μ_{0}qω/(5(√5)πR) The moving charge makes a circular current loop. The current I = dq/dt = q/(period of the circular motion)

AP Physics C — Schedule for E&M Unit 3 (Ch.27, 28)

AP Physics C – For EM Unit 3 Homework

Ch. 27:

OQ1. d

2. qω/2π I = dq/dt or ∆q/∆t

3. 0.00013 m/s J = I/A = nqv_{d}

OQ4. i) a, ii) a

OQ5. c

OQ2. i) e, ii) d

OQ7. a

OQ3. c > a> b >d > e

OQ10. c

OQ12. i) a, ii) b

45. a) 0.66 kWh, b) $0.0726

63. a) Q/(4C), b) Q/4 on C and 3Q/4 on 3C, c) Q^{2}/(32C) on C and 3Q^{2}/(32C) on 3C, d) 3Q^{2}/(8C)

After equilibrium is reached, I = 0, so V_{R} = IR = 0, so C and 3 C reach the same voltage.

Ch. 28:

OQ2. i) d, ii) b

1. a) 6.73 ohms, b) 1.97 ohms

OQ1. a

OQ4. b

OQ14. i) b, ii) a, iii) a, iv) b, v) a, vi) a

OQ15. . i) a, ii) d, iii) a, iv) a, v) d, vi) a

11. R_{1} = 1000 ohms, R_{2} = 2000 ohms, R_{3} = 3000 ohms

OQ12. i) a >d >b =c >e, ii) e> a= b> c =d

24. a) 0.909 A, b) *V _{b} *-

*V*= – 1.82 V Write 1 junction rule equation and 2 loop rule equations.

_{a}OQ7. d

37. a) 0.002 s, b) 0.00018 C, c) 0.000114 a) time constant = RC, c) use (1 – exponential decay) equation

39. a) 0.0616 A, b) 0.235 μC, c) 1.96 A a) Use the exponential decay equation, b) Use the exponential decay equation

63. a) 24.1 μC, b) 16.1 μC, c) 0.0161 A The 2 capacitors are in parallel, so we can treat the two as one capacitor that = the 2 in parallel.

65. a) 240 μC (1– e^{ – t/0.006} ), b) 360 μC (1– e^{ – t/0.006} ) Time constant = R_{eq} C_{eq}

41. a) 1.5 s, b) 1s, c) I_{1} + (I_{2} ) = 200μA + (100 μC ( e^{ – t/1} )) a) Before S is closed, R_{1} and R_{2} are in series. B) After S is closed, the capacitor discharges through R_{2}. c) There is the I_{1} clockwise in left loop and the I_{2} counterclockwise in right loop.

43. a) 6 V, b) 8.29 μs The R in RC is the (4 + 2) ohms and (1 + 8) ohms in parallel.

75. a) Let 12000-ohm be the R_{1 }and 3000-ohm be the R_{3}. I_{R3} = 0, I_{R1} = I_{R2} = 0.000333A, b) 50 μC, c) (278 μA) *e*^{−t/ (0.180 s)} (for *t *> 0), d) 290 ms

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AP Physics C — Schedule for E&M Unit 2 (Ch. 25, 26)

AP Physics C – For EM Unit 2 Homework

Ch. 25

OQ5. a>b=d>c Use: scalar U = kq_{1}q_{2} /r. Must plug in the signs for q.

OQ9. c>a>d>b Use: scalar U = kq_{1}q_{2} /r. Must plug in the signs for q. Add scalar U together for every pair of charges.

24. WK_{req} = ∆U (because no requirement on kinetic energy change)

OQ7. d>c>b>a Use: scalar V = kq/r. Must plug in the sign for q. Add scalar V together.

OQ10. b Same distance from all charges.

23. a) -4.83 m, b) at x= 0.667m and x= -2m V is a scalar.

26. Look at special locations and trend (how E changes with distance).

43. (1/√5 -1) kQ/R V is a scalar.

47. kλ(π+2 ln3) Have to integrate the scalar V.

OQ1. b

OQ4. d Uniform E, so we can use V = Ed and this d has to be parallel to the E. Also: when follow E line, V decreases.

36. E = -slope of the V as a function of x graph.

39. a) (-5 + 6xy) **i** + (3x^{2} – 2z^{2} )**j** – 4yz **k**, b) 7.07N/C** **a)** E _{x} = - ∂**V/

**∂**x, b) Plug in the coordinates to find

**E**’s 3 components, then use Pythagorean theorem to find the magnitude of E.

40. a) larger at A, b) 200N/C downward, c) E lines go outward and are always perpendicular to the equipotential surfaces.

For b), use V = Ed to estimate E.

OQ6. i) a, ii) c

21. a) 4(√2) kQ/a, b) 4(√2) kqQ/a b) WK_{req} = ∆U

OQ11. b

9. 0.3 m/s When q moves, the strings tension does not do any work (no displacement along the direction of tension.) So the K of q only comes from the electric field. Use U = qV and V = Ed (but the d has to be the displacement along the direction of E).

50. a) E = 0 inside conductor, V = V_{surface} = 1.67x10^{6} V, b) E = 5.84x10^{6} V/m radially outward, V = 1.17x10^{6} V, c) E = 1.19x10^{7} V/m radially outward, V = 1.67x10^{6} V

52. a) 450000V, b) 7.51 x10^{-6} C Use E = kq/r^{2} to find q first. Then use V = kq/r to find V.

60. a) –kq/(4a), b) when x>>a, V = -kq/(4.5a), so the difference is about 11%. Add the V scalar by the two charges together.

62. kQ^{2}/(2R) W = ∫Vdq = ∫(kq/R)dq integrate to build the charges from 0 to Q.

65. –λ ln(r_{2} /r_{1})/(2πє_{0}) E is not uniform, so we cannot use V = Ed. We have to integrate V = ∫Edr

72. 3kQ^{2}/(5R)

Ch. 26

2. a) 1μF, b) 100V

7. 4.43x10^{-6} m Have to combine a few equations: q = CV, q = σA, C = є_{0}A/d.

OQ6. b

13. a) 17μF, b) 9V, c) C_{1} has 45 μC, C_{2} has108 μC

14. a) 3.53μF, b) V_{1} = 6.35 V, V_{2} = 2.65 V, c) same q: 31.8μC

20. a) 2C, b) q_{1} > q_{3} > q_{2} , c) V_{1} > V_{2} = V_{3} , d) q_{1} and q_{3} increase, q_{2 }decreases.

24. a) 120μC, b) q_{1} = 80μC, q_{2} = 40μC b) the 2 capacitors share the charges on C_{1} and charges stop flowing between the two capacitors (when things reach static equilibrium), the two capacitors should have the same electric potential.

30. 4470 V

57. 0.00251 m^{3}, 2.51L.

OQ3. a

43. a) 13.3 nC, b) 272 nC

44. a) 3.4, b) nylon, c) between 25 and 85V.

OQ9. a

OQ11. b

36. a) C(∆V)^{2}, b) 4(∆V)/3, c) 4C(∆V)^{2}/3, d) Positive work must be done to pull the plates farther apart, because the 2 oppositely charged plates attract each other. The energy put into the system becomes extra stored U.

49. a) 40μJ, b) 500 V a) WK_{req} = ∆U

CQ6. An ideal voltmeter has infinite resistance. However, a real voltmeter has a finite large resistance. Therefore the capacitor slowly discharges through the voltmeter.

12. mgd tanθ/q Draw force diagram and write force equations for horizontal and vertical directions separately.

41. a) 400μC, b) 2500N/m V = Ed, and each plate’s charge contribute ½ of the E between the plates. To find the electric attractive force between the 2 charged plates, we can use one plate to produce E and place the other plate in this E, so F = qE.

48. 22.5V

59. 0.188 m^{2}

AP Physics C — Schedule for E&M Unit 1: Electric Force, Electric Field, and Gauss’s Law (Ch. 23, 24)

Ch. 23

13. a) 0.951 m, b) Yes, if the 3^{rd} bead is positively charged. a) The 2 electric forces on the 3^{rd} charge must be equal and opposite. b) Try displace the 3^{rd} charge a little bit in a certain way and see if the charge would experience a restoring force or not. Do this for every possible way the 3^{rd}charge can be displaced. If the 3^{rd} charge experiences a restoring force in all possible situations, the equilibrium is stable.

16. 0.299 m Draw a force diagram to show all of the forces acting on one of the 2 charges. The net force on the charge is zero in the x- and the y- directions.

19. kQ^{2} ((1/(2√2)**i** + (2 – 1/(2√2))**j**)/d^{2} We have to add the 2 electric force vectors together. Because one of the 2 forces is slanted, it can be more convenient for us to break that slanted force into x- and y- components. Then we can add the other force to it.

20. a) The net electric force can be written in the “–kx” format when a << d. b) (π/2) √(md^{3}/(kqQ)) , c) 4a√(kqQ/md^{3}) b) use k_{eq} to find ω. c) The midpoint is the equilibrium position. Use ω to find v_{eq} = v_{max}.

24. 2070 N/C, downward mg and electric force must be equal and opposite so they can cancel: mg = qE.

OQ12. a Use Coulomb’s law.

26. 3kq/r^{2} in the –y direction. Add the 3 electric fields produced by the 3 point charges.

29. 1.82 m to the left of the – 2.5μC The electric fields produced by each of the 2 charges at that location must be equal and opposite.

