AP Physics C Mechanics

AP Physics C Mechanics Practice Test: Practice Test 1

Practice Test 1 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 $\vec{F} = (4\hat{i} + 6\hat{j})$ N acts on an object that undergoes a displacement $\vec{d} = (3\hat{i})$ m. A second force $\vec{G}$ also acts on the object. The work done by force $\vec{G}$ on the object is zero, and the net force $\vec{F}_{net}$ on the object is parallel to the displacement $\vec{d}$ . What is the vector $\vec{G}$ ?

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Question 1

$(-4\hat{i})$ N
$(-6\hat{j})$ N (correct answer)
$(-4\hat{i} - 6\hat{j})$ N
$(3\hat{i})$ N

Explanation: This problem combines work-energy concepts with vector addition, testing your understanding of when forces do zero work and how net forces relate to motion direction. Since force $\vec{G}$ does zero work on the object, $\vec{G}$ must be perpendicular to the displacement $\vec{d} = (3\hat{i})$ m. The work formula $W = \vec{F} \cdot \vec{d}$ equals zero only when the dot product is zero, which happens when vectors are perpendicular. Since $\vec{d}$ points purely in the x-direction, $\vec{G}$ can only have a y-component. The key insight is that the net force $\vec{F}_{net}$ is parallel to $\vec{d}$ , meaning $\vec{F}_{net}$ has no y-component. Since $\vec{F}_{net} = \vec{F} + \vec{G}$ , and $\vec{F} = (4\hat{i} + 6\hat{j})$ N has a 6 N component in the y-direction, $\vec{G}$ must cancel this y-component. Therefore, $\vec{G} = (-6\hat{j})$ N. Let's verify: $\vec{F}_{net} = (4\hat{i} + 6\hat{j}) + (-6\hat{j}) = (4\hat{i})$ N, which is indeed parallel to $\vec{d}$ . Choice A) $(-4\hat{i})$ N would reduce the x-component of the net force but wouldn't eliminate the y-component. Choice C) $(-4\hat{i} - 6\hat{j})$ N would make the net force zero, not parallel to $\vec{d}$ . Choice D) $(3\hat{i})$ N would do positive work since it's parallel to the displacement. Remember: when a force does zero work, look for perpendicularity to displacement. When forces must produce motion in a specific direction, focus on which components need to be eliminated through vector addition.

Question 2

A particle is in translational equilibrium. Which of the following statements must be true about the particle?

The particle must be at rest relative to the observer.
The particle's speed must be constant, but its direction of motion may be changing.
The particle's velocity vector must be constant. (correct answer)
The sum of the magnitudes of all forces acting on the particle must be zero.

Explanation: Translational equilibrium means the net force on the particle is zero. According to Newton's First Law, this implies that the particle's acceleration is zero, which means its velocity must be constant. A constant velocity vector means both the speed and the direction of motion are unchanging. Being at rest is a special case where the constant velocity is zero.

Question 3

An object with mass $m$ is projected horizontally with initial velocity $v_0$ on a surface where the only horizontal force is a resistive force given by $F = -kv$ . The velocity of the object as a function of time is $v(t) = v_0 e^{-kt/m}$ . What is the total distance the object travels as it slows to a stop?

$v_0 k / m$
The object never stops, so the distance is infinite.
$v_0 m / k$ (correct answer)
$v_0 m / (2k)$

Explanation: The total distance is the integral of the velocity function from $t=0$ to $t=\infty$ . $x_{total} = \int_0^\infty v(t) dt = \int_0^\infty v_0 e^{-kt/m} dt$ . The integral evaluates to $v_0 [-\frac{m}{k}e^{-kt/m}]_0^\infty = -\frac{mv_0}{k}(0 - 1) = \frac{mv_0}{k}$ .

Question 4

A wheel with moment of inertia $I=1.2\ \text{kg·m}^2$ must speed up from $\omega_0=5.0\ \text{rad/s}$ to $\omega_f=17\ \text{rad/s}$ in $t=4.0\ \text{s}$ under a constant net torque (friction included). Key equations: $\alpha=(\omega_f-\omega_0)/t$ and $\tau=I\alpha$ . Using the given conditions, what is the torque required to achieve this change?

