AP Physics C Mechanics

AP Physics C Mechanics Practice Test: Practice Test 6

Practice Test 6 for AP Physics C Mechanics: real questions and explanations from the Varsity Tutors practice-test pool.

0 / 25 answered

Question 1 of 25

Planet X has mass $M$ and radius $R$ . Planet Y has mass $2M$ and radius $4R$ . An object of mass $m$ is placed on the surface of each planet. What is the ratio of the gravitational force on the object on Planet Y to the force on the object on Planet X, $F_Y / F_X$ ?

Question Navigator

All questions

Question 1

$1/8$ (correct answer)
$1/4$
$1/2$
$2$

Explanation: The gravitational force on the surface of a planet is given by $F = G \frac{M_{planet} m}{R_{planet}^2}$ . For Planet X, $F_X = G \frac{M m}{R^2}$ . For Planet Y, $F_Y = G \frac{(2M) m}{(4R)^2} = G \frac{2Mm}{16R^2} = \frac{1}{8} G \frac{Mm}{R^2}$ . The ratio is $\frac{F_Y}{F_X} = \frac{\frac{1}{8} G \frac{Mm}{R^2}}{G \frac{Mm}{R^2}} = \frac{1}{8}$ .

Question 2

The position of a particle moving in three-dimensional space is given by $\vec{r}(t) = (2t^2)\hat{i} + (\cos(\pi t))\hat{j} + (3t)\hat{k}$ in SI units. What is the speed of the particle at $t = 1$ s?

$(4\hat{i} + 3\hat{k})$ m/s
$\sqrt{14}$ m/s
$5$ m/s (correct answer)
$\sqrt{16 + \pi^4}$ m/s

Explanation: First, find the velocity vector by taking the derivative of the position vector: $\vec{v}(t) = \frac{d\vec{r}}{dt} = (4t)\hat{i} - (\pi\sin(\pi t))\hat{j} + (3)\hat{k}$ . Next, evaluate the velocity vector at $t = 1$ s: $\vec{v}(1) = (4(1))\hat{i} - (\pi\sin(\pi))\hat{j} + 3\hat{k} = 4\hat{i} - 0\hat{j} + 3\hat{k} = (4\hat{i} + 3\hat{k})$ m/s. The speed is the magnitude of this vector: $|\vec{v}(1)| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ m/s.

Question 3

A skater of mass $60\,\text{kg}$ pushes on a rigid wall and accelerates away at $0.80\,\text{m/s}^2$ for $0.50\,\text{s}$ . During the push, the wall exerts a horizontal force on the skater of $F_{\text{wall}\to \text{skater}}=4.8\times10^{1}\,\text{N}$ away from the wall. The skater’s hands remain in contact with the wall only during this interval. What is the relationship between $\vec F_{\text{skater}\to \text{wall}}$ and $\vec F_{\text{wall}\to \text{skater}}$ ?

$\vec F_{\text{skater}\to \text{wall}}$ is larger because the skater accelerates
$\vec F_{\text{skater}\to \text{wall}}=-\vec F_{\text{wall}\to \text{skater}}$ (correct answer)
$\vec F_{\text{skater}\to \text{wall}}=\vec F_{\text{wall}\to \text{skater}}$
They are not a third-law pair because one object is stationary

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 establishes that forces between interacting objects are always equal in magnitude but opposite in direction, regardless of the motion of either object. In the provided scenario, the wall pushes the skater away with 48 N, and by Newton's Third Law, the skater must push the wall with equal force in the opposite direction. Choice B is correct because it correctly states that the skater's force on the wall is the negative (opposite) of the wall's force on the skater, satisfying Newton's Third Law. Choice A is incorrect because it assumes the accelerating object exerts a larger force, which is a common misconception - Newton's Third Law forces are always equal regardless of acceleration. To help students, emphasize that the wall doesn't accelerate not because forces are unequal, but because Earth (to which the wall is attached) has enormous mass. Use examples like a person jumping off Earth to show equal forces produce different accelerations due to different masses.