36. First find the E produced by each charge, then add the E together by adding vectors.

OQ13. c Electric field is a vector.

45. 2.16x10^{7} N/C to the left Have to chop the charges into point charges dq. Each dq provides dE. We have to integrate vector dE. Due to symmetry, we only have to integrate dE’s x-component.

82. First we have to find the E at a distance x from the center of the ring along the x-axis. Then we can find the electric force on the –q at that location: F = qE. Then we have to show that for x<<a, the F matches the –kx format.

49. a) -1/3, b) q_{1} is – and q_{2} is +. The number of E lines coming out of a + charge or going into a – charge is proportional to the amount of the net charge.

57. a) 111ns, b) 0.00567m, c) horizontal: 4.5x10^{5} m/s, vertical: 1.02x10^{5} m/s F = qE = ma

86. a) b) + z direction For example, at the center of the top face, the net E by the top 4 point charges is zero, so we only have to add the E by the lower 4 point charges. Because of symmetry, we only have to add the E_{z}.

Ch. 24

1. a) 1.98x10^{6} Nm^{2}/C, b) zero

4. a) -2340 Nm^{2}/C, b) 2340 Nm^{2}/C, c) 0 We can see that every E line entering the vertical surface comes out of the slanted the surface. That’s why parts a and b have the same amount of flux, but one is - and the other one is +.

OQ1. i) a, ii) b.

11. S_{1}: -Q/є _{0} , S_{2}: 0 , S_{3}: -2Q/є _{0} , S_{4} : 0.

CQ2. a) doubled, b) same, c) same, d) same, e) zero.

CQ3. Zero.

CQ6. +.

OQ4. i) c, ii) b. ii) If we move the charge to a point inside the box but extremely close to the center of one face, then half of the E lines coming out of this charge would go through that particular face.

20. (Q + 6q)/(6є _{0}) Because of symmetry, the flux through each of the 6 faces of the box is the same.

OQ6. i) e, ii) a, iii) c, iv) c, v) a

OQ8. C Inside conductor.

CQ7. a) No. b) The person will experience a shock when the charge on his body neutralizes the charges on the inner surface of the metallic sphere.

CQ8. The large net positive charge can induce charge separation on the other conducting sphere: with large net negative charges at the side near the other sphere and slightly more + charges at the far side, so the attractive force between the 2 spheres can be stronger than the repulsive force.

OQ2. c. The outer conducting shell is a Faraday’s cage. The external electric field induces charge separation on the outer surface of the conducting shell, so the electric field inside the shell is not affected.

OQ7. i) c, ii) e.

OQ9. a) A> B > D > C, b) B = D > A > C.

33. ρr/(2є_{0}), radially outward. Use Gauss’s law and choose a cylinder as Gaussian surface.

53. a) σ/(2є_{0}), b) 3σ/(2є_{0}), c) –σ/(2є_{0}) Use Gauss’s law and choose a cylinder (or a prism with any cross-sectional shape) as Gaussian surface.

56. a) 0, b) σ/є_{0} to the right, c) 0, d): a) 2σ/є_{0} to the left, b) 0, c) σ/є_{0} to the right. Use Gauss’s law and choose a cylinder (or a prism with any cross-sectional shape) as Gaussian surface.

59. σ/(2є_{0}), radially outward Superposition: E = E_{sphere }– E_{disk hole }by subtracting vectors. For finding E_{sphere} , we can use Gauss’s law or Coulomb’s law. For finding E_{disk hole} , we can use Gauss’s law and choose a cylinder (or a prism with any cross-sectional shape) as Gaussian surface. When we look for the E by a disk of charge at a location very very close to the disk, we can treat the disk as a very very large sheet of charge (large compared to the distance to the disk). Therefore, we can use Gauss’s law to find the E_{disk hole} .

60. a) 2kλ/r, radially outward, b) 2k(λ + ρπ(r^{2} – a^{2})) /r, radially outward, c) 2k(λ + ρπ(b^{2} – a^{2})) /r, radially outward. Use Gauss’s law and choose cylinders as Gaussian surfaces.

OQ3. e

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AP Physics C — Schedule for Mechanics Unit 6 (Ch. 12, 15, 13)

Ch. 12

OQ7. If use counterclockwise as + direction: D>C>E>B>A. If only compare the magnitude of torque: A>B=D>E>C=0.

11. Sam: 176N upward, Joe: 274N upward Use proportion or: Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

14. a) Horizontal: friction = (m_{1} g/2 + xm_{2} g/L)cotθ towards the wall, vertical: normal force = (m_{1} + m_{2})g upward Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

b) (m_{1}/2 + dm_{2}/L) cotθ /(m_{1}+m_{2}) μ = Friction/ normal force

15. a) 29.9 N, b) 22.2 N Draw force diagram with each force at the point of application and write force equations in x and y directions separately. For part a, you may look at the entire chain or half a chain. For part b) you will need to look at half of a chain.

39. 0.896m Use proportion or: Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

45. a) F_{g}(L+d)/((2L+d)sinθ), b) F_{x} = F_{g}(L+d)cotθ/(2L+d), F_{y} = F_{g}L/(2L+d) Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

49. a) 5080N, b) 4770N, c) 8260N Draw force diagram with each force at the point of application. Choose a convenient axis and write torque and force equations.

62. a) 120N, b) 0.3

c) 103N at an upper corner 31º above the horizontal, so the force is perpendicular to the diagonal of the rectangular cabinet. To minimize the force, we need to maximize the lever arm.

Ch. 15

5 a) 1.5Hz, b) 0.667s, c) 4m, d) π rad, e) 2.83m f = ω/(2π), etc. To find x at t=0.25s, make sure calculator is in radian mode.

8. a) 2.4s, b) 0.417Hz, c) 2.62 rad/s ω = 2πf

14. a) No mechanical energy is lost in elastic collisions, so the ball rebounds to 4m high each time. The motion repeats, so it’s periodic. b) 1.81s The period is the time it takes the ball to go down and then back up.

c) It’s not SHM because the net force on the ball is not in –kx format.

18. a) 1.26s, b) 0.15m/s, 0.75m/s^{2}, c) x(t) = 0.03cos(5t+π), v(t) = –0.15sin(5t+π), a(t) = –0.75cos(5t+π)

24. 0.153J, b) 0.784 m/s, c) 17.5m/s^{2} E = K + U = U_{max at end pt} = K_{max at equili.}, F = -kx = ma, so a_{max} happens at x_{max}.

66. μ_{s}g /(2πf)^{2} Write F_{net} = ma for the top box so you can find a_{max}. Then use a_{max} = ω^{2} x_{max}.

85. a) 0.5 m/s, b) 0.086m a) v at equilibrium is a max. To find v_{max}, we can use energy conservation or v_{max} = ω x_{max}.

b) It takes ¼ of a period for m_{1} to travel from equilibrium to the right end. During that time, m_{2} travels to the right at constant velocity.

OQ2 C, OQ3 A, OQ4 C, OQ5 D, OQ7 C, OQ8 B

OQ 16 i) B, ii) A, iii) C Use g_{apparent} instead of g.

38. I = mgd/(2πf)^{2}

39. 0.499

42. a) 2.09s I = I_{cm} + Mh^{2} where h is the distance between the c.m. of the meterstick and the pivot.

OQ13. D If the system is released from rest, the system’s c.m. stays at rest while the blocks oscillate. In this case, the 2 blocks are identical, so the c.m. is right in the middle. Because the c.m. does not move, we can treat the system as 2 oscillators one at each side of the c.m. Each oscillator is made of half of a spring and a block. Find the frequency of this oscillator - it is the same as the frequency of the original oscillator.

64. a) 2cm, b) 4s, c) π/2 rad/s, d) π cm/s, e) 4.93 cm/s^{2}, x(t) = 2 sin (πt/2) where x is in cm and t is in seconds.

68. a) Use 1/k_{eq} = 1/k_{1}+ 1/k_{2}.

b) The 2 springs are effectively in parallel – to compress one spring, you have to stretch the other spring by the same amount.

Ch. 13

11. 0.614m/s^{2} g = GM/r^{2}, where r is measured from the center of the earth, not from the surface.

12. 2/3 g = GM/r^{2}, where M = ρ•volume

OQ1. C There is one term between every pair of particles.

49. 2.52×10^{7} m It’s a maximum height problem. We can use conservation of energy. Because the projectile will go up very high (no longer considered “near the surface of the earth”), we need to use – GmM/r instead of mgy for gravitational PE.

38. For orbit speed, write F_{net} = ma. For escape speed, use energy conservation.

OQ7. i) E, ii) C, iii) A

OQ11. E Use Kepler’s law of periods: T^{2 }∝ a^{3} and the semi-major axis a = (max distance + min distance)/2.

CQ5. a) perihelion: when closest to the sun, b) aphelion: when farthest to the sun.

28. a) 6.02 × 10^{24} kg Circular orbit: start with F_{net} = ma.

b) Both the Moon and the Earth orbit around the center of mass of the Earth-Moon system. This center of mass is located close to the center of the Earth, but not at the center of the Earth. Therefore, the Moon’s orbit radius is a little less than the distance between the Moon and the Earth.

72. Circular orbit: start with F_{net} = ma.

CQ9. No. Air resistance causes the satellite to lose mechanical energy. Therefore, the satellite’s orbit gets closer and closer to the earth. With smaller orbit radius, the orbit speed increases.