$\tau=1.8\ \text{N·m}$
$\tau=3.6\ \text{N·m}$ (correct answer)
$\tau=14\ \text{N·m}$
$\tau=0.30\ \text{N·m}$

Explanation: This question tests AP Physics C: Mechanics concepts on rotational kinematics and dynamics, specifically calculating required torque for a desired angular acceleration. The problem requires working backwards from kinematic information to find the necessary torque. For a wheel with I = 1.2 kg·m² accelerating from ω0 = 5.0 rad/s to ωf = 17 rad/s in t = 4.0 s, we first find the angular acceleration. Choice B is correct because α = (ωf - ω0)/t = (17 - 5.0)/4.0 = 12/4.0 = 3.0 rad/s², and then τ = Iα = 1.2 × 3.0 = 3.6 N·m. Choice C at 14 N·m might result from calculation errors or using wrong values. To help students: emphasize the two-step process of finding acceleration first, then torque; practice problems that work backwards from desired motion to required forces/torques; and always verify that calculated values make physical sense.

Question 5

A $1200\,\text{kg}$ car descends a frictionless hill and reaches $25\,\text{m/s}$ at the bottom; what is its translational kinetic energy there?

$3.75\times10^5\,\text{J}$ (correct answer)
$7.50\times10^5\,\text{J}$
$1.88\times10^5\,\text{J}$
$3.00\times10^4\,\text{J}$

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding and calculating translational kinetic energy. Kinetic energy (KE) is the energy an object possesses due to its motion, calculated using KE = 1/2 mv², where m is mass and v is velocity. In this scenario, the car has a mass of 1200 kg and reaches a velocity of 25 m/s at the bottom of the hill, requiring application of the formula to find its kinetic energy. Choice A is correct because it correctly applies the kinetic energy formula: KE = 1/2 × 1200 kg × (25 m/s)² = 1/2 × 1200 × 625 = 375,000 J = 3.75×10⁵ J. Choice B is incorrect because it doubles the correct answer, likely from forgetting the 1/2 factor in the formula. To help students: Emphasize memorizing the complete formula including the 1/2 factor, and practice dimensional analysis to verify units. Create visual aids showing the relationship between mass, velocity squared, and kinetic energy to reinforce the formula's structure.

Question 6

A uniform ladder of mass $m$ and length $L$ leans against a perfectly smooth (frictionless) vertical wall at an angle $\theta$ with the horizontal floor. The floor is rough and provides a static friction force that prevents the ladder from slipping.

Which expression represents the magnitude of the torque produced by the wall's normal force ( $N_w$ ) about the point where the ladder contacts the floor?

$N_w L \sin\theta$ (correct answer)
$N_w L \cos\theta$
$N_w L$
Zero, because the wall is frictionless.

Explanation: The torque is given by $\tau = rF\sin\phi$ , where $r$ is the distance from the pivot to the point of force application, $F$ is the force, and $\phi$ is the angle between the position vector and the force vector. Here, the pivot is the base of the ladder, $r=L$ , and $F = N_w$ . The angle between the ladder (position vector) and the horizontal force $N_w$ is $(180^\circ - \theta)$ . The lever arm of the force $N_w$ about the base is $L\sin\theta$ . Thus, the torque is $N_w (L\sin\theta)$ .

Question 7

Two springs with spring constants $k_1 = 300 \text{ N/m}$ and $k_2 = 500 \text{ N/m}$ are placed end-to-end (in series) and compressed by a total of $0.20 \text{ m}$ from their combined natural length. What is the magnitude of the force exerted by the $k_2$ spring?

