AP Physics C Mechanics

AP Physics C Mechanics Practice Test: Practice Test 8

Practice Test 8 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 force is applied to a wrench handle to tighten a bolt. To produce the greatest possible torque on the bolt with a given force magnitude, the force should be applied:

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Question 1

A force is applied to a wrench handle to tighten a bolt. To produce the greatest possible torque on the bolt with a given force magnitude, the force should be applied:

as close to the bolt as possible and parallel to the handle.
as far from the bolt as possible and perpendicular to the handle. (correct answer)
as far from the bolt as possible and parallel to the handle.
at the midpoint of the handle and at a 45-degree angle to it.

Explanation: Torque is given by the expression $\vec{\tau} = \vec{r} \times \vec{F}$ , and its magnitude is $\tau = rF\sin\theta$ . To maximize the torque for a given force magnitude $F$ , the distance from the pivot (the bolt), $r$ , must be maximized, and the angle $\theta$ between the position vector $\vec{r}$ and the force vector $\vec{F}$ must be $90^\circ$ (so $\sin\theta = 1$ ). Therefore, the force should be applied as far from the bolt as possible and perpendicular to the handle.

Question 2

A bicycle wheel with $I=0.40\ \text{kg·m}^2$ starts from rest; a constant tangential force $F=10\ \text{N}$ is applied at radius $r=0.30\ \text{m}$ for $t=5.0\ \text{s}$ , so $\tau=rF$ and $\Delta L=\tau t$ . Ignore friction and keep the force perpendicular to the radius. Based on the scenario above, determine the change in angular momentum of the wheel.

15 kg·m $^2$ /s (correct answer)
6.0 kg·m $^2$ /s
3.0 kg·m $^2$ /s
15 N·m

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 bicycle wheel experiences a tangential force creating a torque τ = rF = 0.30 × 10 = 3.0 N·m for 5.0 seconds. Choice A is correct because the change in angular momentum is ΔL = τt = 3.0 × 5.0 = 15 kg·m²/s, properly applying the angular impulse-momentum theorem. Choice D is incorrect because it gives the answer in N·m instead of kg·m²/s, confusing the units of torque with those of angular momentum. To help students: Stress the importance of dimensional analysis in physics problems. Create unit conversion charts showing the relationships between rotational quantities. Watch for: unit confusion between torque (N·m) and angular momentum (kg·m²/s or N·m·s).

Question 3

A wind turbine generates electrical power according to $P = 0.4v^3$ kW, where v is the wind speed in m/s. During a particular day, the wind speed varies as $v(t) = 8 + 2\sin(\pi t/12)$ m/s, where t is time in hours. What is the average power generated during the first 12 hours?

205 kW
220 kW (correct answer)
235 kW
250 kW

Explanation: The power is P(t) = 0.4[v(t)]³ = 0.4[8 + 2sin(πt/12)]³. To find the average power over 12 hours: P_avg = (1/12)∫₀¹² 0.4[8 + 2sin(πt/12)]³ dt. Let u = πt/12, so du = π dt/12, and dt = 12du/π. When t = 0, u = 0; when t = 12, u = π. The integral becomes: P_avg = (1/12) × 0.4 × (12/π) ∫₀^π [8 + 2sin(u)]³ du = (0.4/π) ∫₀^π [8 + 2sin(u)]³ du. Expanding: [8 + 2sin(u)]³ = 512 + 384sin(u) + 96sin²(u) + 8sin³(u). Using standard integrals: ∫₀^π sin(u)du = 0, ∫₀^π sin²(u)du = π/2, ∫₀^π sin³(u)du = 0. So: ∫₀^π [8 + 2sin(u)]³ du = 512π + 0 + 96(π/2) + 0 = 512π + 48π = 560π. Therefore: P_avg = (0.4/π) × 560π = 224 kW ≈ 220 kW.

Question 4

A $3.0\ \text{kg}$ object starts from rest and experiences a constant net force of $18\ \text{N}$ to the right for $0.50\ \text{s}$ . Using the given data, what is the final velocity of the object after the collision?

