AP Physics C Mechanics

AP Physics C Mechanics Practice Test: Practice Test 4

Practice Test 4 for AP Physics C Mechanics: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A uniform rod of length $L$ and mass $M$ is pivoted at one end and released from a horizontal position. What is its rotational kinetic energy when it reaches the vertical position? The rotational inertia of a rod about its end is $I = \frac{1}{3}ML^2$ . Use $g$ for the acceleration due to gravity.

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Question 1

$MgL$
$MgL/2$ (correct answer)
$MgL/3$
$MgL/4$

Explanation: This is a conservation of energy problem. The center of mass of the rod is at $L/2$ from the pivot. When the rod is released from a horizontal position, its center of mass falls a vertical distance of $L/2$ . The change in gravitational potential energy is $\Delta U_g = Mg(L/2)$ . This potential energy is converted into rotational kinetic energy. Therefore, the rotational kinetic energy at the vertical position is $K_{rot} = MgL/2$ .

Question 2

A grinding wheel, initially at rest, accelerates with a uniform angular acceleration of $4.0 \, \text{rad/s}^2$ for $5.0 \, \text{s}$ . What is its final angular velocity?

$10 \, \text{rad/s}$
$20 \, \text{rad/s}$ (correct answer)
$40 \, \text{rad/s}$
$50 \, \text{rad/s}$

Explanation: For constant angular acceleration, the final angular velocity is given by the kinematic equation $\omega = \omega_0 + \alpha t$ . Since the wheel starts from rest, $\omega_0 = 0$ . Plugging in the values: $\omega = 0 + (4.0 \, \text{rad/s}^2)(5.0 \, \text{s}) = 20 \, \text{rad/s}$ .

Question 3

A particle of mass $m$ is located inside a uniform solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center ( $r < R$ ). The gravitational force on the particle is $F_g(r) = -G\frac{Mmr}{R^3}$ . If the potential energy is defined to be zero at the center of the sphere ( $U(0)=0$ ), what is the potential energy $U(r)$ of the particle-sphere system?

$\frac{GMmr^2}{2R^3}$ (correct answer)
$-\frac{GMmr^2}{2R^3}$
$G\frac{Mmr}{R^3}$
$\frac{GMm}{R}$

Explanation: The change in potential energy is the negative of the work done by the conservative force. $U(r) - U(0) = -\int_0^r F_g(r') dr' = -\int_0^r (-G\frac{Mmr'}{R^3}) dr' = G\frac{Mm}{R^3} \int_0^r r' dr'$ . Evaluating the integral gives $G\frac{Mm}{R^3} [\frac{1}{2}r'^2]_0^r = G\frac{Mm r^2}{2R^3}$ . Since $U(0)=0$ , we have $U(r) = \frac{GMmr^2}{2R^3}$ .

Question 4

A thin hoop of mass $M$ and radius $R$ has two small beads, each of mass $m$ , attached to its rim at opposite ends of a diameter. What is the total rotational inertia of the hoop-beads system about an axis perpendicular to the plane of the hoop and passing through its center?

$MR^2$
$MR^2 + 2mR^2$ (correct answer)
$(M+2m)R$
$MR^2 + mR^2$

Explanation: Rotational inertias are additive. The total rotational inertia is the sum of the inertia of the hoop and the inertias of the two beads. The hoop's inertia is $I_{hoop} = MR^2$ . Each bead is a point mass at a distance R from the axis, so each has an inertia of $I_{bead} = mR^2$ . The total inertia is $I_{total} = I_{hoop} + I_{bead1} + I_{bead2} = MR^2 + mR^2 + mR^2 = MR^2 + 2mR^2$ .

Question 5

A system consists of three point masses: $m_1 = 2.0\text{ kg}$ at position $x_1 = 0\text{ m}$ , $m_2 = 3.0\text{ kg}$ at position $x_2 = 4.0\text{ m}$ , and $m_3 = 1.0\text{ kg}$ at position $x_3 = 6.0\text{ m}$ . What is the $x$ -coordinate of the center of mass of this system?

