AP Physics C Mechanics

AP Physics C Mechanics Practice Test: Practice Test 5

Practice Test 5 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 thin hoop of mass $M$ and radius $R$ rolls without slipping on a horizontal surface. Its rotational inertia is $I = MR^2$ . What fraction of its total kinetic energy is rotational kinetic energy?

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Question 1

A thin hoop of mass $M$ and radius $R$ rolls without slipping on a horizontal surface. Its rotational inertia is $I = MR^2$ . What fraction of its total kinetic energy is rotational kinetic energy?

$1/4$
$1/3$
$1/2$ (correct answer)
$2/3$

Explanation: The total kinetic energy is $K_{total} = K_{trans} + K_{rot}$ . For a hoop rolling without slipping ( $v=R\omega$ ), $K_{trans} = \frac{1}{2}Mv^2$ and $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(MR^2)(\frac{v}{R})^2 = \frac{1}{2}Mv^2$ . So, $K_{total} = \frac{1}{2}Mv^2 + \frac{1}{2}Mv^2 = Mv^2$ . The fraction that is rotational is $\frac{K_{rot}}{K_{total}} = \frac{\frac{1}{2}Mv^2}{Mv^2} = \frac{1}{2}$ .

Question 2

A horizontal beam of length $4.0\,\text{m}$ is held in static equilibrium by two vertical support cables at its ends. A $300\,\text{N}$ load hangs $1.0\,\text{m}$ from the left end, and the beam’s own weight is negligible. The upward tensions $T_L$ and $T_R$ act at the left and right ends, respectively, and the load’s weight acts downward at its attachment point. Taking torques about the left end, the lever arm for $T_R$ is $4.0\,\text{m}$ and for the load is $1.0\,\text{m}$ . The beam does not rotate, so Newton’s First Law in rotational form implies $\sum \tau_{\text{left}}=0$ . Considering the forces acting on the object, calculate the force needed at a specific point to maintain equilibrium.

$T_R=75\,\text{N}$ upward, from $T_R(4.0)=300(1.0)$ . (correct answer)
$T_R=1200\,\text{N}$ upward, from $T_R(1.0)=300(4.0)$ .
$T_R=300\,\text{N}$ upward, since $\sum F_y=0$ requires $T_R=300$ .
$T_R=75\,\text{N}$ downward, to oppose the load’s downward torque.

Explanation: This question tests understanding of rotational equilibrium and Newton's First Law in rotational form in AP Physics C: Mechanics. Rotational equilibrium occurs when the sum of torques acting on an object is zero, meaning the object is not accelerating rotationally. Newton's First Law in rotational form states that an object at rest will remain so unless acted upon by a net external torque. In this scenario, we have a horizontal beam with a load closer to the left support, requiring us to find the right support tension using torque balance. Choice A is correct because taking torques about the left end: clockwise torque from load = counterclockwise torque from right tension, so 300 N × 1.0 m = TR × 4.0 m, giving TR = 300/4.0 = 75 N upward. Choice C is incorrect because it assumes equal sharing of the load, ignoring that the load is closer to the left support, which must carry more of the weight. To help students: Demonstrate that the support closer to the load carries more weight. Practice choosing convenient pivot points (like one support) to eliminate unknown forces from the torque equation.

Question 3

A mass-spring system oscillates with simple harmonic motion. When the displacement is $\frac{1}{3}$ of the amplitude, the ratio of kinetic energy to potential energy is:

$\frac{1}{9}$
$\frac{1}{8}$
$\frac{8}{1}$ (correct answer)
$\frac{9}{1}$

Explanation: For SHM, $U = \frac{1}{2}kx^2$ and $K = \frac{1}{2}k(A^2 - x^2)$ . When $x = \frac{A}{3}$ : $U = \frac{1}{2}k(\frac{A}{3})^2 = \frac{kA^2}{18}$ and $K = \frac{1}{2}k(A^2 - \frac{A^2}{9}) = \frac{1}{2}k \cdot \frac{8A^2}{9} = \frac{4kA^2}{9}$ . Therefore $\frac{K}{U} = \frac{\frac{4kA^2}{9}}{\frac{kA^2}{18}} = \frac{4kA^2}{9} \cdot \frac{18}{kA^2} = 8$ . Choice A uses $\frac{U}{K}$ instead. Choice B confuses the calculation. Choice D incorrectly squares the amplitude ratio.