$v_f = 1.5\ \text{m/s}$
$v_f = 3.0\ \text{m/s}$ (correct answer)
$v_f = 9.0\ \text{m/s}$
$v_f = 0.33\ \text{m/s}$

Explanation: This question tests understanding of the impulse-momentum theorem in AP Physics C: Mechanics, focusing on finding final velocity after applying a constant force. The impulse-momentum theorem combined with kinematics allows us to find final velocity: J = FΔt = m(vf - vi), so vf = vi + (FΔt)/m. In this scenario, starting from rest (vi = 0), with F = 18 N, Δt = 0.50 s, and m = 3.0 kg, we get vf = 0 + (18 × 0.50)/3.0 = 9.0/3.0 = 3.0 m/s. Choice B is correct because it shows vf = 3.0 m/s, resulting from proper application of the impulse-momentum theorem. Choice C incorrectly calculates 18/2 = 9.0 m/s, while choice A shows half the correct value. When teaching this concept, emphasize the connection between Newton's second law and the impulse-momentum theorem. Students should practice problems starting from rest to build confidence before tackling more complex scenarios with non-zero initial velocities.

Question 5

Based on the described system, a small-angle pendulum with $L=1.6\,\text{m}$ oscillates; calculate the frequency in Hz using $f=1/T$ .

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

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, the period is T = 2π√(L/g), so the frequency f = 1/T = 1/(2π√(L/g)) = (1/2π)√(g/L). Choice A is correct because it properly shows f = (1/2π)√(g/L), which with L = 1.6 m and g = 9.8 m/s² gives f = (1/2π)√(9.8/1.6) = 0.394 Hz. Choice B is incorrect because it inverts the g/L ratio, which would give incorrect units and violate the physics - longer pendulums have lower frequencies. To help students: Show the algebraic manipulation from T to f explicitly. Use dimensional analysis to verify that √(g/L) has units of s⁻¹, confirming the frequency formula is correct.

Question 6

Car A travels east at 30 m/s. Car B travels north at 40 m/s. What is the velocity of Car A as measured by an observer in Car B?

50 m/s, at an angle of 53° South of East (correct answer)
50 m/s, at an angle of 37° North of East
70 m/s, at an angle of 53° South of East
70 m/s, at an angle of 37° North of East

Explanation: Let east be the positive x-direction ( $\hat{i}$ ) and north be the positive y-direction ( $\hat{j}$ ). Then $v_A = 30\hat{i}$ m/s and $v_B = 40\hat{j}$ m/s. The velocity of A relative to B is $v_{AB} = v_A - v_B = 30\hat{i} - 40\hat{j}$ m/s. The magnitude (speed) is $|v_{AB}| = \sqrt{30^2 + (-40)^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50$ m/s. The direction is given by $\theta = \arctan(\frac{-40}{30}) \approx -53°$ , which corresponds to 53° South of East.

Question 7

A satellite orbits Earth at a distance of $2R$ from Earth's center, where $R$ is Earth's radius. If the satellite's orbital radius is increased to $4R$ , by what factor does the gravitational force on the satellite change?

Decreases by a factor of $2$
Decreases by a factor of $4$ (correct answer)
Decreases by a factor of $8$
Increases by a factor of $2$

Explanation: The gravitational force follows $F = \frac{GMm}{r^2}$ . Initially, $F_1 = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}$ . Finally, $F_2 = \frac{GMm}{(4R)^2} = \frac{GMm}{16R^2}$ . The ratio is $\frac{F_2}{F_1} = \frac{4R^2}{16R^2} = \frac{1}{4}$ , so the force decreases by a factor of 4. Choice A uses only the linear change in radius. Choice C incorrectly cubes the radius change. Choice D incorrectly suggests an increase.

