Download KEY Chapter 8 – Rotational Motion Chapter 6 – Work, Energy

AP Physics 1 Test Review
Chapter 8 – Rotational Motion
Chapter 6 – Work, Energy & Power
Chapter 7 – Momentum & Collisions
Physicist: KEY
Chapter 6 – Work, Energy & Power
Multiple Choice:
1. What two factors is work dependent upon?
A. force & energy
B. force & distance
C. distance & energy D. energy & momentum
2. Work is done only if the applied force does which of the following?
A. equals zero
B. acts in the same direction as gravity
C. is greater than the weight D. produces motion
3. The transfer of energy by mechanical means is
A. momentum
B. work
C. effort
D. acceleration
4. A job is done slowly, and an identical job is done quickly. Both jobs require the same amount of
work, but different amounts of
A. energy
B. power
C. both of these
D. none of these
5. An object that has potential energy has this energy because of its
A. speed
B. acceleration
C. momentum
D. location
6. A ball is thrown vertically upward. Which one of the following quantities increases as the ball rises?
A. total mechanical energy
B. potential energy
C. kinetic energy
D. speed
7. A ball is thrown vertically upward. Which one of the following quantities decreases as the ball rises?
A. total mechanical energy
B. potential energy
C. kinetic energy
D. speed
8. A ball is thrown vertically upward. Which one of the following quantities remains constant as the ball
rises?
A. total mechanical energy
B. potential energy
C. kinetic energy
D. speed
9. A box of rivets falls from the top of a skyscraper under construction. As the box passes the 80th
floor, it has a kinetic energy of Ek. By the time it passes the 20th floor its speed has doubled. By
then its kinetic energy is
A. (1/4) Ek
B. (1/2) Ek
C. 4 Ek
D. 2 Ek
10. The total amount of energy in an isolated, closed system
A. is constantly increasing
B. is constantly decreasing
C. remains constant
D. cannot be accurately measured
2
11. The moment of inertia of a solid cylinder about its axis is given by 0.5 MR . If this cylinder rolls without
slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A. 1:1.
B. 1:2.
C. 2:1.
D. 1:3.
12. "The total angular momentum of a system of particles changes when a net external force acts on the
system." This statement is
A. always true.
B. never true.
C. sometimes true. It depends on the force's magnitude.
D. sometimes true. It depends on the force's point of application.
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
Short Answer:
13. 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?
ΔEp = ΔEk
2
Mgh = ½ Mv + Wfriction
2
(40 kg)(9.8 N/kg)(8.0 m) = (1/2)(40 kg)(7.0 m/s) + Wfriction
3136 J = 980 J + Wfriction
Wfriction = 2156 J
14. A jack system will increase the potential energy of a heavy load by 1000 J with a work input of 2000 J. What is the efficiency of
the jack system?
e = (1000 J)/(2000 J) * 100% = 50 %
15. A 25-N force pushes a 25 kg box 2.2 meters across a rough floor. The box is then lifted vertically and placed on a shelf that is 2.2
meters above the floor.
(a) How much total work is done on the box?
(b) How much energy is stored in the box when it is on the shelf?
a)
Wtotal = Wpush + Wlift
Wtotal = Fd + mgh
Wtotal = (25 N)(2.2 m) + (25 kg)(9.8 N/kg)(2.2 m) = 544.5 J
b)
Ep = (25 kg)(9.8 N/kg)(2.2 m) = 489.5 J
16. IF ABCD is a frictionless track, what is the speed of the object on section CD? Use an energy bar chart in your response.
ΔEp = ΔEk
2
MgΔh = ½ MΔv
Δv = √(2gΔh) = √(2)(9.8 N/kg)(0.35 m) = 2.62 m/s
17. What is the speed of the pendulum shown at the bottom of its swing? Use an energy bar chart in your response.
ΔEp = ΔEk
2
MgΔh = ½ MΔv
Δv = √(2gΔh) = √(2)(9.8 N/kg)(0.20 m) = 1.98 m/s
18. A ball of mass 2.00-kg is dropped from a height of 1.5 m (from the ground) onto a massless spring (the spring has an equilibrium
length of 0.50 m). The ball compresses the spring by an amount of 0.20 m by the time it comes to a stop. Calculate the spring
constant of the spring, k.