42. GmM/(12R_{E}) WK_{nonconservative} = ∆E

43. a) 8.5× 10^{8} J WK_{nonconservative} = ∆E = ∆U, because the object does not have K up there.

b) 2.71× 10^{9} J WK_{nonconservative} = ∆E = ∆(K + U)

21. 1.26× 10^{32} kg Circular orbit: start with F_{net} = ma.

74. Circular orbit: start with F_{net} = ma.

AP Physics C — Schedule for Mechanics Unit 5 (Ch. 10, 11)

AP Physics C – For Mechanics Unit 5 Homework

Answers and hints for Ch. 10 problems: Try to do the problems without the hints first. It’s more fun that way!!

OQ2. b The centripetal acceleration is not zero.

OQ3. b = e > a = d > c = 0 a_{T} = rα and α is a constant. a_{r} = v^{2}/r = (rω)^{ 2}/r.

3. a) 5 rad, 10 rad/s, 4 rad/s^{2} , b) 53 rad, 22 rad/s, 4 rad/s^{2} ω = dθ/dt, α = dω/dt

17. a) 25 rad/s, b) 39.8 rad/s^{2}, c) 0.628 s ω = v/r. Need to convert rev. to radians.

70. a) ω(t) = ω + At + Bt^{2}/2, b) θ(t) = ωt + At^{2}/2 + Bt^{3}/6 ω = ∫αdt, θ = ∫ωdt .

18. a) 0.605 m/s ω = v/r, need to convert cm to m and rev/min to rad/s.

b) 17.3 rad/s The bike wheel has the same ω as the rear sprocket. The rear sprocket has the same v as the front sprocket because they are connected by a chain.

c) 5.82 m/s v = rω of the rear wheel.

d) 1.39 m/s The length of the pedal cranks being 17.5 cm long.

24. a (√(1 + π^{2} ))/g Friction is the only horizontal force to provide the horizontal acceleration. There are 2 horizontal acceleration components: a_{T} and a_{R}.

OQ6. i) a You want the mass to be close to the axis.

ii) c You want the mass to be far away from the axis.

OQ11. e I ∝ mr^{2}.

39. a) 5.8 kgm^{2} , b) the height of the door If you look at the top view of the door (looking down at the door from above), the door is just like a thin rod rotating about one of its ends.

43. 11 mL^{2}/12 Method 1: You can shift the axis to the center of mass of the 3-rod structure first and then shift the axis back and use the parallel-axis theorem to find the I. Method 2: You can look at one rod at a time. There are 3 rods. Look at one rod at a time: the one on x-axis is a thin rod rotating about one end, the one on y-axis is like a point mass, and the one on z-axis is a little more complicated. We can first find the I_{cm} of the z-axis rod and then shift the axis by L/2 and use the parallel-axis theorem to find the I.

27. clockwise 3.55 Nm

28. clockwise 168 Nm Lever arm is the distance between the line of force and the axis.

OQ7. a When the 50N weight accelerates downward, the tension in the string is less than the weight hung under.

29. 21.5 N Use τ_{net} = I α. The net torque is produced by the torque by T_{u} and T_{lower}. Those 2 tensions produce torques against each other.

46. 13 MR^{2}ω^{2}/24 K_{rot }= ½ I ω^{2} . You have to be careful when you find I. I = I_{cam} + I_{shaft}. I_{shaft} is just the I of a solid uniform cylinder rotating about its center axis. The I_{cam} is the I of a solid uniform cylinder with a cylindrical hole, so I_{cam} = I_{cylinder }– I_{hole}. For the I_{cylinder}, you can find I_{cm}first and then shift the axis from the c.m. to the actual axis using the parallel-axis theorem. Also note that the mass of the cam with hole is M, so the mass without hole is not M (it is 4M/3).

32. a) make sure that you draw each force from the point of application.

b) 0.309 m/s^{2}, c) Horizontal portion: T_{1} = 7.67 N, slanted portion: T_{2} = 9.22 N Write F_{net} = ma for each of the 2 boxes. Write τ_{net} = Iα for the pulley. Replace α with a/R. In order for you to be able to stack the 3 equations together and add them, you will need to divide by R on both sides of the τ_{net} = Iα equation, so you have force in newtons on the left side of the equation, so you can add it to the forces in the other 2 equations. When you add the 3 equations together, the internal forces T_{1} and T_{2} should cancel, so you can find acceleration conveniently.

37. 0.312 Use kinematics to find α and then write τ_{net} = Iα.

OQ9. i) a The basketball also has rotational KE.

ii) c p = mv

iii) a The basketball has more KE to turn into mgy.

OQ12. Less than Just look a small piece of mass (point mass) m attached to a light rod a distance R to the axis. When the rod is released from the horizontal position and swings down, we can use conservation of energy to find the ω at the bottom of the swing: mgR = ½ Iω^{2} = ½ (mR)^{2}ω^{2}, so we get ω^{ }∝ √(1/R), so ω is smaller the bigger the R. If the rod is not light, then we would consider all the mass on the rod that has smaller “R” than the “R” of the mass attached at the end. Therefore, the mass added at the end of the rod makes the ω at the bottom of the swing less than before.

56. Use conservation of energy, the lost mgy of the hanging mass is turned into KE of hanging mass and the KE_{rot} of the spool.

57. a) 2(√(Rg/3)) Lost mgy (y = R) is turned into KE_{rot }of the disk. Be careful that the I of the disk is not I_{cm}.

b) 4(√(Rg/3)) v = rω, and the lowest point is 2R from the axis.

c) √(Rg)

61. a) 2gsinθ/3 The disk does both translational motion and rotational motion, so we need to write both the F_{net} = ma and τ_{net} = Iα and for the disk.

b) hoop: gsinθ/2, c) tanθ/3

78. a) Mg/3, b) 2g/3 The disk does both translational motion and rotational motion, so we need to write both the F_{net} = ma and τ_{net} = Iα and for the disk.

b) – 2/3 fractional change = change/ initial value = (K_{f} – K_{i})/K_{i}

c) (√(4gh/3)) The disk does constant acceleration motion, so you may use kinematics equations to find the final speed. Or, you may use conservation of energy to find it: lost mgy = KE_{translational} + KE_{rotational}.

81. (√(10(R-r)(1 – cosθ)/(7r^{2}))) Use conservation of energy. Be careful when you look for the height change. The radius of the circular motion of the center of mass of the sphere is (R-r).

82. a), b) The disk does both translational motion and rotational motion, so we need to write both the F_{net} = ma and τ_{net} = Iα and for the disk.

c) (√(8Fd/(3M)) Use kinematics equations.

83. a) 2.7 R First use forces to find the minimum speed for the sphere to make the loop at the top of the circle, then use energy conservation to find the starting height.

b) 20mg/7 to the left and 5mg/7 upward We have to look at both the a_{c} and a_{T}. The normal force provides the ma_{c} = mv^{2}/R. We can use energy conservation to find the speed of the sphere at P. (v_{P} = (√(20gR/7))) The a_{T }is provided by the upward friction and downward mg. So write the force equation for a_{T }as well as the τ_{net} = Iα for the sphere, use the 2 equations to find the a_{T}. m times a_{T }gives use the net force in the vertical direction.

73. a) (√(3g/L)) Use conservation of energy. Be careful when you look for the height change.

b) 3g/2L The rod is rotating about a fixed axis, so there is only rotational motion about that axis. There is no translational motion. So we only have to write τ_{net} = Iα. Be careful that the I is not the I_{cm}.

c) –(3g/2) **i** – (3g/4) **j** We have to look at both the a_{c} and a_{T}.

d) –(3Mg/2) **i** + (Mg/4) **j ** If we look at the entire rod, there are only 2 external forces acting on the rod: Mg and the force from the pivot. Together these 2 forces provide the acceleration we found in part c).

79. (√((2mgdsinθ + kd^{2})/(I + mR^{2}))) Use conservation of energy. Height change is dsinθ.

85. a) (√(3gh/4)) Use conservation of energy. The rod does not have a fix axis for rotation, so it’s easier if we treat the motion of the rod as: rotation about its center of mass and translational motion of the center of mass. So lost mgy = KE_{translational} + KE_{rotational}.

b) (√(3gh/4)) Use conservation of energy. This time the rod rotates about a fixed axis at one of its ends, so it’s easier if we treat the motion of the rod as: only rotation about its end and no translational motion. So lost mgy = KE_{rotational}. Be careful that the I in this case is not the I_{cm}.

92. a) Use conservation of energy. The lost E is friction times d.

b) 2Mg(sinθ – μcosθ)/(2M + m) Since it is constant acceleration motion, we can use kinematics equations to find a.

Ch. 11

10. a) No, b) **B** x **A** is perpendicular to both **B** and **A**. When 2 vectors are perpendicular to each other, their dot-product is zero. So we can find the dot product of (2, -3, 4) and (4, 3, -1). Since this dot-product is not zero, they are not perpendicular to each other.

12. – 22**k** kgm^{2}/s For a point mass, L = mvr_{┴}. And we can use right-hand-rule to find the direction for L. Or, we can use the formula in section 11.1.

13. (mxv_{y} – myv_{x})**k** For a point mass, L = mvr_{┴}. And we can use right-hand-rule to find the direction for L. Or, we can use the formula in section 11.1.

29. a) 1.57 x 10^{8} kgm^{2} We want the centripetal acceleration to equal to g. This will give us the final ω. Then we can use L = Iω.

b) 6260 s We can use τ_{ave} ∆t = ∆L.