$2.3\text{ m}$
$2.7\text{ m}$ (correct answer)
$3.3\text{ m}$
$4.0\text{ m}$

Explanation: The center of mass is calculated using $x_{cm} = \frac{\sum m_i x_i}{\sum m_i} = \frac{(2.0)(0) + (3.0)(4.0) + (1.0)(6.0)}{2.0 + 3.0 + 1.0} = \frac{0 + 12.0 + 6.0}{6.0} = \frac{18.0}{6.0} = 2.7\text{ m}$ . Choice A uses an incorrect calculation. Choice C represents the simple average of positions without mass weighting. Choice D is the position of the middle mass, not the center of mass.

Question 6

Two blocks, $M_1$ and $M_2$ , are in contact on a frictionless horizontal surface. A horizontal force $F$ is applied to $M_1$ , causing both blocks to accelerate. What forces should be shown on a free-body diagram for block $M_2$ ?

Only the applied force $F$ , gravity, and the normal force from the surface.
Gravity, the normal force from the surface, and a contact force exerted by $M_1$ on $M_2$ . (correct answer)
Gravity, the normal force from the surface, the applied force $F$ , and a contact force from $M_1$ on $M_2$ .
Gravity and the normal force from the surface, as the horizontal forces are internal to the system.

Explanation: A free-body diagram shows forces exerted on a specific object. For block $M_2$ , the forces are: gravity (down), the normal force from the horizontal surface (up), and the contact force from block $M_1$ pushing it (horizontally). The applied force $F$ acts directly on $M_1$ , not $M_2$ .

Question 7

A 0.60 kg cart is attached to a spring with $k=240\ \text{N/m}$ and oscillates on a frictionless track about $x=0$ . It is pulled to $x=+0.070\ \text{m}$ and released from rest at $t=0$ , so $A=0.070\ \text{m}$ and $\phi=0$ . The motion is $x(t)=A\cos(\omega t+\phi)$ with $\omega=\sqrt{k/m}$ . The acceleration is $a(t)=-\omega^2 x(t)$ and $a_{\max}=\omega^2 A$ . All quantities are in SI units and amplitude remains constant. Find the maximum acceleration magnitude $a_{\max}$ .

$a_{\max}=28\ \text{m/s}^2$ (correct answer)
$a_{\max}=7.0\ \text{m/s}^2$
$a_{\max}=0.28\ \text{m/s}^2$
$a_{\max}=4.0\ \text{m/s}^2$

Explanation: This question tests the ability to represent and analyze simple harmonic motion (SHM) in the context of AP Physics C: Mechanics. SHM is characterized by periodic motion where the restoring force is proportional to displacement, commonly modeled by mass-spring systems or pendulums. In this scenario, a 0.60 kg cart on a spring with k=240 N/m oscillates with amplitude A=0.070 m, and we need the maximum acceleration amax=ω²A. Choice A is correct because ω=√(k/m)=√(240/0.60)=20 rad/s, giving amax=(20)²×0.070=400×0.070=28 m/s². Choice B is incorrect due to a common error where students might calculate amax=ωA instead of ω²A, getting 20×0.070=1.4 m/s² (though this doesn't match choice B exactly). To help students: Emphasize that acceleration in SHM is proportional to displacement with factor -ω², making maximum acceleration occur at maximum displacement. Practice problems should include finding acceleration at various positions to reinforce the a=-ω²x relationship.

Question 8

A particle moves along a helical path described by the parametric equations $x(t) = a\cos(\omega t)$ , $y(t) = a\sin(\omega t)$ , and $z(t) = bt$ , where $a$ , $b$ , and $\omega$ are positive constants. What is the magnitude of the particle's acceleration?

$\omega^2\sqrt{a^2 + b^2}$
$\omega\sqrt{a^2 + b^2}$
$\sqrt{a^2\omega^4 + b^2}$
$a\omega^2$ (correct answer)