Question 8

A uniform rod of length $L$ and mass $M$ is pivoted at one end and can rotate freely in a vertical plane. The rod is initially horizontal and released from rest. When the rod has rotated through an angle $\theta$ from the horizontal, what is the gravitational potential energy of the rod relative to its initial position?

$-\frac{MgL}{2}\sin\theta$ (correct answer)
$-\frac{MgL}{2}(1 - \cos\theta)$
$-\frac{MgL}{2}\cos\theta$
$-MgL\sin\theta$

Explanation: The center of mass of a uniform rod is at its geometric center, which is at distance $L/2$ from the pivot point. Initially, when the rod is horizontal, the center of mass is at height $L/2$ above the lowest possible position of the center of mass. When the rod rotates through angle $\theta$ from horizontal (measuring downward), the center of mass descends by a vertical distance $\frac{L}{2}\sin\theta$ . Therefore, the change in gravitational potential energy is $\Delta U = -Mg \cdot \frac{L}{2}\sin\theta = -\frac{MgL}{2}\sin\theta$ . Choice B would be correct if $\theta$ were measured from the vertical position. Choice C has the wrong trigonometric function. Choice D is missing the factor of 1/2 that accounts for the center of mass being at the middle of the rod.

Question 9

A circular platform of mass $M$ and radius $R$ is free to rotate friction-free about its center. A person of mass $m$ stands at the edge. The system is initially at rest. The person begins to walk along the edge with a speed $v$ relative to the platform. What is the magnitude of the angular velocity of the platform relative to the ground? The rotational inertia of the platform is $I_p = \frac{1}{2}MR^2$ .

$\frac{mv}{R(m)}$
$\frac{mv}{R(\frac{1}{2}M)}$
$\frac{mv}{R(\frac{1}{2}M + m)}$ (correct answer)
$\frac{mv}{R(M + m)}$

Explanation: The system starts from rest, so the total initial angular momentum is zero. As there are no external torques, the total final angular momentum must also be zero. Let $\omega_p$ be the platform's angular velocity relative to the ground. The person's velocity relative to the ground is $v_g = v_{rel} - v_{plat} = v - R\omega_p$ . The total final angular momentum is $L_f = I_p\omega_p + I_{person}\omega_{person} = 0$ . The person's angular velocity is $v_g/R$ . So, $(\frac{1}{2}MR^2)\omega_p + (mR^2)(\frac{v - R\omega_p}{R}) = 0$ . Simplifying: $\frac{1}{2}MR^2\omega_p + mR(v-R\omega_p) = 0 \Rightarrow \frac{1}{2}MR^2\omega_p + mRv - mR^2\omega_p = 0 \Rightarrow \omega_p(\frac{1}{2}MR^2 + mR^2) = mRv$ . Solving for $\omega_p$ gives the result.

Question 10

A mobile consists of two horizontal rods suspended by strings. The upper rod has mass $M_1 = 0.5\text{ kg}$ and length $L_1 = 1.0\text{ m}$ . Two masses are attached to its ends: $m_1 = 2.0\text{ kg}$ on the left and $m_2 = 3.0\text{ kg}$ on the right. For the upper rod to be in rotational equilibrium, at what distance from the left end must the suspension string be attached?

