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AP Physics C Mechanics

AP Physics C Mechanics Practice Test: Practice Test 2

Practice Test 2 for AP Physics C Mechanics: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Two objects, a rubber ball and a clay ball of identical mass, are thrown with the same speed toward a brick wall. The rubber ball bounces back with nearly the same speed, while the clay ball sticks to the wall. Which statement correctly compares the impulse delivered by the wall to each ball?

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Question 1

Two objects, a rubber ball and a clay ball of identical mass, are thrown with the same speed toward a brick wall. The rubber ball bounces back with nearly the same speed, while the clay ball sticks to the wall. Which statement correctly compares the impulse delivered by the wall to each ball?

  1. The clay ball experiences a greater impulse because it undergoes a perfectly inelastic collision.
  2. The rubber ball experiences a greater impulse because its change in momentum is larger. (correct answer)
  3. Both balls experience the same impulse because their mass and initial speed are identical.
  4. The clay ball experiences a greater impulse because the contact time with the wall is longer.

Explanation: Impulse equals the change in momentum (Δp\Delta pΔp). The clay ball's momentum changes from mvmvmv to 0, so ∣Δpclay∣=mv|\Delta p_{clay}| = mv∣Δpclay​∣=mv. The rubber ball's momentum changes from mvmvmv to approximately −mv-mv−mv, so ∣Δprubber∣≈2mv|\Delta p_{rubber}| \approx 2mv∣Δprubber​∣≈2mv. Since the rubber ball has a larger change in momentum, it experiences a greater impulse.

Question 2

The potential energy of a particle undergoing one-dimensional simple harmonic motion is given by the function U(x)=(2.0 J/m2)x2U(x) = (2.0 \text{ J/m}^2)x^2U(x)=(2.0 J/m2)x2. If the total mechanical energy of the particle is 8.0 J, what is the amplitude of the oscillation?

  1. 0.5 m
  2. 1.0 m
  3. 2.0 m (correct answer)
  4. 4.0 m

Explanation: The total mechanical energy EEE is equal to the maximum potential energy, which occurs at the amplitude, x=Ax=Ax=A. So, E=U(A)E = U(A)E=U(A). We have 8.0 J=(2.0 J/m2)A28.0 \text{ J} = (2.0 \text{ J/m}^2)A^28.0 J=(2.0 J/m2)A2. Solving for A2A^2A2 gives A2=8.0/2.0=4.0 m2A^2 = 8.0/2.0 = 4.0 \text{ m}^2A2=8.0/2.0=4.0 m2. Taking the square root gives the amplitude A=2.0A = 2.0A=2.0 m.

Question 3

A roller coaster car of mass mmm travels along a frictionless track with several hills and valleys. Which of the following statements about the work done by the normal force, WNW_NWN​, exerted by the track on the car is correct?

  1. WNW_NWN​ is always positive because the normal force must support the car's weight.
  2. WNW_NWN​ is always zero because the normal force is always perpendicular to the car's velocity. (correct answer)
  3. WNW_NWN​ is positive on uphill sections and negative on downhill sections to assist the motion.
  4. WNW_NWN​ depends on the car's speed because a higher speed requires a larger normal force in curves.

Explanation: Work is defined by the dot product of force and displacement, W=∫F⃗⋅ds⃗W = \int \vec{F} \cdot d\vec{s}W=∫F⋅ds. The normal force is, by definition, always perpendicular to the surface of the track. The instantaneous displacement (and thus velocity) of the car is always tangent to the track. Since the normal force and displacement vectors are always perpendicular, their dot product is zero, and the normal force does no work.

Question 4

A satellite in elliptical orbit obeys Kepler’s second law; what happens to its speed as it approaches perigee?

  1. Speed decreases because angular momentum must decrease near Earth.
  2. Speed increases because angular momentum conservation requires larger vvv at smaller rrr. (correct answer)
  3. Speed stays constant because gravity does no work in orbit.
  4. Speed increases because gravitational potential energy increases at smaller rrr.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). Kepler's second law states that a satellite sweeps out equal areas in equal times, which is a consequence of angular momentum conservation. As a satellite in elliptical orbit approaches perigee (closest point to Earth), its distance r from Earth decreases while angular momentum L = mrv remains constant. Choice B is correct because it accurately applies angular momentum conservation: as r decreases at perigee, v must increase to maintain constant L = mrv. Choice D is incorrect because it misstates the energy relationship - gravitational potential energy actually becomes more negative (decreases) at smaller r. To help students: Emphasize the connection between Kepler's second law and angular momentum conservation, use area-sweeping diagrams to visualize the law, and practice calculating speeds at apogee and perigee using L conservation. Show energy conservation as a separate constraint.