Explanation: When analyzing motion along a parametric path like this helix, you need to find acceleration by taking the second derivative of the position vector with respect to time. Starting with the position vector $\vec{r}(t) = a\cos(\omega t)\hat{i} + a\sin(\omega t)\hat{j} + bt\hat{k}$ , find the velocity by differentiating: $\vec{v}(t) = -a\omega\sin(\omega t)\hat{i} + a\omega\cos(\omega t)\hat{j} + b\hat{k}$ Then differentiate again to get acceleration: $\vec{a}(t) = -a\omega^2\cos(\omega t)\hat{i} - a\omega^2\sin(\omega t)\hat{j} + 0\hat{k}$ Notice that the $z$ -component acceleration is zero because the $z$ -velocity is constant ( $b$ ). The particle moves at constant speed vertically while accelerating horizontally due to circular motion. The magnitude is: $|\vec{a}| = \sqrt{(-a\omega^2\cos(\omega t))^2 + (-a\omega^2\sin(\omega t))^2} = a\omega^2\sqrt{\cos^2(\omega t) + \sin^2(\omega t)} = a\omega^2$ Choice A ( $\omega^2\sqrt{a^2 + b^2}$ ) incorrectly includes the constant vertical velocity $b$ in the acceleration calculation. Choice B ( $\omega\sqrt{a^2 + b^2}$ ) makes the same mistake and uses $\omega$ instead of $\omega^2$ . Choice C ( $\sqrt{a^2\omega^4 + b^2}$ ) also incorrectly includes $b$ and has the wrong power structure. The correct answer is D: $a\omega^2$ . Key takeaway: In helical motion problems, remember that constant-velocity components contribute zero to acceleration. Only the circular motion components (here, the $x$ and $y$ parts) contribute to the centripetal acceleration.

Question 9

A person stands on a moving walkway that travels at $1.5 \text{ m/s}$ . The person walks forward on the walkway at $2.0 \text{ m/s}$ relative to the walkway for $10 \text{ s}$ , then turns around and walks backward at the same speed relative to the walkway for another $10 \text{ s}$ . What is the person's displacement relative to the ground after the entire $20 \text{ s}$ journey?

$20 \text{ m}$ in the forward direction
$35 \text{ m}$ in the forward direction
$30 \text{ m}$ in the forward direction (correct answer)
$10 \text{ m}$ in the forward direction

Explanation: When you encounter relative motion problems involving moving platforms, you need to carefully track velocities and apply vector addition. The key is understanding that the person's velocity relative to the ground equals their velocity relative to the walkway plus the walkway's velocity relative to the ground. For the first 10 seconds, the person walks forward at 2.0 m/s relative to the walkway, which moves at 1.5 m/s forward. Their ground velocity is $2.0 + 1.5 = 3.5 \text{ m/s}$ forward. Distance covered: $3.5 \times 10 = 35 \text{ m}$ forward. For the next 10 seconds, the person walks backward at 2.0 m/s relative to the walkway. Since they're moving opposite to the walkway's direction, their ground velocity is $-2.0 + 1.5 = -0.5 \text{ m/s}$ (backward). Distance: $0.5 \times 10 = 5 \text{ m}$ backward. Total displacement: $35 - 5 = 30 \text{ m}$ forward, confirming answer C. Answer A (20 m) likely comes from ignoring the walkway's motion entirely and just considering the person's relative motion. Answer B (35 m) represents only the first segment, forgetting about the return journey. Answer D (10 m) might result from incorrectly calculating the backward segment or confusing the time intervals. Remember: in relative motion problems, always establish a reference frame and use vector addition. The walkway continues moving throughout the entire journey, affecting the person's ground displacement even when they walk backward relative to it.

Question 10

A rocket in space ejects exhaust gases at a rate of 100 kg/s with a velocity of 3000 m/s relative to the rocket. At a particular instant, the rocket has a mass of 5000 kg and is accelerating at $60 \text{ m/s}^2$ . What is the magnitude of the force that the exhaust gases exert on the rocket according to Newton's third law?

600,000 N
300,000 N (correct answer)
900,000 N
200,000 N

Explanation: When you encounter rocket propulsion problems, you're dealing with Newton's third law and the momentum principle. The key insight is that the force exerted by exhaust gases on the rocket equals the rate of momentum change of the expelled gases. To find this force, use the relationship $F = \frac{dm}{dt} \times v_{exhaust}$ , where $\frac{dm}{dt}$ is the mass flow rate and $v_{exhaust}$ is the exhaust velocity relative to the rocket. Given: mass flow rate = 100 kg/s and exhaust velocity = 3000 m/s $F = 100 \text{ kg/s} \times 3000 \text{ m/s} = 300,000 \text{ N}$ This is the thrust force the exhaust gases exert on the rocket, making B correct. Now for the distractors: Choice A (600,000 N) likely comes from incorrectly doubling the thrust force, perhaps by misapplying Newton's third law. Choice C (900,000 N) might result from adding the rocket's weight to the thrust, but remember we're in space where gravitational effects are negligible. Choice D (200,000 N) could come from calculation errors or using incorrect formulas. Notice that the rocket's mass (5000 kg) and acceleration (60 m/s²) are red herrings in this problem. While these could help you find the net force on the rocket, the question specifically asks for the exhaust force according to Newton's third law. Remember: In rocket problems, focus on the momentum transfer rate of the exhaust. The thrust force depends only on how fast mass is ejected and at what velocity, not on the rocket's current state of motion.