$22.5 \text{ N}$
$60.0 \text{ N}$
$37.5 \text{ N}$ (correct answer)
$100 \text{ N}$

Explanation: When you encounter springs connected in series, remember that they share the same force but have different compressions. This is a fundamental principle that distinguishes series from parallel spring arrangements. To solve this problem, you need to find how the total compression distributes between the two springs. Since the springs are in series, the force in each spring must be equal: $F_1 = F_2 = F$ . Using Hooke's law ( $F = kx$ ), you can write $k_1 x_1 = k_2 x_2$ , where $x_1$ and $x_2$ are the individual compressions. Since $x_1 + x_2 = 0.20 \text{ m}$ and $300x_1 = 500x_2$ , you can solve: $x_1 = \frac{500}{300}x_2 = \frac{5}{3}x_2$ . Substituting: $\frac{5}{3}x_2 + x_2 = 0.20$ , so $\frac{8}{3}x_2 = 0.20$ , giving $x_2 = 0.075 \text{ m}$ . The force in the $k_2$ spring is: $F_2 = k_2 x_2 = 500 \times 0.075 = 37.5 \text{ N}$ . Answer A (22.5 N) incorrectly uses the compression of the $k_1$ spring with the wrong spring constant. Answer B (60.0 N) appears to use an incorrect distribution of compression. Answer D (100 N) mistakenly assumes the entire compression applies to the $k_2$ spring. Study tip: For series springs, always remember that forces are equal but compressions differ inversely with spring constants. The stiffer spring (higher k) compresses less, while the softer spring (lower k) compresses more.

Question 8

Which of the following lists contains only vector quantities?

Velocity, acceleration, displacement, and force. (correct answer)
Speed, force, momentum, and electric charge.
Distance, work, power, and kinetic energy.
Temperature, mass, time, and electric potential.

Explanation: Vector quantities are defined as physical quantities that have both magnitude and direction. Velocity, acceleration, displacement, and force all fit this definition. The other choices contain scalar quantities: speed, distance, work, power, kinetic energy, electric charge, temperature, mass, time, and electric potential are all described by magnitude only.

Question 9

Two projectiles are launched simultaneously from the same point. Projectile A is launched at angle $30°$ with initial speed $v_A$ , and projectile B is launched at angle $60°$ with initial speed $v_B$ . If both projectiles have the same range, what is the ratio $\frac{v_A}{v_B}$ ?

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

Explanation: The range formula is $R = \frac{v^2\sin(2\theta)}{g}$ . For projectile A: $R_A = \frac{v_A^2\sin(60°)}{g} = \frac{v_A^2\frac{\sqrt{3}}{2}}{g}$ . For projectile B: $R_B = \frac{v_B^2\sin(120°)}{g} = \frac{v_B^2\frac{\sqrt{3}}{2}}{g}$ since $\sin(120°) = \sin(60°) = \frac{\sqrt{3}}{2}$ . Setting $R_A = R_B$ : $\frac{v_A^2\frac{\sqrt{3}}{2}}{g} = \frac{v_B^2\frac{\sqrt{3}}{2}}{g}$ , which simplifies to $v_A^2 = v_B^2$ , so $v_A = v_B$ and $\frac{v_A}{v_B} = 1$ . This makes physical sense because complementary angles ( $30°$ and $60°$ are complementary since $30° + 60° = 90°$ ) give the same range for the same initial speed. Choice B comes from incorrectly using $\frac{\sin(60°)}{\sin(30°)} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$ . Choice C results from the reciprocal calculation with a factor error. Choice D is the reciprocal of choice B.

Question 10

An electric motor operates at constant mechanical power output of 1.5 kW while lifting a load vertically. Initially, the load has a mass of 100 kg and is lifted at constant velocity. After 10 seconds, an additional 50 kg mass is suddenly attached to the load. Assuming the motor maintains the same power output, what is the new constant velocity of the combined load after the transient effects have died down?