Question 5

A 1500 kg car moves at 10 m/s and then speeds up to 25 m/s on level ground. Calculate the change in translational kinetic energy.

  1. 4.50×105 J4.50\times10^5\ \text{J}4.50×105 J
  2. 6.25×105 J6.25\times10^5\ \text{J}6.25×105 J
  3. 3.94×105 J3.94\times10^5\ \text{J}3.94×105 J (correct answer)
  4. 2.81×105 J2.81\times10^5\ \text{J}2.81×105 J

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding and calculating changes in translational kinetic energy. The change in kinetic energy is calculated as ΔKE = KE_f - KE_i = ½mv_f² - ½mv_i². In this scenario, the 1500 kg car speeds up from 10 m/s to 25 m/s, so ΔKE = ½(1500)(25²) - ½(1500)(10²) = ½(1500)(625 - 100) = ½(1500)(525) = 393,750 J ≈ 3.94×10⁵ J. Choice C is correct because it properly calculates the difference between final and initial kinetic energies. Choice B might calculate only the final kinetic energy without subtracting the initial, while choice A might use an incorrect velocity difference. To help students: Emphasize that kinetic energy change requires calculating both initial and final values separately before subtracting. Practice problems involving velocity changes to reinforce that KE is proportional to v², not v.

Question 6

A uniform sphere of mass MMM and radius RRR is in rotational equilibrium when suspended by a string attached to a point on its surface. A small mass mmm is then attached to the opposite point on the sphere's surface. What is the angle θ\thetaθ that the string makes with the vertical when the system reaches new equilibrium?

  1. sin⁡θ=mR(M+m)R=mM+m\sin\theta = \frac{mR}{(M + m)R} = \frac{m}{M + m}sinθ=(M+m)RmR​=M+mm​
  2. tan⁡θ=mR(M+m)R=mM+m\tan\theta = \frac{mR}{(M + m)R} = \frac{m}{M + m}tanθ=(M+m)RmR​=M+mm​
  3. tan⁡θ=2mR(M+m)R=2mM+m\tan\theta = \frac{2mR}{(M + m)R} = \frac{2m}{M + m}tanθ=(M+m)R2mR​=M+m2m​ (correct answer)
  4. sin⁡θ=2mR(M+m)R=2mM+m\sin\theta = \frac{2mR}{(M + m)R} = \frac{2m}{M + m}sinθ=(M+m)R2mR​=M+m2m​

Explanation: When you encounter rotational equilibrium problems involving suspended objects, you need to analyze both the forces and torques acting on the system. The key insight is identifying the pivot point and applying the condition that net torque equals zero. Initially, the sphere hangs vertically because its center of mass aligns directly below the suspension point. When mass mmm is added to the opposite side, the system's center of mass shifts, requiring the sphere to tilt to reach a new equilibrium position. For the new equilibrium, choose the suspension point as your pivot. The sphere's weight (M+m)g(M+m)g(M+m)g acts downward through the system's center of mass, which is now displaced from the sphere's geometric center due to the added mass mmm. The horizontal displacement of the center of mass from the sphere's original center is mRM+m\frac{mR}{M+m}M+mmR​ (using the center of mass formula). When the string makes angle θ\thetaθ with the vertical, the total weight creates a restoring torque. The horizontal distance from the suspension point to the line of action of the total weight is Rsin⁡θ+mRM+mR\sin\theta + \frac{mR}{M+m}Rsinθ+M+mmR​. For equilibrium, this torque must equal the torque from the shifted center of mass. Setting up the torque balance: (M+m)g(Rsin⁡θ)=mg⋅2Rcos⁡θ(M+m)g(R\sin\theta) = mg \cdot 2R\cos\theta(M+m)g(Rsinθ)=mg⋅2Rcosθ (where 2R2R2R is the diameter). Simplifying gives tan⁡θ=2mM+m\tan\theta = \frac{2m}{M+m}tanθ=M+m2m​. Answer A uses sine instead of tangent. Answer B omits the factor of 2 from the diameter. Answer D uses sine with the correct coefficient but wrong trigonometric function. Study tip: In rotational equilibrium problems, always identify your pivot point first, then carefully track the perpendicular distances for each torque term.