2
Mghi = Mghf + ½ kΔx
2
Mghi - Mghf = ½ kΔx
2
2(Mghi - Mghf) = kΔx
2
(2(Mghi - Mghf))/( Δx ) = k
k = 1180 N/m
2
19. A wheel of moment of inertia of 5.00 kg∙m starts from rest and accelerates under a constant torque of 3.00 N∙m for 8.00 s.
What is the wheel's rotational kinetic energy at the end of 8.00 s?
τ = Iα
2
3.00 N∙m = (5.00 kg∙m ) α
α = 0.600 rad/s²
ωf = ωi + αt = (0.600 rad/s²)(8.00 s) = 4.80 rad/s
2
2
Ek= ½ Iω = (0.5)(5.00 kg∙m )(4.80 rad/s)² = 57.6 J
20. A solid sphere of mass 1.0 kg and radius 0.010 m rolls with a speed of 10 m/s. How high up an inclined plane can it climb before
coming to rest?
The moment of inertia of solid sphere is:
2
I = (2/5)mr
Law of Conservation of Energy:
2
2
0.5mv + 0.5Iω = mgh
We also know that v = rω
We get then:
2
2
2
h = (1/(mg))*0.5*[m + (2/5)m(r )/(r )]*v
with:
m = 1 kg
g = 9.81 N/kg
r = 0.01 m
v = 10 m/s
h = (1/(9.81))*0.5*[1 + 0.4]*10
2
h = (50/9.81)*[1.4] = 7.14 m
Chapter 7 – Momentum & Collisions
Multiple Choice:
21. The greatest change in momentum will be produced by a:
A. large force acting over a long time
B. small force acting over a short time
C. large force acting over a short time
D. small force acting over a long time
22. Impulse can be represented by
A. Δv/Δt
B. FΔt
C. mv
D. none of these
23. The momentum change of an object is equal to the
A. force acting on the object
B. velocity change of the object
C. impulse acting on the object
D. mass of the object times the force acting on it
24. A Ping-Pong gun fires a Ping-Pong ball out of its barrell. Compared to the impulse on the ball, the
impulse on the gun is
A. larger
B. smaller
C. the same
Use the diagram below to answer the next three questions (25-27).
Two blocks, A and B travelling towards each other with equal momenta on a frictionless surface,
are separated by a spring of negligible mass. The mass of block A is twice that of block B.
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
25. Compared with the total momentum of the blocks before the collision, the total momentum after
the collision is:
A. the same
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
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
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
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
B. the same as the original velocity
C. one-half the original velocity
D. zero
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
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 Etotal = Ep.
At bottom of ramp Etotal = Ek.
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.
vf =√ (2ad) = √[(2)(9.8)(1.5)] = 5.4 m/s
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?
Favg= J/Δt = (570 N•s) / (0.4 s) = 1420 N
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
35. Two dynamics carts, on a frictionless surface, are blasted apart by the spring between them.
If cart B moves off at + 5.0 m/s, what is the velocity of cart A? Use a momentum bar chart in your response.
Cart A
Cart B
BEFORE
AFTER
Momentum Vector Diagram
pA + pB = pA’ + pB’
0 = pA’ + pB’
pA’ = -pB’
mAvA = -mBvB
vA = (-mBvB)/mA
vA = - [(2.8 kg)(5.0 m/s)]/(1.16 kg) = 121 m/s
36. An arrow weighing 0.10 kg travelling at 80 m/s is shot into a straw target suspended by a rope. You are given that the target
weighs 7.0 kg and that the arrow sticks inside the target. Calculate the velocity of the target immediately after being pierced by the
arrow. Use a momentum bar chart in your response.