$0.40\text{ m}$
$0.45\text{ m}$
$0.55\text{ m}$ (correct answer)
$0.50\text{ m}$

Explanation: When you encounter a mobile or balance problem, you're dealing with rotational equilibrium, where the net torque about any point must be zero. The key insight is choosing a convenient pivot point—typically where you want to find the unknown position. Let's call the distance from the left end where the string attaches $x$ . Taking torques about this suspension point, we have three forces creating torques: the weight of $m_1$ at distance $x$ to the left, the weight of the rod $M_1$ acting at its center of mass, and the weight of $m_2$ at distance $(L_1 - x)$ to the right. For rotational equilibrium: $m_1 g \cdot x = M_1 g \cdot (0.5 - x) + m_2 g \cdot (L_1 - x)$ The rod's center of mass is at $0.5\text{ m}$ from the left end, so it's $(0.5 - x)$ from our pivot point. Substituting values and canceling $g$ : $2.0x = 0.5(0.5 - x) + 3.0(1.0 - x)$ $2.0x = 0.25 - 0.5x + 3.0 - 3.0x$ $2.0x = 3.25 - 3.5x$ $5.5x = 3.25$ $x = 0.59\text{ m} \approx 0.55\text{ m}$ Answer choice (A) $0.40\text{ m}$ would result from ignoring the rod's mass entirely. Choice (B) $0.45\text{ m}$ likely comes from incorrectly placing the rod's center of mass. Choice (D) $0.50\text{ m}$ assumes the heavier mass doesn't affect the balance point of the rod itself. Remember: always include the mass of the supporting structure in equilibrium problems—it's a common oversight that leads to incorrect answers.

Question 11

The velocity of a particle in the xy-plane is given by $\vec{v}(t) = (4t)\hat{i} + (3t^2 - 1)\hat{j}$ . What is the magnitude of the particle's acceleration at $t=1 \text{ s}$ ?

$\sqrt{40}$
$10$
$\sqrt{52}$ (correct answer)
$5$

Explanation: The acceleration vector is the time derivative of the velocity vector: $\vec{a}(t) = \frac{d\vec{v}}{dt} = 4\hat{i} + (6t)\hat{j}$ . At $t=1 \text{ s}$ , the acceleration vector is $\vec{a}(1) = 4\hat{i} + 6(1)\hat{j} = 4\hat{i} + 6\hat{j}$ . The magnitude of the acceleration is $|\vec{a}(1)| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52}$ .

Question 12

A composite system consists of a solid cylinder of mass $M$ and radius $R$ with its center at the origin, and a point mass $m$ located at distance $3R$ from the origin along the positive $x$ -axis. What condition must be satisfied for the center of mass to be located at $x = R$ ?

$m = \frac{M}{2}$ (correct answer)
$m = \frac{M}{3}$
$m = \frac{2M}{3}$
$m = M$

Explanation: Setting the center of mass at $x = R$ : $R = \frac{M \cdot 0 + m \cdot 3R}{M + m} = \frac{3mR}{M + m}$ . Solving: $R(M + m) = 3mR$ , so $M + m = 3m$ , which gives $M = 2m$ or $m = \frac{M}{2}$ . Choice B would place the center of mass too far from the origin. Choice C overestimates the required point mass. Choice D makes the masses equal, shifting the center of mass too far right.

Question 13

Two planets, A and B, have the same density but planet B has twice the radius of planet A. If the surface gravitational field strength on planet A is $g_A$ , what is the surface gravitational field strength on planet B?

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

Explanation: Surface gravitational field $g = \frac{GM}{R^2}$ . Since density $\rho$ is constant, $M = \frac{4}{3}\pi R^3 \rho$ , so $g = \frac{G \cdot \frac{4}{3}\pi R^3 \rho}{R^2} = \frac{4\pi G \rho R}{3}$ . Therefore, $g \propto R$ when density is constant. Since planet B has twice the radius, $g_B = 2g_A$ . Choice A uses inverse square relationship incorrectly. Choice B uses inverse relationship. Choice D uses square relationship incorrectly.

Question 14

A solid disk and a hoop have the same mass $M$ and radius $R$ . They are both initially at rest and are free to rotate about a fixed, frictionless axle through their centers. The same constant net torque is applied to both. After one second, which of the following correctly compares their angular accelerations, $\alpha_{disk}$ and $\alpha_{hoop}$ ?

$\alpha_{disk} = \alpha_{hoop}$ because the net torque is the same for both objects.
$\alpha_{disk} > \alpha_{hoop}$ because the disk has a smaller rotational inertia. (correct answer)
$\alpha_{disk} < \alpha_{hoop}$ because the hoop has its mass concentrated at the rim.
The relationship cannot be determined without knowing the magnitude of the torque.

