Everything below prints as one AP AP Physics 1 practice paper set: papers A & B with their answer keys, plus the full-length study package exam. Use the Download PDF / Print button (or Cmd/Ctrl+P) to save it.
Paper A
AP Physics 1 — Practice Paper A
Original unofficial practice questions · paper A · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
For an object in free fall (no air), acceleration is
A. 9.8 m/s² downwardB. zeroC. 9.8 m/s² upwardD. depends on massNewton's Third Law states that forces come in
A. equal and opposite pairsB. equal same-direction pairsC. unequal opposite pairsD. single forcesWork done on an object equals
A. change in kinetic energyB. change in momentumC. force × timeD. mass × velocityMomentum p of a mass m moving at v is
A. mvB. mv²C. maD. ½mv²A projectile's horizontal velocity is
A. constantB. increasingC. decreasingD. zeroThe period of a simple pendulum depends on
A. lengthB. massC. amplitude (large)D. both mass and lengthIf net force is zero, an object's velocity is
A. constantB. zeroC. increasingD. decreasingKinetic energy of a 2 kg object moving at 3 m/s is
A. 9 JB. 6 JC. 18 JD. 3 JHeat flows spontaneously from
A. hot to coldB. cold to hotC. in both directionsD. high pressure to low pressureCentripetal force on a 1 kg mass moving at 2 m/s in a circle of radius 1 m is
A. 4 NB. 2 NC. 1 ND. 8 NA volt is a unit of
A. electric potentialB. chargeC. currentD. resistanceImpulse equals change in
A. momentumB. energyC. massD. accelerationTwo resistors in series have
A. the same currentB. the same voltageC. equal powerD. equal resistanceSnell's law relates angles in refraction to
A. indices of refractionB. wavelengths onlyC. frequenciesD. intensitiesKinetic friction is generally
A. less than static frictionB. greater than static frictionC. equal to static frictionD. zeroSection II — Free Response
A 3 kg block slides from rest down a frictionless ramp of height 4 m. (a) Find speed at the bottom using energy. (b) If friction did work -30 J, find the new speed.
8 points · rubric: (a) energy conservation 4 pts; (b) work-energy with W=-30 J 4 pts.
A 0.5 kg ball moving at 4 m/s hits a wall and rebounds at 3 m/s. Find the impulse delivered to the ball.
5 points · rubric: Impulse = Δp 2 pts, sign/direction 2 pts, magnitude 1 pt.
Explain the difference between mass and weight and compute the weight of a 5 kg object on Earth (g=9.8).
4 points · rubric: Concept 2 pts, calculation 2 pts.
A 12 V battery drives 4 Ω and 2 Ω resistors in series. Find total current and power in the 4 Ω resistor.
6 points · rubric: Series R 2 pts, Ohm's law 2 pts, power 2 pts.
Answer Key
1. 9.8 m/s² downward — Constant g.
2. equal and opposite pairs — Action-reaction pairs.
3. change in kinetic energy — Work-energy theorem.
4. mv — Definition.
5. constant — No horizontal acceleration (ideal).
6. length — T = 2π√(L/g).
7. constant — Newton's 1st law.
8. 9 J — ½mv² = 9 J.
9. hot to cold — Second law.
10. 4 N — F = mv²/r = 4 N.
11. electric potential — Potential difference.
12. momentum — Impulse-momentum theorem.
13. the same current — Series current is common.
14. indices of refraction — n₁sinθ₁ = n₂sinθ₂.
15. less than static friction — Static is larger (stiction).
Free response — rubric notes
1. (a) energy conservation 4 pts; (b) work-energy with W=-30 J 4 pts. · model: (a) mgh = ½mv² → v=√(2gh)=√78.4≈8.9 m/s. (b) ½mv² = mgh - 30 → v≈ 7.3 m/s.
2. Impulse = Δp 2 pts, sign/direction 2 pts, magnitude 1 pt. · model: Δp = m(v_f - v_i) = 0.5(-3 - 4) = -3.5 N·s; impulse magnitude 3.5 N·s away from wall.
3. Concept 2 pts, calculation 2 pts. · model: Mass is amount of matter (kg), weight is gravitational force (W=mg) = 49 N.