Question 4

A particle starts from rest at the origin and moves along the x-axis with an acceleration $a_x(t) = C t^{1/2}$ , where $C$ is a positive constant. Which of the following represents the particle's position $x(t)$ as a function of time?

$\frac{2C}{3}t^{3/2}$
$\frac{C}{2}t^{-1/2}$
$\frac{4C}{15}t^{5/2}$ (correct answer)
$C t^{3/2}$

Explanation: To find position from acceleration, we must integrate twice. First, find velocity: $v_x(t) = \int a_x(t) dt = \int C t^{1/2} dt = C \frac{t^{3/2}}{3/2} + v_0$ . Since the particle starts from rest, $v_0=0$ , so $v_x(t) = \frac{2C}{3}t^{3/2}$ . Next, find position: $x(t) = \int v_x(t) dt = \int \frac{2C}{3}t^{3/2} dt = \frac{2C}{3} \frac{t^{5/2}}{5/2} + x_0$ . Since it starts at the origin, $x_0=0$ . Thus, $x(t) = \frac{4C}{15}t^{5/2}$ .

Question 5

The graph of angular velocity $\omega$ versus time $t$ for a rotating rigid body is a straight line with a positive slope, passing through the origin. What does this indicate about the body's motion?

The body has a constant positive angular velocity.
The body is undergoing a constant positive angular acceleration, starting from rest. (correct answer)
The body is undergoing an angular acceleration that increases linearly with time.
The body has a constant positive angular displacement from its starting point.

Explanation: The angular acceleration $\alpha$ is the slope of the $\omega$ versus $t$ graph. A straight line with a positive slope indicates a constant positive angular acceleration. Since the line passes through the origin, the initial angular velocity at $t=0$ is zero, meaning the body started from rest.

Question 6

An object starts from rest and experiences an acceleration given by $a(t) = A\cos(\frac{\pi}{2}t)$ , where $A$ is a constant. What is the object's change in velocity during the time interval from $t=0$ to $t=1$ s?

$0$
$A$
$\frac{\pi A}{2}$
$\frac{2A}{\pi}$ (correct answer)

Explanation: The change in velocity is the definite integral of acceleration over the time interval. $\Delta v = \int_0^1 a(t) dt = \int_0^1 A\cos(\frac{\pi}{2}t) dt$ . The integral of $\cos(kt)$ is $\frac{1}{k}\sin(kt)$ . So, $\Delta v = A [\frac{2}{\pi}\sin(\frac{\pi}{2}t)]_0^1 = \frac{2A}{\pi}[\sin(\frac{\pi}{2}) - \sin(0)] = \frac{2A}{\pi}[1 - 0] = \frac{2A}{\pi}$ .

Question 7

A wheel starts from rest and rotates with constant angular acceleration $\alpha$ . After rotating through an angle $\theta_1$ , its angular velocity is $\omega_1$ . After rotating through an additional angle $\theta_2$ , its angular velocity becomes $\omega_2$ . Which expression correctly relates these quantities?

$\omega_2^2 = \omega_1^2 + 2\alpha\theta_2$ (correct answer)
$\omega_2^2 = \omega_1^2 + 2\alpha(\theta_1 + \theta_2)$
$\omega_2 = \omega_1 + \alpha\sqrt{2\theta_2}$
$\omega_2^2 = 2\alpha(\theta_1 + \theta_2)$

Explanation: This problem requires careful application of rotational kinematic equations. The wheel starts from rest and after angle $\theta_1$ , has angular velocity $\omega_1$ . Then after an additional angle $\theta_2$ , it has angular velocity $\omega_2$ . For the second phase of motion (from $\omega_1$ to $\omega_2$ through angle $\theta_2$ ), we use $\omega_f^2 = \omega_i^2 + 2\alpha\theta$ , which gives $\omega_2^2 = \omega_1^2 + 2\alpha\theta_2$ . Choice B incorrectly uses the total angle from the start. Choice C incorrectly takes the square root of the angle term. Choice D ignores the initial angular velocity $\omega_1$ for the second phase.