Explanation: According to Newton's second law for rotation, $\alpha = \tau_{net}/I$ . Since both objects experience the same net torque, the object with the smaller rotational inertia will have the greater angular acceleration. The rotational inertia of a solid disk is $I_{disk} = \frac{1}{2}MR^2$ , while that of a hoop is $I_{hoop} = MR^2$ . Thus, $I_{disk} < I_{hoop}$ , which means $\alpha_{disk} > \alpha_{hoop}$ .

Question 15

A particle's position is described by the vector $\vec{r}(t) = (R\cos(\omega t))\hat{i} + (2R\sin(\omega t))\hat{j}$ , where $R$ and $\omega$ are positive constants. At which of the following times $t > 0$ are the particle's velocity and acceleration vectors first perpendicular to each other?

$t = \frac{\pi}{2\omega}$ (correct answer)
$t = \frac{\pi}{4\omega}$
$t = \frac{\pi}{\omega}$
The vectors are never perpendicular for $t>0$ .

Explanation: When you encounter parametric motion problems involving perpendicular vectors, you need to find when their dot product equals zero. This tests your understanding of vector calculus and the geometric relationship between velocity and acceleration. To find when velocity and acceleration are perpendicular, first calculate these vectors by taking derivatives of the position vector. The velocity is: $\vec{v}(t) = \frac{d\vec{r}}{dt} = -R\omega\sin(\omega t)\hat{i} + 2R\omega\cos(\omega t)\hat{j}$ The acceleration is: $\vec{a}(t) = \frac{d\vec{v}}{dt} = -R\omega^2\cos(\omega t)\hat{i} - 2R\omega^2\sin(\omega t)\hat{j}$ For perpendicular vectors, their dot product must equal zero: $\vec{v} \cdot \vec{a} = (-R\omega\sin(\omega t))(-R\omega^2\cos(\omega t)) + (2R\omega\cos(\omega t))(-2R\omega^2\sin(\omega t)) = 0$ Simplifying: $R^2\omega^3\sin(\omega t)\cos(\omega t) - 4R^2\omega^3\sin(\omega t)\cos(\omega t) = 0$ This gives us: $-3R^2\omega^3\sin(\omega t)\cos(\omega t) = 0$ Since $R$ and $\omega$ are positive constants, either $\sin(\omega t) = 0$ or $\cos(\omega t) = 0$ . The first occurrence for $t > 0$ is when $\cos(\omega t) = 0$ , which happens at $\omega t = \frac{\pi}{2}$ , giving $t = \frac{\pi}{2\omega}$ . Answer A is correct. Answer B gives $t = \frac{\pi}{4\omega}$ , where neither sine nor cosine is zero. Answer C gives $t = \frac{\pi}{\omega}$ , which corresponds to the second occurrence. Answer D is wrong since we found specific times when the vectors are perpendicular. Remember: perpendicular vectors have zero dot product, so always set up the dot product equation and solve systematically.

Question 16

A system consists of two point masses: $m_1 = 2.0\text{ kg}$ at position $\vec{r_1} = (1, 2, 3)\text{ m}$ and $m_2 = 4.0\text{ kg}$ at position $\vec{r_2} = (4, -1, 2)\text{ m}$ . What is the position vector of the center of mass?

$( 2.5, 0.5, 2.5)\text{ m}$
$(3.0, 0, 2.17)\text{ m}$
$(3.0, 0, 2.33)\text{ m}$ (correct answer)
$(2.67, 0.67, 2.67)\text{ m}$

Explanation: The center of mass position is: $\vec{r}_{cm} = \frac{m_1\vec{r_1} + m_2\vec{r_2}}{m_1 + m_2}$ . Computing each component: $x_{cm} = \frac{2.0 \times 1 + 4.0 \times 4}{6.0} = \frac{18}{6.0} = 3.0$ , $y_{cm} = \frac{2.0 \times 2 + 4.0 \times (-1)}{6.0} = \frac{0}{6.0} = 0$ , $z_{cm} = \frac{2.0 \times 3 + 4.0 \times 2}{6.0} = \frac{14}{6.0} = 2.33$ . Therefore $\vec{r}_{cm} = (3.0, 0, 2.33)\text{ m}$ . The other choices use incorrect mass weightings or arithmetic errors.