OQ1. b L = constant

OQ8. d L = constant if τ_{net} = 0, p = constant if F_{net} = 0.

CQ3. To increase the moment of inertia of the walker, so it’s harder to change the walker’s state of rotational motion.

CQ6. The net torque (produced by gravitational force) on the c.m. of the “rider + motorcycle” system is zero, so the system’s angular momentum is constant. so if the wheels gain angular momentum in one direction, the rest of the system must gain angular momentum in the opposite direction to keep the total L in the system constant.

CQ8. It can still have mvr_{┴}, if r_{┴ }is not zero.

CQ9. a) I would increase, but only very slightly. b) Since L = constant, increase in I means decrease in ω. Therefore, the revolution period would increase, but only by a small fraction of a second.

30. a) I_{1}ω_{1}/ (I_{1 }+ I_{2}) It’s a collision problem with a fixed axle, so the linear momentum is not conserved; however, the angular momentum is conserved.

b) I_{1}/(I_{1 }+ I_{2})

31. 7.14 rev/min It’s a collision problem with a fixed axle, so the linear momentum is not conserved; however, the angular momentum is conserved, so: L_{i} = L_{f}. We can treat the child as a point mass.

33. a) No. Some calories (chemical PE) from the woman is turned into mechanical energy.

b) No. The turntable axle’s force and mg on the system do not cancel, so the net force on the system is not zero.

c) Yes. The net torque on the system is zero.

d) 0.36 rad/s counterclockwise. L_{i} = L_{f}.

e) 99.9 J It’s the KE gain.

39. a) 2mv_{i} d/((M + 2m)R^{2}) L_{i} = L_{f}

b) No, it’s a completely inelastic collision. Some KE is lost to heat in this case.

c) No. The fixed axle’s force does not cancel with mg. So the net force on the system is not zero.

49. a) 7md^{2}/3 Find I for each particle and then add all the I together.

b) mgd counterclockwise. There are three mg providing torque. Add the torque together to get net torque – remember that torque is a vector. Or: we can use the c.m. to represent the gravitational force of all 3 pieces of mass and find the torque produced by that gravitational force.

c) 3g/(7d) Use τ_{net} = Iα.

d) 2g/7 upward a_{T} = rα and a_{c} = v^{2}/r.

e) mgd That’s when the system has the lowest gravitational PE (when c.m. is the lowest), and the lost PE at that moment is the KE_{max}.

f) √(6g/(7d)) KE_{max} = KE_{rot} because the system is rotating about a fixed axis. There is no KE_{translational}.

53. a) r_{i}v_{i}/r L_{i} = L_{f}.

b) m(r_{i}^{2}v_{i}^{2})/r^{3} Tension provides the centripetal force for the puck.

c) ½ mv_{i}^{2}((r_{i}^{2}/r^{2}) – 1)) The work is turned into KE gain.

56. a) Mvd L_{total} = L_{1} + L_{2}

b) Mv^{2} K_{total} = K_{1} + K_{2}

c) Mvd L_{i} = L_{f}.

d) 2v L_{i} = L_{f}.

e) 4Mv^{2}, f) 3Mv^{2} The calories used is converted into the KE gain.

61. a) ω_{i }/3 That’s when v_{f} = Rω_{f}. Write force equation to find a and torque equation to find α. Because both a and α are constants, we can use kinematics equations to find the v_{f} and ω_{f}. Set v_{f} = Rω_{f} and find ω_{f}. (Note: When you write these equations, you can include μ. You will find that μ will cancel, so it does not show up in your answer for ω_{f }.)

c) R ω_{i} /(9μg) The disk starts with v_{i} = 0, so ∆v = v_{f} - v_{i} = Rω_{f} . And F_{ave net} ∆t = ∆p = m ∆v, and F_{ave net} = F_{fr} = μmg.

d) R^{2} ω_{i}^{2} /(18μg) Can use kinematics equations.

63. 4(√(ga((√2) – 1)/3)) This problem involves 2 parts: a collision (L_{i} = L_{f}) and then the cube “swings up” about its lower right edge. So we have to do it as a 2-part problem: first the cube’s L_{i} = L_{f}, then we have conservation of energy for the swinging part. In order for the block to “swing” over the edge, the block’s c.m. has to get beyond the edge of the table – i.e. rising to a height that is half of the square’s diagonal – a height = a√2. Using formula on p. 304 and the parallel-axis-theorem, you should find that the I of the cube about one edge to be 8Ma^{2}/3. (If you have trouble finding the correct I, you may wish to just use I = 8Ma^{2}/3 to work on the problem.)

AP Physics C — Schedule for Mechanics Unit 4 (Ch. 9)

Answers and hints for **Ch. 10 problems**:

OQ2. b The centripetal acceleration is not zero.

OQ3. b = e > a = d > c = 0 a_{T} = rα and α is a constant. a_{r} = v^{2}/r = (rω)^{ 2}/r.

3. a) 5 rad, 10 rad/s, 4 rad/s^{2} , b) 53 rad, 22 rad/s, 4 rad/s^{2} ω = dθ/dt, α = dω/dt

17. a) 25 rad/s, b) 39.8 rad/s^{2}, c) 0.628 s ω = v/r. Need to convert rev. to radians.

70. a) ω(t) = ω + At + Bt^{2}/2, b) θ(t) = ωt + At^{2}/2 + Bt^{3}/6 ω = ∫αdt, θ = ∫ωdt .

Answers and hints for Ch. 9 problems:

48. (11.7cm, 13.3cm) Because we can easily find the location of c.m. of a uniform rectangle, it can be convenient for us to divide this sheet metal into 3 rectangles. For each rectangle, we can use the c.m. of the rectangle as the location for this piece and use the area to represent the mass of this piece. Then we can find the center of mass of the 3 pieces combined using the center of mass equation. It’s a 2-d problem, so do the x and y separately.

50. 0.00673nm below the center of the oxygen Because the molecule is symmetric left and right to the dotted line in Fig. P9.50, we know that the c.m. has to be on the dotted line. Therefore, we can treat this problem as a 1-d problem with 2 pieces of mass: the oxygen (mass 16u) at the top and the combination of the 2 hydrogen (mass 2u) at the mid-point between the 2 H-atoms. Then we can just use proportion to find the location of the c.m. of the H_{2}O.

53. 0.7m If the friction between the boat and the water is negligible, the net external force on the system is zero, and the c.m. of the system should stay at rest the entire time. So, M∆**X**_{c.m.}= m_{1}∆**x**_{1} + m_{2}∆**x**_{2} = m_{1}∆**x**_{1 rel to 2} + (m_{1} + m_{2})_{ }∆**x**_{2} = 0. Use m_{1} for Juliet and m_{2} for (Romeo + boat). For “m_{1}∆**x**_{1 rel to 2} + (m_{1} + m_{2})_{ }∆**x**_{2}” part, because we have to consider that the boat carries Juliet with it, hence the (m_{1} + m_{2}).

OQ1. b>e>a=d>c The Frisbee “collides” with you on frictionless surface, so p_{i} = p_{f}, but that’s only true in the horizontal direction. So focus on the horizontal motion. The more horizontal velocity (and therefore, momentum) the Frisbee keeps in its original direction, the less momentum (and therefore, speed) you get in that direction. Note: For d, I guess the authors meant to say that the Frisbee ends up going vertically upward as viewed by an observer on the ground.

OQ2. a) No. It’s completely inelastic collision. b) No. When cart #1 rolled down, the net force on the boxcar is not zero. c) Yes because there is no friction. d) Yes, because the net force on the boxcar-Earth system is zero. e) No. See part a. f) Yes. See part c.

CQ4. a) No. b) No. The ball cannot end with negative KE. c) Yes. Momentum is a vector, so the ball can end with negative p (e.g. if it turns around).

3. 7N **F**_{net} = d**p**/dt or Impulse by net force = Δ**p = F**_{ave net} Δt

5. 3263N toward homeplate and 3988N downward Look at the ball and use Impulse by net force on the ball: **J** = Δ**p = F**_{ave net} Δt. The force on the bat is the same amount as but in opposite direction to the force on the ball. In this case, we can ignore the mg of the ball because the force from the bat is so much stronger.

7. a) *v _{pi} *= −

*m*+

_{g}v_{gp}/(m_{g}*m*p = constant, so

_{p})*p*because we have to consider that the plank carries the girl with it, hence the

_{i}= 0 = p_{f}= m_{g}v_{gp}+ (m_{g}+ m_{p}) v_{pi}*(m*.

_{g}+ m_{p})

b) *v _{gp}+*

*v*=

_{pi}*v*−

_{gp}*m*+

_{g}v_{gp}/(m_{g}*m*The girl gets two velocities: one by herself relative to the plank, the other by the plank. Therefore: her net velocity relative to the ice is

_{p})*v*

_{gi}= v_{gp}+*v*.

_{pi}11. a) 6m/s to the left Use *p _{i} = p_{f}*.

b) 8.4 J Energy is conserved: energy stored in the spring is turned into the KE of the 2 blocks.

c) spring , d) The cord is burned before the blocks begin to move, therefore the cord’s forces do not have any displacement, so the cord’s forces do not do any work on the boxes. I.e. no energy is transferred from the cord to the boxes.

e) Yes. Because the net force on the whole system (consisting the 2 boxes) is zero, and that is the condition for momentum conservation. The spring’s forces on the two boxes are equal and opposite, so the net force by the spring on the 2-box system is zero.