Question 8

A uniform solid sphere of mass $M$ and radius $R$ rolls without slipping down an inclined plane of angle $\theta$ . At the bottom of the incline, it encounters a horizontal surface with coefficient of kinetic friction $\mu_k$ . If the sphere was released from rest at height $h$ above the horizontal surface, what is the distance it travels on the horizontal surface before coming to rest?

$\frac{5h}{7\mu_k}$ (correct answer)
$\frac{h}{\mu_k}$
$\frac{2h}{3\mu_k}$
$\frac{7h}{5\mu_k}$

Explanation: Using energy conservation on the incline: $Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}Mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2 = \frac{7}{10}Mv^2$ . So $v = \sqrt{\frac{10gh}{7}}$ . On the horizontal surface, friction does work: $\mu_k Mgd = \frac{7}{10}Mv^2 = \frac{7}{10}M(\frac{10gh}{7}) = Mgh$ . Therefore $d = \frac{h}{\mu_k} \cdot \frac{5}{7} = \frac{5h}{7\mu_k}$ . Choice B neglects rotational energy. Choice C uses wrong moment of inertia. Choice D inverts the fraction.

Question 9

An object is dropped from rest from a height $H$ and simultaneously another object is launched horizontally with an initial speed $v_0$ from the same height $H$ . Which object reaches the horizontal ground first, assuming negligible air resistance?

The object dropped from rest reaches the ground first because it travels a shorter distance.
The object launched horizontally reaches the ground first because it has a greater initial speed.
Both objects reach the ground at the same time. (correct answer)
The answer depends on the value of the initial horizontal speed $v_0$ .

Explanation: The horizontal and vertical components of motion are independent. For both objects, the initial vertical velocity is zero, and they both fall the same vertical distance $H$ under the same acceleration due to gravity, $g$ . The time to fall depends only on the vertical motion, which is identical for both. Therefore, they reach the ground at the same time.

Question 10

Based on the scenario, how does the velocity vector change over time for uniform circular motion?

Magnitude constant; direction rotates; $\Delta\vec v$ points inward (correct answer)
Magnitude increases; direction constant; $\Delta\vec v$ is tangent
Magnitude constant; direction constant; $\Delta\vec v=\vec 0$
Magnitude decreases; direction rotates; $\Delta\vec v$ points outward

Explanation: This question tests AP Physics C kinematics, specifically motion in two or three dimensions for uniform circular motion velocity vectors. The motion requires understanding how velocity vectors behave in circular motion - constant magnitude but continuously changing direction. In this scenario, we analyze how the velocity vector evolves during uniform circular motion. Choice A is correct because it accurately describes that velocity magnitude stays constant while direction rotates, and the change in velocity (Δv) points inward toward the center, creating centripetal acceleration. Choice C is incorrect because it suggests the velocity vector doesn't change at all, which would mean no acceleration and thus no circular motion. To help students: Use vector diagrams showing velocity at different points around the circle. Demonstrate how subtracting consecutive velocity vectors yields an inward-pointing Δv. Watch for: confusion between speed (scalar) and velocity (vector) and misunderstanding how vector subtraction works.

Question 11

A yo-yo of mass $m$ is modeled as a solid cylinder of radius $R$ with a string wrapped around an inner axle of radius $r$ . It is released from rest and falls as the string unwinds. The rotational inertia of the yo-yo is $I$ . The tension in the string is $T$ . Which pair of equations correctly describes its translational and rotational motion?