Question 7

A block is attached to a horizontal spring with spring constant kkk. The block is pulled from its equilibrium position at x=0x=0x=0 to a position x=Ax=Ax=A. What is the work done by the spring on the block during this displacement?

  1. 12kA2\frac{1}{2}kA^221​kA2
  2. −12kA2-\frac{1}{2}kA^2−21​kA2 (correct answer)
  3. −kA2-kA^2−kA2
  4. kA2kA^2kA2

Explanation: The force exerted by the spring is Fs=−kxF_s = -kxFs​=−kx. The work done by the spring is the integral of this force over the displacement: Ws=∫0A(−kx)dx=−k[x22]0A=−12kA2W_s = \int_0^A (-kx) dx = -k\left[\frac{x^2}{2}\right]_0^A = -\frac{1}{2}kA^2Ws​=∫0A​(−kx)dx=−k[2x2​]0A​=−21​kA2. The work is negative because the spring force opposes the direction of displacement.

Question 8

Rotational inertia is the physical quantity that quantifies an object's...

  1. resistance to a change in its rotational motion. (correct answer)
  2. resistance to a change in its translational motion.
  3. total angular momentum while it is rotating at a constant speed.
  4. kinetic energy stored in its rotation for a given angular velocity.

Explanation: Rotational inertia (III) is the rotational analog of mass (inertia). Just as mass measures an object's resistance to a change in its linear velocity (i.e., its resistance to linear acceleration), rotational inertia measures an object's resistance to a change in its angular velocity (i.e., its resistance to angular acceleration).

Question 9

A spacecraft with I=80 kg\cdotpm2I=80\ \text{kg·m}^2I=80 kg\cdotpm2 must reach ω=0.60 rad/s\omega=0.60\ \text{rad/s}ω=0.60 rad/s from rest; during a short burn, external torques are negligible except for the thrusters, so J=ΔL=IΔωJ=\Delta L=I\Delta\omegaJ=ΔL=IΔω. Assume the torque direction matches the desired rotation. Considering the described system, calculate the angular impulse applied to the spacecraft during the interval.

  1. 48 N·m·s (correct answer)
  2. 133 N·m·s
  3. 48 N·m
  4. 0.48 N·m·s

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and is conserved in a closed system. Angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the spacecraft needs to change from rest (ωi = 0) to ωf = 0.60 rad/s, requiring a change in angular momentum. Choice A is correct because the angular impulse J = ΔL = IΔω = I(ωf - ωi) = 80 × (0.60 - 0) = 48 N·m·s. Choice B is incorrect because it might result from a calculation error or misunderstanding of the initial conditions, possibly assuming a different initial angular velocity. To help students: Emphasize careful reading of initial conditions (starting from rest means ωi = 0). Practice problems with various initial states to build pattern recognition. Watch for: assumptions about initial conditions and ensure students identify when objects start from rest versus already in motion.

Question 10

A parachutist of mass 70 kg reaches terminal velocity when the drag force equals 686 N. If the drag coefficient and air density remain constant, what would be the terminal velocity of the same parachutist carrying an additional 30 kg of equipment?

  1. The terminal velocity increases by a factor of 100/70=1.20\sqrt{100/70} = 1.20100/70​=1.20 (correct answer)
  2. The terminal velocity increases by a factor of 70/100=0.84\sqrt{70/100} = 0.8470/100​=0.84
  3. The terminal velocity decreases by a factor of 100/70=1.20\sqrt{100/70} = 1.20100/70​=1.20
  4. The terminal velocity decreases by a factor of 70/100=0.84\sqrt{70/100} = 0.8470/100​=0.84

Explanation: At terminal velocity, mg=Fd=12ρCdAv2mg = F_d = \frac{1}{2}\rho C_d A v^2mg=Fd​=21​ρCd​Av2. For constant air density, drag coefficient, and area: v∝mgv \propto \sqrt{mg}v∝mg​. The ratio is v2/v1=m2/m1=100/70=1.20v_2/v_1 = \sqrt{m_2/m_1} = \sqrt{100/70} = 1.20v2​/v1​=m2​/m1​​=100/70​=1.20. Terminal velocity increases because more weight requires higher speed to generate equal drag. Choice B inverts the mass ratio. Choices C and D incorrectly suggest velocity decreases with increased mass.

Question 11

A physical pendulum consists of a rigid object of mass MMM that oscillates about a pivot point a distance ddd from its center of mass. It is released from rest at a small maximum angular displacement θmax\theta_{max}θmax​. Which of the following changes will increase the total mechanical energy of the pendulum-Earth system?

  1. Decreasing the mass MMM of the pendulum while keeping ddd and θmax\theta_{max}θmax​ constant.
  2. Moving the pivot point closer to the center of mass, decreasing ddd.
  3. Increasing the mass MMM of the pendulum while keeping ddd and θmax\theta_{max}θmax​ constant. (correct answer)
  4. Performing the experiment on a planet with a smaller acceleration due to gravity.

Explanation: The total mechanical energy is determined by the maximum gravitational potential energy. The maximum height of the center of mass above its equilibrium position is h=d(1−cos⁡θmax)h = d(1-\cos\theta_{max})h=d(1−cosθmax​). The total energy is E=Mgh=Mgd(1−cos⁡θmax)E = Mgh = Mgd(1-\cos\theta_{max})E=Mgh=Mgd(1−cosθmax​). To increase EEE, one must increase MMM, ggg, ddd, or θmax\theta_{max}θmax​. Increasing the mass MMM while other factors are constant will increase the total energy.

Question 12

Two hockey pucks slide on frictionless ice. Puck A (mass 2kg) moves east at 8 m/s, and puck B (mass 3kg) moves north at 6 m/s. They collide and stick together. What is the magnitude of their velocity immediately after the collision?

  1. 4.0 m/s
  2. 5.2 m/s (correct answer)
  3. 6.4 m/s
  4. 7.0 m/s

Explanation: This is a perfectly inelastic collision in two dimensions. Using conservation of momentum: In the x-direction: px=(2)(8)+(3)(0)=16p_x = (2)(8) + (3)(0) = 16px​=(2)(8)+(3)(0)=16 kg⋅m/s. In the y-direction: py=(2)(0)+(3)(6)=18p_y = (2)(0) + (3)(6) = 18py​=(2)(0)+(3)(6)=18 kg⋅m/s. Total mass after collision is 5 kg. Final velocity components: vx=16/5=3.2v_x = 16/5 = 3.2vx​=16/5=3.2 m/s, vy=18/5=3.6v_y = 18/5 = 3.6vy​=18/5=3.6 m/s. Magnitude: v=(3.2)2+(3.6)2=10.24+12.96=23.2=4.82v = \sqrt{(3.2)^2 + (3.6)^2} = \sqrt{10.24 + 12.96} = \sqrt{23.2} = 4.82v=(3.2)2+(3.6)2​=10.24+12.96​=23.2​=4.82 m/s ≈ 5.2 m/s.

Question 13

A block slides down a frictionless inclined plane of height hhh and angle θ\thetaθ. At the bottom, it encounters a rough horizontal surface with coefficient of kinetic friction μk\mu_kμk​ and slides a distance ddd before coming to rest. What is the total work done by friction on the block during its entire motion?

  1. −μkmgd-\mu_k mg d−μk​mgd (correct answer)
  2. −mgh-mgh−mgh
  3. −μkmgdcos⁡θ-\mu_k mg d \cos\theta−μk​mgdcosθ
  4. −μkmg(d+hcot⁡θ)-\mu_k mg(d + h\cot\theta)−μk​mg(d+hcotθ)

Explanation: Friction only acts on the rough horizontal surface, not on the frictionless incline. The work done by friction is Wf=−fk⋅d=−μkmgdW_f = -f_k \cdot d = -\mu_k mg dWf​=−fk​⋅d=−μk​mgd. Choice B represents the total energy dissipated but doesn't specify work by friction alone. Choice C incorrectly applies the incline angle to the horizontal motion. Choice D incorrectly assumes friction acts along the entire path including the incline.

Question 14

Two cars move in opposite directions: Car A travels east at 20 m/s20\ \text{m/s}20 m/s and Car B travels west at 15 m/s15\ \text{m/s}15 m/s. A reference frame is the viewpoint used to assign signs and directions to velocities. Relative motion compares one object’s velocity as measured from another object’s frame. This helps estimate closing speeds in traffic and collision analysis. Based on the scenario described, what is the relative speed between the two cars?

  1. 5 m/s5\ \text{m/s}5 m/s
  2. 35 m/s35\ \text{m/s}35 m/s (correct answer)
  3. 20 m/s20\ \text{m/s}20 m/s
  4. 15 m/s15\ \text{m/s}15 m/s

Explanation: This question tests AP Physics C Mechanics skills: understanding reference frames and relative motion. A reference frame is a perspective from which motion is measured. Relative motion denotes how an object moves in relation to a chosen frame. In this scenario, Car A travels east at 20 m/s and Car B travels west at 15 m/s, moving in opposite directions, and we need their relative speed. Choice B is correct because when objects move in opposite directions, their relative speed is the sum of their speeds: 20 + 15 = 35 m/s. Choice A is incorrect because it subtracts the speeds (20 - 15 = 5 m/s), which would only be correct if both cars moved in the same direction. To help students: Emphasize that opposite-direction motion means the cars approach each other, so speeds add. Use the concept of closing speed in head-on scenarios.

Question 15

A balloon is released with internal pressure 1.3×105 Pa1.3\times10^{5}\,\text{Pa}1.3×105Pa while ambient pressure is 1.0×105 Pa1.0\times10^{5}\,\text{Pa}1.0×105Pa; escaping air jets backward and a force sensor estimates the balloon exerts Fballoon→air=5.0×10−1 NF_{\text{balloon}\to\text{air}}=5.0\times10^{-1}\,\text{N}Fballoon→air​=5.0×10−1N backward on the air as it accelerates out. What is the relationship between the force the balloon exerts on the air and the force the air exerts on the balloon?

  1. The air exerts 5.0×10−1 N5.0\times10^{-1}\,\text{N}5.0×10−1N forward on the balloon. (correct answer)
  2. The air exerts 5.0×10−1 N5.0\times10^{-1}\,\text{N}5.0×10−1N backward on the balloon.
  3. The air exerts a smaller force because the air has less mass.
  4. The reaction force is the balloon’s weight, not an air force.

Explanation: This question tests understanding of Newton's Third Law, which states that for every action, there is an equal and opposite reaction. Newton's Third Law involves pairs of forces that are equal in magnitude but opposite in direction, acting on different objects. In the provided scenario, the balloon exerts 0.5 N backward on the escaping air, so by Newton's Third Law, the air must exert 0.5 N forward on the balloon. Choice A is correct because it correctly identifies that the air exerts 0.5 N forward on the balloon, forming an action-reaction pair with the balloon's backward push on the air. Choice C is incorrect because it assumes the force depends on mass - Newton's Third Law guarantees equal magnitudes regardless of the masses involved. To help students, emphasize that action-reaction pairs always have equal magnitudes even when the objects have very different masses. The different accelerations result from the different masses, not different forces.

Question 16

In a laboratory on Earth, a student measures the mass of a block to be mgm_gmg​ using a spring scale and mim_imi​ using an inertial balance. The experiment is then moved to a spaceship accelerating at a=ga=ga=g in deep space. How will the new measurements, mg′m'_gmg′​ and mi′m'_imi′​, compare to the original measurements?

  1. mg′=mgm'_g = m_gmg′​=mg​ and mi′=mim'_i = m_imi′​=mi​ (correct answer)
  2. mg′=0m'_g = 0mg′​=0 and mi′=mim'_i = m_imi′​=mi​
  3. mg′=mgm'_g = m_gmg′​=mg​ and mi′=0m'_i = 0mi′​=0
  4. mg′=0m'_g = 0mg′​=0 and mi′=0m'_i = 0mi′​=0

Explanation: Mass is an intrinsic property of an object. The spring scale measures apparent weight, and in the accelerating spaceship, the apparent gravity is the same as on Earth, so it will measure the same gravitational mass. The inertial balance measures resistance to acceleration (inertial mass), which is also an intrinsic property. Therefore, both measurements will yield the same results.

Question 17

A physical pendulum has a certain period of oscillation. If the mass of the pendulum is doubled, but its shape, size, and pivot point remain exactly the same, what is the effect on its period?

  1. The period is halved because it is inversely proportional to mass.
  2. The period is doubled because it is directly proportional to mass.
  3. The period remains the same because the effects of mass in inertia and torque cancel. (correct answer)
  4. The period increases by a factor of 2\sqrt{2}2​ because it is proportional to the square root of mass.

Explanation: The period of a physical pendulum is T=2πI/(mgd)T = 2\pi\sqrt{I/(mgd)}T=2πI/(mgd)​. The rotational inertia III is directly proportional to the mass mmm (e.g., I=βmR2I = \beta m R^2I=βmR2 for some constant β\betaβ). Therefore, the mass mmm in the numerator (within III) cancels with the mass mmm in the denominator, making the period independent of mass.

Question 18

A particle moves along the x-axis with a velocity given by vx(t)=4t3−2tv_x(t) = 4t^3 - 2tvx​(t)=4t3−2t. If the particle is at x=1x=1x=1 at time t=1t=1t=1, what is its position at time t=2t=2t=2?

  1. 121212
  2. 131313 (correct answer)
  3. 111111
  4. 141414

Explanation: Position is the integral of velocity. x(t)=∫vx(t)dt=∫(4t3−2t)dt=t4−t2+Cx(t) = \int v_x(t) dt = \int (4t^3 - 2t) dt = t^4 - t^2 + Cx(t)=∫vx​(t)dt=∫(4t3−2t)dt=t4−t2+C. Use the condition x(1)=1x(1)=1x(1)=1 to find CCC. 1=(1)4−(1)2+C1 = (1)^4 - (1)^2 + C1=(1)4−(1)2+C, which gives 1=0+C1 = 0 + C1=0+C, so C=1C=1C=1. The position function is x(t)=t4−t2+1x(t) = t^4 - t^2 + 1x(t)=t4−t2+1. At t=2t=2t=2, the position is x(2)=(2)4−(2)2+1=16−4+1=13x(2) = (2)^4 - (2)^2 + 1 = 16 - 4 + 1 = 13x(2)=(2)4−(2)2+1=16−4+1=13.

Question 19

An ice skater rotates about a vertical axis with Ii=3.0 kg\cdotpm2I_i=3.0\ \text{kg·m}^2Ii​=3.0 kg\cdotpm2 and ωi=2.0 rad/s\omega_i=2.0\ \text{rad/s}ωi​=2.0 rad/s. The skater pulls in arms to If=1.2 kg\cdotpm2I_f=1.2\ \text{kg·m}^2If​=1.2 kg\cdotpm2 with negligible external torque, so L=IωL=I\omegaL=Iω is conserved. Considering the described system, what is the final angular velocity of the system after the skater pulls in arms?

  1. 0.80 rad/s
  2. 5.0 rad/s (correct answer)
  3. 3.2 rad/s
  4. 7.2 rad/s

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is conserved in a closed system when no external torque acts, which is the key principle in this ice skater problem. In this problem, the ice skater changes their moment of inertia by pulling in their arms, with negligible external torque, so angular momentum remains constant. Choice B is correct because it accurately applies conservation of angular momentum: Li = Lf, so Iiωi = Ifωf, giving ωf = Iiωi/If = (3.0 × 2.0)/1.2 = 5.0 rad/s. Choice C is incorrect because it appears to use an incorrect calculation, possibly confusing the ratio of moments of inertia. To help students: Emphasize that angular momentum conservation is analogous to linear momentum conservation but involves I and ω. Encourage students to identify when external torque is negligible (like this skating problem). Watch for: confusion about when angular momentum is conserved versus when angular velocity is constant, and ensure understanding that decreasing I increases ω.

Question 20

Based on the scenario, a 600 kg600\,\text{kg}600kg boat has drag Fd=kv2F_d=kv^2Fd​=kv2 with k=100 N s2/m2k=100\,\text{N\,s}^2/\text{m}^2k=100Ns2/m2; calculate terminal speed if engine thrust is constant 1600 N1600\,\text{N}1600N.

  1. 4.0 m/s4.0\,\text{m/s}4.0m/s (correct answer)
  2. 16 m/s16\,\text{m/s}16m/s
  3. 40 m/s40\,\text{m/s}40m/s
  4. 2.7 m/s2.7\,\text{m/s}2.7m/s

Explanation: This question tests understanding of resistive forces in translational dynamics, specifically finding terminal velocity when thrust balances drag force. Resistive forces, such as water drag, oppose motion and at terminal speed, the propulsive thrust exactly balances the drag force for zero acceleration. In this scenario, the boat reaches terminal speed when thrust equals drag: 1600 = kv², so v = √(1600/k) = √(1600/100) = √16 = 4.0 m/s. Choice A is correct because it accurately calculates terminal speed using the force balance equation, demonstrating understanding of equilibrium conditions. Choice C is incorrect because it appears to use v = √(F×k) instead of v = √(F/k), a common algebraic manipulation error. To help students: Emphasize setting up equilibrium equations correctly, practice algebraic manipulation of quadratic relationships, and verify units throughout the calculation process.

Question 21

A wind turbine rotor with I=2.0×104 kg\cdotpm2I=2.0\times10^4\ \text{kg·m}^2I=2.0×104 kg\cdotpm2 spins at ωi=0.80 rad/s\omega_i=0.80\ \text{rad/s}ωi​=0.80 rad/s. The wind drops and net torque becomes τ=−2.0×103 N\cdotpm\tau=-2.0\times10^3\ \text{N·m}τ=−2.0×103 N\cdotpm for t=4.0 st=4.0\ \text{s}t=4.0 s, using ΔL=τt\Delta L=\tau tΔL=τt and L=IωL=I\omegaL=Iω. Considering the described system, what is the final angular velocity of the system after 4.0 seconds?

  1. 0.60 rad/s (correct answer)
  2. 1.0 rad/s
  3. 0.40 rad/s
  4. 8.0×103^33 kg·m2^22/s

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding angular momentum and angular impulse. Angular momentum is a measure of the rotation of an object and angular impulse is the change in angular momentum due to a torque applied over time. In this problem, the wind turbine rotor experiences a negative net torque as the wind drops, which decreases its angular momentum and slows its rotation. Choice A is correct because it accurately calculates the final angular velocity: ΔL = τt = -2.0×10³ × 4.0 = -8.0×10³ kg·m²/s, then ωf = (Li + ΔL)/I = (2.0×10⁴ × 0.80 - 8.0×10³)/(2.0×10⁴) = (1.6×10⁴ - 8.0×10³)/(2.0×10⁴) = 0.60 rad/s. Choice B is incorrect because it appears to use an incorrect calculation, possibly making an error with the scientific notation. To help students: Emphasize careful handling of scientific notation in multi-step problems. Encourage students to verify that negative torque reduces angular velocity. Watch for: calculation errors with large numbers and scientific notation, and ensure understanding that wind turbines slow down when wind decreases.

Question 22

In a circular orbit, what energy exchange occurs if a satellite is boosted from LEO to a higher circular orbit?

  1. Kinetic energy increases and gravitational potential energy decreases.
  2. Kinetic energy decreases while gravitational potential energy increases (less negative). (correct answer)
  3. Both kinetic and gravitational potential energies decrease.
  4. Both kinetic and gravitational potential energies remain constant.

Explanation: This question tests understanding of the motion of orbiting satellites within the context of energy and momentum in rotating systems (AP Physics C: Mechanics). When a satellite moves to a higher orbit, both its kinetic and potential energies change, but total mechanical energy increases due to work done by the boosting force. In a higher orbit, the satellite moves more slowly (v = √(GM/r) decreases with increasing r), so kinetic energy K = ½mv² decreases. Choice B is correct because it accurately states that kinetic energy decreases while gravitational potential energy U = -GMm/r becomes less negative (increases) as r increases. Choice A is incorrect because it reverses the energy changes - kinetic energy actually decreases in higher orbits. To help students: Emphasize that orbital speed decreases with altitude, practice calculating both K and U for different orbits, and show that total energy E = -GMm/2r becomes less negative (increases) for higher orbits. Use energy bar charts to visualize the trade-off between kinetic and potential energy.

Question 23

A 12.0 kg12.0\,\text{kg}12.0kg box is pulled up a 30∘30^\circ30∘ incline at constant speed by a rope parallel to the plane; μk=0.15\mu_k=0.15μk​=0.15. Consider the scenario described above, with kinetic friction opposing the motion and g=9.8 m/s2g=9.8\,\text{m/s}^2g=9.8m/s2. The normal force is perpendicular to the incline and weight is vertical. No air resistance acts. Calculate the frictional force given the conditions.

  1. 15.3 N15.3\,\text{N}15.3N up the plane
  2. 15.3 N15.3\,\text{N}15.3N down the plane (correct answer)
  3. 17.6 N17.6\,\text{N}17.6N down the plane
  4. 14.7 N14.7\,\text{N}14.7N down the plane
  5. 18 N18\,\text{N}18N down the plane

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams help visualize all forces acting on an object, including weight, normal force, friction, and applied forces, which is essential for analyzing motion on inclined planes. In this scenario, a 12.0 kg box moves up a 30° incline at constant speed, meaning the net force is zero and all forces balance. The problem asks specifically for the frictional force, which opposes motion (pointing down the plane). Choice B is correct because the normal force equals mg cos(30°) = 101.8 N, and the kinetic friction force equals μₖN = 0.15 × 101.8 = 15.3 N down the plane. Choice A is incorrect because it has the right magnitude but wrong direction - friction must oppose motion, so it points down the plane, not up. To help students: Emphasize that friction always opposes relative motion between surfaces, and at constant velocity, the net force must be zero. Practice identifying the direction of friction based on the direction of motion or impending motion. Watch for: students confusing the direction of friction, mixing up static and kinetic friction, or forgetting that constant speed means zero acceleration.

Question 24

An object of mass m has a momentum of magnitude p. Which of the following expressions correctly represents its translational kinetic energy, K?

  1. K=p2mK = \frac{p}{2m}K=2mp​
  2. K=p2mK = \frac{p^2}{m}K=mp2​
  3. K=2p2mK = \frac{2p^2}{m}K=m2p2​
  4. K=p22mK = \frac{p^2}{2m}K=2mp2​ (correct answer)

Explanation: Kinetic energy is K=12mv2K = \frac{1}{2}mv^2K=21​mv2 and momentum is p=mvp = mvp=mv. We can express velocity as v=p/mv = p/mv=p/m. Substituting this into the kinetic energy equation gives K=12m(pm)2=12mp2m2=p22mK = \frac{1}{2}m(\frac{p}{m})^2 = \frac{1}{2}m\frac{p^2}{m^2} = \frac{p^2}{2m}K=21​m(mp​)2=21​mm2p2​=2mp2​.

Question 25

Two billiard balls collided head-on elastically: ball 1 (0.16 kg0.16\,\text{kg}0.16kg) moved right at +2.0 m/s+2.0\,\text{m/s}+2.0m/s and ball 2 (0.16 kg0.16\,\text{kg}0.16kg) moved left at −1.0 m/s-1.0\,\text{m/s}−1.0m/s, with negligible external impulse. Determine the velocity of the second object after the collision.

  1. v2f=+2.0 m/sv_{2f}=+2.0\,\text{m/s}v2f​=+2.0m/s (correct answer)
  2. v2f=−2.0 m/sv_{2f}=-2.0\,\text{m/s}v2f​=−2.0m/s
  3. v2f=+1.0 m/sv_{2f}=+1.0\,\text{m/s}v2f​=+1.0m/s
  4. v2f=−1.0 m/sv_{2f}=-1.0\,\text{m/s}v2f​=−1.0m/s

Explanation: For elastic collisions between equal masses, velocities exchange. Initial: ball 1 at +2.0 m/s, ball 2 at -1.0 m/s. Since masses are equal (0.16 kg each), in an elastic collision the velocities swap: ball 1 takes ball 2's initial velocity (-1.0 m/s) and ball 2 takes ball 1's initial velocity (+2.0 m/s). Therefore, v2f=+2.0v_{2f} = +2.0v2f​=+2.0 m/s. Choice A is correct. This is a special case of elastic collisions that students should memorize. Watch for sign errors or attempting complex calculations when the simple velocity exchange rule applies.