Question 11

A hollow spherical shell of mass $M$ and inner radius $a$ and outer radius $b$ has uniform density. A point mass $m$ is located at the geometric center of the shell. What is the gravitational force on the point mass?

$\frac{GMm}{a^2}$
$\frac{GMm}{b^2}$
$\frac{GMm}{(\frac{a+b}{2})^2}$
$0$ (correct answer)

Explanation: According to the shell theorem, a spherically symmetric mass distribution exerts zero net gravitational force on a point mass located at its center. This is because every element of mass in the shell has a corresponding element on the opposite side, and these paired elements exert equal and opposite forces on the central point mass, resulting in zero net force. Choices A, B, and C incorrectly apply gravitational force formulas using different characteristic distances of the shell.

Question 12

A mass $m$ is attached to two identical springs, each with spring constant $k$ , arranged in parallel (both springs attached between the mass and a fixed wall). The mass is displaced from equilibrium by distance $x_0$ and released. At the moment when the displacement is $\frac{x_0}{2}$ , what is the ratio of kinetic energy to elastic potential energy?

$\frac{1}{3}$
$\frac{1}{2}$
$\frac{3}{1}$ (correct answer)
$\frac{2}{1}$

Explanation: When you encounter problems involving springs in parallel and energy conservation, remember that parallel springs act as a single spring with combined stiffness, and mechanical energy remains constant throughout the motion. Since the two springs are in parallel, they share the load equally, giving an effective spring constant of $k_{eff} = 2k$ . You can solve this using conservation of mechanical energy: $E_{total} = KE + PE = constant$ . At the initial position ( $x_0$ ), all energy is potential: $E_{total} = \frac{1}{2}(2k)x_0^2 = kx_0^2$ . At position $x = \frac{x_0}{2}$ , the potential energy is: $PE = \frac{1}{2}(2k)\left(\frac{x_0}{2}\right)^2 = \frac{1}{2}(2k)\frac{x_0^2}{4} = \frac{kx_0^2}{4}$ . Using energy conservation: $KE = E_{total} - PE = kx_0^2 - \frac{kx_0^2}{4} = \frac{3kx_0^2}{4}$ . Therefore: $\frac{KE}{PE} = \frac{\frac{3kx_0^2}{4}}{\frac{kx_0^2}{4}} = 3 = \frac{3}{1}$ , confirming answer C. Answer A ( $\frac{1}{3}$ ) incorrectly inverts the ratio. Answer B ( $\frac{1}{2}$ ) likely comes from forgetting about the parallel spring configuration and using the wrong effective spring constant. Answer D ( $\frac{2}{1}$ ) might result from calculation errors in the energy expressions or misapplying the displacement value. Study tip: For spring-mass energy problems, always identify the effective spring constant first (series: $\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}$ ; parallel: $k_{eff} = k_1 + k_2$ ), then apply conservation of mechanical energy systematically.

Question 13

An object moves at a constant speed in a circular path. Which of the following statements about the work done on the object is correct?

The net force does positive work, increasing the kinetic energy.
The net force does negative work, decreasing the kinetic energy.
The net force does zero work, and the kinetic energy remains constant. (correct answer)
The work done depends on the displacement for one revolution.

Explanation: The net force on an object in uniform circular motion is the centripetal force, which is always directed towards the center of the circle. The object's instantaneous displacement is always tangent to the circle. Therefore, the centripetal force is always perpendicular to the displacement. The work done by a force is given by $W = Fd\cos\theta$ . Since the angle $θ$ between the force and displacement is 90°, the work done is zero. By the work-energy theorem, zero net work means the kinetic energy is constant, which is consistent with constant speed.

Question 14

Using the given parameters, a pendulum of length $L=2.0\,\text{m}$ swings at small angles; determine the time for one full oscillation.