15. a) x√(k/m) Use conservation of energy.

b) x√(km) Impulse by net force = Δ**p.**

c) No. Same work no matter what mass. Use conservation of energy.

17. a) 0.096 s Use kinematics equations for constant acceleration motion.

b) 365000N Impulse by net force = Δ**p = F**_{ave net} Δt

c) 265 m/s^{2} = 26.5g, d) No. The 11-kg toddler will seem to weigh 11×26.5 = 296 kg = 641 lb.

19. a) 12**i **Ns Impulse by net force = Δ**p = F**_{ave net} Δt = area of **F**_{net} as a function of time graph.

b) 4.8**i** m/s, c) 2.8**i** m/s Impulse by net force = Δ**p = **m Δ**v**.

d) 2.4**i** N Impulse by net force = Δ**p = F**_{ave net} Δt

20. a) 981 Ns, up Impulse by a force **= F** Δt = area of **F** as a function of time graph = ∫**F** dt

b) 3.43 m/s, down Use kinematics equations for constant acceleration or conservation of energy.

c) 3.83 m/s, up Impulse by net force **= **Impulse by (m**g** + **F**_{on her by platform}) = Δ**p**

d) 0.748 m Use kinematics equations for constant acceleration or conservation of energy.

21. 16.5 N The scale reads the weight of (the bucket + water inside) + the force related to the impulse of falling water. For the falling water part, consider **F**_{ave net} = Δ**p/**Δt = m Δ**v/**Δt = (m/Δt)Δ**v**.

61. 15 N The additional force is the force related to the impulse of the shooting water.

67. -260**i** N It’s 2-dim case, so we need to look at one dimension at a time. There is only momentum change in the x direction, so the force is only in the x-direction. Find components of **v _{i} **and

**v**

_{f}in the x-direction. Then do

**F**

_{ave net x}= Δ

**p**Δt.

_{x}/OQ3. i) c Newton’s 3^{rd} law of motion.

ii) a K = ½ mv^{2} = p^{2}/(2m) ∝ (1/m)

OQ7. a K = ½ mv^{2} = p^{2}/(2m) ∝ p^{2}

OQ8. d K = ½ mv^{2} = p^{2}/(2m) ∝ p^{2}

OQ11. b First use momentum conservation to find the speed of the block immediately after the impact. Then the KE is lost to friction.

OQ13. a Same F and same d, so the work done by the force are the same, so the ∆K are the same. Both particles start from rest, so K_{f} are the same.

OQ14. d Both the fall and bounce are interaction between the ball and the earth. The net external force on the ball + earth system is zero during the fall and the bounce.

CQ11. No **F**_{ave net} Δt = Δ**p**

CQ12. No WK_{Fnet} = ΔK

29. a) 4.85 m/s Kinematics or conservation of energy.

b) 8.41 m The basketball rebounds at the same speed and collides elastically with the tennis ball coming down toward it at the same speed. Find the final velocity of the tennis ball and then find the max. height.

30. (4M√(gL))/m Because the M is attached to a stiff rod (not a string), barely making the loop means the speed of the M at the top of the loop is zero. Then use conservation of energy to find the speed of the M at the bottom of the loop after the bullet has emerged. Then use conservation of momentum for the collision part to find the speed of the bullet before collision.

32. ((m+M)√(2μgd))/m For the collision part: p_{i} = p_{f}. For the sliding part: lost KE is taken away by friction.

33. 0.556 m For the sliding part: use conservation of energy to find the speed of m_{1} just before the collision. The blocks never touch à the collision is elastic. So use v_{approaching} = v_{separating} and p_{i} = p_{f} for the collision part. Find the velocity of m_{1} at the end of the collision and then use conservation of energy to find how high it rises.

34. a) 2.24 m/s to the right Use p_{i} = p_{f}. All 3 carts have the same velocity at the end.

b) No.

73. v_{m} = 2v_{i}, v_{3m} = 0 Use v_{approaching} = v_{separating} and p_{i} = p_{f} for the elastic collision.

35. a) v_{x} = 0.929 m/s, v_{y} = 0.53 m/s 2-dim case, so separate the x and y. Find the final v_{x} and v_{y} of the 0.2-kg puck. Then write p_{i} = p_{f} for x and y directions separately to find the final v_{x} and v_{y} of the 0.3-kg puck.

37. (3**i **– 1.2**j**) m/s Write p_{i} = p_{f} for x and y directions separately to find the final v_{x} and v_{y}.

57. a) The cart “suddenly” comes to a stop, so the top of the cart suddenly stops. However, the particle still has the v_{i} at that moment. So the particle swings up: use conservation of energy for this part.

b) at the lowest point The connection between the cart and the particle is the string. A string’s tension is along the direction of the string. So the tension has no horizontal component when the string is vertical.

69. a) 1.33 m/s to the right Use p_{i} = p_{f}.

b) 235 N to the left F_{fr} = μ_{k} F_{N}

c) 0.68 s Impulse by net force** = F**_{ave net} Δt = Δ**p = **m Δ**v**. In this case, the net force changing the person’s momentum is the friction.

d) Person: 160Ns to the left, cart: 160 Ns to the right.

e) 1.82 m For the person: F_{net} = F_{fr} = 235 = ma = 60a. We can find a (to the left), and we know v_{i} = 4 and v_{f} = 1.33, so we can find ∆x.

f) 0.454 m For the cart: F_{net} = F_{fr} = 235 = ma = 120a. We can find a (to the right), and we know v_{i} =0 and v_{f} = 1.33, so we can find ∆x.

g) -427 J, h) 107 J, i) (427 – 107) J is lost to friction.

72. a) There are 2 parts in this event: the collision and then the rise to maximum height. We can use conservation of momentum for the collision part and then conservation of energy for the rise to maximum height part.

b) ((m+M)√(2gh))/m

75. a) (m_{1}v_{1} + m_{2}v_{2})/( m_{1} + m_{2}) This problem shows a classic way to model elastic collision. During the collision, some KE is turned into the PE stored in the spring. At the end of the collision, the spring is relaxed again. All the energy stored in the spring during the collision is back in KE. So KE_{i} = KE_{f} à it’s an elastic collision. During this collision, when the spring has the maximum compression, the relative velocity between the 2 blocks is zero. Therefore at that moment the 2 blocks have the same velocity – kind of like the 2 blocks are stuck together at that moment.

80. a) 0.667 m/s to the left It’s like an explosion problem: one piece, the block and wedge combination) breaks into 2 pieces. So we can use p_{i} = p_{f}, although the momentum is only conserved in the horizontal direction.

b) 0.952 m The block and wedge has no KE at the beginning. The KE they gain at the end comes from the lost mgy of the block.

82. ((M+m)√(gd^{2}/(2h)))/m We know information at the end, so we can go backwards: Use projectile motion kinematics to find the speed of the (m+M) at the end of the collion. Then we can use p_{i} = p_{f} to find the initial speed of the bullet.

85. 0.403 They both swing down from the same height, so they meet at the lowest point with horizontal velocities of the same magnitude but in opposite directions. They collide and stick together, so we can use p_{i} = p_{f} to find their velocity immediately after the collision. Then we can use conservation of energy for their swing up part.

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**Physics is Phun Demonstration Lecture: Celebrating Light" **is geared toward a high school audience and aim to both educate and entertain! Friday, October 16th @ 7:00 PM to about 8:30PM, Saturday, October 17th @ 7:00 PM to about 8:30PM (The same lecture on both days.) To earn bonus points, a student should turn in his/her stamped notes on the lecture before Tuesday, Oct. 20.

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AP Physics C — Schedule for Mechanics Unit 3 (Ch. 7, 8)

Hints for Ch. 7 Try to do the problems without the hints first. It’s more fun that way!!

1. a) 1590 J WK=Fd cos<

b) 1590 J Since friction stays the same, new F still has the same horizontal (parallel to d) component as before.

2. a) 0.0328 J WK=Fd cos<

b) -0.0328 J a = 0.

OQ 5. a>b=e>d>c cos<

CQ4. a) 90º ≥ θ > 0, b) 180º ≥ θ > 90º cos<

9. 16 For **A**-**B**, do each component separately. For scalar product, remember that cos 90º = 0

11. a) 16 J Work is a scalar product.

14. a) 24 J, b) -3 J, c) 21 J Area.

23. a) When one tray is removed, the force compressing the springs is reduces by the weight of one tray. If the springs are Hookes law springs, F = kx means ∆F = ∆(kx) = k ∆x. Same ∆F gives the same ∆x. We just have to make the ∆x for the weight of one tray equal to the thickness of the tray.

b) 316 N/m There are 4 springs in parallel supporting the trays.

c) 45.3cm by 35.6cm.

28. a) 9000J WK = ∫Fdx

OQ4. C WK = Fdcos<

CQ3. Sometimes. It’s true if the initial K is zero.

CQ8. a) Not necessarily. If the force does no work, no K change, ie. no speed change. (WK = 0 when F ┴ d, ie. F ┴ v, because v is always tangent to the path. So if F ┴ v, this F only provides centripetal (or radial) acceleration. No tangential acceleration means no speed change. )

b) Yes. F = ma. If F is not zero, a is not zero, so velocity has to change.

38. a) 23400N, opposite to the bullet’s v WK_{Fnet} = F_{net} d cos< = ∆K

b) 0.000191s Can use kinematics.

25. a) Draw force diagram and write force equation. Because the particle is being pulled at constant speed, there is no tangential acceleration. There is only centripetal acceleration.

b) mgR Use conservation of energy.

45. a) 125 J For the first segment of the path, only the x-component of the force does work. For the 2nd segment, only y-component of the force does work.

b), c) 125J

47. A/(r^{2}) F = -dU/dr

49. F_{x} = 7 – 9x^{2}y, F_{y} = - 3x^{3 }F_{x } = -∂U/∂x, F_{y } = -∂U/∂y

50. a) Ax^{2}/2 – Bx^{3}/3 U = - ∫Fdx

b) 2.5A – 6.33 B, c) - 2.5A + 6.33 B There is no non-conservative forces doing work, so E = constant, so ∆E = ∆K + ∆U = 0

52. a) A: 0, B: +, C: 0, D: -, E: 0 F = -dU/dx = - slope

b) A, E: unstable, C: stable, c)

OQ14. d U = ½ kx^{2} ∝ x^{2} , when x à2, U à4, so the extra U we need to provide is 4 – 1 = 3 times the old value.

CQ11. 2k When you pull on a spring with a force F, the entire spring has the same tension F. If a force F stretches the original spring by x, the same F would stretch half of the spring by ½ x.

59. 0.299 m/s Lost K turns into U in the spring. U also equals to the energy given to the spring by the car, i.e. the work done on the spring by the car. And WK = area of the F as a function of x graph. Or we can also find U by finding the energy stored in both springs.

60. The ball can only go up the incline by 0.881 m along the incline. That is only about half of the height of a professional basketball player. Need to convert k from N/cm to N/m and convert d from cm to m. E_{i }= ½ k x^{2} = mgy_{f} = mg (distance • sin θ)

64. E_{i }= ½ mv^{2} + mg (d + x) sin θ = E_{f} = ½ kx^{2}

Hints for Ch. 8

OQ1. a E_{i} = ½ kx^{2} = E_{f} = ½ mv^{2} , since you pull the slingshot for the same amount, E_{i }is the same for the pebble and the bean. This means mv^{2}à 1.

8. a) √(2(m_{1} – m_{2})gh/(m_{1} + m_{2})) Write E_{i} = E_{f}. Notice that one mass has mgy at the beginning and the other has mgy at the end. At the end, both boxes move at the same speed.

b) 2m_{1}h/(m_{1} + m_{2}) After m_{1} strikes the floor, m_{2} still has an upward velocity, so it keeps going up to max. h. before coming down.

11. √(8gh/15) The blocks have equal mass, and A is pulled up by 1T while B is pulled up by 2T. Therefore, block A will go down and B go up. Also, when A goes down by 1m, B goes up by ½ m causing their separation to be 1.5 meters. This means a separation of h means A goes down by 2h/3 and B goes up by h/3. Write E_{i} = E_{f}.

24. a) 0.381 m Write E_{i} = E_{f}. Note that the initial mgy has a y that is (1.2 + x) above the final position. You will get a quadratic equation from E_{i}= E_{f}. This means you will get 2 solutions for x, one +, one –. The answer is the + solution, of course. But the – solution also has a special meaning. What do you think it means?

13. v^{2}/(2μ_{k}g) ∆E = WK_{nonconservative force} or lost E = amount of energy take away by friction.

15. a) 0.791 m/s Write E_{i} = E_{f}.

b) 0.531 m/s ∆E = WK_{nonconservative force} or lost E = amount of energy take away by friction.

33. $145 (Power in kW) (time in h) = (energy used in kWh)

41. a) 10200 W P = WK/t = Fv, in this case, a = 0, so F = tension in cable = mg sinθ.

b) 10600 W In this case, T – mg sinθ = ma.

c) 5820000 J It’s easier to use conservation of energy.

45. h + d^{2}/(4h) Since the information at the end is given, we can work backwards to find the initial height H. First do the projectile part to find the speed at which the child goes off the slide. Then use conservation of energy to find H.

46. a) 2.49 m/s It’s convenient to use conservation of energy to find the speed of the blocks the moment m_{2} strikes the floor. After m_{2} strikes the floor, m_{1 }continues to move along the table at constant velocity until it goes off the table.

b) 5.45 m/s Can use conservation of energy to find landing speed. Look at only m_{1} and start from the moment it goes off the table to the moment before it lands.

c) 1.23 m It would be the ∆x of m_{1}’s projectile motion.

d) No, e) m_{2} loses its K when it lands.

47. a) K = 2 + 24t^{2} + 72t^{4} v = dx/dt

b) a = 12t, F_{net }= 48t a = dv/dt

c) P = 48t + 288t^{3} P = Fv

d) WK = 1250 J Can do WK = ∫Pdt, but it’s easier to use conservation of energy: WK = ∆K.

64. 1.24 m/s Write E_{i} = E_{f}. Note that block m_{2} will go down by h = 0.2 m, while m_{1} will have a vertical height increase of h sinθ. The 2 blocks have the same speed.

68. a) E_{i} = E_{f} because the peg’s force on the string does no work.

b) In order for the ball to complete the circle, the ball has to have at least a minimum speed that can be found if you write the force equation at the top of the circle. Then write E_{i} = E_{f}. v^{2} cannot be negative, so d has to be at least 3L/5.

84. a) For this problem, in order for 3m to be the correct answer, I think you will have to make the corner of the table frictionless – kind of like having a frictionless pulley at the corner for the chain to hang down from it, so the hanging part of the chain does not contribute to the normal force on the part of the table with friction. Then: Let the linear mass density of the chain be λ. I.e. every meter of the chain has a mass of λ. λ = Mass/Length. If the amount hanging over the edge has a length x, the mass of this hanging part would be x λ. OR: You may say the total mass of the chain is M and the hanging part has a mass that is a fraction x/L out of M.

University of Maryland College Park, Department of Materials Science and Engineering:

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AP Physics C — Schedule for Mechanics Unit 2 (Ch. 5, 6)

Answers and hints for Ch. 5:

CQ13. The scale reading oscillates up and down – more than (m_{weightlifter }g + m_{barbell} g) when his arms + barbell have an upward acceleration and less than (m_{weightlifter }g + m_{barbell} g) when his arms + barbell have a downward acceleration. When he throws the upward, the scale would read more than (m_{weightlifter }g + m_{barbell} g) when he accelerates the barbell upward. Once he lets go of the barbell, the scale would just read his weight.

CQ14. The sack moves up with the athlete whether the athlete is speeding up or slowing down. If you draw a force diagram for each of them, you will see that you have exactly the same force diagrams and same amount of forces acting on them. So they have the same a = F_{net}/m. They also both start from rest.

CQ22. a) Both students slide toward each other with B experiencing twice the force of A. b) Same as a). c) Same as a). d) Both students slide toward each other with B experiencing the force as A.

28. (If use g = 10 m/s^{2}) a) 50N, b) 50N, c) 100N, d) 25N

32. 112N Find accelerations x- and y- components. Then find the magnitude of the acceleration.

51. (If use g = 10 m/s^{2}) a) 720N, b) 828N, c) 720N, d) 662N

4. 8.71N x-components cancel.

26. a) 3 forces: one vertical (mg), one horizontal (magnetic force), one slanted (T) b) 1.03N, c) 0.805N to the right. Find components of the slanted force. Then write F_{net} = ma for x- and y-directions separately.

86. a) T = f/(2sinθ), b) 410N Draw forces for the midpoint of the cable. Take advantage of the symmetry: i.e. make one of your axis along f. Find components of the slanted force. Then write F_{net} = ma for x- and y-directions separately.

29. a) 7m/s^{2} to the right, b) 21N, c) 14N a) Can look all 3 boxes as one system. b) Can look at 1kg + 2kg combined. c) Can look at the 2kg box.

78. a) Note: In this problem the sagging of the rope is ignored. So even though there is no vertical force to cancel with the mg of the rope, the rope is drawn straight and horizontal. b) a = F/(m_{b} + m_{r}), c) T = F m_{b} /(m_{b} + m_{r}), d) For a massless string, the tension would be the same throughout.

42. b) 3.57 m/s^{2}, c) 26.7 N, d) 7.14 m/s. For m_{2}, only need to draw forces along the incline.

45. (If use g = 10 m/s^{2}) a) 20N, b) If F_{x} is a 80N pushing force to the left (or: F_{x} = - 80N), c) It’s a straight line with + slope for -80N 10) to 100N. (a = (F_{x} – 20)/10). For -100N to -80N, the line bents at -80N, the equation becomes: a = F_{x}/8.) b) The tension becomes zero when m_{1} is in free fall.

46. a) a_{2} = 2a_{1}, b) T_{1} = 2 m_{1}m_{2} g/(4m_{2} + m_{1}), T_{2} = m_{1}m_{2} g/(4m_{2} + m_{1}), and c) a_{1} = m_{1} g/(4 m_{2} + m_{1}), a_{2} = 2m_{1} g/(4 m_{2} + m_{1}) a) When P_{2} moves 1 m, m_{2}moves by 2 meters. b) Draw force diagram and write force equation for each box. Also T_{1} = 2T_{2}.

OQ1. d) Max friction also doubles.

OQ5. b) F_{net} = ma

OQ8. d) Can use F_{net} = ma and kinematics or use work-energy theorem.

