Ap Physics Worksheet With Answers Page 4
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825. Compared with the total momentum of the blocks before the collision, the total momentum after
the collision is:
A. the same
A
B
C
D
B. one-half as great
C. twice as great
D. four times as great
26. After the collision, the magnitude of the momentum of block A compared with that of block B is
A. one-half as great
B. twice as great
A
B
C
D
C. four times as great
D. the same
27. After the collision, the magnitude of the velocity of block A compared with that of block B is
A. the same
B. twice as great
A
B
C
D
C. one-half as great
D. four times as great
28. A moving freight car runs into an identical car at rest on the track. The cars couple together.
Compared to the velocity of the first car before the collision, the velocity of the combined cars
after the collision is
A. twice as great
A
B
C
D
B. the same
C. one-half as great
D. zero
29. Two gliders having the same mass and speed move toward each other on an air track and stick
together. After the collision, the velocity of the gliders is
A. twice the original velocity
A
B
C
D
B. the same as the original velocity
C. one-half the original velocity
D. zero
30. A 40.0-kg object initially at rest slides down from the top of a ramp. The ramp is raised 8.0 m at one end. The object reaches the
bottom of the ramp with a speed of 7.0 m/s. How much of the original potential energy was converted to forms of energy other than
kinetic? (THIS QUESTION SHOULD BE INCLUDED WITH CHAPTER 6 QUESTIONS)
At top of ramp E
= Ep.
total
At bottom of ramp E
= Ek.
total
ΔE = Ep - Ek
2
ΔE = mgh – ½ mv
ΔE = 2156 J
31. A steadily increasing force, acting in one direction, is applied to a body of mass 4.0 kg which is initially at rest. Below is the graph
of the force versus time.
32. What is the speed of the object at the end of the first 4 s ?
J = area under the graph = ½ (20.0 N)(4.0 s) = 40.0 N•s
J = Δp
J = mΔv = 40.0 N•s
Δv = (40.0 N•s)/(4.0 kg) = 10.0 m/s
33. a) Calculate the impulse suffered by a 105 kg man who lands on firm ground after jumping from a height of 1.5 m.
v
=√ (2ad) = √[(2)(9.8)(1.5)] = 5.4 m/s
f
J = Δp = mΔv = (105)(5.4) = 570 N•s
b) What force would be exerted on the man if he bent his knees and absorbed the fall in 0.4 s?
F
= J/Δt = (570 N•s) / (0.4 s) = 1420 N
avg
34. An impulse of 120.0 N•s is applied to a mass of 6.2 kg. What change in velocity would this cause?
J = Δp = mΔv
Δv = J/m = (120 N•s)/(6.2 kg) = 19 m/s
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