$mg-T=ma$ and $TR = I\alpha$
$mg-T=ma$ and $Tr = I\alpha$ (correct answer)
$T-mg=ma$ and $Tr = I\alpha$
$mg-T=ma$ and $T(R-r) = I\alpha$

Explanation: For the translational motion of the center of mass, Newton's second law gives the net force as $F_{net} = mg - T$ , so $mg - T = ma$ , where $a$ is the downward acceleration. For the rotational motion, the tension $T$ provides a torque about the center of mass. The lever arm is the radius of the inner axle, $r$ . Therefore, the torque is $\tau = Tr$ . Applying Newton's second law for rotation gives $Tr = I\alpha$ . The no-slip condition is $a = r\alpha$ .

Question 12

A hollow cylinder and a solid cylinder, both with the same mass $M$ and radius $R$ , are released simultaneously from rest at the top of an inclined plane. Both roll without slipping. When the solid cylinder has traveled a distance $d$ down the incline, what distance has the hollow cylinder traveled?

$\frac{2d}{3}$
$\frac{3d}{4}$ (correct answer)
$\frac{4d}{5}$
$d$

Explanation: For rolling without slipping, $a = \frac{g\sin\theta}{1 + \frac{I}{MR^2}}$ . For solid cylinder: $I = \frac{1}{2}MR^2$ , so $a_s = \frac{2g\sin\theta}{3}$ . For hollow cylinder: $I = MR^2$ , so $a_h = \frac{g\sin\theta}{2}$ . Using $s = \frac{1}{2}at^2$ , when solid travels distance $d$ : $t = \sqrt{\frac{3d}{g\sin\theta}}$ . In this time, hollow travels: $s_h = \frac{1}{2} \cdot \frac{g\sin\theta}{2} \cdot \frac{3d}{g\sin\theta} = \frac{3d}{4}$ . Choice A uses wrong moment ratios. Choice C confuses with sphere values. Choice D assumes equal accelerations.

Question 13

A block of mass $m$ is dropped from a height $h$ above the top of a vertical spring with spring constant $k$ . The block sticks to the spring and compresses it. Which equation must be solved to find the maximum compression $x$ of the spring?

$mgh = \frac{1}{2}kx^2$
$mg(h+x) = \frac{1}{2}kx^2$ (correct answer)
$mgh = \frac{1}{2}k(h+x)^2$
$mgx = \frac{1}{2}kx^2 - mgh$

Explanation: We apply conservation of mechanical energy between the initial point (height $h$ above the spring) and the final point (maximum compression $x$ ). The total vertical distance the block falls is $h+x$ . This loss in gravitational potential energy is converted into elastic potential energy in the spring. Initial energy (relative to max compression point) is $E_i = mg(h+x)$ . Final energy is $E_f = \frac{1}{2}kx^2$ . Setting $E_i = E_f$ gives $mg(h+x) = \frac{1}{2}kx^2$ .

Question 14

A hockey puck slides on a sheet of frictionless ice at a constant speed of $10 \text{ m/s}$ . What is the net force acting on the puck?

A force with a magnitude equal to the puck's weight, acting perpendicular to the ice.
A constant horizontal force in the direction of the puck's velocity to maintain the speed.
A horizontal force that is directly proportional to the puck's speed.
Zero. (correct answer)

Explanation: The puck is moving with a constant velocity (constant speed and direction). According to Newton's First Law, if an object's velocity is constant, the net force acting on it must be zero. The vertical forces (gravity and normal force) cancel, and there is no horizontal force in the absence of friction or propulsion.

Question 15

A heavy crate rests motionless on a horizontal floor. According to Newton's First Law, what can be concluded about the forces acting on the crate?

The gravitational force is the only force acting on the crate, but it is not strong enough to cause motion.
The gravitational force and the normal force are the only forces, and they happen to be equal and opposite.
The vector sum of all forces acting on the crate is zero, maintaining its state of rest. (correct answer)
A static friction force is the primary force preventing the crate from starting to move on its own.