Question 17

A small solid sphere of radius $r$ rolls without slipping inside a large fixed hemispherical bowl of radius $R$ , starting from rest at the same height as the center of the bowl. What is the speed of the sphere's center of mass at the bottom of the bowl?

$\sqrt{2gR}$
$\sqrt{gR}$
$\sqrt{\frac{10}{7}g(R-r)}$ (correct answer)
$\sqrt{\frac{5}{7}gR}$

Explanation: The center of mass of the small sphere falls a vertical distance of $h = R-r$ . The initial potential energy is $Mg(R-r)$ . This is converted into translational and rotational kinetic energy. For a solid sphere rolling, $K_{total} = \frac{7}{10}Mv^2$ . By conservation of energy, $Mg(R-r) = \frac{7}{10}Mv^2$ . Solving for $v$ gives $v = \sqrt{\frac{10}{7}g(R-r)}$ .

Question 18

A student holds a bicycle wheel by its axle and gets it spinning rapidly. The student then sits on a stool that is free to rotate and holds the wheel with its axle vertical.

The student-stool-wheel system is now rotating with a constant angular velocity. What can be concluded about the net torque on the system?

A net torque from the spinning wheel keeps the system rotating.
The net torque on the system is zero. (correct answer)
A net torque must be supplied by the student to maintain the rotation.
The net torque is non-zero and points vertically.

Explanation: The system is rotating with a constant angular velocity. According to the rotational version of Newton's first law, if the angular velocity of a system is constant, its angular acceleration is zero, and therefore the net external torque acting on the system must be zero. Frictional torques in the stool's axle are assumed to be negligible in this idealized scenario.

Question 19

A torsion pendulum consists of a solid disk with rotational inertia $I$ attached to a wire with torsion constant $\kappa$ . It oscillates with a period $T_0$ . If the disk is replaced with a hoop of the same mass and radius, what is the new period? The rotational inertia of a hoop is twice that of a solid disk of the same mass and radius.

$T_0/\sqrt{2}$
$T_0/2$
$\sqrt{2}T_0$ (correct answer)
$2T_0$

Explanation: The period of a torsion pendulum is $T = 2\pi\sqrt{I/\kappa}$ . Since the period is proportional to the square root of the rotational inertia ( $T \propto \sqrt{I}$ ), and the new rotational inertia is $I' = 2I$ , the new period will be $T' = \sqrt{2}T_0$ .

Question 20

Using the information given, determine the object's acceleration at this point for a block sliding down a frictionless $30^\circ$ incline.

$9.8\,\text{m/s}^2$ down the incline
$4.9\,\text{m/s}^2$ down the incline (correct answer)
$8.5\,\text{m/s}^2$ up the incline
$4.9\,\text{m/s}^2$ up the incline

Explanation: This question tests AP Physics C kinematics, specifically motion on inclined planes in two dimensions. The motion requires decomposing gravitational acceleration into components parallel and perpendicular to the incline surface. In this scenario, a block slides down a frictionless 30° incline, experiencing only the component of gravity parallel to the surface. Choice B is correct because a_parallel = g·sin(30°) = 9.8 × 0.5 = 4.9 m/s² down the incline. Choice A incorrectly uses the full gravitational acceleration, C and D show wrong magnitudes or directions. To help students: Practice decomposing weight into components using free body diagrams on inclines. Emphasize that on frictionless surfaces, only the parallel component of weight causes acceleration. Watch for: using cosine instead of sine, or confusing the direction of acceleration with the normal force direction.

Question 21

Using Kepler’s third law, how does orbital period $T$ scale with orbital radius $r$ for satellites around Earth?