1.0 m/s (correct answer)
1.2 m/s
1.5 m/s
2.0 m/s

Explanation: Initially, with m₁ = 100 kg at constant velocity v₁: P = F₁v₁ = m₁gv₁, so v₁ = P/(m₁g) = 1500/(100×9.8) = 1.53 m/s. After the additional mass is added, the total mass is m₂ = 150 kg. At the new constant velocity v₂: P = F₂v₂ = m₂gv₂, so v₂ = P/(m₂g) = 1500/(150×9.8) = 1.02 m/s ≈ 1.0 m/s. Choice B (1.2 m/s) might result from using an incorrect mass ratio or calculation error. Choice C (1.5 m/s) would result from not accounting for the increased mass properly. Choice D (2.0 m/s) would result from incorrectly thinking the velocity increases when mass increases.

Question 11

The position of a particle is given by the function $x(t)$ . Which of the following mathematical expressions represents the particle's instantaneous acceleration at time $t$ ?

$\int x(t) dt$
$\frac{dx}{dt}$
$\frac{d^2x}{dt^2}$ (correct answer)
$\frac{d}{dt} \left( \frac{1}{2} \left( \frac{dx}{dt} \right)^2 \right)$

Explanation: Instantaneous velocity is the first derivative of position with respect to time, $v(t) = \frac{dx}{dt}$ . Instantaneous acceleration is the first derivative of velocity with respect to time, $a(t) = \frac{dv}{dt}$ . Substituting the expression for velocity gives $a(t) = \frac{d}{dt} \left( \frac{dx}{dt} \right) = \frac{d^2x}{dt^2}$ .

Question 12

An object is moving to the right with a constant velocity. If a single, constant force is then applied to the object, which of the following is NOT a possible resulting motion?

The object continues moving right but its speed increases.
The object's path curves, and it begins to move upwards and to the right.
The object instantaneously stops and remains at rest. (correct answer)
The object continues moving right but its speed decreases.

Explanation: A net force causes acceleration, which means the velocity must change. However, velocity is the integral of acceleration, so it must change continuously over time. An object cannot instantaneously stop unless an infinite force is applied over an infinitesimal time. The object would first have to decelerate to zero velocity.

Question 13

A position vector $\vec{r}$ makes an angle of $\theta$ with the positive x-axis and has magnitude $r$ . The vector is then rotated clockwise by angle $\phi$ to create a new vector $\vec{r}'$ . Which expression correctly represents the x-component of $\vec{r}'$ ?

$r\cos(\theta - \phi)$ (correct answer)
$r\cos(\theta + \phi)$
$r\sin(\theta - \phi)$
$r\sin(\theta + \phi)$

Explanation: When a vector is rotated clockwise by angle $\phi$ , the new angle with respect to the positive x-axis becomes $\theta - \phi$ (since clockwise rotation decreases the angle). The x-component of the rotated vector is $r'_x = r\cos(\theta - \phi)$ . Choice B represents counterclockwise rotation. Choices C and D incorrectly use sine instead of cosine for the x-component.

Question 14

A solid disk and a spoked wheel have the same mass $M$ and the same outer radius $R$ . Both are accelerated from rest to the same final angular velocity $\omega$ . Which object has more rotational kinetic energy, and why?

The solid disk, because its mass is more uniformly distributed, leading to a more efficient rotation.
The spoked wheel, because more of its mass is located at a larger radius, giving it a larger rotational inertia. (correct answer)
They have the same rotational kinetic energy, because their mass, radius, and angular velocity are identical.
They have the same rotational kinetic energy, because the work required to accelerate them is the same.

Explanation: Rotational kinetic energy is $K_{rot} = \frac{1}{2}I\omega^2$ . Since both objects have the same mass $M$ and are accelerated to the same angular velocity $\omega$ , the one with the larger rotational inertia $I$ will have more kinetic energy. Rotational inertia depends on how mass is distributed relative to the axis of rotation. The spoked wheel has most of its mass concentrated at the outer radius, while the solid disk's mass is distributed throughout. Therefore, the spoked wheel has a larger rotational inertia and thus more rotational kinetic energy.

