2025 AP AP Physics C: Mechanics Practice Paper with Answers

A. i.
Initial momentum of Glider 1: $p_{1,i} = (3m)(+2v_0) = +6mv_0$
Initial momentum of Glider 2: $p_{2,i} = (m)(-4v_0) = -4mv_0$
Total initial momentum of the two-glider system: $p_{\text{sys}} = p_{1,i} + p_{2,i} = +6mv_0 - 4mv_0 = +2mv_0$
Since the track is frictionless and no net external horizontal force acts on the system, linear momentum is conserved, so the total momentum after the collision is also $p_{\text{sys}} = +2mv_0$ (in the $+x$-direction).

ii.
By conservation of momentum, the final velocity $v_f$ of the combined gliders is:
$p_{\text{sys}} = (3m + m)v_f = 4mv_f = +2mv_0 \implies v_f = +\frac{1}{2}v_0$

The change in momentum of Glider 2 is:
$\Delta p_2 = p_{2,f} - p_{2,i} = m\left(+\frac{1}{2}v_0\right) - (-4mv_0) = +\frac{9}{2}mv_0$

Using the impulse-momentum theorem:
$J = \int_{0}^{T} F(t)\,dt = \Delta p_2$
$\int_{0}^{T} b(Tt - t^2)\,dt = \frac{9}{2}mv_0$
$b\left[ \frac{Tt^2}{2} - \frac{t^3}{3} \right]_0^T = \frac{9}{2}mv_0$
$b\left( \frac{T^3}{2} - \frac{T^3}{3} \right) = b\frac{T^3}{6} = \frac{9}{2}mv_0$
$b = \frac{6 \cdot 9 m v_0}{2 T^3} = \frac{27mv_0}{T^3}$

B.
Applying conservation of linear momentum to the system for the second scenario:
$\sum p_i = \sum p_f$
$(3m)(v_1) + (m)(-3v_0) = (3m + m)(+2v_0)$
$3mv_1 - 3mv_0 = 4m(2v_0) = 8mv_0$
$3mv_1 = 11mv_0$
$v_1 = \frac{11}{3}v_0$

Part A(i) [2 points]:
- 1 point for correctly determining the initial momentum of Glider 2 as $-4mv_0$ (or magnitude $4mv_0$ in the $-x$-direction).
- 1 point for correctly determining the momentum of the two-glider system before and after the collision as $+2mv_0$ (or magnitude $2mv_0$ in the $+x$-direction).

Part A(ii) [5 points]:
- 1 point for a multistep derivation that begins with the integral form of impulse ($J = \int F\,dt = \Delta p$) or the differential form of Newton's second law ($F = \frac{dp}{dt}$).
- 1 point for correctly finding the change in momentum of Glider 2 as $+\frac{9}{2}mv_0$ (or the change in momentum of Glider 1 as $-\frac{9}{2}mv_0$).
- 1 point for substituting the given force function $F(t) = bt(T - t)$ into the integral.
- 1 point for evaluating the definite integral with correct limits from $0$ to $T$ to obtain $\frac{bT^3}{6}$.
- 1 point for the correct final expression for $b = \frac{27mv_0}{T^3}$.

Part B [3 points]:
- 1 point for starting with a statement of conservation of linear momentum ($\sum p_i = \sum p_f$).
- 1 point for correctly setting the total final momentum equal to $+8mv_0$ or $(3m+m)(2v_0)$.
- 1 point for arriving at the correct final expression $v_1 = \frac{11}{3}v_0$.

A.
Total mechanical energy of the system when released from rest at \(x = x_0\):
\[ E_{\text{total}} = \frac{1}{2} k x_0^2 \]
At position \(x = \frac{1}{2} x_0\):
- Elastic potential energy:
\[ U_s = \frac{1}{2} k \left(\frac{1}{2} x_0\right)^2 = \frac{1}{4} \left(\frac{1}{2} k x_0^2\right) = \frac{1}{4} E_{\text{total}} \]
- Total kinetic energy:
\[ K_{\text{total}} = E_{\text{total}} - U_s = \frac{3}{4} E_{\text{total}} \]
Since the cylinder rolls without slipping, \(v = \omega R\):
\[ K_{\text{trans}} = \frac{1}{2} M v^2 \]
\[ K_{\text{rot}} = \frac{1}{2} I \omega^2 = \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{4} M v^2 = \frac{1}{2} K_{\text{trans}} \]
Thus, \(K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{3}{2} K_{\text{trans}}\).
Therefore:
\[ K_{\text{trans}} = \frac{2}{3} K_{\text{total}} = \frac{2}{3} \left(\frac{3}{4} E_{\text{total}}\right) = \frac{1}{2} E_{\text{total}} \]
\[ K_{\text{rot}} = \frac{1}{3} K_{\text{total}} = \frac{1}{3} \left(\frac{3}{4} E_{\text{total}}\right) = \frac{1}{4} E_{\text{total}} \]
Comparing the ratios:
\[ U_s : K_{\text{trans}} : K_{\text{rot}} = 1 : 2 : 1 \]
The bar heights must be drawn with positive heights in this exact \(1:2:1\) ratio.

B.
From the conservation of mechanical energy between release at \(x = x_0\) and equilibrium \(x = 0\):
\[ E_0 = E_f \]
\[ \frac{1}{2} k x_0^2 = K_{\text{trans, max}} + K_{\text{rot, max}} \]
Substitute \(K_{\text{trans}} = \frac{1}{2} M v_{\text{max}}^2\) and \(K_{\text{rot}} = \frac{1}{2} I \omega_{\text{max}}^2\):
\[ \frac{1}{2} k x_0^2 = \frac{1}{2} M v_{\text{max}}^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v_{\text{max}}}{R}\right)^2 \]
\[ \frac{1}{2} k x_0^2 = \frac{1}{2} M v_{\text{max}}^2 + \frac{1}{4} M v_{\text{max}}^2 = \frac{3}{4} M v_{\text{max}}^2 \]
Solving for \(v_{\text{max}}\):
\[ v_{\text{max}}^2 = \frac{2 k x_0^2}{3 M} \implies v_{\text{max}} = x_0 \sqrt{\frac{2k}{3M}} \]

C.
The kinetic energy \(K(t)\) satisfies:
1. At \(t = 0\), the cylinder is released from rest, so \(K(0) = 0\).
2. Kinetic energy is always non-negative (\(K \ge 0\)).
3. The function oscillates periodically, reaching zero whenever the cylinder reaches turning points (spaced by \(\frac{1}{2} T\)).
4. Due to damping, each successive peak has a smaller maximum amplitude than the preceding peak.

D.
Feature: The period \(T\) of oscillation of the position-time curve will be greater (or the oscillation frequency will be smaller).
Justification: For rolling without slipping, total kinetic energy is \(K = \frac{1}{2} M_{\text{eff}} v^2\), where \(M_{\text{eff}} = M + \frac{I}{R^2}\). For the solid cylinder, \(M_{\text{eff, cyl}} = \frac{3}{2} M\), yielding \(T = 2\pi \sqrt{\frac{3M}{2k}}\). For the thin hoop, \(I = M R^2\), so \(M_{\text{eff, hoop}} = 2M\), giving \(T = 2\pi \sqrt{\frac{2M}{k}}\). Since the hoop has a greater rotational inertia, it has greater effective inertia, increasing the period of oscillation.

Part A (3 points):
- Point A1: For drawing bars for \(U_s\), \(K_{\text{trans}}\), and \(K_{\text{rot}}\) with positive heights.
- Point A2: For drawing the bar for \(K_{\text{rot}}\) with a height equal to the height of the bar drawn for \(U_s\).
- Point A3: For drawing the bar for \(K_{\text{trans}}\) with a height twice that of \(U_s\) (or \(K_{\text{rot}}\)).

Part B (4 points):
- Point B1: For a multistep derivation that begins with a statement of conservation of mechanical energy: \(E_i = E_f\) or \(\Delta E = 0\).
- Point B2: For correctly expressing total kinetic energy as the sum of translational and rotational kinetic energy: \(K_{\text{total}} = \frac{1}{2}M v^2 + \frac{1}{2}I \omega^2\).
- Point B3: For substituting \(I = \frac{1}{2}MR^2\) and the rolling without slipping condition \(\omega = \frac{v}{R}\) into the energy equation.
- Point B4: For obtaining the correct final expression for \(v_{\text{max}}\) in terms of given quantities: \(v_{\text{max}} = x_0 \sqrt{\frac{2k}{3M}}\).

Part C (3 points):
- Point C1: For sketching a curve that starts at \(K = 0\) at \(t = 0\) and is always greater than or equal to zero.
- Point C2: For sketching a periodic curve with zeros regularly spaced at intervals representing \(\frac{1}{2} T\).
- Point C3: For sketching successive peaks that decrease in height over time (damped behavior).

Part D (2 points):
- Point D1: For identifying that the period of the position graph increases (or the frequency decreases, or maximum velocity decreases).
- Point D2: For a valid justification linking the larger rotational inertia of the hoop (\(I_{\text{hoop}} > I_{\text{cyl}}\)) to greater effective inertia and thus a longer oscillation period (or lower angular frequency).