Arrow
Target
BEFORE
AFTER
Momentum Vector Diagram
p(arrow) + p(target) = p(arrow)’ + p(target)’
p(arrow) +0 = p(arrow)’ + p(target)’
p(arrow) +0 = p(arrow + target)’
p(arrow + target)’ = p(arrow)
m(arrow + target)v(arrow + target)’ = m(arrow)v(arrow)
v(arrow + target)’ = m(arrow)v(arrow)/ m(arrow + target)
v(arrow + target)’ = (0.10 kg)(80 m/s)/(7.1 kg)
v(arrow + target)’ = 1.13 m/s
Note: arrow is embedded in target; combined unit has
same velocity
Chapter 8 – Rotational Motion
37. A point object moves from point A to point B along a circular path with a
radius R. What is the size of the angle θ?
A. 1 rad B. 2 rad
C. 3 rad D. 4 rad
38. A solid disk with a radius R rotates at a constant rate ω. Which of the following
points has the greatest angular displacement?
A. A B. B
C. C D. All points have the same angular displacement
39. A solid disk with a radius R rotates at a constant rate ω. Which of the following
points has the greatest linear velocity?
A. A B. B
C. C D. All points have the same linear velocity
40. A solid disk with a radius R rotates at a constant rate ω. Which of the following represents the
period of rotations?
A. 2πω B. 𝜔2𝜋 C. 2𝜋/𝜔 D. 2ω
41. An object starts from rest and accelerates at a constant rate α in a circular path with a radius R.
The radius describes an angle θ after time t. Which of the following represents the angular velocity as
a function of θ?
A. 2𝛼𝜃 B. 2𝜃/𝛼 C. √(2𝛼𝜃) D. √(2𝜃/𝛼)
42. An object starts from rest and accelerates at a constant rate in a circular path. After a certain time
t, the object reaches the angular velocity ω. How many revolutions did it make during time t?
A. 4𝜋𝜔 B. 4𝜔/𝜋𝑡 C. 𝜔𝑡/4𝜋 D. 4𝜔𝑡
43. Torque is the rotational analog of which of the following quantities?
A. kinetic energy B. linear momentum
C. acceleration
D. force
44. Two wheels are fixed to each other and are free to rotate about a
frictionless axis through their concentric centers. Four forces are
exerted tangentially to the wheels. The magnitude of the net torque is:
A. zero B. FR
C. 2 FR D. 4 FR
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
A
B
C
D
45. In which of the following diagrams is the torque about point O equal in
magnitude to the torque about point X in the diagram to the right?
46. Which of the following must be true for the below system of two concentric
massless pulleys and two masses to be at equilibrium?
A. m1 = m2
B. m1R1 = m2R2
2
C. m1R2 = m2 R1
2
2
D. m1R1 = m2R2
47. In translational motion, mass is a measure of an object’s acceleration in response to a force.
What is the rotational analog of mass?
A. torque
B. moment of inertia
C. angular acceleration
D. angular momentum
48. When a solid object rotates with a constant angular acceleration, which of the following is true?
A. All points on the object rotate with the same centripetal acceleration.
B. The net torque applied to the object must be zero.
C. The net torque applied to the object must be a constant.
D. All points on the object rotate with the same tangential acceleration.
49. A baseball player swings his bat with his arms fully extended. If his arms are pulled in closer to his
body, which of the following choices correctly describes the impact of his motion on his swing’s
angular momentum and kinetic energy?
Angular Momentum
Kinetic Energy
A.
increases
increases
B.
increases
remains constant
C.
remains constant
increases
D.
remains constant
remains constant
50. An ice skater performs a fast spin by pulling her outstretched arms down and close to her body.
What happens to her angular momentum with respect to the axis of rotation?
A. increases
B. decreases
C. stays the same
D. depends on her initial rotational velocity
51. The system below rotates with an angular velocity, ω. If the mass of
the rod supports is negligible, what is the ratio of the angular momentum
of the two upper spheres to the two lower spheres?