Question 15

A 0.50 kg block on a frictionless track is attached to a spring with $k=200\ \text{N/m}$ and equilibrium at $x=0$ . It is pulled to $x=+0.080\ \text{m}$ and released from rest at $t=0$ , so $A=0.080\ \text{m}$ and $\phi=0$ . The angular frequency is $\omega=\sqrt{k/m}$ and the motion is $x(t)=A\cos(\omega t+\phi)$ with $x$ in meters and $t$ in seconds. The velocity is $v(t)=-A\omega\sin(\omega t+\phi)$ and the acceleration is $a(t)=-\omega^2 x(t)$ . The restoring force is $F_s=-kx$ . Assume SHM holds for all times. What is the period $T$ of the oscillation?

$T=0.16\ \text{s}$
$T=0.31\ \text{s}$ (correct answer)
$T=3.1\ \text{s}$
$T=1.6\ \text{s}$

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 0.50 kg block attached to a spring with k=200 N/m undergoes SHM with angular frequency ω=√(k/m)=√(200/0.50)=20 rad/s. Choice B is correct because the period T=2π/ω=2π/20≈0.314 s, which rounds to 0.31 s. Choice D is incorrect due to a common error where students might calculate T=2π√(m/k) but forget the 2π factor, getting approximately 0.16 s instead. To help students: Emphasize the importance of remembering the complete period formula T=2π/ω and the relationship between angular frequency and spring-mass parameters. Practice problems should include variations in mass and spring constant to reinforce the inverse relationship between period and angular frequency.

Question 16

A block slides at a constant velocity across a horizontal, frictionless surface. A constant vertical force of magnitude F is then applied to the block. How much work is done by this new vertical force as the block continues to travel a horizontal distance $d$ ?

$Fd$
$-Fd$
Zero (correct answer)
It cannot be determined without the mass of the block.

Explanation: Work is done only when there is a component of force in the direction of displacement. The applied force is vertical, while the block's displacement is horizontal. Since the force and displacement vectors are perpendicular to each other, the dot product of the two is zero, and no work is done by the vertical force.

Question 17

A non-uniform bar of length $L$ and mass $M$ has its center of mass at a distance $L/4$ from one end, End A. The bar is to be balanced on a single pivot.

Where must the pivot be placed for the bar to be in static equilibrium?

At a distance of $L/4$ from End A. (correct answer)
At a distance of $L/2$ from End A.
At a distance of $3L/4$ from End A.
The pivot position depends on the orientation of the bar.

Explanation: For the bar to be in static equilibrium under its own weight, the pivot must provide a normal force that exactly opposes the gravitational force, and there must be zero net torque. To achieve zero net torque from gravity, the pivot must be placed directly under the center of mass. This way, the lever arm for the gravitational force is zero, resulting in zero torque.

Question 18

A boat of mass $m$ starts from rest at $t=0$ . A constant force $F_{app}$ is applied, and it experiences a resistive force $F_r = -kv$ . The boat's speed $v$ as a function of time $t$ is given by which expression?

$v(t) = \frac{F_{app}}{k}(1 - e^{-kt/m})$ (correct answer)
$v(t) = \frac{F_{app}}{k} e^{-kt/m}$
$v(t) = \frac{F_{app}}{m}t - \frac{k}{2m}t^2$
$v(t) = \frac{F_{app}}{m}(1 - e^{-kt/m})$

Explanation: The differential equation for the boat's motion is $m \frac{dv}{dt} = F_{app} - kv$ . This equation is mathematically analogous to that of a falling object, with the constant gravitational force $mg$ replaced by the constant applied force $F_{app}$ . The solution, with the initial condition $v(0)=0$ , is therefore $v(t) = \frac{F_{app}}{k}(1 - e^{-kt/m})$ , where $F_{app}/k$ is the terminal velocity.

Question 19

A diver performs a somersault after jumping from a diving board. They initially leave the board with their body extended, then tuck into a compact shape, and finally extend their body again before entering the water. Neglecting air resistance, which of the following correctly describes the changes in their rotational inertia $I$ and angular momentum $L$ about their center of mass during this process?

$I$ decreases, then increases; $L$ remains approximately constant. (correct answer)
$I$ remains constant; $L$ decreases, then increases.
Both $I$ and $L$ decrease, then increase.
Both $I$ and $L$ remain approximately constant.

Explanation: The only significant external force on the diver during flight is gravity, which acts on the center of mass. Therefore, there is no net external torque about the diver's center of mass. This means their angular momentum $L$ is conserved and remains approximately constant. When the diver tucks, they bring their mass closer to the axis of rotation, decreasing their rotational inertia $I$ . When they extend their body, they move mass away from the axis, increasing $I$ .

Question 20

A particle moves in a one-dimensional potential given by $U(x) = x^4 - 8x^2 + 12$ . If the particle has a total mechanical energy of $E = 4$ J, which of the following is a turning point of its motion?

$x = 0$ m
$x = \sqrt{6}$ m
$x = 3$ m
$x = 2$ m (correct answer)

Explanation: Turning points occur where the kinetic energy is zero, so the total energy equals the potential energy: $E = U(x)$ . We set $4 = x^4 - 8x^2 + 12$ . This gives the equation $x^4 - 8x^2 + 8 = 0$ . Let $y = x^2$ , then $y^2 - 8y + 8 = 0$ . Using the quadratic formula: $y = \frac{8 \pm \sqrt{64 - 32}}{2} = \frac{8 \pm \sqrt{32}}{2} = 4 \pm 2\sqrt{2}$ . Since $y = x^2 \geq 0$ , both solutions are valid: $x^2 = 4 + 2\sqrt{2} \approx 6.83$ or $x^2 = 4 - 2\sqrt{2} = 4$ . The turning points are at $x = \pm 2$ m and $x = \pm\sqrt{4 + 2\sqrt{2}}$ m. Of the choices given, $x = 2$ m is a turning point.

Question 21

A mass attached to a vertical spring oscillates with amplitude $A$ . Taking the equilibrium position as the reference for gravitational potential energy, at what displacement from equilibrium is the total mechanical energy (including gravitational potential energy) equal to twice the elastic potential energy?

$\pm \frac{A}{\sqrt{2}}$ (correct answer)
$\pm \frac{A}{2}$
$\pm \frac{A\sqrt{3}}{2}$
$\pm A\sqrt{2}$

Explanation: Total mechanical energy $E = \frac{1}{2}kA^2$ . Elastic potential energy at displacement $x$ is $U_e = \frac{1}{2}kx^2$ . When $E = 2U_e$ : $\frac{1}{2}kA^2 = 2 \cdot \frac{1}{2}kx^2$ , so $A^2 = 2x^2$ , giving $x = \pm \frac{A}{\sqrt{2}}$ . Note that gravitational PE is zero at equilibrium by the problem setup. Choice B uses $A^2 = 4x^2$ . Choice C comes from incorrectly using energy ratios. Choice D inverts the square root.

Question 22

The potential energy of a particle moving along the x-axis is given by the function $U(x) = -4x^3 + 6x^2$ J, where x is in meters. What is the force on the particle at $x = -1$ m?

-24 N
24 N (correct answer)
-10 N
10 N

Explanation: The conservative force is related to the potential energy by $F(x) = -\frac{dU}{dx}$ . Differentiating $U(x)$ gives $\frac{dU}{dx} = -12x^2 + 12x$ . Therefore, the force is $F(x) = -(-12x^2 + 12x) = 12x^2 - 12x$ . Evaluating at $x = -1$ m gives $F(-1) = 12(-1)^2 - 12(-1) = 12(1) + 12 = 24$ N.

Question 23

Consider two oscillating systems: System X has restoring force $F_X = -kx$ and System Y has restoring force $F_Y = -kx^3$ where $k > 0$ in both cases. A physics student argues that both systems exhibit simple harmonic motion because both have restoring forces that oppose displacement. Which analysis correctly evaluates this argument?