4. Series R 2 pts, Ohm's law 2 pts, power 2 pts. · model: R_total = 6 Ω, I = 12/6 = 2 A, P = I²R = 16 W.
Paper B
AP Physics 1 — Practice Paper B
Original unofficial practice questions · paper B · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
For an object in free fall (no air), acceleration is
A. depends on massB. zeroC. 9.8 m/s² downwardD. 9.8 m/s² upwardNewton's Third Law states that forces come in
A. equal same-direction pairsB. unequal opposite pairsC. single forcesD. equal and opposite pairsWork done on an object equals
A. change in momentumB. change in kinetic energyC. mass × velocityD. force × timeMomentum p of a mass m moving at v is
A. ½mv²B. maC. mvD. mv²A projectile's horizontal velocity is
A. decreasingB. constantC. increasingD. zeroThe period of a simple pendulum depends on
A. amplitude (large)B. both mass and lengthC. lengthD. massIf net force is zero, an object's velocity is
A. zeroB. decreasingC. increasingD. constantKinetic energy of a 2 kg object moving at 3 m/s is
A. 3 JB. 9 JC. 18 JD. 6 JHeat flows spontaneously from
A. cold to hotB. in both directionsC. hot to coldD. high pressure to low pressureCentripetal force on a 1 kg mass moving at 2 m/s in a circle of radius 1 m is
A. 2 NB. 8 NC. 1 ND. 4 NA volt is a unit of
A. resistanceB. electric potentialC. chargeD. currentImpulse equals change in
A. accelerationB. energyC. massD. momentumTwo resistors in series have
A. equal resistanceB. equal powerC. the same currentD. the same voltageSnell's law relates angles in refraction to
A. wavelengths onlyB. intensitiesC. indices of refractionD. frequenciesKinetic friction is generally
A. greater than static frictionB. zeroC. equal to static frictionD. less than static frictionSection II — Free Response
A 3 kg block slides from rest down a frictionless ramp of height 4 m. (a) Find speed at the bottom using energy. (b) If friction did work -30 J, find the new speed.
8 points · rubric: (a) energy conservation 4 pts; (b) work-energy with W=-30 J 4 pts.
A 0.5 kg ball moving at 4 m/s hits a wall and rebounds at 3 m/s. Find the impulse delivered to the ball.
5 points · rubric: Impulse = Δp 2 pts, sign/direction 2 pts, magnitude 1 pt.
Explain the difference between mass and weight and compute the weight of a 5 kg object on Earth (g=9.8).
4 points · rubric: Concept 2 pts, calculation 2 pts.
A 12 V battery drives 4 Ω and 2 Ω resistors in series. Find total current and power in the 4 Ω resistor.
6 points · rubric: Series R 2 pts, Ohm's law 2 pts, power 2 pts.
Answer Key
1. 9.8 m/s² downward — Constant g.
2. equal and opposite pairs — Action-reaction pairs.
3. change in kinetic energy — Work-energy theorem.
4. mv — Definition.
5. constant — No horizontal acceleration (ideal).
6. length — T = 2π√(L/g).
7. constant — Newton's 1st law.
8. 9 J — ½mv² = 9 J.
9. hot to cold — Second law.
10. 4 N — F = mv²/r = 4 N.
11. electric potential — Potential difference.
12. momentum — Impulse-momentum theorem.
13. the same current — Series current is common.
14. indices of refraction — n₁sinθ₁ = n₂sinθ₂.
15. less than static friction — Static is larger (stiction).
Free response — rubric notes
1. (a) energy conservation 4 pts; (b) work-energy with W=-30 J 4 pts. · model: (a) mgh = ½mv² → v=√(2gh)=√78.4≈8.9 m/s. (b) ½mv² = mgh - 30 → v≈ 7.3 m/s.
2. Impulse = Δp 2 pts, sign/direction 2 pts, magnitude 1 pt. · model: Δp = m(v_f - v_i) = 0.5(-3 - 4) = -3.5 N·s; impulse magnitude 3.5 N·s away from wall.
3. Concept 2 pts, calculation 2 pts. · model: Mass is amount of matter (kg), weight is gravitational force (W=mg) = 49 N.
4. Series R 2 pts, Ohm's law 2 pts, power 2 pts. · model: R_total = 6 Ω, I = 12/6 = 2 A, P = I²R = 16 W.
Full-length study package exam
AP Physics 1 — Full Practice Exam
Time: 3 hours (90 minutes Section I + 90 minutes Section II)
Calculator and equation sheet allowed. Use g = 9.8 m/s² unless otherwise noted.