OQ12. d) F_{net} = ma = 0

OQ13. a), c), d)

52. a) 14.7m, b) 10000kg and 20000kg

57. μ_{s} = 0.727, μ_{k} = 0.577 Look at forces along the incline.

58. a) 4.18, b) Applying any higher engine power would only make the car skid and therefore turning the friction to the smaller kinetic friction. If the driver keeps at that higher engine power, the wheels will spin faster (while skidding), lifting front end more and perhaps flip over.

60. a) The free-body diagram should include 4 forces. b) 55.2° c) 167N Find components for the slanted force and write one F_{net} = ma for each of the x and y directions.

66. a) between 31.7 N and 48.6 N. b) If P is too big, the block will slide up the wall. If P is too small, the block will slide down the wall. Draw force diagram – considering the 2 conditions described in answer for part b) separately. Find components for the slanted force and write one F_{net} = ma for each of the x and y directions.

69. 0.06 m Since we want to know the motion of the mug, let’s draw force and write force equation for the mug, so we can find the mug’s acceleration. Because the mug move during the pull of the tablecloth, the tablecloth has to move more than 30 cm relative to the table. However, we know how far the tablecloth has to move relative to the mug (30cm), we can use that to find out how much time it takes by looking at the relative motion between the mug and the tablecloth. We can then use this time to find the distance traveled by the mug within that time.

70. a) There are 4 forces acting on the 5kg and 6 forces on the 10kg. b) T = 9.8 N, a = 0.58 m/s^{2}. Since the 2 blocks have different accelerations, we should draw force and write force equation for the blocks separately.

81. a) b) 0.408 m/s^{2}, c) 83.3 N For b), look at Nick and the chair as one system. c) To find the force between Nick and the chair, we would have to look at Nick and the chair separately. Since there is only one unknown we want to find, we can choose to look at either Nick or the chair.

82. a) Since the child is lighter than Nick + chair, the child will acceleration upward. a = 0.426 m/s^{2} and T = 459 N. b) In this case, a = 0 and T = the weight of Nick + chair = 480N. This is more than 250N in 81 and the 459N in part a). a) This scenario is like the Atwood machine – a pulley with one mass hanging on each side.

85. a) b) T_{1} = T_{2} = T_{3} = Mg/2, T_{4} = 3Mg/2, and T_{5} = Mg, c) Mg/2 a = 0 for every object.

91. a) The cushion starts from rest and the acceleration is a slanted constant acceleration (downward g and a horizontal constant amount too), so the cushion will accelerate along the straight line that is slanted like the constant acceleration. b) No. See answer for a). c) 1.63 m from the base of the building. d) Parabola c) We can still look at the forces and kinematics one direction at a time – vertical and horizontal directions separately. d) The object starts with an initial velocity that is in a different direction than its constant acceleration (like a projectile), so its path is parabolic. The parabola is tilted.

92. a) a_{1} = 2a_{2} because when m_{2} goes down by 1 meter, the string has to increase by 1 m on each side of the pulley above m_{2}, so m_{1} has to move by 2 meters. b) a_{2} = 12.7N/(1.3kg + 4m_{1}) downward. c) a_{2} would be almost g downward: i.e. m_{2} would be almost like a falling object. d) a_{2} would approach 0, because the tension on m_{1} is a limited force. If m_{1} approaches infinity, the acceleration will have to approach to zero. e) 6.37N, f) Yes. See previous answers.

93. (M + m_{1} + m_{2}) m_{1} g/m_{2} m_{1} has no vertical acceleration. This is the information we can use to find the tension in the string. Knowing the tension, we will be able to find the acceleration of m_{2}. Then we can look at the whole 3-box system to find F.

Ch. 6

OQ1. a) A>C=D>B=E=0, b) A: north, B: west, C: south, c) A: west, B: nonexistent, C: east.

OQ4. a) Yes: A: to the right and downward: perpendicular to the string. b) Yes, A. See answer to part a) for acceleration. c) No. d) Yes, B: It’s to the right between (but not including) the direction along the string and the direction of the rightward tangent.

OQ5. b)

OQ6. a) No, because v = 0: turning point. b) Yes.

OQ7. i) c), ii) b): the angle has to do with acceleration (not velocity).

CQ4. a) The object moves in a circle at constant speed. b) The objet moves along a straight line at changing speed. a) Only radial acceleration and no tangential acceleration. b) Only tangential acceleration and no radial acceleration.

9. a) static friction, b) 0.085 Write force equation for the horizontal direction.

16. a) 24900N, b) 12.1 m/s b) almost lose contact, so normal force is zero.

42. Draw force diagram and write force equation: The acceleration is horizontal. It’s more convenient to choose an upright coordinate system. Find components for any forces that are slanted in your coordinate system.

54. a) m_{2}g, b) m_{2}g, c) √(m_{2}gR/m_{1}), d) m_{2} will go down and the radius of the puck’s circular motion will decrease. Because of angular momentum conservation, the puck will also speed up. e) m_{2} will go up and the radius of the puck’s circular motion will increase. Because of angular momentum conservation, the puck will also slow down.

63. 12.8 N Draw force diagram and write force equation: The acceleration is horizontal. In this case, either upright coordinate system or the slanted one will give us 2 slanted vectors in a coordinate system. So for either coordinate system you choose, you will have to find components for 2 vectors.

64. a) 2.63 m/s^{2}, b) 201 m, c) 17.7 m/s This problem is just like prob. 42. Draw force diagram and write force equation: The acceleration is horizontal. It’s more convenient to choose an upright coordinate system. Find components for any forces that are slanted in your coordinate system.

20. a) 3.6 m/s^{2} to the right, b) T = 0, c) Both observers in parts c and d will see the same mg and normal force. An observer in the car may think that in addition to the tension read by the spring scale there must also be a (fictitious) force of Ma to left acting on the m. d) An observer on the ground would think that the tension read by the spring scale is the only horizontal force acting on the m.

21. a) 17°, b) 5.12 N This problem is just like prob. 42 and 64. Draw force diagram and write force equation: The acceleration is horizontal. It’s more convenient to choose an upright coordinate system. Find components for any forces that are slanted in your coordinate system.

24. 2(vt – L)/[(g + a) t^{2}] Use kinematics and forces. Because of the acceleration of the elevator, the normal force on the backpack is not the same as its mg.

51. a = g(cosϕtanθ – sinϕ) Draw force diagram and write force equation. It can be convenient to choose your coordinate system tilted in the same way as the acceleration. You will have 2 slanted forces in this coordinate system. Be careful that the 2 forces slant at different angles.

AP Physics C — Schedule for Mechanics Unit 1 (Ch. 2, 4)

AP Physics C – For Mechanics Unit 1 Homework

Answers and hints for Ch. 2 problems: Try to do the problems without the hints first. It’s more fun that way!!

OQ11. b To obtain a(t), we need to first obtain v(t) from the x vs. t graph.

7. a) -2.4 m/s, b) about -3.8 m/s, c) about 4s a) ave. v = ∆x/∆t, c) v = slope of x vs. t graph, so v = 0 when the slope of the graph is zero.

11. a) 5m, b) 4875s Hint: Both hare and tortoise run at constant speed if you ignore the waiting time of the hare.

Solution: Total running time for hare = 1000m/(8m/s) = 125s, Total running time for tortoise = 1000m/(0.2m/s) = 5000s. So the hare has to wait 5000-125= 4875 s. At first the hare runs for 0.8km/(8m//s) = 100 s, so the hare must begin to run again 125-100 = 25s before the race ends. That means the tortoise must be (0.2m/s)(25s) = 5 m from finish line.

14. 13400m/s^{2} ave. a = ∆v/∆t. Pay attention to the direction of the velocities.

39. a) against (i.e. not equal), b) for (i.e. equal) Hint: The glider goes through constant acceleration motion, so its ave. v = (v_{0} + v) /2.

Solution: Since the glider goes through constant acceleration motion, its velocity changes at a constant rate, it reaches the (v_{0} + v) /2 half way through time. I.e. it reaches the ave. v half way through time. This is not the same as the v half way through space, because the glider speeds up, so it takes less time to travel the second half of the space than the time it takes to travel the first half of the space.

79. 1.6m/s^{2} From problem 39, we know that for constant acceleration motion, the average velocity is the velocity half way through time, and we have enough information to find the average v when each car goes by. So we can find the velocities of the cars when they pass Liz half way in time.

17. a) 1.3 m/s^{2}, b) at t = 3 s, 2 m/s^{2}, c) at t = 6s and t = about 10.3 s to 12 s, d) 8s, -1.5 m/s^{2} a) Use definition of ave. a. b) steepest part of the graph, c) where slope =0, d) steepest negative slope.

19. a) at 10s: 20m/s, at 20s: 5m/s, b) 263m a) use ∆v = area of graph, b) Can plot a v vs. t graph and then find area of graph: Since we are looking for distance, we have to add +∆x and the absolute value of -∆x together. Another way to do this is to look at the motion in 3 separate segments (0 to 10s, 10 to 15s, and 15 to 20s), and use kinematics equations to find the displacement traveled in each segment. In this case the object does not turn around, otherwise we have to be careful to pay attention to when the object turns around in order to find the correct distance traveled

21. a) 2m, b) -3m/s, c) -2 m/s^{2} a) Plug in t. b) Take time derivative and then plug in t. c) Take one more time derivative and then plug in t.

