**.)**

__If you do a Google search for the textbook, you will be able to find a pdf file of the textbook__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< 2. a) 0.0328 J WK=Fd
cos< OQ 5. a>b=e>d>c cos< CQ4. a) 90º > θ ≥ 0, b) 180º ≥ θ > 90º cos< 9. 16 For
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. 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. ) 38. a) 23400N, opposite to the bullet’s v WK 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. 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. 47. A/(r 49. F 50. a)
Ax 52. a)
A: 0, B: +, C: 0, D: -, E: 0 F
= -dU/dx = - slope OQ14.
d U = ½ kx 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 64. E
Hints for Ch. 8 OQ1. a E 8. a)
√(2(m 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 24. a)
0.381 m Write E 13. v 15. a)
0.791 m/s Write E 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. 45. h
+ d 46. a)
2.49 m/s It’s
convenient to use conservation of energy to find the speed of the blocks the
moment m 47. a)
K = 2 + 24t 64.
1.24 m/s Write E 68. a)
E
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 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∆ OQ1. b>e>a=d>c The Frisbee “collides” with you
on frictionless surface, so p 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 5. 3263N toward homeplate and
3988N downward Look at
the ball and use Impulse
by net force on the ball: 7. a) m+ _{g} v_{gp}
/(m_{g} m p = constant, so _{p})p_{i}
= 0 = p_{f} = m_{g} v_{gp} + (m_{g} +
mbecause
we have to consider that the plank carries the girl with it, hence the _{p}) v_{pi} (m_{g}
+ m. _{p}) b)
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 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. 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 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 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 32. 112N Find accelerations x- and y- components. Then find the magnitude of the acceleration. 51. (If use g = 10 m/s 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 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 29. a) 7m/s 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 42. b) 3.57 m/s 45. (If use g = 10 m/s 46. a) a OQ1. d) Max friction also doubles. OQ5. b)
F OQ8. d)
Can use F OQ12. d)
F OQ13. a), c), d) 52. a) 14.7m, b) 10000kg and 20000kg 57. μ 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 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 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 81. a)
b) 0.408 m/s 82. a) Since the child is
lighter than Nick + chair, the child will acceleration upward. a = 0.426 m/s 85. a)
b) T 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 93. (M + m 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 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 20. a) 3.6 m/s 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 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: 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. 14. 13400m/s 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 79. 1.6m/s 17. a) 1.3 m/s 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 24. 160ft Use the v 29. -16 cm/s 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 OQ14. b) The 2 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 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 54. a) (h/t) + (gt/2), b) (h/t) –(gt/2) 55. (if use g = 9.8 m/s 56. a) v 85. (if use g = 9.8 m/s Ch. 4 OQ1. e) What is the definition of ave. a? OQ4. d) Use the v 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 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 29. a) (0, 50m), b) (18m/s,
0), c) Constant acceleration motion with downward a = - g = - 9.8 m/s 15. 53.1º Start with an initial velocity v 21. d(tanθ 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 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
Summer Assignment –
2018 AP Physics C Review Schedule:
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 22. a) 1.79×10 73. a) 1.04×10 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 24. a) 6.86 × 10 29. 244000V/m Undeflected: so F 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. 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 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 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 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 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 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 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 OQ12. i) a >d >b =c >e, ii) e> a= b> c =d 24. a) 0.909 A, 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 41. a) 1.5 s, b) 1s, c) I 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 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 OQ9. c>a>d>b Use: scalar U = kq 24. WK 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) ∂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 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 52. a) 450000V, b) 7.51 x10 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 65. –λ ln(r 72. 3kQ Ch. 26 2. a) 1μF, b) 100V 7. 4.43x10 OQ6. b 13. a) 17μF, b) 9V, c) C 14. a) 3.53μF, b) V 20. a) 2C, b) q 24. a) 120μC, b) q 30. 4470 V 57. 0.00251 m 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) 49. a) 40μJ, b) 500 V a) WK 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
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 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 20. a) The net electric force can be written in
the “–kx” format when a << d. b)
(π/2) √(md 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 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 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 57. a) 111ns, b) 0.00567m,
c) horizontal: 4.5x10 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 Ch. 24 1. a) 1.98x10 4. a) -2340 Nm OQ1. i) a, ii) b. 11. S 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 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 53. a) σ/(2є 56. a) 0, b) σ/є 59. σ/(2є 60. a) 2kλ/r, radially
outward, b) 2k(λ + ρπ(r
OQ3. e
High school seniors who demonstrate exceptional leadership, drive, integrity, and citizenship are invited to apply for the 2018 GE-Reagan Foundation Scholarship Program. This program annually provides college-bound students with $10,000 renewable scholarships – up to $40,000 total per recipient – and supports them as they lead and serve in college and beyond. Download a program flyer here. Deadline: January 4, 2018 For those with financial needs: Horatio Alger Association scholarship programs. For 4 year colleges, the application is due 10/25: https://scholars.horatioalger.org/scholarships/about-our-scholarship-programs/national-scholarships/ and https://scholars.horatioalger.org/scholarships/about-our-scholarship-programs/state-scholarships/ For community colleges, the application will open in March and apparently they usually don't get enough applications for this program: https://scholars.horatioalger.org/scholarships/about-our-scholarship-programs/technical/ 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 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 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 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. 18. a) 1.26s, b) 0.15m/s,
0.75m/s 24. 0.153J, b) 0.784 m/s, c)
17.5m/s 66. μ 85. a) 0.5 m/s, b) 0.086m a) v at equilibrium is a max.