The argument is correct because both restoring forces are directed toward equilibrium and both systems will oscillate periodically about $x = 0$
The argument is incorrect because only System X satisfies $a \propto -x$ ; System Y has $a \propto -x^3$ , violating the SHM definition (correct answer)
The argument is correct because both systems conserve energy and exhibit periodic motion with well-defined equilibrium positions at the origin
The argument is incorrect because System Y's period depends on amplitude, while SHM requires the period to be independent of amplitude

Explanation: Simple harmonic motion specifically requires acceleration proportional to displacement: $a = -\omega^2 x$ . System X: $F = ma = -kx$ , so $a = -\frac{k}{m}x$ (SHM). System Y: $F = ma = -kx^3$ , so $a = -\frac{k}{m}x^3$ (not SHM). The defining characteristic of SHM is linear restoring force, not just any restoring force. Choice A confuses restoring force with SHM definition. Choice C lists properties that both motions share but misses the key distinction. Choice D mentions a consequence rather than the fundamental definition.

Question 24

A mass $m$ is attached to a spring with spring constant $k$ and undergoes simple harmonic motion. At the moment when the displacement is $x = \frac{A}{2}$ (where $A$ is the amplitude), the speed is $v = \frac{A\sqrt{3}}{2}\omega$ , where $\omega$ is the angular frequency. What is the period of this motion?

The given conditions are physically impossible
$\pi\sqrt{\frac{m}{k}}$
$4\pi\sqrt{\frac{m}{k}}$
$2\pi\sqrt{\frac{m}{k}}$ (correct answer)

Explanation: When you encounter simple harmonic motion problems, always start with the fundamental energy relationship. The total mechanical energy in SHM remains constant and equals the maximum potential energy: $E = \frac{1}{2}kA^2$ . At any position, this energy splits between kinetic and potential: $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$ . Let's check if the given conditions are physically consistent by substituting the values. With $x = \frac{A}{2}$ and $v = \frac{A\sqrt{3}}{2}\omega$ : $\frac{1}{2}kA^2 = \frac{1}{2}m\left(\frac{A\sqrt{3}}{2}\omega\right)^2 + \frac{1}{2}k\left(\frac{A}{2}\right)^2$ $\frac{1}{2}kA^2 = \frac{1}{2}m \cdot \frac{3A^2\omega^2}{4} + \frac{1}{2}k \cdot \frac{A^2}{4}$ $\frac{1}{2}kA^2 = \frac{3mA^2\omega^2}{8} + \frac{kA^2}{8}$ Dividing by $\frac{A^2}{2}$ : $k = \frac{3m\omega^2}{4} + \frac{k}{4}$ Solving: $\frac{3k}{4} = \frac{3m\omega^2}{4}$ , so $k = m\omega^2$ This gives us $\omega = \sqrt{\frac{k}{m}}$ , which is exactly the standard formula for SHM angular frequency. Therefore, the period is $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$ . Answer choice A is wrong because the conditions are physically possible, as we just verified. Choice B gives half the correct period, while choice C gives twice the correct period—both likely stem from errors in the $2\pi$ factor. Study tip: Always verify that given SHM conditions satisfy energy conservation before solving. This catches impossible scenarios and confirms your approach is correct.

Question 25

A car travels around a banked curve with banking angle $\theta = 25°$ . The coefficient of static friction between the tires and road is $\mu_s = 0.40$ . For a curve of radius 200 m, what is the maximum speed the car can travel without slipping?

$25 \text{ m/s}$
$32 \text{ m/s}$
$38 \text{ m/s}$ (correct answer)
$45 \text{ m/s}$

Explanation: For maximum speed, friction acts down the incline. The centripetal force equation becomes: $mg\sin\theta + \mu_s mg\cos\theta = \frac{mv^2}{r}$ . Solving: $v = \sqrt{gr(\sin\theta + \mu_s\cos\theta)} = \sqrt{9.8 \times 200 \times (\sin25° + 0.40\cos25°)} = 38 \text{ m/s}$ . Choice A ignores the banking angle contribution. Choice B uses only the banking term without friction. Choice D incorrectly adds the terms under separate square roots.

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