$T=\pi\sqrt{\dfrac{L}{g}}\,\text{s}$
$T=2\pi\sqrt{\dfrac{L}{g}}\,\text{s}$ (correct answer)
$T=2\pi\sqrt{\dfrac{g}{L}}\,\text{s}$
$T=2\pi\sqrt{\dfrac{m}{k}}\,\text{s}$

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 simple pendulum at small angles, the period T is determined by T = 2π√(L/g), where L is the pendulum length and g is gravitational acceleration. Choice B is correct because it shows the standard pendulum formula T = 2π√(L/g), which with L = 2.0 m and g = 9.8 m/s² gives T = 2π√(2.0/9.8) = 2.84 s. Choice A is incorrect because it's missing the factor of 2 in front of π, which would give half the actual period. To help students: Derive the pendulum formula from the restoring torque τ = -mgL sin θ ≈ -mgLθ for small angles. Practice calculating periods for different pendulum lengths to build intuition about the √L dependence.

Question 15

A fan blade of radius $0.25$ m is rotating at $120$ rad/s. It is then turned off and decelerates uniformly, coming to rest in $8.0$ s. What is the magnitude of the total linear acceleration of a point on the tip of the blade at the instant it is turned off ( $t=0$ )?

$3.75 \text{ m/s}^2$
$3600 \text{ m/s}^2$ (correct answer)
$3604 \text{ m/s}^2$
$14400 \text{ m/s}^2$

Explanation: At $t=0$ , the initial angular velocity is $\omega_0 = 120$ rad/s. The angular acceleration is constant: $\alpha = (\omega_f - \omega_0) / t = (0 - 120 \text{ rad/s}) / 8.0 \text{ s} = -15 \text{ rad/s}^2$ . The tangential acceleration is $a_T = R|\alpha| = (0.25 \text{ m})(15 \text{ rad/s}^2) = 3.75 \text{ m/s}^2$ . The centripetal acceleration at $t=0$ is $a_c = R\omega_0^2 = (0.25 \text{ m})(120 \text{ rad/s})^2 = 3600 \text{ m/s}^2$ . The total acceleration is $a = \sqrt{a_T^2 + a_c^2} = \sqrt{(3.75)^2 + (3600)^2} \approx \sqrt{14 + 12960000} \approx 3600 \text{ m/s}^2$ . The centripetal component is much larger than the tangential component, so the total acceleration is very close to the centripetal acceleration.

Question 16

Two identical balls collide head-on. Ball 1 initially moves at 6 m/s to the right, and ball 2 initially moves at 2 m/s to the left. After collision, ball 1 moves at 1 m/s to the left. Assuming the collision occurs along a straight line, what can be concluded about this collision?

The collision is elastic because momentum is conserved throughout the interaction
The collision is inelastic because the coefficient of restitution is less than unity (correct answer)
The collision violates conservation of momentum and is therefore physically impossible
The collision is perfectly inelastic because one ball reverses its direction of motion

Explanation: Let's check momentum conservation first. Initial: $p_i = m(6) + m(-2) = 4m$ . After collision, ball 1 has velocity -1 m/s. From momentum conservation: $4m = m(-1) + mv_2$ , so $v_2 = 5$ m/s (to the right). Coefficient of restitution: $e = \frac{|v_2 - v_1|}{|u_1 - u_2|} = \frac{|5 - (-1)|}{|6 - (-2)|} = \frac{6}{8} = 0.75 < 1$ . Since $e < 1$ , the collision is inelastic but not perfectly inelastic. Choice A is wrong because momentum conservation doesn't determine if collision is elastic. Choice C is wrong because momentum is conserved. Choice D is wrong because perfectly inelastic means $e = 0$ .

Question 17

A particle's velocity vector is $\vec{v} = (-3.0 \hat{i} + 4.0 \hat{j})$ m/s. What is the approximate angle of the velocity vector, measured counterclockwise from the positive x-axis?

53.1°
126.9° (correct answer)
233.1°
306.9°

Explanation: The vector has a negative x-component and a positive y-component, placing it in the second quadrant. The reference angle relative to the negative x-axis can be found using the arctangent: $\alpha = \arctan{\left|\frac{v_y}{v_x}\right|} = \arctan{\left|\frac{4.0}{-3.0}\right|} \approx 53.1^\circ$ . The angle measured counterclockwise from the positive x-axis is $180^\circ - \alpha = 180^\circ - 53.1^\circ = 126.9^\circ$ . Distractor A is only the reference angle. The other distractors correspond to angles in the third and fourth quadrants.