Explanation: Since the crate is at rest, its velocity is constant (zero). Newton's First Law states that for an object to have a constant velocity, the net force (the vector sum of all forces) acting on it must be zero. Other forces besides gravity and the normal force could be present, but the net effect of all forces must be zero.

Question 16

Three identical point masses $m$ are fixed at the vertices of an equilateral triangle with side length $s$ .

What is the rotational inertia of the system about an axis that passes through one of the masses and is perpendicular to the plane of the triangle?

$ms^2$
$2ms^2$ (correct answer)
$3ms^2$
$\frac{3}{2}ms^2$

Explanation: Let the axis pass through mass 1. Its distance from the axis is $r_1 = 0$ . The other two masses, mass 2 and mass 3, are at a distance $s$ from mass 1. The total rotational inertia is the sum of the individual inertias: $I = \sum m_i r_i^2 = m(0)^2 + m(s)^2 + m(s)^2 = 0 + ms^2 + ms^2 = 2ms^2$ .

Question 17

A uniform rod of mass $M$ and length $L$ is pivoted at one end and oscillates as a physical pendulum. The rotational inertia of the rod about its end is $\frac{1}{3}ML^2$ . What is the period of the rod for small-amplitude oscillations?

$2\pi\sqrt{\frac{L}{g}}$
$2\pi\sqrt{\frac{2L}{3g}}$ (correct answer)
$2\pi\sqrt{\frac{L}{3g}}$
$2\pi\sqrt{\frac{L}{2g}}$

Explanation: The period of a physical pendulum is given by $T = 2\pi\sqrt{\frac{I}{mgd}}$ . For a uniform rod pivoted at one end, the rotational inertia is $I = \frac{1}{3}ML^2$ and the distance from the pivot to the center of mass is $d = L/2$ . Substituting these values gives $T = 2\pi\sqrt{\frac{\frac{1}{3}ML^2}{Mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}$ .

Question 18

A gyroscope with its axis tilted precesses with a constant angular speed about a vertical axis. The gravitational force exerts a torque on the gyroscope, causing the precession. What is the work done by this gravitational torque during one complete precession cycle?

Zero, because the torque vector is always perpendicular to the angular displacement of precession. (correct answer)
Positive, because the torque is required to maintain the precession against dissipative forces.
Negative, because the potential energy of the gyroscope's center of mass does not change.
It cannot be determined without knowing the gyroscope's spin and precession speeds.

Explanation: The work done by a torque is given by $W = \int \vec{\tau} \cdot d\vec{\theta}$ . For a precessing gyroscope, the gravitational torque vector is horizontal. The angular displacement vector for the precession is along the vertical axis of precession. Since the torque vector is always perpendicular to the angular displacement vector, their dot product is zero, and the work done is zero. This torque changes the direction of the angular momentum but not its magnitude.

Question 19

A uniform rod of length $L$ and mass $M$ is pivoted at its center. A force $F$ is applied perpendicularly to the rod at one end, and a force $2F$ is applied perpendicularly at the other end, in the opposite direction. The rotational inertia of the rod about its center is $I = \frac{1}{12}ML^2$ . What is the magnitude of the initial angular acceleration of the rod?

$\frac{18F}{ML}$ (correct answer)
$\frac{12F}{ML}$
$\frac{6F}{ML}$
$\frac{36F}{ML}$

Explanation: Both forces produce torques in the same rotational direction. The lever arm for each force is $L/2$ . The net torque is $\tau_{net} = F(L/2) + 2F(L/2) = \frac{3}{2}FL$ . Using $\tau_{net} = I\alpha$ , we have $\frac{3}{2}FL = (\frac{1}{12}ML^2)\alpha$ . Solving for $\alpha$ gives $\alpha = \frac{3FL/2}{ML^2/12} = \frac{18F}{ML}$ .

Question 20

A flywheel in the shape of a uniform disk has moment of inertia $I$ and is initially rotating with angular velocity $\omega_0$ . A constant braking torque $\tau_b$ is applied until the flywheel stops. During the braking process, what is the average power dissipated?