Question 4

An object of mass $m$ is released from rest at the top of a frictionless ramp of height $h$ and angle $\theta$ . What is the work done by the normal force on the object as it slides down the entire length of the ramp?

$mgh$
Zero (correct answer)
$-mgh$
$mg \cos\theta \cdot \frac{h}{\sin\theta}$

Explanation: The normal force exerted by the ramp on the object is, by definition, perpendicular to the surface of the ramp. The object's displacement is parallel to the surface of the ramp. Since the normal force vector is always perpendicular to the displacement vector, the dot product is zero, and the work done by the normal force is zero.

Question 5

A rope of negligible mass passes over an ideal pulley. A 4.0 kg mass hangs from one end, and a 6.0 kg mass hangs from the other end. When the system is released, what is the magnitude of the force that the 4.0 kg mass exerts on the rope?

59 N
39 N
47 N (correct answer)
49 N

Explanation: When you encounter an Atwood machine problem like this, you're dealing with connected objects where Newton's second law must be applied to the entire system, then to individual components to find internal forces. Since the rope is massless and the pulley is ideal, the tension is uniform throughout the rope. The 6.0 kg mass will accelerate downward while the 4.0 kg mass accelerates upward at the same rate. First, find the system's acceleration by applying Newton's second law to the entire system: the net force is $(6.0 - 4.0)g = 2.0g = 19.6 \text{ N}$ downward, and the total mass is $10.0 \text{ kg}$ . Therefore, $a = \frac{19.6}{10.0} = 1.96 \text{ m/s}^2$ . Now analyze the 4.0 kg mass individually. Two forces act on it: weight (39.2 N downward) and tension (upward). Since it accelerates upward at 1.96 m/s², Newton's second law gives: $T - mg = ma$ , so $T = m(g + a) = 4.0(9.8 + 1.96) = 47.1 \text{ N}$ . By Newton's third law, the mass exerts an equal and opposite 47 N force on the rope. Choice A (59 N) incorrectly assumes the entire weight difference creates tension. Choice B (39 N) represents just the weight of the 4.0 kg mass, ignoring acceleration. Choice D (49 N) likely results from calculation errors in determining the acceleration. Strategy tip: In Atwood machine problems, always find the system acceleration first, then analyze individual objects. Remember that tension in an accelerating system differs from the weight of hanging objects due to the additional force needed for acceleration.

Question 6

A projectile is fired from the edge of a cliff at angle $\theta$ above the horizontal with initial speed $v_0$ . The projectile lands at a horizontal distance $d$ from the launch point and at a vertical distance $h$ below the launch point. If the projectile were instead fired horizontally with the same initial speed from the same point, at what horizontal distance from the launch point would it land?

$d\cos\theta$
$\frac{d\cos\theta}{\sin\theta + \sqrt{\sin^2\theta + \frac{2gh}{v_0^2}}}$
$v_0\sqrt{\frac{2h}{g}}$ (correct answer)
$\frac{v_0^2\cos\theta}{g}\sqrt{\frac{2gh}{v_0^2\sin^2\theta + 2gh}}$

Explanation: For horizontal launch, the time of flight is determined by the vertical motion: $h = \frac{1}{2}gt^2$ , so $t = \sqrt{\frac{2h}{g}}$ . The horizontal distance is then $d_{horizontal} = v_0 t = v_0\sqrt{\frac{2h}{g}}$ . This is independent of the original trajectory parameters $\theta$ and $d$ because horizontal launch only depends on the initial horizontal speed and the fall height. Choice A incorrectly tries to relate it to the original range through a cosine factor. Choice B attempts a complex relationship involving the original angle and trajectory, which is unnecessary for horizontal launch. Choice D over-complicates the relationship and includes terms from the angled launch that don't affect the horizontal launch scenario.

Question 7

Two masses $m_1=2.0\,\text{kg}$ and $m_2=2.5\,\text{kg}$ are connected by a light rope over a frictionless pulley. Consider the scenario described above, with $m_2$ moving downward and the system accelerating. Take $g=9.8\,\text{m/s}^2$ and assume the rope is massless. Tension is the same on both sides. Determine the tension in the rope.

$22.1\,\text{N}$
$4.9\,\text{N}$
$24.5\,\text{N}$
$21.8\,\text{N}$ (correct answer)
$19.6\,\text{N}$

Explanation: This question tests AP Physics C: Mechanics skills, specifically understanding forces and free-body diagrams. Free-body diagrams for Atwood machines show that tension is the same throughout a massless rope, and both masses experience the same magnitude of acceleration in opposite directions. In this scenario, masses of 2.0 kg and 2.5 kg are connected over a frictionless pulley, with the heavier mass accelerating downward. Choice D is correct because the system acceleration is a = (m₂ - m₁)g/(m₁ + m₂) = (0.5)(9.8)/4.5 = 1.09 m/s², and the tension T = m₁(g + a) = 2.0(9.8 + 1.09) = 21.8 N, which also equals m₂(g - a) as a check. Choice E is incorrect because it represents the weight of the lighter mass (19.6 N), showing the common error of assuming tension equals weight when the system accelerates. To help students: Set up separate free-body diagrams and equations for each mass, use the constraint that both accelerations have the same magnitude, and solve the system of equations. Verify the answer by checking that the same tension satisfies both force equations. Watch for: students assuming tension equals one of the weights, using different accelerations for each mass, or making sign errors in the force equations.

Question 8

A ceiling fan blade rotating at $10 \, \text{rad/s}$ is switched off and slows down with a constant angular acceleration of magnitude $2.0 \, \text{rad/s}^2$ . Through what angle does the blade rotate before coming to rest?

$5.0 \, \text{rad}$
$12.5 \, \text{rad}$
$25 \, \text{rad}$ (correct answer)
$50 \, \text{rad}$

Explanation: Use the rotational kinematic equation $\omega^2 = \omega_0^2 + 2\alpha\Delta\theta$ . The final angular velocity is $\omega = 0$ , the initial angular velocity is $\omega_0 = 10 \, \text{rad/s}$ , and the angular acceleration is $\alpha = -2.0 \, \text{rad/s}^2$ (negative because it's slowing down). Solving for $\Delta\theta$ : $0^2 = (10)^2 + 2(-2.0)\Delta\theta$ . This gives $0 = 100 - 4.0\Delta\theta$ , so $4.0\Delta\theta = 100$ , and $\Delta\theta = 25 \, \text{rad}$ .

Question 9

A ball is attached to a string and moves in a vertical circle. When the ball is at the side of the circle (horizontal position), the string makes an angle of 15° with the vertical. If the radius of the circular path is 0.6 m, what is the ball's speed at this position?

$1.3 \text{ m/s}$ (correct answer)
$2.0 \text{ m/s}$
$2.7 \text{ m/s}$
$3.5 \text{ m/s}$

Explanation: At the horizontal position, the centripetal force is provided by the horizontal component of tension: $T\sin(15°) = \frac{mv^2}{r}$ . The vertical component balances weight: $T\cos(15°) = mg$ . Dividing these equations: $\tan(15°) = \frac{v^2}{gr}$ . Solving: $v = \sqrt{gr\tan(15°)} = \sqrt{9.8 \times 0.6 \times 0.268} = 1.3 \text{ m/s}$ . Choice B uses $\sin(15°)$ instead of $\tan(15°)$ . Choice C uses $\cos(15°)$ . Choice D uses the full radius without the trigonometric factor.

Question 10

A mass on a spring oscillates vertically. When the mass is at position $y = +0.10$ m above equilibrium, its acceleration is $a = -4.0$ m/s² (downward). When the mass is at position $y = -0.05$ m below equilibrium, its acceleration is $a = +2.0$ m/s² (upward). What can be concluded about this motion?

This is simple harmonic motion with spring constant $k = 40$ N/m, assuming the mass equals 1.0 kg based on the given data
This is not simple harmonic motion because the ratio of acceleration to displacement differs between the two positions given
This is simple harmonic motion with angular frequency $\omega = 2\sqrt{10}$ rad/s because acceleration is proportional to displacement from equilibrium (correct answer)
This is not simple harmonic motion because the acceleration values suggest the equilibrium position is not where $y = 0$

Explanation: When analyzing oscillatory motion, you need to determine whether the acceleration is proportional to displacement from equilibrium. For simple harmonic motion, the relationship is $a = -\omega^2 y$ , where the acceleration is always directed toward equilibrium. Let's check if this relationship holds by calculating the ratio $a/y$ at both positions. At $y = +0.10$ m, $a = -4.0$ m/s², so $a/y = -4.0/0.10 = -40$ s⁻². At $y = -0.05$ m, $a = +2.0$ m/s², so $a/y = +2.0/(-0.05) = -40$ s⁻². Since this ratio is constant, we have $-\omega^2 = -40$ , giving us $\omega = \sqrt{40} = 2\sqrt{10}$ rad/s. Choice C correctly identifies this as simple harmonic motion with the right angular frequency. The acceleration is indeed proportional to displacement, confirming SHM. Choice A makes an unjustified assumption about the mass being 1.0 kg, which isn't given in the problem. Even if true, the spring constant calculation would require knowing the mass explicitly. Choice B incorrectly claims the acceleration-to-displacement ratio differs between positions. As shown above, this ratio is actually constant at both locations. Choice D misinterprets the data by suggesting the equilibrium position isn't at $y = 0$ . The fact that acceleration points toward $y = 0$ in both cases (downward when above, upward when below) confirms that $y = 0$ is indeed equilibrium. Study tip: Always check if $a/y$ is constant when determining whether motion is simple harmonic. If the ratio is constant and negative, you have SHM with $\omega = \sqrt{|a/y|}$ .

Question 11

A student is analyzing the forces on a block on an inclined plane. According to the course framework, which convention should be followed when drawing the forces on the free-body diagram?

Draw the components of the gravitational force instead of the gravitational force vector itself.
Draw all force vectors originating from the dot representing the object, without showing components. (correct answer)
Draw force vectors originating from their points of application on the object's surface.
Draw the net force vector in addition to the individual force vectors.

Explanation: The standard convention for AP Physics free-body diagrams is to represent the object as a dot and draw all individual external forces as vectors originating from that dot. Force components are used for calculations but should not be drawn on the primary free-body diagram itself. The net force is the vector sum of the individual forces and is also not drawn on the diagram.

Question 12

A particle moves in the xy-plane such that its x-coordinate is given by $x(t) = R\cos(\omega t)$ and its y-coordinate is given by $y(t) = R\sin(\omega t)$ , where $R$ and $\omega$ are positive constants. Which statement correctly describes the magnitude of the particle's acceleration vector?

It is zero because the particle's speed is constant.
It increases with time as the particle covers more distance.
It is constant and directed toward the origin. (correct answer)
It is constant and directed away from the origin.

Explanation: This is uniform circular motion. The position vector is $\vec{r}(t) = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}$ . Differentiating twice gives the acceleration vector: $\vec{v}(t) = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j}$ and $\vec{a}(t) = -R\omega^2\cos(\omega t)\hat{i} - R\omega^2\sin(\omega t)\hat{j}$ . This can be written as $\vec{a}(t) = -\omega^2 \vec{r}(t)$ . The magnitude of the acceleration is $|\vec{a}| = \omega^2 |\vec{r}| = \omega^2 R$ , which is constant. The direction is opposite to the position vector $\vec{r}$ , meaning it is always directed toward the origin.

Question 13

An object of mass $m$ is released from rest in a fluid that exerts a resistive force $F_r = kv$ , where $v$ is the object's speed. What is the initial acceleration of the object immediately after it is released?

0
$g - k/m$
$g$ (correct answer)
$mg/k$

Explanation: Immediately after release, at $t=0$ , the object's speed is $v=0$ . Therefore, the resistive force $F_r = kv = 0$ . The only force acting on the object is gravity, $mg$ . By Newton's second law, $F_{net} = mg = ma$ , so the initial acceleration is $a = g$ .

Question 14

A compound pendulum consists of a uniform disk of radius $R$ and mass $M$ that can pivot about a horizontal axis passing through a point on its rim. If the disk oscillates with small amplitude, what is the length of the equivalent simple pendulum that would have the same period?