Question 18

A rigid body with rotational inertia $I$ experiences a net torque $\tau$ , resulting in an angular acceleration $\alpha$ . If the mass of the body is doubled, with its shape and dimensions remaining identical, what is the new angular acceleration for the same net torque $\tau$ ?

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

Explanation: Rotational inertia $I$ is directly proportional to mass $M$ (e.g., $I=kMR^2$ ). If the mass is doubled while shape and dimensions are unchanged, the new rotational inertia $I'$ will be $2I$ . From Newton's second law for rotation, $\alpha = \tau/I$ . The new angular acceleration $\alpha'$ will be $\alpha' = \tau/I' = \tau/(2I) = \frac{1}{2}(\tau/I) = \frac{1}{2}\alpha$ .

Question 19

Two identical uniform rods, each of mass $M$ and length $L$ , are initially at rest. Rod A lies along the x-axis with its center at the origin. Rod B lies along the y-axis with its center also at the origin, forming a cross. A brief torque is applied about the z-axis (perpendicular to both rods) to the entire system, giving it an angular impulse $J$ . What is the final angular momentum of rod A about the origin?

$\frac{J}{2}$ (correct answer)
$\frac{J}{4}$
$\frac{ML^2J}{12I_{total}}$ where $I_{total}$ is the total moment of inertia
$\frac{J\sqrt{2}}{4}$

Explanation: Both rods have the same moment of inertia about the z-axis: $I = \frac{1}{12}ML^2$ . The total moment of inertia is $I_{total} = 2 \cdot \frac{1}{12}ML^2 = \frac{1}{6}ML^2$ . The angular impulse gives the system angular velocity $\omega = \frac{J}{I_{total}} = \frac{6J}{ML^2}$ . Each rod rotates with this same angular velocity, so rod A's angular momentum is $L_A = I_A\omega = \frac{1}{12}ML^2 \cdot \frac{6J}{ML^2} = \frac{J}{2}$ . Choice B would be correct if there were 4 identical rods. Choice C is unnecessarily complicated and doesn't simplify correctly. Choice D introduces an incorrect $\sqrt{2}$ factor.

Question 20

A uniform solid sphere of mass $M$ and radius $R$ rolls without slipping down an incline that makes an angle $\theta$ with the horizontal. What is the magnitude of the linear acceleration of the sphere's center of mass? The rotational inertia of a solid sphere is $I = \frac{2}{5}MR^2$ .

$g \sin\theta$
$\frac{2}{3} g \sin\theta$
$\frac{5}{7} g \sin\theta$ (correct answer)
$\frac{1}{2} g \sin\theta$

Explanation: The net force down the incline is $Mg\sin\theta - f = Ma$ . The torque causing rotation is $\tau = fR = I\alpha$ . For rolling without slipping, $a = R\alpha$ . Substituting for $f$ and $\alpha$ : $f = I\alpha/R = (\frac{2}{5}MR^2)(a/R)/R = \frac{2}{5}Ma$ . Now substitute this into the force equation: $Mg\sin\theta - \frac{2}{5}Ma = Ma$ . This simplifies to $Mg\sin\theta = \frac{7}{5}Ma$ , so $a = \frac{5}{7}g\sin\theta$ .

Question 21

A wind turbine rotor with $I=1.0\times10^3\ \text{kg·m}^2$ spins at $\omega_i=2.0\ \text{rad/s}$ ; the wind drops and the net torque becomes $\tau=-200\ \text{N·m}$ for $t=5.0\ \text{s}$ , so $\Delta L=\tau t$ and $L=I\omega$ . Neglect other torques. Based on the scenario above, what is the final angular velocity of the system after 5.0 seconds?