$\frac{\tau_b \omega_0}{2}$ (correct answer)
$\frac{I \omega_0^2}{2t_{stop}}$ where $t_{stop} = \frac{I\omega_0}{\tau_b}$
$\frac{\tau_b^2}{2I\omega_0}$
$\frac{I\omega_0^3}{2\tau_b}$

Explanation: The angular deceleration is $\alpha = \tau_b/I$ , so the time to stop is $t_{stop} = \omega_0/\alpha = I\omega_0/\tau_b$ . The initial kinetic energy is $KE_i = \frac{1}{2}I\omega_0^2$ . The average power is $P_{avg} = \frac{\text{Energy dissipated}}{\text{Time}} = \frac{\frac{1}{2}I\omega_0^2}{\frac{I\omega_0}{\tau_b}} = \frac{\tau_b\omega_0}{2}$ . Alternatively, since power $P = \tau\omega$ and $\omega$ decreases linearly from $\omega_0$ to 0, the average angular velocity is $\omega_0/2$ , giving $P_{avg} = \tau_b \cdot \frac{\omega_0}{2}$ . Choice B gives the same result when simplified. Choice C has wrong units. Choice D has wrong units and incorrect relationship.

Question 21

Two satellites collided elastically in space: A ( $200\,\text{kg}$ ) moved at $+5.0\,\text{m/s}$ and B ( $200\,\text{kg}$ ) moved at $-2.0\,\text{m/s}$ , with negligible external impulse. Determine the velocity of the second object after the collision.

$v_{2f}=+5.0\,\text{m/s}$ (correct answer)
$v_{2f}=-5.0\,\text{m/s}$
$v_{2f}=+2.0\,\text{m/s}$
$v_{2f}=-2.0\,\text{m/s}$

Explanation: Elastic collision between equal-mass satellites. Initial: satellite A (200 kg) at +5.0 m/s, satellite B (200 kg) at -2.0 m/s. For elastic collisions between equal masses, velocities exchange: satellite A takes B's velocity (-2.0 m/s) and satellite B takes A's velocity (+5.0 m/s). Therefore, $v_{2f} = +5.0$ m/s. Choice A is correct. This velocity exchange rule for equal masses simplifies calculations and helps avoid algebraic errors in elastic collision problems.

Question 22

Three particles are arranged in a straight line. Particle A has mass $2m$ at position $x = 0$ , particle B has mass $m$ at position $x = L$ , and particle C has mass $3m$ at position $x = 2L$ . What is the distance from particle A to the center of mass of the system?

$\frac{4L}{3}$
$\frac{5L}{3}$
$\frac{7L}{6}$ (correct answer)
$\frac{4L}{3}$

Explanation: The center of mass position is: $x_{cm} = \frac{2m \cdot 0 + m \cdot L + 3m \cdot 2L}{2m + m + 3m} = \frac{0 + L + 6L}{6m} = \frac{7L}{6}$ . The distance from particle A (at $x = 0$ ) to the center of mass is $\frac{7L}{6} - 0 = \frac{7L}{6}$ . Choice A represents an incorrect average. Choice B uses wrong mass weighting. Choice D duplicates choice A with calculation error.

Question 23

A 1000 kg drag racing car starts from rest while its engine delivers constant power of 80 kW for 5.0 s, neglecting losses. What kinetic energy does it gain?

$4.0\times10^5\ \text{J}$ (correct answer)
$1.6\times10^4\ \text{J}$
$8.0\times10^4\ \text{J}$
$2.0\times10^5\ \text{J}$

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding and calculating translational kinetic energy from power and time relationships. Power is the rate of energy transfer, so the total energy delivered equals power multiplied by time: E = P×t. In this scenario, the engine delivers 80 kW (80,000 W) for 5.0 seconds, and neglecting losses means all this energy converts to kinetic energy: KE = (80,000 W)(5.0 s) = 400,000 J = 4.0×10⁵ J. Choice A is correct because it properly calculates the total energy delivered as kinetic energy gain. Choice B might incorrectly divide by time again, while choices C and D show calculation errors. To help students: Emphasize the relationship between power, energy, and time (P = E/t). Practice problems involving energy conversions and power calculations to reinforce unit consistency and the meaning of 'neglecting losses'.