Section I: Multiple-Choice Questions (50 questions, 90 minutes)
Questions 1–3 refer to the following.
A car initially moving at 15 m/s brakes uniformly and comes to rest in 6 seconds.
1. What is the car's acceleration during braking? (A) −90 m/s² (B) −2.5 m/s² (C) 2.5 m/s² (D) 90 m/s²
2. How far does the car travel while braking? (A) 25 m (B) 45 m (C) 90 m (D) 180 m
3. Which graph best represents the car's velocity versus time during braking? (A) A horizontal line (B) A line with positive slope (C) A line with negative slope starting from positive v (D) A parabola opening downward
4. A ball is thrown horizontally from a cliff and lands 60 m from the base. If the cliff is 45 m high, what was the ball's initial horizontal speed? (A) 10 m/s (B) 14 m/s (C) 20 m/s (D) 28 m/s
5. An object's position-time graph is a straight line with a negative slope. The object is: (A) At rest (B) Moving with constant positive velocity (C) Moving with constant negative velocity (D) Accelerating
6. Two forces act on a 3 kg object: 8 N east and 5 N west. The object's acceleration is: (A) 1.0 m/s² east (B) 4.3 m/s² east (C) 1.0 m/s² west (D) 4.3 m/s² west
7. A 5 kg box sits on a surface with μs = 0.4 and μk = 0.3. A 30 N horizontal force is applied. The friction force on the box is: (A) 0 N (B) 14.7 N (C) 19.6 N (D) 30 N
8. An elevator accelerates upward at 3 m/s². A 50 kg person in the elevator experiences a normal force of: (A) 150 N (B) 340 N (C) 490 N (D) 640 N
9. A 4 kg block is pushed across a floor by a 20 N horizontal force against a kinetic friction of 8 N. The block's acceleration is: (A) 2.0 m/s² (B) 3.0 m/s² (C) 5.0 m/s² (D) 7.0 m/s²
10. Which of the following is a Newton's Third Law pair? (A) The normal force and weight of a book on a table (B) A person pushing on a wall and the wall pushing on the person (C) The applied force and friction on a sliding box (D) Tension and weight on a hanging object
11. A 1200 kg car rounds a 50 m radius curve at 20 m/s on a flat road. The required centripetal force is: (A) 240 N (B) 480 N (C) 4800 N (D) 9600 N
12. If the distance between two masses is doubled, the gravitational force between them becomes: (A) Half (B) One-fourth (C) One-third (D) Double
13. A satellite in a circular orbit moves to a higher orbit. What happens to its speed? (A) Increases (B) Decreases (C) Stays the same (D) Depends on satellite mass
14. A ball on a string in a vertical circle has minimum tension at the: (A) Top of the circle (B) Bottom of the circle (C) Midpoint going up (D) Tension is the same everywhere
15. A 10 N force pushes a box 3 m in the direction of the force. The work done is: (A) 3.3 J (B) 10 J (C) 13 J (D) 30 J
16. A spring with k = 200 N/m is compressed 0.1 m. The potential energy stored is: (A) 0.5 J (B) 1.0 J (C) 2.0 J (D) 10 J
17. A 3 kg object at a height of 10 m has a gravitational potential energy of: (take reference at ground level) (A) 30 J (B) 98 J (C) 294 J (D) 300 J
18. A 2 kg ball is dropped from 20 m. Neglecting air resistance, its speed at 10 m is: (A) 7.0 m/s (B) 9.9 m/s (C) 14 m/s (D) 20 m/s
19. A motor does 5000 J of work in 10 s. Its power output is: (A) 50 W (B) 250 W (C) 500 W (D) 5000 W
20. A 0.5 kg ball moving at 8 m/s has a momentum of: (A) 2.0 kg·m/s (B) 4.0 kg·m/s (C) 8.0 kg·m/s (D) 16 kg·m/s
21. A 1000 kg car moving at 10 m/s brakes to rest in 5 s. The average braking force is: (A) 500 N (B) 1000 N (C) 2000 N (D) 5000 N
22. A 6 kg object moving at 4 m/s collides with a 2 kg object at rest. They stick together. The final speed is: (A) 1.0 m/s (B) 2.0 m/s (C) 3.0 m/s (D) 4.0 m/s
23. In a perfectly elastic collision between two objects, which quantities are conserved? (A) Momentum only (B) Kinetic energy only (C) Both momentum and kinetic energy (D) Neither
24. A force of 40 N acts on a 2 kg object for 0.1 s. The impulse is: (A) 2.0 N·s (B) 4.0 N·s (C) 8.0 N·s (D) 20 N·s
25. A spring-mass system has a period of 1.0 s. If the mass is quadrupled, the new period is: (A) 0.25 s (B) 0.5 s (C) 2.0 s (D) 4.0 s
26. A pendulum has a period of 2.0 s on Earth. On the Moon (g ≈ 1.6 m/s²), the period is approximately: (A) 2.0 s (B) 2.8 s (C) 4.9 s (D) 5.6 s
27. At the equilibrium position of a spring-mass oscillator, the: (A) Acceleration and velocity are both maximum (B) Acceleration is maximum, velocity is zero (C) Acceleration is zero, velocity is maximum (D) Acceleration and velocity are both zero
28. A uniform rod of length L and mass M is pivoted at one end. Its moment of inertia is: (A) ¹⁄₁₂ML² (B) ¹⁄₃ML² (C) ½ML² (D) ML²
29. A force of 20 N is applied perpendicular to a wrench 0.3 m from the bolt. The torque is: (A) 3.0 N·m (B) 6.0 N·m (C) 6.7 N·m (D) 60 N·m
30. A figure skater spinning with arms extended pulls them in. What happens to her angular speed? (A) Decreases (B) Stays the same (C) Increases (D) Depends on her mass
31. A solid disk and a hoop of the same mass and radius roll down the same incline without slipping. Which has the greater speed at the bottom? (A) The hoop (B) The disk (C) Both the same (D) Depends on the angle
32. The rotational kinetic energy of an object is given by: (A) ½mv² (B) ½Iω² (C) Iα (D) τθ
33. A projectile is launched at 45° with speed v. If the launch angle is increased to 60° (same speed), the maximum height: (A) Decreases (B) Increases (C) Stays the same (D) Becomes zero
34. An object in uniform circular motion has: (A) Constant velocity and constant acceleration (B) Changing velocity and constant acceleration magnitude (C) Constant velocity and changing acceleration (D) Changing velocity and zero acceleration
35. A box is pulled across a rough floor at constant speed. The work done by the applied force equals: (A) Zero (B) The change in kinetic energy (C) The work done by friction (D) The work done by gravity
36. In a perfectly inelastic collision, which is true? (A) Objects bounce off each other with the same speeds (B) Maximum kinetic energy is conserved (C) Objects stick together and momentum is conserved (D) Objects stick together and kinetic energy is conserved
37. A pendulum's period depends on: (A) Mass and length (B) Length and g (C) Mass and amplitude (D) Amplitude and g
38. An object in free fall near Earth's surface has an acceleration of: (A) 0 m/s² (B) 9.8 m/s² upward (C) 9.8 m/s² downward (D) Varies with mass
39. Which force does NO work on an object sliding along a horizontal surface? (A) Applied force (B) Friction (C) Normal force (D) Air resistance
40. The area under a force-time graph represents: (A) Work (B) Impulse (C) Momentum (D) Kinetic energy
41. A 5 kg object is at rest. A net force of 10 N acts on it for 3 s. What is its final velocity? (A) 1.5 m/s (B) 6.0 m/s (C) 15 m/s (D) 30 m/s
42. Two identical springs in parallel (side by side) supporting a mass have an effective spring constant of: (A) k/2 (B) k (C) 2k (D) 4k
43. For an object in rotational equilibrium: (A) ΣF = 0 only (B) Στ = 0 only (C) Both ΣF = 0 and Στ = 0 (D) Neither is required
44. A satellite in circular orbit has a period T and orbital radius r. If r is increased to 4r, the new period is: (A) 2T (B) 4T (C) 8T (D) 16T
45. The normal force on an object on an inclined plane equals: (A) mg (B) mg sin θ (C) mg cos θ (D) mg tan θ
46. A ball is thrown straight up. At the highest point, the ball's: (A) Velocity and acceleration are both zero (B) Velocity is zero, acceleration is g downward (C) Velocity is nonzero, acceleration is zero (D) Velocity and acceleration are both g
47. An object's speed doubles. Its kinetic energy: (A) Doubles (B) Quadruples (C) Increases by a factor of 8 (D) Stays the same
48. A wheel rotates from rest with angular acceleration 2 rad/s². After 5 s, its angular displacement is: (A) 5 rad (B) 10 rad (C) 25 rad (D) 50 rad
49. The center of mass of a system moves according to: (A) The net internal force (B) The net external force (C) The total internal torque (D) The sum of all velocities
50. A mass m on a spring with constant k oscillates with amplitude A. The total energy is: (A) ½kA (B) kA² (C) ½kA² (D) ½mv²
Section II: Free-Response Questions (5 questions, 90 minutes)
Question 1: Experimental Design
A student wants to determine the acceleration due to gravity using a simple pendulum.
The student uses a pendulum bob of mass 0.100 kg attached to a string of length 1.00 m. The pendulum is displaced by a small angle and released. The student measures the time for 20 complete oscillations to be 40.2 s.
(a) Calculate the experimental value of g from this data.
(b) The student wants to graphically determine g. The student collects data for pendulums of various lengths and measures the period for each. What should the student graph (which variables on which axes) to obtain a straight line from which g can be determined? Explain your reasoning.
(c) The student claims that using a heavier bob would give a more accurate result for g. Evaluate this claim using physics principles.
(d) Describe how the student could modify the experiment to improve the accuracy of the period measurement. Identify one source of experimental error and describe how to reduce it.
(e) If the experiment were performed on the Moon (g ≈ 1.6 m/s²) with the same 1.00 m pendulum, predict the time for 20 oscillations.
Question 2: Qualitative/Quantitative Translation
A 2.0 kg block is released from rest at the top of a curved frictionless ramp of height 3.0 m. At the bottom of the ramp, the block slides onto a rough horizontal surface (μk = 0.4) for a distance d before coming to rest.
(a) Calculate the speed of the block at the bottom of the ramp.
(b) Calculate the distance d the block slides on the rough surface.
(c) On a single set of axes, sketch a graph of the block's kinetic energy as a function of position from the top of the ramp (x = 0) to the point where it stops. Label the key positions (bottom of ramp, stopping point) with numerical values.
(d) Suppose the ramp has a small amount of friction. Explain qualitatively how this would affect the speed at the bottom of the ramp and the distance d on the horizontal surface.
Question 3: Short Answer
Two blocks (m₁ = 3 kg and m₂ = 5 kg) are connected by a light string over a frictionless, massless pulley. Block m₁ rests on a rough horizontal surface (μk = 0.2), and block m₂ hangs vertically. The system is released from rest.
(a) Draw free-body diagrams for each block.
(b) Calculate the acceleration of the system.
(c) Calculate the tension in the string.
Question 4: Short Answer
A 0.15 kg baseball is thrown horizontally at 30 m/s toward a batter. The batter hits the ball, and it leaves the bat at 40 m/s horizontally in the opposite direction. The contact time between bat and ball is 0.002 s.
(a) Calculate the magnitude of the average force exerted on the ball by the bat.
(b) Is the momentum of the ball conserved during the collision? Explain.
(c) Explain why the batter follows through with the swing (keeps the bat moving after contact). Use the impulse-momentum theorem in your explanation.
Question 5: Short Answer
A solid disk (M = 4.0 kg, R = 0.20 m) is free to rotate about a frictionless axle through its center. A light string is wrapped around the disk and a 1.0 kg mass hangs from the free end. The system is released from rest.
(a) Calculate the moment of inertia of the disk.
(b) Calculate the linear acceleration of the hanging mass.
(c) Calculate the angular speed of the disk after the mass has fallen 0.5 m.
END OF EXAM
Answer Key & Rubric
AP Physics 1 — Full Practice Exam: Answer Key and Detailed Solutions
Section I: Multiple-Choice Answers
1. B. a = (v − v₀)/t = (0 − 15)/6 = −2.5 m/s². Negative indicates deceleration.
2. B. Δx = ½(v₀ + v)t = ½(15 + 0)(6) = 45 m. Alternatively: Δx = v₀t + ½at² = 15(6) + ½(−2.5)(36) = 90 − 45 = 45 m.
3. C. Uniform deceleration from positive velocity produces a straight line with negative slope on a v-t graph.
4. C. Time to fall: 45 = ½(9.8)t² → t² = 9.184 → t = 3.03 s. Horizontal: v = Δx/t = 60/3.03 = 19.8 m/s ≈ 20 m/s.
5. C. A straight line on a position-time graph means constant velocity. Negative slope means negative velocity (moving in the negative direction).
6. A. Net force = 8 − 5 = 3 N east. a = F/m = 3/3 = 1.0 m/s² east.
7. D. Maximum static friction: μsFn = 0.4(5)(9.8) = 19.6 N. Since the applied force (30 N) exceeds 19.6 N, the box moves, and kinetic friction applies. Wait — the box IS moving, so kinetic friction: Ff = μkFn = 0.3(49) = 14.7 N. The friction force is 14.7 N (B). But the question asks for friction force once the box is moving: since 30 > 19.6, the box accelerates. Friction = μkFn = 14.7 N. Answer: B.
8. D. Fn − mg = ma → Fn = m(g + a) = 50(9.8 + 3) = 50(12.8) = 640 N.
9. B. Net force = 20 − 8 = 12 N. a = 12/4 = 3.0 m/s².
10. B. A Newton's Third Law pair involves two objects and two forces that are equal and opposite. The person pushes on the wall, and the wall pushes on the person — two objects, two forces. In option A, both forces act on the book. In C, applied force and friction are not a Third Law pair (different origins). In D, tension and weight on a hanging object both act on the same object.
11. D. Fc = mv²/r = 1200(20)²/50 = 1200(400)/50 = 1200(8) = 9600 N.
12. B. F ∝ 1/r². If r doubles, F becomes 1/4 of original.
13. B. v = √(GM/r). Larger r → smaller v.
14. A. At the top: T + mg = mv²/r → T = mv²/r − mg. At the bottom: T − mg = mv²/r → T = mv²/r + mg. The bottom always has greater tension. Minimum at the top.
15. D. W = Fd cos θ = 10(3)cos 0° = 10(3)(1) = 30 J.
16. B. PEs = ½kx² = ½(200)(0.1)² = ½(200)(0.01) = 1.0 J.
17. C. PEg = mgh = 3(9.8)(10) = 294 J.
18. C. Energy conservation from 20 m to 10 m: mg(20) = ½mv² + mg(10) → g(10) = ½v² → v = √(2g × 10) = √(196) = 14 m/s.
19. C. P = W/t = 5000/10 = 500 W.
20. B. p = mv = 0.5(8) = 4.0 kg·m/s.
21. C. Impulse = Δp = mΔv = 1000(0 − 10) = −10000 N·s. F = Δp/Δt = −10000/5 = −2000 N. Magnitude: 2000 N.
22. C. Conservation of momentum: 6(4) + 2(0) = (6 + 2)v → 24 = 8v → v = 3.0 m/s.
23. C. In a perfectly elastic collision, both momentum and kinetic energy are conserved.
24. B. Impulse = FΔt = 40(0.1) = 4.0 N·s.
25. C. T = 2π√(m/k). If m → 4m: T → 2T = 2(1.0) = 2.0 s.
26. C. T = 2π√(L/g). On Earth: 2.0 = 2π√(L/9.8) → L = 9.8(2.0/2π)² = 0.994 m. On Moon: T_moon = 2π√(0.994/1.6) = 2π√(0.621) = 2π(0.788) = 4.95 s ≈ 4.9 s.
27. C. At equilibrium, the spring force is zero, so acceleration is zero. Velocity is maximum at equilibrium.
28. B. A uniform rod pivoted at one end: I = ¹⁄₃ML².
29. B. τ = rF sin 90° = 0.3(20)(1) = 6.0 N·m.
30. C. I decreases (mass moves closer to axis), so ω increases (L = Iω conserved).
31. B. The disk has less rotational inertia (I = ½MR² vs. MR²), so more energy goes into translational KE, giving it greater speed.
32. B. KE_rot = ½Iω².
33. B. Maximum height = (v sin θ)²/(2g). At 60°, sin 60° > sin 45°, so height increases.
34. B. Velocity changes direction (so it changes), but the magnitude of centripetal acceleration (v²/r) is constant since speed is constant in UCM.
35. C. At constant speed, net work = 0 (W_net = ΔKE = 0). So work by applied force + work by friction = 0. Work by applied force = −(work by friction) = |work by friction|. Since friction does negative work, the applied force does positive work equal in magnitude to the friction's negative work. C is correct.
36. C. In a perfectly inelastic collision, objects stick together, momentum is conserved, and kinetic energy is not conserved.
37. B. T = 2π√(L/g) — depends on length and gravitational acceleration, not mass or amplitude.
38. C. In free fall, acceleration = g = 9.8 m/s² downward (ignoring air resistance).
39. C. The normal force is perpendicular to the horizontal displacement, so cos 90° = 0, and no work is done.
40. B. The area under a force-time graph equals impulse (FΔt = J = Δp).
41. B. J = FΔt = 10(3) = 30 N·s. Δp = 30 = m(v − 0) = 5v → v = 6.0 m/s.
42. C. Two identical springs in parallel share the load. Each stretches half as much as a single spring would for the same force. Effective k = k₁ + k₂ = 2k.
43. C. Full equilibrium requires both translational (ΣF = 0) and rotational (Στ = 0) equilibrium.
44. C. T ∝ r^(3/2). If r → 4r: T → 4^(3/2) × T = 8T.
45. C. The normal force equals the component of weight perpendicular to the surface: mg cos θ.
46. B. At the highest point, velocity is zero (momentarily), but acceleration is still g downward (gravity still acts).
47. B. KE = ½mv². If v doubles, KE = ½m(2v)² = 4(½mv²). KE quadruples.
48. C. θ = ω₀t + ½αt² = 0 + ½(2)(25) = 25 rad.
49. B. The center of mass accelerates according to the net external force: ΣF_ext = M_total × a_cm.
50. C. Total energy in SHM = maximum PE = ½kA².
Section II: Free-Response Answers
Question 1: Experimental Design (Pendulum)
(a) Period for one oscillation: T = 40.2/20 = 2.01 s. Using T = 2π√(L/g): 2.01 = 2π√(1.00/g) (2.01/2π)² = 1.00/g (0.320)² = 1.00/g 0.1024 = 1.00/g g = 1.00/0.1024 = 9.77 m/s²
(b) Square the pendulum period equation: T² = 4π²L/g.
Rearranging: T² = (4π²/g)L
This is in the form y = mx, where:
- y = T² (dependent variable, on vertical axis)
- x = L (independent variable, on horizontal axis)
- slope = 4π²/g
The student should graph T² vs. L to obtain a straight line through the origin. From the slope m_slope = 4π²/g, the student can calculate g = 4π²/m_slope.
(c) The student's claim is incorrect. The period of a simple pendulum T = 2π√(L/g) is independent of mass. A heavier bob does not change the period and therefore cannot improve the accuracy of the g measurement. The bob's mass does not appear in the formula. A heavier bob might reduce the effect of air resistance slightly, but for small angles and typical lab conditions, this effect is negligible.
(d) Source of error: Reaction time in starting/stopping the timer can introduce random error in the period measurement.
Improvement: Time 20 or more oscillations (which the student already does) and divide by the number of oscillations. This divides the reaction-time error by the number of oscillations. Additionally, the student could use a photogate timer or motion sensor for more precise timing, or use video analysis.
(e) On the Moon: T = 2π√(L/g_moon) = 2π√(1.00/1.6) = 2π√(0.625) = 2π(0.7906) = 4.968 s. Time for 20 oscillations = 20 × 4.968 = 99.4 s.
Question 2: Qualitative/Quantitative Translation (Energy)
(a) Using energy conservation on the frictionless ramp: mgh = ½mv² (2.0)(9.8)(3.0) = ½(2.0)v² 58.8 = v² v = 7.67 m/s
(b) On the rough surface, kinetic energy converts to thermal energy: ½mv² = Ff × d = μk × mg × d ½(2.0)(7.67²) = 0.4 × 2.0 × 9.8 × d ½(2.0)(58.8) = 7.84d 58.8 = 7.84d d = 7.50 m
(c) The KE graph should show:
- At x = 0 (top of ramp, height = 3.0 m): KE = 0 J
- At x = bottom of ramp: KE = 58.8 J (vertical jump from the ramp curve to 58.8 J; note: KE increases gradually along the ramp as PE decreases)
- Actually, on the ramp, KE increases linearly with position (since PE decreases linearly with height, and on a straight ramp, height decreases linearly with position). From the top to the bottom of the ramp: KE increases from 0 to 58.8 J.
- On the horizontal surface: KE decreases linearly from 58.8 J to 0 J over the distance d = 7.50 m.
- The graph looks like a triangle (ramp up) followed by a triangle (linear decrease on the flat surface).
Key values to label: (0, 0), (bottom of ramp, 58.8 J), (bottom of ramp + 7.50 m, 0 J).
(d) If the ramp has friction:
- Some mechanical energy is lost to friction on the ramp, so the block arrives at the bottom with less speed than 7.67 m/s (less KE than 58.8 J).
- Since the block has less KE at the bottom, it will stop in a shorter distance d on the horizontal surface (less KE to dissipate via friction).
Question 3: Short Answer (Atwood-like with Friction)
(a)
- Block m₁ (on table): Tension T to the right (toward pulley), kinetic friction Ff = μk(m₁g) to the left, weight m₁g down, normal force Fn = m₁g up.
- Block m₂ (hanging): Tension T upward, weight m₂g downward.
(b) For m₁ (horizontal, right is positive): T − μk(m₁g) = m₁a For m₂ (vertical, down is positive): m₂g − T = m₂a
Adding equations: m₂g − μk(m₁g) = (m₁ + m₂)a a = (m₂g − μk(m₁g))/(m₁ + m₂) a = (5.0 × 9.8 − 0.2 × 3.0 × 9.8)/(3.0 + 5.0) a = (49 − 5.88)/8 a = 43.12/8 a = 5.39 m/s²
(c) From the equation for m₂: T = m₂(g − a) = 5.0(9.8 − 5.39) = 5.0(4.41) = 22.1 N.
Check with m₁: T = m₁a + μk(m₁g) = 3.0(5.39) + 0.2(3.0)(9.8) = 16.17 + 5.88 = 22.05 N ≈ 22.1 N ✓
Question 4: Short Answer (Impulse)
(a) Let the initial direction (toward the batter) be positive.
- Initial momentum: p_i = mv_i = 0.15(30) = 4.5 kg·m/s
- Final momentum: p_f = mv_f = 0.15(−40) = −6.0 kg·m/s
- Change in momentum: Δp = p_f − p_i = −6.0 − 4.5 = −10.5 kg·m/s
- Force: F = Δp/Δt = −10.5/0.002 = −5250 N
- Magnitude of the average force: 5250 N (direction is opposite to the initial pitch, i.e., away from the batter).
(b) No, the momentum of the ball is not conserved. Momentum is conserved for a SYSTEM when the net external force is zero. The ball alone experiences an external force from the bat, so its momentum changes. The momentum of the ball + bat SYSTEM is approximately conserved (if we neglect external forces during the brief collision), but the ball's individual momentum is not.
(c) Following through increases the contact time Δt between the bat and the ball. By the impulse-momentum theorem: J = FΔt = Δp. For a desired change in momentum Δp (to send the ball at high speed), a larger Δt means a smaller average force is needed. But more importantly for the batter, following through ensures the bat maintains force on the ball throughout a longer contact, maximizing the impulse delivered. If the batter stopped the swing at contact, the bat would decelerate and the contact time would be very short, reducing the total impulse.
Question 5: Short Answer (Rotation)
(a) Moment of inertia of a solid disk: I = ½MR² = ½(4.0)(0.20)² = ½(4.0)(0.04) = 0.080 kg·m².
(b) For the hanging mass (Newton's Second Law, down positive): mg − T = ma ... (i)
For the disk (rotational): TR = Iα = I(a/R) [since a = αR for no slipping] T = Ia/R² = (I/R²)a ... (ii)
Substituting (ii) into (i): mg − (I/R²)a = ma mg = ma + (I/R²)a = a(m + I/R²) a = mg/(m + I/R²) a = (1.0)(9.8)/(1.0 + 0.080/0.04) a = 9.8/(1.0 + 2.0) a = 9.8/3.0 a = 3.27 m/s²
(c) Using energy conservation: The mass falls 0.5 m, so gravitational PE converts to translational KE of the mass plus rotational KE of the disk: mgh = ½mv² + ½Iω² Since ω = v/R: mgh = ½mv² + ½I(v/R)² = ½mv² + ½(I/R²)v² = v²(m/2 + I/(2R²)) (1.0)(9.8)(0.5) = v²(0.5 + 0.080/0.08) = v²(0.5 + 1.0) = 1.5v² 4.9 = 1.5v² v² = 3.267 v = 1.81 m/s
Angular speed: ω = v/R = 1.81/0.20 = 9.05 rad/s
Alternatively, using kinematics: v² = 2aΔy = 2(3.27)(0.5) = 3.27 v = 1.81 m/s ω = v/R = 1.81/0.20 = 9.05 rad/s
END OF ANSWER KEY