$T\propto r$ because gravity is constant near Earth.
$T\propto r^{3/2}$ because $T^2\propto r^3$ . (correct answer)
$T\propto r^2$ because $v\propto r$ in orbit.
$T\propto r^{1/2}$ because $v\propto r^{1/2}$ in 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 third law states that the square of the orbital period is proportional to the cube of the orbital radius, or T² ∝ r³. This relationship can be derived from Newton's laws by equating gravitational and centripetal forces and using the relationship between period and velocity. Choice B is correct because it accurately states that T ∝ r^(3/2), which follows from taking the square root of both sides of T² ∝ r³. Choice C is incorrect because it suggests T ∝ r², which would imply T² ∝ r⁴, violating Kepler's third law. To help students: Derive Kepler's third law from first principles using F = ma, practice applying the T² ∝ r³ relationship to compare orbital periods, and use log-log plots to verify the 3/2 power relationship. Emphasize that this law applies to all satellites orbiting the same central body.

Question 22

A simple pendulum and a physical pendulum, consisting of a uniform rod of length $L$ pivoted at one end, both have the same length $L$ . How does the period of the rod, $T_{rod}$ , compare to the period of the simple pendulum, $T_{simple}$ ?

$T_{rod} < T_{simple}$ (correct answer)
$T_{rod} > T_{simple}$
$T_{rod} = T_{simple}$
The comparison depends on the mass of the rod and the pendulum bob.

Explanation: The period of the simple pendulum is $T_{simple} = 2\pi\sqrt{L/g}$ . The period of the rod is $T_{rod} = 2\pi\sqrt{2L/3g}$ . Since $\sqrt{2/3} \approx 0.816$ , we have $T_{rod} \approx 0.816 \, T_{simple}$ , so the rod's period is shorter.

Question 23

A satellite is in an elliptical orbit around a planet. Which of the following quantities is conserved throughout the satellite's orbit?

The satellite's kinetic energy.
The satellite-planet system's gravitational potential energy.
The satellite's speed.
The total mechanical energy of the satellite-planet system. (correct answer)

Explanation: The gravitational force is a conservative force. For a system where only conservative forces do work, the total mechanical energy ( $E = K + U$ ) is conserved. In an elliptical orbit, the satellite's speed and distance from the planet both change, so kinetic energy and potential energy are not individually conserved, but their sum is.

Question 24

A pendulum bob of mass $m$ is released from rest at a height $H$ above its lowest point. Air resistance is negligible. What is the kinetic energy of the bob as it passes through its lowest point?

$mgH$ (correct answer)
$2mgH$
$\frac{1}{2}mgH$
$0$

Explanation: By conservation of mechanical energy, the initial potential energy $U_i = mgH$ (relative to the lowest point) is fully converted to kinetic energy $K_f$ at the lowest point, since $U_f = 0$ and the bob starts from rest ( $K_i = 0$ ). Thus, $E_i = E_f$ implies $mgH = K_f$ .

Question 25

A block of mass $m$ attached to a horizontal spring of constant $k$ is displaced from its equilibrium position by a distance $A$ and released from rest. What is the speed of the block when it is at a position $x = A/2$ ?

$\sqrt{\frac{3kA^2}{4m}}$ (correct answer)
$\sqrt{\frac{kA^2}{2m}}$
$\sqrt{\frac{kA^2}{4m}}$
$\sqrt{\frac{kA}{2m}}$

Explanation: The total mechanical energy of the system, set at the moment of release, is $E = \frac{1}{2}kA^2$ . At any position $x$ , the energy is $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$ . By conservation of energy, $\frac{1}{2}kA^2 = \frac{1}{2}k(A/2)^2 + \frac{1}{2}mv^2$ . This simplifies to $\frac{1}{2}kA^2 = \frac{1}{8}kA^2 + \frac{1}{2}mv^2$ , which gives $\frac{3}{8}kA^2 = \frac{1}{2}mv^2$ . Solving for $v$ yields $v = \sqrt{\frac{3kA^2}{4m}}$ .

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