Question 24

Two skaters collide elastically: $m_1=50\,\text{kg}$ with $v_{1i}=+4.0\,\text{m/s}$ and $m_2=70\,\text{kg}$ with $v_{2i}=0$ . Afterward $v_{2f}=+3.0\,\text{m/s}$ . What is the final velocity of skater 1?

$-0.20\,\text{m/s}$ (correct answer)
$+0.20\,\text{m/s}$
$+1.0\,\text{m/s}$
$-1.0\,\text{m/s}$

Explanation: This question tests the understanding of linear momentum conservation in AP Physics C: Mechanics, specifically in elastic collisions where both momentum and kinetic energy are conserved. In elastic collisions, knowing the masses, initial velocities, and one final velocity allows us to calculate the other final velocity using momentum conservation. In this problem, a 50 kg skater moving at +4.0 m/s collides elastically with a stationary 70 kg skater, and after collision the second skater moves at +3.0 m/s. Choice A (-0.20 m/s) is correct because applying momentum conservation: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f gives us (50)(4.0) + (70)(0) = (50)v₁f + (70)(3.0), which simplifies to 200 = 50v₁f + 210, yielding v₁f = -10/50 = -0.20 m/s. Choice C (+1.0 m/s) is incorrect because it would not conserve momentum, as the total final momentum would exceed the initial momentum. To help students: In elastic collisions, always verify both momentum and energy conservation. The negative final velocity indicates the first skater rebounds backward, which is physically reasonable when a lighter object collides with a heavier stationary object.

Question 25

A particle moves in a conservative force field where the potential energy is given by $U(x) = ax^2 - bx^3$ , where $a$ and $b$ are positive constants. The particle starts from rest at $x = 0$ . What is the maximum value of $x$ the particle can reach if its total mechanical energy is $E$ ?

The positive root of $ax^2 - bx^3 - E = 0$
$\sqrt{\frac{E}{a}}$
$\sqrt[3]{\frac{E}{b}}$
The positive root of $ax^2 - bx^3 = E$ (correct answer)

Explanation: When you encounter problems involving conservative forces and potential energy, think about energy conservation. The key insight is that a particle can only reach positions where its kinetic energy remains non-negative, since $KE = \frac{1}{2}mv^2 \geq 0$ . Since the particle starts from rest at $x = 0$ , its initial kinetic energy is zero and its initial potential energy is $U(0) = 0$ . Therefore, the total mechanical energy is $E = KE_0 + U_0 = 0 + 0 = E$ , which means this energy $E$ must be supplied to the system initially. At any position $x$ , energy conservation gives us: $E = KE + U(x) = \frac{1}{2}mv^2 + ax^2 - bx^3$ . For the particle to physically reach position $x$ , we need $KE \geq 0$ , which means $E - U(x) \geq 0$ , or $E \geq ax^2 - bx^3$ . The maximum value of $x$ occurs when the kinetic energy just reaches zero, meaning all energy is potential energy. At this turning point: $E = U(x) = ax^2 - bx^3$ , which gives us $ax^2 - bx^3 = E$ . Choice A incorrectly subtracts $E$ instead of setting the potential energy equal to $E$ . Choice B assumes the cubic term is negligible and only considers the quadratic term $ax^2 = E$ . Choice C makes the opposite error, ignoring the quadratic term and setting $bx^3 = E$ . Choice D correctly represents the condition where all mechanical energy equals potential energy at the turning point. Remember: turning points in conservative force problems occur where kinetic energy equals zero, so total energy equals potential energy.

Opening subject page...