24. 160ft Use the v^{2} equation for constant acceleration motion. v_{0}^{2} is proportional to ∆x.

29. -16 cm/s^{2}

33. a) 35s, b) 15.7m/s Either plot a v vs. t graph or use kinematics equations and work through the 3 segments of the motion one at a time.

38. a) 2.56m, b) -3m/s a) What’s special about turning point? b) When x = 0 again.

59. a) –(1×10^{8})t + 3×10^{5} m/s^{2} , b) 3×10^{-3}s, c) 450 m/s a) a = dv/dt, b) make a = 0 and solve for t, c) plug the previous answer into the v(t) equation.

OQ14. b) The 2^{nd} ball’s motion is like the reverse of the 1^{st} ball’s. And they meet at the same time, so it must be half way in time.

OQ15. b) Can use kinematics or energy conservation.

OQ16. e)

OQ17. a) A=C=E>B=D, b) D>A=B=C=E

50. 7.96s (if use g = 9.8m/s^{2}) Find h(t=2) to find initial height. Take time derivative of h to find v(t) and plug in t = 2 to find initial velocity of the mailbag.

52. 0.6s When the two balls meet, they are at the same position at the same time. So if we use the same coordinate system for both balls and write one y(t) for each ball. The balls would meet when y_{1}(t) = y_{2}(t). Solve for t for time. (Of course, you can also write the upward displacement of 1^{st} ball as a function of t. Write distance fallen by 2^{nd} ball as a function of time. Then add these 2 together to equal to 15m. Solve for t.) .

54. a) (h/t) + (gt/2), b) (h/t) –(gt/2)

55. (if use g = 9.8 m/s^{2}) b) 0.782s, a) 7.82 m b) Look at vertical motion.a) Look at horizontal motion.

56. a) v_{i} + gt, b) gt^{2}/2, c): a) The absolute value of (v_{i} - gt), b) gt^{2}/2 In this case the speed is the absolute value of the velocity. To find the distance between the helicopter and the package, we can find the difference in position of the two. So if we use the same coordinate system for both the helicopter and the package to write one y(t) for each. The absolute value of the difference in their y(t) would be the distance between the two.

85. (if use g = 9.8 m/s^{2}) a) 26.4 m, b) 28.2 m: 6.82% error a) Let t be the time it takes for rock to hit water, then the time it takes for sound to reach man is (2.4 –t). Use t to express the distance traveled by rock and the distance traveled by the sound. These 2 distances should be equal which gives us a quadratic equation. Solve the quadratic equation to find t. Then use t to find depth of the well. b) Treat 2.4 s as the time it takes for the rock to fall.

Ch. 4

OQ1. e) What is the definition of ave. a?

OQ4. d) Use the v^{2} equation.

OQ7. d) First find out by what factor the time changes. Then use ∆x = vt.

OQ11. a) T = 2πr/v

OQ14. d)

2. 2.5 m/s Which component of the velocity represents the velocity of the shadow?

3. a) (1**i** + 0.75**j**) m/s, b) v = (1**i** + 0.5**j**) m/s, speed = 1.12m/s, c) parabola a) b) Do the x- and y- components separately. c) because y(x) is a second degree equation.

CQ1. Parabola The spacecraft has no acceleration in one direction and has a constant acceleration in another direction.

CQ2. For the first loop, the velocity is always tangent to the circle and with increasing magnitude. The acceleration slants forward and towards the center of the circle. For the second loop, the velocity is always tangent to the circle and has a constant magnitude (no smaller than the largest velocity vector in the first loop). The acceleration always points toward the center of the circle and has constant magnitude. For acceleration, first look at tangential and radial accelerations separately.

CQ3. You can determine the average velocity, but not the instantaneous velocity.

CQ4. a) Follow a straight path. b) Follow a straight path (a = constant = 0), or follow a circular path (a = non-zero constant magnitude)

22. a) ∆x = 1.18 m, so the space is wide enough for a walkway. But a very tall person will have to walk carefully close to the wall to avoid getting his/her head wet. b) 0.491 m/s a) Projectile shot horizontally. b) ∆x = v_{x} t. When the height fallen changes by a factor of 1/12, the ∆y = 0 + ½ at^{2}changes by a factor of 1/12, so t^{2} changes by a factor of 1/12. Since we also want ∆x to change by a factor of 1/12, we can find the factor by which v_{x} has to change.

29. a) (0, 50m), b) (18m/s, 0), c) Constant acceleration motion with downward a = - g = - 9.8 m/s^{2}. d) Constant velocity motion with v = 18 m/s, e) v_{x} = v_{i }, v_{y} = -gt, f) x = v_{i}t_{ }, y = y_{0} - (1/2)gt^{2}, g) 3.19s, h) 36.1m/s, 60.1º below the horizontal.

15. 53.1º Start with an initial velocity v_{0} and initial angle θ. (Since v_{0 }does not matter, we can also say the initial velocity is 1m/s or 100m/s. It won’t matter.) Find max. height and ∆x in terms of v_{0}, θ, and g. Then make ∆x = (3 times the max. height). v_{0 }should cancel. Solve for the angle.

21. d(tanθ_{i} ) – (gd^{2} )/(2v_{i}^{2}cos^{2 }θ_{i}) Just like a normal projectile motion problem. Separate the horizontal and vertical sides. Use the horizontal side to find t and then use that t on the vertical side to find h.

23. a) The ball clears the crossbar by 0.89m. b) While falling. a) It’s just like problem 21. b) Using the initial velocity and the initial angle, we can find the time it takes the ball to reach maximum height. Compare this time to the time found in part a).

49. a) The acceleration has a downward component of 9.8 m/s^{2} and a southward component of 2.5 m/s^{2}. I.e. the acceleration is 10.1 m/s^{2}, 14.3º to the south from the vertical. b) a downward acceleration of 9.8 m/s^{2}. c) The bolt starts from rest and has a constant acceleration of 10.1 m/s^{2}, 14.3º to the south from the vertical. So its trajectory is a straight line that goes 14.3º to the south from the vertical. d) The bolt has an initial horizontal velocity that equals to the train’s velocity at the moment the bolt leaves the ceiling. So the bolt is like a projectile shot horizontally. The trajectory is parabolic.

53. 15.3m Since the observer standing on the ground observes the ball shot straight up, the ball must have been thrown with a horizontal velocity that equals to 10m/s backwards. Using trig, we can find the vertical component of the ball’s velocity which can be used to calculate for the ball’s max. height.

55. 54.4 m/s^{2} Radial acceleration is the same as the centripetal acceleration.

__AP PHYSICS C SYLLABUS (2015-2016)__

Textbook: Physics for Scientists and Engineers by Serway, Jewett. 9th edition.

Supplemental material: AP Physics C Exam Problems binder book.

__MECHANICS:__

**Unit 1:** Ch. 2, 4: 1-dim kinematics and free fall, the limit, derivatives & integration, ectors, 2-dim kinematics

**Unit 2:** Ch. 5, 6: Newton’s laws, friction and the dynamics of circular motion

**Unit 3:** Ch. 7, 8: Work and energy, Conservation of energy

**Unit 4:** Ch. 9: Center of mass, Conservation of momentum, Impulse and collisions

**Unit 5:** Ch.10, 11: Rotational Kinematics, Rotational Dynamics, Conservation of angular momentum

**Unit 6:** Ch. 12, 13, 15: Static Equilibrium, Oscillations, Gravitation and Satellite motion

__ELECTRICITY AND MAGNETISM:__

**Unit 1:** Ch. 23, 24: Electrostatics, Coulomb’s law of electrostatics, Electric field, Gauss’s law of electrostatics

**Unit 2:** Ch. 25, 26: Electric potential, Capacitors and dielectrics

**Unit 3:** Ch. 27, 28: Current and resistance, resistivity, DC Circuits, Resistance-capacitance circuit

**Unit 4:** Ch. 29, 30: Magnetic field, Biot-Savart law, Ampere’s law

**Unit 5:** Ch. 31, 32, a little bit of 34: Faraday’s law of induction, Inductance, Inductance-resistance circuits, Inductance-capacitance circuits, Maxwell’s equations

**Summer homework for the 2015-16 school year:**

AP Physics C – Summer Assignment – Due on the first day of school: August 25, 2015

**Students are allowed to use a calculator and the equation sheets (p.67 to 69) on the entire AP Physics 1, AP Physics 2, AP Physics C: Electricity & Magnetism, and AP Physics C: Mechanics Exams including both the multiple-choice and free response sections. **

Date

In class

Homework

3/31/2015

Tue

Kinematics

**Review Kinematics and do **MCh (Multiple Choice): kin.& F(2pg), **AP Prob. 96-2, 92-1**: Count 23 pieces of paper (i.e. 46 pages) from the back of the AP problem book to find the problems.

4/1

Forces.

05-1, 07-1

4/2

Forces.

Solving differential equations.

Mechanics: 08-1 (solve diff. eq.) and 14-2 (solve diff. eq.): They are not in the AP Problems binder, but you can find old AP Exam Problems online.

4/13 Mon

Work & Energy.

89-1 (near end of AP Problems book), 14-1 (online), MCh.:wk&E(2pg)

4/14

Review.

Review for quiz.

4/20

Mon

4/21 Parcc

SHM.

Gravitation.

** ** Parcc (Please report to the auditorium.)

13-1, 94-3

MCh: SHM & G (2pg)