3.0 rad/s
1.0 rad/s (correct answer)
0.0 rad/s
2.2 rad/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 wind turbine experiences a negative torque of -200 N·m for 5.0 seconds, changing its angular momentum by ΔL = τt = -200 × 5.0 = -1000 N·m·s. Choice B is correct because the initial angular momentum is Li = Iωi = 1000 × 2.0 = 2000 N·m·s, the final angular momentum is Lf = Li + ΔL = 2000 + (-1000) = 1000 N·m·s, and the final angular velocity is ωf = Lf/I = 1000/1000 = 1.0 rad/s. Choice C is incorrect because it would mean the turbine stops completely, which would require exactly -2000 N·m·s of impulse, not -1000 N·m·s. To help students: Practice problems involving partial reduction of angular velocity versus complete stopping. Emphasize checking if the final result makes physical sense. Watch for: assumptions that negative torque always brings objects to rest, rather than just reducing angular velocity.

Question 22

An object moves along the x-axis with a velocity given by $v(t) = 3t^2 + 2t$ , where $v$ is in m/s and $t$ is in seconds. What is its displacement during the interval from $t=1$ s to $t=3$ s?

$36$ m
$34$ m (correct answer)
$33$ m/s
$2$ m

Explanation: Displacement is the integral of the velocity function over the time interval. $\Delta x = \int_{1}^{3} v(t) dt = \int_{1}^{3} (3t^2 + 2t) dt = [t^3 + t^2]_{1}^{3} = (3^3 + 3^2) - (1^3 + 1^2) = (27 + 9) - (1 + 1) = 36 - 2 = 34$ m.

Question 23

How does conservation of angular momentum affect satellite speed if orbital radius decreases due to a brief inward impulse?

Speed decreases so that $L=mrv$ remains constant.
Speed increases so that $L=mrv$ remains constant. (correct answer)
Speed is unchanged because gravity cancels angular momentum.
Speed increases because potential energy increases as $r$ decreases.

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). Angular momentum L = mrv is conserved for a satellite when no external torques act, which is true for central forces like gravity. When orbital radius r decreases due to an inward impulse, conservation of angular momentum requires that velocity v must increase to keep L constant. Choice B is correct because it accurately applies conservation of angular momentum: as r decreases, v must increase proportionally so that the product mrv remains constant. Choice A is incorrect because it states speed decreases, which would violate angular momentum conservation when radius decreases. To help students: Emphasize that gravity exerts no torque about the central body, practice applying L = mrv for different orbital scenarios, and use the ice skater analogy where pulling arms inward increases rotation speed. Have students calculate specific velocity changes for given radius changes.

Question 24

A block of mass $m$ slides from rest down a frictionless incline of height $h$ . It then moves onto a rough horizontal surface with a coefficient of kinetic friction $\mu_k$ . How far does the block slide on the horizontal surface before coming to rest?

$\frac{h}{\mu_k}$ (correct answer)
$\frac{gh}{\mu_k}$
$\mu_k h$
$\frac{2gh}{\mu_k}$

Explanation: On the incline, conservation of energy gives the block's kinetic energy at the bottom as $K = mgh$ . On the horizontal surface, the work done by friction, $W_f = -f_k d = -\mu_k mgd$ , must equal the change in kinetic energy, which is $0 - K = -mgh$ . Setting $-\mu_k mgd = -mgh$ and solving for $d$ gives $d = h/\mu_k$ .

Question 25

A square plate of side length L is pivoted at its center. Four forces of equal magnitude F are applied. Which of the following applications produces the greatest magnitude of torque about the pivot?

Force applied at a corner, perpendicular to the line connecting the center to that corner. (correct answer)
Force applied at the midpoint of an edge, perpendicular to that edge.
Force applied at a corner, directed parallel to an adjacent edge of the plate.
Force applied at a corner, directed toward the center of the plate.

Explanation: The magnitude of the torque is $\tau = rF\sin\theta$ . We want to maximize $r \sin\theta$ . In case A, the distance to a corner is $r = \sqrt{(L/2)^2 + (L/2)^2} = L/\sqrt{2}$ , and the force is perpendicular ( $\sin\theta=1$ ), so $\tau_A = (L/\sqrt{2})F \approx 0.707LF$ . In case B, the distance is $r=L/2$ and the force is perpendicular, so $\tau_B = (L/2)F = 0.5LF$ . In case C, the distance is $r = L/\sqrt{2}$ but the angle is not 90 degrees; the torque is $LF/2 = 0.5LF$ . In case D, the force points toward the pivot, so $\theta=180^\circ$ and the torque is zero. Comparing the magnitudes, case A produces the greatest torque.

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