To find v OQ2 C, OQ3 A, OQ4 C, OQ5 D, OQ7 C, OQ8 B OQ 16 i) B, ii) A, iii) C Use g 38. I = mgd/(2πf) 39. 0.499 42. a) 2.09s I = I 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 68. a) Use 1/k Ch. 13 11. 0.614m/s 12. 2/3 g = GM/r OQ1. C There is one term between every pair of particles. 49. 2.52×10 38. For orbit speed, write F OQ7. i) E, ii) C, iii) A OQ11. E Use Kepler’s law of periods: T CQ5. a) perihelion: when closest to the sun, b) aphelion: when farthest to the sun. 28. a) 6.02 × 10 72. Circular orbit: start with F 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 43. a) 8.5× 10 21. 1.26× 10
74. Circular
orbit: start with F 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 3. a) 5 rad, 10 rad/s, 4
rad/s 17. a) 25 rad/s, b) 39.8
rad/s 70. a) ω(t) = ω + At + Bt 18. a) 0.605 m/s ω = v/r, need to convert cm
to m and rev/min to rad/s. 24. a (√(1 + π OQ6. i) a You want the mass to be close to
the axis. OQ11. e I ∝ mr 39. a) 5.8 kgm 43. 11 mL 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 46. 13 MR 32. a) make sure that you draw each force from
the point of application. 37. 0.312 Use kinematics to find α and then
write τ OQ9. i) a The basketball
also has rotational KE. 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 56. Use
conservation of energy, the lost mgy of the hanging mass is turned into KE of
hanging mass and the KE 57. a) 2(√(Rg/3)) Lost mgy (y = R) is turned
into KE 61. a) 2gsinθ/3 The disk does both
translational motion and rotational motion, so we need to write both the F 78. a) Mg/3, b) 2g/3 The disk does both
translational motion and rotational motion, so we need to write both the F 81. (√(10(R-r)(1 – cosθ)/(7r 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 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. 73. a) (√(3g/L)) Use conservation of energy.
Be careful when you look for the height change. 79. (√((2mgdsinθ + kd 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 92. a) Use conservation of energy.
The lost E is friction times d. Ch. 11 10. a) No, b) 12. – 22 13. (mxv 29. a) 1.57 x 10 OQ1. b L = constant OQ8. d L = constant
if τ 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 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 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 33. a) No. Some calories
(chemical PE) from the woman is turned into mechanical energy. 39. a) 2mv 49. a) 7md 53. a) r 56. a) Mvd L 61. a) ω
63. 4(√(ga((√2) – 1)/3)) This problem involves 2
parts: a collision (L 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 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∆ OQ1. b>e>a=d>c The Frisbee “collides” with you
on frictionless surface, so p 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 5. 3263N toward homeplate and
3988N downward Look at
the ball and use Impulse
by net force on the ball: 7. a) m+ _{g} v_{gp}
/(m_{g} m p = constant, so _{p})p_{i}
= 0 = p_{f} = m_{g} v_{gp} + (m_{g} +
mbecause
we have to consider that the plank carries the girl with it, hence the _{p}) v_{pi} (m_{g}
+ m. _{p}) b)
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 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 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 c) 265 m/s
19. a) 12 b) 4.8 d) 2.4 20. a) 981 Ns, up Impulse by a force 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 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 61. 15 N The additional force is the force related to the impulse of the shooting water. 67. -260 v_{f} in the
x-direction. Then do F_{ave net
x} = ΔpΔt._{x}/OQ3. i) c Newton’s
3 ii) a K
= ½ mv OQ7. a K
= ½ mv OQ8. d K
= ½ mv 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 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 CQ12. No WK 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 33. 0.556 m For
the sliding part: use conservation of energy to find the speed of m 34. a) 2.24 m/s to the right Use p b) No. 73. v 35. a) v 37. (3 57. a) The
cart “suddenly” comes to a stop, so the top of the cart suddenly stops.
However, the particle still has the v 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 b) 235 N to the left F c) 0.68 s Impulse by net force d) Person: 160Ns to the left, cart: 160 Ns to the right. e) 1.82 m For
the person: F f) 0.454 m For
the cart: F 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 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 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 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
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< 2. a) 0.0328 J WK=Fd
cos< OQ 5. a>b=e>d>c cos< CQ4. a) 90º ≥ θ > 0, b) 180º ≥ θ > 90º cos< 9. 16 For
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. 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. ) 38. a) 23400N, opposite to the bullet’s v WK 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. 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. 47. A/(r 49. F 50. a)
Ax 52. a)
A: 0, B: +, C: 0, D: -, E: 0 F
= -dU/dx = - slope OQ14.
d U = ½ kx 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 64. E
Hints for Ch. 8 OQ1. a E 8. a)
√(2(m 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 24. a)
0.381 m Write E 13. v 15. a)
0.791 m/s Write E 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. 45. h
+ d 46. a)
2.49 m/s It’s
convenient to use conservation of energy to find the speed of the blocks the
moment m 47. a)
K = 2 + 24t 64.
1.24 m/s Write E 68. a)
E
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.
The QuestBridge National College Match is a college and scholarship application process that helps outstanding low-income high school seniors gain admission and full four-year scholarships to the nation's most selective colleges. For academically outstanding high school seniors from families earning less than $65,000/year for a family of four. The deadline for students to apply is September 27, 2017.On November 18th, 2017 from 10am-3:30 pm the Biomedical Engineering Society (BMES) at the University of Maryland will be hosting a "Crash Course" event where high school students are invited to come learn more about bioengineering. We hope to help students develop an idea for what it means to be an engineer by introducing some basic skills they may need as college engineering students. There will be 3 workshops throughout the day that focus on Computer Aided Design (CAD) in conjunction with 3D printing, Coding (Python), and bioengineering lab techniques (DNA Gel Electrophoresis). This event, which will be held in the Kim Engineering Building at the University of Maryland, College Park. Lunch and Prizes will also be provided. Link for Registration: ter.ps/bmesccAP 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 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 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 32. 112N Find accelerations x- and y- components. Then find the magnitude of the acceleration. 51. (If use g = 10 m/s 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 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 29. a) 7m/s 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 42. b) 3.57 m/s 45. (If use g = 10 m/s 46. a) a OQ1. d) Max friction also doubles. OQ5. b)
F OQ8. d)
Can use F OQ12. d)
F OQ13. a), c), d) 52. a) 14.7m, b) 10000kg and 20000kg 57. μ 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 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 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 81. a)
b) 0.408 m/s 82. a) Since the child is
lighter than Nick + chair, the child will acceleration upward. a = 0.426 m/s 85. a)
b) T 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 93. (M + m 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 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 20. a) 3.6 m/s 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 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: 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. 14. 13400m/s 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 79. 1.6m/s 17. a) 1.3 m/s 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 24. 160ft Use the v 29. -16 cm/s 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 OQ14. b) The 2 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 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 54. a) (h/t) + (gt/2), b) (h/t) –(gt/2) 55. (if use g = 9.8 m/s 56. a) v 85. (if use g = 9.8 m/s Ch. 4 OQ1. e) What is the definition of ave. a? OQ4. d) Use the v 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 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 29. a) (0, 50m), b) (18m/s,
0), c) Constant acceleration motion with downward a = - g = - 9.8 m/s 15. 53.1º Start with an initial velocity v 21. d(tanθ 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 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 |