PGCPS Letter about AP Exams 12_9182017.pdf : Please click on link to read the letter. 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^{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. 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) (1i + 0.75j) m/s, b) v = (1i + 0.5j) 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 = nonzero 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. For 20172018 AP Physics C Students: According to the feedback provided by the class of 2017 AP Physics C students, I should still give you the hefty summer assignment. (Sorry, but they think it's important for you to do all the summer assignment.) The class of 2017 students also said that I should still cover all the topics and not to remove the most challenging problems. However, they recommended that we spend less time during the school year on Mechanics, so we can spend more time on Electricity & Magnetism. Below are the summer assignments: ALL students have summer assignments, but those who took AP Physics 2 have a little less to do. 20172018: AP Physics C – Summer Assignment – Notes and work are due on the first day of school: Sept. 6, 2017