A. 1:4
B. 1:2
C. 2:1
D. 4:1
A
B
C
D
A
B
C
D
A
B
C
D
52. A rod rotates about a pivot at its center at 2 rad/s. Its angular velocity increases uniformly to 14 rad/s in 3 s. Find the rod’s
angular acceleration over this time period.
ωf = ωi + 2αt
α = (ωf - ωi)/2t
2
α = (14 rad/s – 2 rad/s)/(2*3 s) = 2 rad/s
2
53. A bicycle wheel of radius 40.0 cm and angular velocity of 10.0 rad/s starts accelerating at 80.0 rad/s . (a) What is the tangential
acceleration of the wheel at this time point? (b) What is the centripetal acceleration of the wheel at this time point?
2
2
(a) atan = rα = (0.40 m)(80.0 rad/s ) = 32.0 m/s
2
2
2
(b) ar = ac = v /r = ω r = (10.0 rad/s)2(0.40 m) = 40 m/s
54. A record on a turntable is spinning and coming to rest. At t = 0 s, the angular velocity of the record is 20.0 rad/s with a constant
2
acceleration of – 5.0 rad/s . How long does it take for the record to come to rest?
ωf = ωi + 2αt
t = (ωf - ωi)/2 α
2
t = (0 rad/s – 20.0 rad/s)/2*(-5.0 rad/s ) = 2.0 S
55. As shown to the right, a wooden square of side length 1.0 m is on a horizontal tabletop and is
free to rotate about its center axis. The square is subject to two forces and rotates.
Find the net torque of the system.
The moment arm, r, from the contact point of the force to the center axis is 0.707 m. This is
found by using the Pythagorean Theorem and the fact that the ‘halfway’ distance horizontally
and vertically to the center point is 0.5 m.
τnet = τ4N + τ8N
o
o
τnet = -(4 N)(0.707 m)(sin 135 ) + (8 N)(0.707 m)(sin 135 ) { τ4N is negative since its direction is CW}
τnet = 2 Nm
56. Two masses of mass 10.0 and 6.0 kg are hung from massless strings at the end of a light rod. The rod is virtually weightless. A
pivot (fulcrum) is placed off center and the system is free to rotate.
If the 6.0 kg mass is 4.0 m away from the pivot, how far away is the 10.0 kg mass if the system is not rotating?
Στ = 0 (if not rotating)
τ10kg + τ6kg = 0
τ10kg =- τ6kg
o
o
(10 kg)(9.8 N/kg)(x)(sin 90 ) = (6 kg)(9.8 N/kg)(4 m)(sin 90 )
x = (6 kg)(4 m)/(10 kg) = 2.4 m
57. A solid rubber ball, of radius 1.2 cm and mass 0.50 kg, rolls on a flat surface. Its angular velocity is 4.0 rad/s. What is the ball’s
rotational kinetic energy?
2
2
2
I = 2/5 mr = 2/5(0.50 kg)(0.012 m) = 0.0000288 kg•m
2
2
2
Ek = ½ Iω = ½(0.0000288 kg•m )(4.0 rad/s) = 0.0000576 J
58. Gina rolls a bowling ball of mass 7 kg and radius 10.9 cm down a lane with a velocity of 6 m/s. (a) Find the rotational kinetic
energy of the bowling ball, assuming it does not slip. (b) Find the total kinetic energy of the bowling ball.
(a) To find the rotational kinetic energy of the bowling ball, you need to know its moment of inertia and its angular velocity.
Assume the bowling ball is a solid sphere to find its moment of inertia.
Next, find the ball’s angular velocity.
Finally, solve for the rotational kinetic energy of the bowling ball.
(b) The total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy of the bowling ball.
59. Angelina spins on a rotating pedestal with an angular velocity of 8 rad/s.. Bob throws her an exercise ball, which increases her
2
2
moment of inertia from 2 kg·m to 2.5 kg·m . What is Angelina’s angular velocity after catching the exercise ball? (Neglect any
external torque from the ball.)
Since there are no external torques, you know that the initial spin angular momentum must equal the final spin angular
momentum, and can therefore solve for Angelina’s final angular velocity: