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AP Physics C Mechanics

AP Physics C Mechanics Practice Test: Practice Test 3

Practice Test 3 for AP Physics C Mechanics: real questions and explanations from the Varsity Tutors practice-test pool.

0 / 25 answered

Question 1 of 25

A 90 kg load is lifted 6.0 m using an ideal block-and-tackle with mechanical advantage 2, so the rope tension is $T=\tfrac{mg}{2}$ while the load rises at constant speed. Because two rope segments support the load, the free end of the rope must be pulled 12 m to raise the load 6.0 m. The pull takes 8.0 s, and friction is negligible. Use $g=9.8\,\text{m/s}^2$ . In the given system, what is the average power output of the system described?

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Question 1

$6.6\times10^{2}\,\text{W}$ (correct answer)
$1.3\times10^{3}\,\text{W}$
$3.3\times10^{2}\,\text{W}$
$5.5\times10^{2}\,\text{W}$

Explanation: This question tests power in a mechanical advantage system. With a block-and-tackle system having mechanical advantage 2, the rope tension is half the load weight, but the rope must be pulled twice the distance. Given: mass m = 90 kg, load lift height h = 6.0 m, rope pull distance = 12 m, time t = 8.0 s, g = 9.8 m/s². The weight is W = mg = 90 × 9.8 = 882 N. The rope tension is T = mg/2 = 441 N. The work done by pulling the rope is W = T × rope distance = 441 × 12 = 5292 J. This equals the gravitational potential energy gained: mgh = 90 × 9.8 × 6.0 = 5292 J ✓. The power is P = W/t = 5292/8.0 = 661.5 W ≈ 6.6 × 10² W. Choice A is correct. The mechanical advantage doesn't change the work required (energy is conserved), but it does change the force-distance trade-off.

Question 2

A uniform solid sphere rolls down a curved track and then moves along a horizontal loop-the-loop of radius $R$ . If the sphere is released from rest at height $h$ above the bottom of the track, what is the minimum value of $h$ required for the sphere to maintain contact with the track at the top of the loop?

$3R$
$\frac{5R}{2}$
$\frac{7R}{2}$
$\frac{27R}{10}$ (correct answer)

Explanation: This problem combines rotational motion with circular motion dynamics. When you see a rolling object going through a loop, you need to consider both translational and rotational kinetic energy, plus the condition for maintaining contact. At the top of the loop, the minimum condition for contact occurs when the normal force equals zero, meaning gravity alone provides the centripetal force: $mg = \frac{mv_{top}^2}{R}$ , so $v_{top} = \sqrt{gR}$ . Using energy conservation from release point to loop top: The initial potential energy $mgh$ converts to kinetic energy (both translational and rotational) plus potential energy at height $2R$ : $mgh = \frac{1}{2}mv_{top}^2 + \frac{1}{2}I\omega^2 + mg(2R)$ For a solid sphere, $I = \frac{2}{5}mr^2$ and $v = \omega r$ , so $\frac{1}{2}I\omega^2 = \frac{1}{5}mv^2$ . Therefore: $mgh = \frac{1}{2}mv_{top}^2 + \frac{1}{5}mv_{top}^2 + 2mgR = \frac{7}{10}mv_{top}^2 + 2mgR$ Substituting $v_{top}^2 = gR$ : $h = \frac{7gR}{10g} + 2R = \frac{7R}{10} + 2R = \frac{27R}{10}$ Choice A ( $3R$ ) ignores rotational energy entirely. Choice B ( $\frac{5R}{2}$ ) incorrectly uses the condition for a point mass. Choice C ( $\frac{7R}{2}$ ) applies the rolling energy ratio incorrectly. Remember: For rolling objects in loops, always account for rotational kinetic energy using $\frac{1}{2}I\omega^2$ , and know the moment of inertia for common shapes—solid spheres have $I = \frac{2}{5}mr^2$ .

Question 3

A satellite in an elliptical orbit follows Kepler’s laws: (1) orbits are ellipses with Earth at a focus, (2) equal areas in equal times, and (3) $T^2\propto a^3$ . The equal-areas law is a consequence of constant angular momentum $L=mrv_\perp$ when gravity (a central force) exerts zero torque about Earth’s center. How does conservation of angular momentum affect the orbit of a satellite?

It forces the semi-major axis to stay constant even if energy changes.
It implies faster motion near perigee and slower motion near apogee. (correct answer)
It makes orbital speed constant everywhere along an ellipse.
It requires an external torque from Earth’s rotation to maintain orbit.

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 (equal areas in equal times) is a direct consequence of angular momentum conservation, as gravitational force exerts zero torque about the central body. In an elliptical orbit, the perpendicular component of velocity v⊥ varies inversely with distance r to maintain constant angular momentum L = mrv⊥. Choice B is correct because conservation of angular momentum requires the satellite to move faster when closer to Earth (perigee) and slower when farther away (apogee), as rv⊥ must remain constant. Choice C is incorrect as it claims constant speed throughout the ellipse, which would violate both energy conservation and angular momentum conservation in a varying gravitational field. To help students: Use visual demonstrations of equal area sweeping, derive Kepler's second law from L = constant, and have students calculate velocities at different points in elliptical orbits. Emphasize the distinction between constant angular momentum and varying linear speed.

Question 4

Two spheres, A and B, have the same translational kinetic energy. The mass of sphere A is four times the mass of sphere B ( $m_A = 4m_B$ ). What is the ratio of the speed of sphere A to the speed of sphere B, $v_A/v_B$ ?

1/4
1/2 (correct answer)
2
4

Explanation: We are given that $K_A = K_B$ . Therefore, $\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2$ . Substituting $m_A = 4m_B$ gives $(4m_B) v_A^2 = m_B v_B^2$ . The $m_B$ terms cancel, leaving $4v_A^2 = v_B^2$ . Taking the square root of both sides gives $2v_A = v_B$ , so the ratio $v_A/v_B = 1/2$ .

Question 5

A gyroscope has spin angular momentum magnitude $L=0.40\ \text{kg·m}^2/\text{s}$ . Gravity exerts a torque of magnitude $\tau=0.080\ \text{N·m}$ about the pivot, producing steady precession; assume the spin magnitude stays constant. Key equation: $\Omega=\tau/L$ . Using the given conditions, determine the precession rate $\Omega$ .

$\Omega=0.20\ \text{rad/s}$ (correct answer)
$\Omega=5.0\ \text{rad/s}$
$\Omega=0.032\ \text{rad/s}$
$\Omega=2.0\ \text{rad/s}$

Explanation: This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically gyroscopic precession. Precession occurs when a torque acts perpendicular to a spinning object's angular momentum, causing the spin axis to rotate. For a gyroscope with angular momentum L = 0.40 kg·m²/s experiencing torque τ = 0.080 N·m, the precession rate is found using the gyroscopic equation. Choice A is correct because Ω = τ/L = 0.080/0.40 = 0.20 rad/s. Choice B at 5.0 rad/s would result from inverting the fraction or other calculation errors. To help students: explain that precession is different from ordinary rotation; emphasize that the precession equation Ω = τ/L applies when spin angular momentum is much larger than precession angular momentum; and use demonstrations or videos to visualize gyroscopic motion.

Question 6

A $1500\ \text{kg}$ car moving east at $12\ \text{m/s}$ is brought to rest by a collision in $0.30\ \text{s}$ . Using the given data, determine the change in momentum of the car.

$\Delta p = +1.8\times10^4\ \text{kg·m/s}$
$\Delta p = -1.8\times10^4\ \text{kg·m/s}$ (correct answer)
$\Delta p = -5.0\times10^3\ \text{kg·m/s}$
$\Delta p = -6.0\times10^4\ \text{kg·m/s}$

Explanation: This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating change in momentum during a collision that brings a car to rest. Change in momentum equals mass times the change in velocity, Δp = m(vf - vi), where careful attention to direction is essential. In this scenario, the 1500 kg car moving east at 12 m/s comes to rest (vf = 0), so Δp = 1500(0 - 12) = -18,000 kg·m/s = -1.8×10^4 kg·m/s. Choice B is correct because it shows Δp = -1.8×10^4 kg·m/s, with the negative sign indicating the westward direction of the momentum change (opposite to initial motion). Choice A incorrectly shows positive change, while choices C and D show incorrect magnitudes. When teaching momentum change, emphasize that coming to rest means final velocity is zero, and the change in momentum opposes the initial direction of motion. Students should practice problems involving objects brought to rest to reinforce sign conventions.

Question 7

Based on the described system, a vertical spring-mass oscillator has $m=0.50\,\text{kg}$ , $k=80\,\text{N/m}$ , and amplitude $0.03\,\text{m}$ about equilibrium; calculate the frequency of the given system in Hz.

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ (correct answer)
$f=2\pi\sqrt{\dfrac{m}{k}}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{m}{k}}$
$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}$

Explanation: This question tests understanding of the frequency and period of simple harmonic motion in AP Physics C: Mechanics. In SHM, the period is the time it takes to complete one full oscillation, and frequency is the number of oscillations per unit time. They are inversely related (f = 1/T). For a mass-spring system, whether vertical or horizontal, the frequency f is given by f = (1/2π)√(k/m), where k is the spring constant and m is mass. Choice A is correct because it properly applies the frequency formula f = (1/2π)√(k/m), which gives oscillations per second (Hz) for the mass-spring system. Choice C is incorrect because it has m/k under the square root instead of k/m, which would give incorrect units and violate the physics - stiffer springs (larger k) should produce higher frequencies. To help students: Emphasize that vertical and horizontal springs use the same frequency formula. Practice unit analysis to verify that √(k/m) has units of 1/time.

Question 8

A uniform disk of mass $M$ and radius $R$ is pivoted at a point on its rim. A horizontal force $F$ is applied at the opposite point on the rim. For the disk to remain in rotational equilibrium, what must be the relationship between $F$ and the other parameters?

$F = \frac{Mg}{2}$ (correct answer)
$F = Mg$
$F = \frac{Mg}{4}$
$F = 2Mg$

Explanation: The disk is pivoted at a point on its rim, so the center of mass is at distance $R$ from the pivot. The gravitational torque about the pivot is $\tau_g = MgR$ (taking clockwise as positive). The applied force $F$ acts at the opposite point on the rim, which is at distance $2R$ from the pivot, creating torque $\tau_F = F(2R)$ in the opposite direction. For equilibrium: $MgR = F(2R)$ , so $F = \frac{Mg}{2}$ . Choice B assumes the lever arm for $F$ is $R$ . Choice C uses incorrect geometry. Choice D assumes $F$ acts at distance $R/2$ .

Question 9

A wheel of radius $R$ and moment of inertia $I$ about its center is initially at rest on a horizontal surface. A horizontal force $F$ is applied to the axle for time $t$ . If the coefficient of static friction between the wheel and surface is $\mu_s$ , and the wheel rolls without slipping throughout the motion, what is the angular acceleration of the wheel?

$\alpha = \frac{FR}{I + MR^2}$ (correct answer)
$\alpha = \frac{FR}{I}$
$\alpha = \frac{F}{MR}$
$\alpha = \frac{(F - \mu_s Mg)R}{I}$

Explanation: For rolling without slipping, $a = \alpha R$ . Applying Newton's second law to translation: $F - f = Ma = M\alpha R$ , where $f$ is the friction force. For rotation about the center: $fR = I\alpha$ . From the rotational equation, $f = \frac{I\alpha}{R}$ . Substituting into the translational equation: $F - \frac{I\alpha}{R} = M\alpha R$ . Solving for $\alpha$ : $F = \alpha(MR + \frac{I}{R}) = \alpha\frac{MR^2 + I}{R}$ , so $\alpha = \frac{FR}{I + MR^2}$ . Choice B ignores the constraint of rolling motion and treats it as pure rotation. Choice C treats it as pure translation. Choice D incorrectly assumes kinetic friction opposes motion, but static friction here enables rolling.

Question 10

Car A, of mass 1000 kg, travels east at 20 m/s. Car B, of mass 1500 kg, travels west at 10 m/s. Both speeds are measured relative to the ground.

What is the total translational kinetic energy of the two-car system as measured by an observer stationary on the ground?

5,000 J
125,000 J
275,000 J (correct answer)
1,125,000 J

Explanation: The total kinetic energy is the scalar sum of the individual kinetic energies. $K_A = \frac{1}{2}m_A v_A^2 = \frac{1}{2}(1000 \text{ kg})(20 \text{ m/s})^2 = 200,000 \text{ J}$ . $K_B = \frac{1}{2}m_B v_B^2 = \frac{1}{2}(1500 \text{ kg})(10 \text{ m/s})^2 = 75,000 \text{ J}$ . The total kinetic energy is $K_{total} = K_A + K_B = 200,000 \text{ J} + 75,000 \text{ J} = 275,000 \text{ J}$ .

Question 11

A simple pendulum of length $L$ oscillates with small amplitude in a region where the gravitational acceleration is $g$ . If the pendulum bob is replaced with one having twice the mass and the length is increased by a factor of 4, what is the ratio of the new period to the original period?

$\frac{1}{2}$
$1$
$2$ (correct answer)
$4$

Explanation: The period of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$ , which is independent of the mass of the bob. When the length increases by a factor of 4, the new period becomes $T_{new} = 2\pi\sqrt{\frac{4L}{g}} = 2 \cdot 2\pi\sqrt{\frac{L}{g}} = 2T_{original}$ . The mass change has no effect on the period. Choice A results from incorrectly thinking the period decreases with length. Choice B assumes both mass and length changes cancel out. Choice D incorrectly applies the length factor directly without taking the square root.

Question 12

A ball is thrown from ground level at $v_0=18\,\text{m/s}$ and $\theta=50^\circ$ above the horizontal, with $g=9.8\,\text{m/s}^2$ downward and no air resistance. Take $t=0$ at launch and analyze motion during the first $2.0\,\text{s}$ . The only force during flight is gravity, so acceleration is constant and vertical. Initial position is $\vec r_0=\langle 0,0\rangle\,\text{m}$ , and $v_x$ remains constant. Use component kinematics to model the motion. Based on the scenario described, describe the motion of the object using kinematic equations.

$x(t)=(v_0\cos\theta)t,\; y(t)=(v_0\sin\theta)t-\tfrac12gt^2$ (correct answer)
$x(t)=\tfrac12(v_0\cos\theta)t^2,\; y(t)=(v_0\sin\theta)t-gt^2$
$x(t)=(v_0\cos\theta)t-\tfrac12gt^2,\; y(t)=(v_0\sin\theta)t$
$x(t)=(v_0\cos\theta)t,\; y(t)=(v_0\sin\theta)t-\tfrac12g^2t^2$

Explanation: This question tests AP Physics C: Mechanics skills in representing motion through parametric equations for projectile motion. Projectile motion requires separate kinematic equations for horizontal and vertical components, with constant horizontal velocity and constant vertical acceleration. In this scenario, a ball is thrown at an angle, requiring component analysis with x(t) for horizontal motion and y(t) for vertical motion. Choice A is correct because horizontal motion has constant velocity: x(t)=(v₀cosθ)t with no acceleration term, while vertical motion includes initial velocity and gravitational acceleration: y(t)=(v₀sinθ)t-½gt². Choice B is incorrect because it incorrectly includes ½ in the horizontal equation, suggesting acceleration where none exists, and omits the ½ factor in the vertical equation. To help students: Emphasize that horizontal motion has zero acceleration in projectile problems. Create tables showing initial conditions and accelerations for each component, and practice deriving position equations from first principles.

Question 13

A horizontal disk with rotational inertia $I_0$ is rotating freely with angular velocity $\omega_0$ about a vertical axis. A small piece of putty of mass $m$ is dropped vertically and sticks to the disk at a distance $r$ from the axis. What is the final angular velocity of the disk-putty system?

$\frac{I_0 \omega_0}{I_0 + mr^2}$ (correct answer)
$\frac{I_0 \omega_0}{I_0 - mr^2}$
$\frac{(I_0 + mr^2) \omega_0}{I_0}$
$\omega_0$

Explanation: Since the putty is dropped vertically, it carries no initial horizontal momentum and thus no initial angular momentum about the disk's axis. The forces during the collision are internal to the disk-putty system, so the total angular momentum is conserved. The initial angular momentum is $L_i = I_0 \omega_0$ . The final rotational inertia of the system is $I_f = I_0 + I_{putty} = I_0 + mr^2$ . The final angular momentum is $L_f = I_f \omega_f = (I_0 + mr^2)\omega_f$ . Setting $L_i = L_f$ gives $I_0 \omega_0 = (I_0 + mr^2)\omega_f$ . Solving for $\omega_f$ yields the correct answer.

Question 14

A uniform chain of mass $M$ and length $L$ hangs vertically from one end. A portion of length $x$ (measured from the bottom) is then lifted and placed horizontally on a frictionless table while the remaining portion of length $(L-x)$ continues to hang vertically. What is the position of the center of mass of the entire chain, measured vertically downward from the table surface?

$\frac{L-x}{2}$
$\frac{L^2 - x^2}{2L}$
$\frac{(L-x)^2}{2L}$ (correct answer)
$\frac{L^2 + x^2}{2L}$

Explanation: When you encounter center of mass problems involving objects with different orientations, you need to treat each portion separately and then combine them using the weighted average principle. The chain has two distinct portions: a horizontal section of length $x$ on the table, and a vertical hanging section of length $(L-x)$ . Since the chain is uniform, its linear mass density is $\lambda = M/L$ . For the horizontal portion (mass $\frac{Mx}{L}$ ), the center of mass is at table level, contributing zero to the vertical position. For the hanging portion (mass $\frac{M(L-x)}{L}$ ), the center of mass is at the midpoint of the hanging length, which is $\frac{L-x}{2}$ below the table. Using the center of mass formula: $y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}$ $y_{cm} = \frac{\frac{Mx}{L} \cdot 0 + \frac{M(L-x)}{L} \cdot \frac{L-x}{2}}{M} = \frac{M(L-x)^2}{2LM} = \frac{(L-x)^2}{2L}$ Choice A gives $\frac{L-x}{2}$ , which is just the center of mass of the hanging portion alone, ignoring the horizontal section entirely. Choice B yields $\frac{L^2-x^2}{2L} = \frac{(L-x)(L+x)}{2L}$ , which incorrectly weights the calculation. Choice D gives $\frac{L^2+x^2}{2L}$ , suggesting the portions somehow add constructively rather than being weighted by their masses. Remember: in center of mass problems, always identify each distinct portion, find their individual centers of mass, then use the weighted average based on their masses.

Question 15

The total angular momentum of a system of particles is conserved under which of the following conditions?

The net external force on the system is zero, ensuring that no work is done on the system.
The net external torque on the system about a particular axis is zero. (correct answer)
The total kinetic energy of the system is constant, preventing any changes in rotational speed.
The moment of inertia of the system remains constant throughout its motion.

Explanation: The condition for the conservation of angular momentum is that the net external torque acting on the system is zero. This is the rotational analogue of Newton's second law, $\vec{\tau}_{net, ext} = d\vec{L}/dt$ . If $\vec{\tau}_{net, ext} = 0$ , then $d\vec{L}/dt = 0$ , which means $\vec{L}$ is constant (conserved). Zero net external force conserves linear momentum, not necessarily angular momentum. Constant kinetic energy is not a prerequisite, and the moment of inertia can change while angular momentum is conserved.

Question 16

A block is pulled up a rough inclined plane at constant velocity by a force $\vec{F}$ applied parallel to the incline. The incline makes angle $\theta$ with the horizontal, and the coefficient of kinetic friction is $\mu_k$ . In the free-body diagram analysis, which expression correctly represents the relationship between the applied force and other forces?

$F = mg\sin\theta + \mu_k mg\cos\theta$ because friction opposes motion up the incline (correct answer)
$F = mg\sin\theta - \mu_k mg\cos\theta$ because friction aids motion up the incline
$F = mg\cos\theta + \mu_k mg\sin\theta$ due to the perpendicular force components
$F = \mu_k mg\cos\theta - mg\sin\theta$ when friction exceeds the gravitational component

Explanation: At constant velocity, the net force is zero, so the applied force must balance both the component of weight down the incline ( $mg\sin\theta$ ) and the friction force opposing motion ( $\mu_k N = \mu_k mg\cos\theta$ ). Both oppose the applied force, so $F = mg\sin\theta + \mu_k mg\cos\theta$ . Choice B incorrectly subtracts friction, suggesting it aids upward motion. Choice C confuses sine and cosine components. Choice D has the wrong sign relationship and would only apply if friction somehow overcame gravity.

Question 17

A projectile is launched from the surface of a planet of mass $M$ and radius $R$ . What is the minimum total mechanical energy required for the projectile-planet system so that the projectile can escape the planet's gravitational pull? Assume the potential energy of the system is zero at infinite separation.