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

Explanation: For a physical pendulum, the equivalent simple pendulum length is $L_{eq} = \frac{I}{md}$ , where $I$ is the moment of inertia about the pivot, $m$ is the mass, and $d$ is the distance from pivot to center of mass. For a disk pivoting at its rim, $d = R$ . The moment of inertia about the rim is $I = I_{center} + MR^2 = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}$ . Therefore, $L_{eq} = \frac{3MR^2/2}{M \cdot R} = \frac{3R}{2}$ . Choice A uses only the center-of-mass moment of inertia. Choice B assumes the equivalent length equals the distance to center of mass. Choice D incorrectly doubles the radius without proper calculation.

Question 15

An object of mass $m$ is placed inside a uniform solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center, where $r < R$ . The magnitude of the gravitational force on the object is proportional to which of the following quantities?

$r$ (correct answer)
$r^2$
$1/r$
$1/r^2$

Explanation: The gravitational force inside a uniform solid sphere is given by the expression $F_g = G \frac{Mmr}{R^3}$ . Since $G, M, m,$ and $R$ are constants for this situation, the force $F_g$ is directly proportional to the distance $r$ from the center.

Question 16

An object of mass m is moving east at a constant speed v. The object then makes a sharp turn and moves north at the same constant speed v. What is the change in the object's translational kinetic energy?

$-\frac{1}{2}mv^2$
Zero (correct answer)
$\frac{1}{2}mv^2$
$mv^2$

Explanation: Translational kinetic energy ( $K = \frac{1}{2}mv^2$ ) is a scalar quantity that depends on the magnitude of the velocity (speed), not its direction. Since the speed v remains the same, the kinetic energy is unchanged. Therefore, the change in kinetic energy is zero.

Question 17

A drag car has translational kinetic energy $2.0\times10^5\,\text{J}$ at speed $20\,\text{m/s}$ ; determine the car’s mass from these values.

$500\,\text{kg}$
$1000\,\text{kg}$ (correct answer)
$200\,\text{kg}$
$4000\,\text{kg}$

Explanation: This question tests AP Physics C: Mechanics skills, specifically rearranging the kinetic energy formula to solve for mass. Kinetic energy is given by KE = 1/2 mv², which can be rearranged to m = 2KE/v² when solving for mass. In this scenario, a drag car has kinetic energy 2.0×10⁵ J at speed 20 m/s, requiring calculation of the car's mass. Choice B is correct because rearranging the formula gives: m = 2 × (2.0×10⁵ J) / (20 m/s)² = 4.0×10⁵ / 400 = 1000 kg. Choice A incorrectly uses half this value, possibly from forgetting to multiply by 2 when rearranging the formula. To help students: Practice algebraic manipulation of the kinetic energy formula to solve for different variables. Emphasize checking units and using dimensional analysis to verify calculations are set up correctly.

Question 18

A geostationary satellite orbits Earth at radius $r_{geo}$ . A spy satellite is placed in a circular orbit at radius $r = \frac{r_{geo}}{4}$ . How many times does the spy satellite orbit Earth while the geostationary satellite completes one orbit?

$4$
$8$ (correct answer)
$16$
$64$

Explanation: Using Kepler's third law: $\frac{T_{spy}^2}{T_{geo}^2} = \frac{r_{spy}^3}{r_{geo}^3} = \frac{(r_{geo}/4)^3}{r_{geo}^3} = \frac{1}{64}$ . Therefore $T_{spy} = \frac{T_{geo}}{8}$ . In one geostationary period, the spy satellite completes $\frac{T_{geo}}{T_{spy}} = 8$ orbits. Choice A incorrectly uses linear scaling. Choice C incorrectly uses quadratic scaling. Choice D uses the cube of the radius ratio directly.

Question 19

A vector $\vec{F}$ can be written as $\vec{F} = F_x\hat{i} + F_y\hat{j}$ where $F_x = 12$ N and $F_y = -5$ N. If this vector is rotated counterclockwise by $90°$ about the origin, what are the components of the resulting vector $\vec{F}'$ ?

$F'_x = 5$ N, $F'_y = 12$ N (correct answer)
$F'_x = -5$ N, $F'_y = -12$ N
$F'_x = -12$ N, $F'_y = 5$ N
$F'_x = 12$ N, $F'_y = 5$ N

Explanation: When a vector is rotated counterclockwise by 90°, the transformation is: $\vec{F}' = -F_y\hat{i} + F_x\hat{j}$ . With $F_x = 12$ N and $F_y = -5$ N, we get: $F'_x = -F_y = -(-5) = 5$ N and $F'_y = F_x = 12$ N. Choice B results from incorrectly applying $F'_x = F_y$ and $F'_y = -F_x$ . Choice C comes from using the clockwise rotation formula. Choice D results from simply changing the sign of $F_y$ without proper rotation.

Question 20

Vector $\vec{A}$ has a magnitude of 8 units, and vector $\vec{B}$ has a magnitude of 5 units. The vector $\vec{C}$ is defined as $\vec{C} = \vec{A} - \vec{B}$ . Which of the following is a possible magnitude for vector $\vec{C}$ ?

2
12 (correct answer)
14
0

Explanation: When working with vector subtraction, you need to consider that the magnitude of the resultant vector depends on both the magnitudes of the individual vectors and the angle between them. For $\vec{C} = \vec{A} - \vec{B}$ , the magnitude of $\vec{C}$ can range from the absolute difference to the sum of the magnitudes. The key insight is that vector subtraction $\vec{A} - \vec{B}$ is equivalent to $\vec{A} + (-\vec{B})$ , where $-\vec{B}$ has the same magnitude as $\vec{B}$ but points in the opposite direction. Using the triangle inequality, the magnitude of $\vec{C}$ must satisfy: $|\vec{A}| - |\vec{B}| \leq |\vec{C}| \leq |\vec{A}| + |\vec{B}|$ With $|\vec{A}| = 8$ and $|\vec{B}| = 5$ , this gives us: $3 \leq |\vec{C}| \leq 13$ The minimum occurs when vectors $\vec{A}$ and $\vec{B}$ point in the same direction (making $\vec{A}$ and $-\vec{B}$ opposite), and the maximum occurs when they point in opposite directions (making $\vec{A}$ and $-\vec{B}$ parallel). Answer B (12) falls within this valid range of 3 to 13 units, making it correct. Answer A (2) is too small—it's less than the minimum possible value of 3. Answer C (14) exceeds the maximum possible value of 13. Answer D (0) is impossible since the minimum magnitude is 3 when the vectors have different magnitudes. Remember: For any vector operation involving magnitudes, always check whether your answer falls within the physically possible range determined by the triangle inequality.

Question 21

A 4.0 kg block sliding at 6.0 m/s on a frictionless surface collides elastically with a 2.0 kg block moving at 3.0 m/s in the opposite direction. After the collision, the 2.0 kg block moves at 7.0 m/s in its original direction. A student calculates that the 4.0 kg block's final velocity should be 1.0 m/s in its original direction. Which statement best describes this result?

The calculation is correct because momentum is conserved and the collision satisfies elastic collision conditions for these masses and velocities
The calculation violates conservation of momentum; the 4.0 kg block should have final velocity 2.0 m/s in its original direction
The calculation violates conservation of kinetic energy; this collision cannot be elastic with the given final velocity of the 2.0 kg block (correct answer)
The calculation is incorrect because both momentum and energy conservation are violated; the collision scenario is physically impossible

Explanation: Check both conservation laws. Taking the 4.0 kg block's initial direction as positive: Initial momentum = $(4.0)(6.0) + (2.0)(-3.0) = 24 - 6 = 18\text{ kg·m/s}$ . Final momentum = $(4.0)(1.0) + (2.0)(7.0) = 4 + 14 = 18\text{ kg·m/s}$ . Momentum is conserved. Initial KE = $\frac{1}{2}(4.0)(6.0)^2 + \frac{1}{2}(2.0)(-3.0)^2 = 72 + 9 = 81\text{ J}$ . Final KE = $\frac{1}{2}(4.0)(1.0)^2 + \frac{1}{2}(2.0)(7.0)^2 = 2 + 49 = 51\text{ J}$ . Kinetic energy is not conserved (81 J ≠ 51 J), so this cannot be an elastic collision. The student's calculation satisfies momentum conservation but violates energy conservation for an elastic collision.

Question 22

Based on the scenario, how does the velocity vector change over time for a projectile under uniform gravity with no air resistance?

$v_x$ decreases linearly; $v_y$ constant
$v_x$ constant; $v_y$ decreases linearly (correct answer)
Both $v_x$ and $v_y$ decrease linearly
Speed remains constant; direction changes only

Explanation: This question tests AP Physics C kinematics, specifically understanding velocity components in projectile motion under uniform gravity. The motion in two dimensions requires recognizing that gravity acts only vertically, leaving horizontal motion unaffected. In this scenario with no air resistance, only gravity acts on the projectile, providing constant downward acceleration. Choice B is correct because vₓ remains constant (no horizontal forces) while vᵧ decreases linearly due to constant gravitational acceleration (vᵧ = v₀ᵧ - gt). Choice A reverses the components, C incorrectly suggests both components change, and D misunderstands that while speed changes, velocity components change predictably. To help students: Emphasize the independence of horizontal and vertical motion in projectile problems. Practice analyzing force diagrams to identify which components experience acceleration. Watch for: confusing speed with velocity or forgetting that only vertical motion is affected by gravity.

Question 23

A disk of radius $2R$ rotates from rest about its center with a constant angular acceleration $\alpha$ . Consider point A at radius $R$ and point B at radius $2R$ . After a non-zero time $t$ , what is the ratio of the magnitude of the total linear acceleration of point B to that of point A, $a_B / a_A$ ?

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

Explanation: For any point at radius $r$ , the tangential acceleration is $a_T = r\alpha$ and the centripetal acceleration is $a_c = r\omega^2$ . After time $t$ , the angular velocity is $\omega = \alpha t$ . The magnitude of the total acceleration is $a = \sqrt{a_T^2 + a_c^2} = \sqrt{(r\alpha)^2 + (r\omega^2)^2} = \sqrt{r^2\alpha^2 + r^2\alpha^2t^4} = r\alpha\sqrt{1 + \alpha^2t^4}$ . Since $\alpha$ and $t$ are the same for both points, the total acceleration is directly proportional to the radius $r$ . Therefore, the ratio $a_B / a_A = (2R) / R = 2$ .

Question 24

A simple pendulum consists of a bob of mass $m$ attached to a string of length $L$ . The pendulum is pulled back to a maximum angle $\theta_{max}$ with the vertical and released from rest. What is the total mechanical energy of the pendulum-Earth system with respect to the lowest point of the swing?

$E = mgL$
$E = mgL(1 - \cos\theta_{max})$ (correct answer)
$E = \frac{1}{2}mgL\theta_{max}^2$
$E = mgL\sin\theta_{max}$

Explanation: The total mechanical energy is conserved. At the maximum angular displacement, the bob is momentarily at rest, so its kinetic energy is zero. The total energy is equal to the gravitational potential energy at that point. The height of the bob above the lowest point is $h = L - L\cos\theta_{max} = L(1 - \cos\theta_{max})$ . Therefore, the total energy is $E = mgh = mgL(1 - \cos\theta_{max})$ .

Question 25

Two cars, A and B, start at the same point. Car A travels north at a constant speed $v$ . Car B starts from rest and accelerates east with a constant acceleration $a$ . What is the velocity of car B relative to car A at time $t$ ?

$at\hat{i} - v\hat{j}$ (correct answer)
$at\hat{i} + v\hat{j}$
$\sqrt{(at)^2 + v^2}$
$\frac{1}{2}at^2\hat{i} - vt\hat{j}$

Explanation: Let the eastward direction be represented by the unit vector $\hat{i}$ and the northward direction by $\hat{j}$ . The velocity of car A is constant: $v_A = v\hat{j}$ . The velocity of car B at time $t$ is given by $v_B = v_{B0} + at = 0 + at\hat{i} = at\hat{i}$ . The velocity of car B relative to car A is $v_{BA} = v_B - v_A = at\hat{i} - v\hat{j}$ .

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