Zero (correct answer)
$\frac{GMm}{R}$
$-\frac{GMm}{R}$
$\frac{GMm}{2R}$

Explanation: For the projectile to escape, it must be able to reach an infinite distance from the planet with a non-negative kinetic energy (i.e., its speed must be real). At infinite separation ( $r \to \infty$ ), the potential energy $U = -GMm/r$ approaches zero. The minimum condition for escape is that the kinetic energy also approaches zero at infinite separation. By conservation of mechanical energy, the total energy $E = K + U$ must be constant. If $K \to 0$ and $U \to 0$ as $r \to \infty$ , the total mechanical energy of the system must be zero.

Question 18

A $2.0\ \text{kg}$ cart moves right at $6.0\ \text{m/s}$ and experiences a constant leftward force of $10\ \text{N}$ for $0.50\ \text{s}$ . Using the given data, determine the change in momentum of the cart.

$\Delta p = +5.0\ \text{kg·m/s}$
$\Delta p = -5.0\ \text{kg·m/s}$ (correct answer)
$\Delta p = -20\ \text{kg·m/s}$
$\Delta p = -10\ \text{kg·m/s}$

Explanation: This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on calculating change in momentum from applied force. The impulse-momentum theorem states that impulse (J = FΔt) equals the change in momentum (Δp), where force and momentum are vector quantities. In this scenario, the cart experiences a leftward force of 10 N for 0.50 s, creating an impulse of J = (-10)(0.50) = -5.0 N·s, which equals the change in momentum. Choice B is correct because Δp = -5.0 kg·m/s, with the negative sign indicating the leftward direction opposing the initial rightward motion. Choice A incorrectly shows positive change, while choices C and D show incorrect magnitudes. When teaching this concept, emphasize that the change in momentum depends only on the impulse (force × time), not on the object's initial velocity. Students should practice identifying force directions and applying consistent sign conventions throughout their calculations.

Question 19

A parachutist falling toward Earth opens their parachute and reaches a constant terminal velocity. At this point, what is the relationship between the magnitude of the gravitational force $F_g$ and the magnitude of the upward air drag force $F_d$ ?

$F_g > F_d$
$F_g < F_d$
$F_g = F_d$ (correct answer)
The relationship cannot be determined without knowing the parachutist's mass.

Explanation: The parachutist is moving at a constant terminal velocity. According to Newton's First Law, this means the net force acting on the parachutist is zero. The forces are the downward gravitational force $F_g$ and the upward drag force $F_d$ . For the net force to be zero, these forces must be equal in magnitude and opposite in direction. Thus, $F_g = F_d$ .

Question 20

An object's position is given by the function $x(t) = 2t^3 - 15t^2 + 36t - 8$ . At which of the following times is the object momentarily at rest?

$t=1.5 \text{ s}$ and $t=6 \text{ s}$
$t=2.5 \text{ s}$ only
$t=2 \text{ s}$ and $t=3 \text{ s}$ (correct answer)
$t=6 \text{ s}$ only

Explanation: The object is momentarily at rest when its velocity is zero. We find the velocity function by differentiating the position function: $v(t) = \frac{dx}{dt} = 6t^2 - 30t + 36$ . Set $v(t) = 0$ and solve for $t$ : $6(t^2 - 5t + 6) = 0$ , which factors to $6(t-2)(t-3) = 0$ . The solutions are $t=2 \text{ s}$ and $t=3 \text{ s}$ .

Question 21

Two point masses, $m_1 = 2.0$ kg and $m_2 = 3.0$ kg, are located at $x_1 = -1.0$ m and $x_2 = 2.0$ m, respectively, on the x-axis. What is the rotational inertia of this system about the y-axis?

$5.0 \, \text{kg} \cdot \text{m}^2$
$8.0 \, \text{kg} \cdot \text{m}^2$
$14 \, \text{kg} \cdot \text{m}^2$ (correct answer)
$17 \, \text{kg} \cdot \text{m}^2$

Explanation: The rotational inertia of a system of point masses is given by the sum $I = \sum m_i r_i^2$ , where $r_i$ is the perpendicular distance of each mass from the axis of rotation. Here, the axis is the y-axis, so the distances are the absolute values of the x-coordinates. $I = m_1 r_1^2 + m_2 r_2^2 = (2.0 \, \text{kg})(-1.0 \, \text{m})^2 + (3.0 \, \text{kg})(2.0 \, \text{m})^2 = 2.0 \, \text{kg} \cdot \text{m}^2 + 12.0 \, \text{kg} \cdot \text{m}^2 = 14 \, \text{kg} \cdot \text{m}^2$ .

Question 22

A $1500\,\text{kg}$ car descends a frictionless hill from rest and reaches $30\,\text{m/s}$ ; how much work is done on the car by gravity?

$3.38\times10^5\,\text{J}$
$6.75\times10^5\,\text{J}$ (correct answer)
$1.35\times10^6\,\text{J}$
$4.50\times10^4\,\text{J}$

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding the work-energy theorem and calculating translational kinetic energy. The work-energy theorem states that the work done on an object equals its change in kinetic energy, and KE = 1/2 mv². In this scenario, a 1500 kg car descends from rest to reach 30 m/s, requiring calculation of the work done by gravity. Choice B is correct because the work done equals the change in kinetic energy: W = ΔKE = KEf - KEi = 1/2 × 1500 kg × (30 m/s)² - 0 = 1/2 × 1500 × 900 = 675,000 J = 6.75×10⁵ J. Choice A incorrectly uses half the correct value, possibly from an arithmetic error. To help students: Reinforce that work done by gravity equals the change in kinetic energy for frictionless motion. Practice energy conservation problems to show the equivalence of gravitational potential energy loss and kinetic energy gain.

Question 23

An object's momentum along a straight line is plotted as a function of time. The graph is a straight line passing through the points $(t=1 \text{ s}, p=5 \text{ kg⋅m/s})$ and $(t=4 \text{ s}, p=14 \text{ kg⋅m/s})$ . What is the constant net force acting on the object?

$3$ N (correct answer)
$4.5$ N
$9$ N
$19$ N

Explanation: The net force is the rate of change of momentum, $F_{net} = \frac{dp}{dt}$ . For a linear momentum-time graph, this is the slope of the line. $F_{net} = \frac{\Delta p}{\Delta t} = \frac{14 \text{ kg⋅m/s} - 5 \text{ kg⋅m/s}}{4 \text{ s} - 1 \text{ s}} = \frac{9 \text{ kg⋅m/s}}{3 \text{ s}} = 3$ N.

Question 24

Two vectors $\vec{P}$ and $\vec{Q}$ are added to form a resultant vector $\vec{R} = \vec{P} + \vec{Q}$ . If the magnitudes satisfy the relation $|\vec{P}| + |\vec{Q}| = |\vec{R}|$ , what must be true about the vectors $\vec{P}$ and $\vec{Q}$ ?

They are perpendicular to each other.
They are in opposite directions.
They are in the same direction. (correct answer)
One of the vectors must be the zero vector.

Explanation: The magnitude of the sum of two vectors is equal to the sum of their individual magnitudes only in the specific case where the vectors are parallel and point in the same direction (an angle of $0^\circ$ between them). In all other cases, due to the triangle inequality, the magnitude of the resultant vector will be less than the sum of the individual magnitudes.

Question 25

Based on the scenario, calculate $K_\text{rot}$ for a flywheel with $I=2.5\ \text{kg·m}^2$ spinning at $\omega=20\ \text{rad/s}$ .

$250\ \text{J}$
$500\ \text{J}$ (correct answer)
$1000\ \text{J}$
$50\ \text{J}$

Explanation: This question tests AP Physics C understanding of rotational kinetic energy in rotating systems. Rotational kinetic energy is calculated using the formula K_rot = (1/2)Iω², where I is the moment of inertia and ω is the angular velocity. In this scenario, we calculate K_rot for a flywheel with I = 2.5 kg·m² and ω = 20 rad/s. Choice B is correct because K_rot = (1/2)(2.5)(20)² = (1/2)(2.5)(400) = 500 J. Choice C is incorrect as it omits the factor of 1/2, giving 1000 J instead of 500 J. To help students: Always remember the factor of 1/2 in the rotational kinetic energy formula. Create mnemonics or visual aids to distinguish between formulas that include 1/2 and those that don't.