### Part A(i)
Experimental Procedure:
1. Measure the radius \(r\) of the hub using calipers/ruler.
2. Hang a block of known mass \(m\) from the string and place the motion sensor directly below it on the floor.
3. Release the block from rest and use the motion sensor/software to record the position-time or velocity-time data to find the constant downward acceleration \(a\).
4. Repeat the drop multiple times for the same mass \(m\) and calculate the average acceleration to minimize random error.
5. Repeat this procedure for several different hanging masses \(m\).

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### Part A(ii)
Apply Newton's second law to the hanging mass:
$$\sum F_y = mg - T = ma \implies T = m(g - a)$$

Apply Newton's second law for rotation to the wheel:
$$\sum \tau = T r = I \alpha$$

Since the string unwinds without slipping, the linear and angular accelerations are related by:
$$a = r \alpha \implies \alpha = \frac{a}{r}$$

Substitute \(T\) and \(\alpha\) into the torque equation:
$$m(g - a) r = I\left(\frac{a}{r}\right) \implies m(g - a) = \frac{I}{r^2} a$$

Rearranging gives:
$$\frac{1}{a} = \frac{1}{g} + \left(\frac{I}{g r^2}\right) \frac{1}{m} \quad \text{or} \quad m(g - a) = \left(\frac{I}{r^2}\right) a$$

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### Part B(i)
One convenient choice of variables to plot is:
- Vertical axis: \(m(g - a)\) (in \(\text{N}\)) with \(g = 9.8\text{ m/s}^2\)
- Horizontal axis: \(a\) (in \(\text{m/s}^2\))

*(Alternatively, plot \(1/a\) on the vertical axis vs. \(1/m\) on the horizontal axis, or \(T r\) vs. \(a/r\).)*

Calculating \(m(g - a)\) for each trial:
- For \(m = 0.10\text{ kg}\), \(a = 0.096\text{ m/s}^2\): \(m(g-a) = 0.10(9.8 - 0.096) = 0.970\text{ N}\)
- For \(m = 0.20\text{ kg}\), \(a = 0.190\text{ m/s}^2\): \(m(g-a) = 0.20(9.8 - 0.190) = 1.922\text{ N}\)
- For \(m = 0.30\text{ kg}\), \(a = 0.280\text{ m/s}^2\): \(m(g-a) = 0.30(9.8 - 0.280) = 2.856\text{ N}\)
- For \(m = 0.40\text{ kg}\), \(a = 0.380\text{ m/s}^2\): \(m(g-a) = 0.40(9.8 - 0.380) = 3.768\text{ N}\)
- For \(m = 0.50\text{ kg}\), \(a = 0.465\text{ m/s}^2\): \(m(g-a) = 0.50(9.8 - 0.465) = 4.668\text{ N}\)

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### Part B(ii)
- Axes: Horizontal axis scaled from \(0\) to \(0.50\text{ m/s}^2\); vertical axis scaled from \(0\) to \(5.0\text{ N}\).
- Points plotted accurately: \((0.096, 0.970)\), \((0.190, 1.922)\), \((0.280, 2.856)\), \((0.380, 3.768)\), \((0.465, 4.668)\).
- Best-fit line drawn smoothly through the data points.

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### Part C
Calculate the slope from two points on the best-fit line:
$$\text{Slope} = \frac{4.67 - 0.97}{0.465 - 0.096} = \frac{3.70}{0.369} \approx 10.0\text{ kg}$$

From the linear relationship \(m(g - a) = \left(\frac{I}{r^2}\right) a\):
$$\text{Slope} = \frac{I}{r^2} \implies I = (\text{Slope}) \times r^2$$
$$I = (10.0\text{ kg})(0.040\text{ m})^2 = 10.0 \times 0.0016 = 0.016\text{ kg}\cdot\text{m}^2$$

Part A(i): 2 points
- 1 point for describing a valid experimental procedure that measures the acceleration of the falling mass across varying hanging mass values.
- 1 point for describing a valid method to reduce experimental uncertainty (e.g., conducting multiple trials per mass to obtain an average acceleration).

Part A(ii): 2 points
- 1 point for setting up correct Newton's second law equations for both translational and rotational motion with consistent sign convention (\(mg - T = ma\) and \(Tr = I\alpha\)).
- 1 point for substituting \(\alpha = a/r\) and correctly combining equations to relate \(m\), \(a\), \(g\), \(r\), and \(I\).

Part B(i): 1 point
- 1 point for identifying appropriate variables to plot that yield a linear relationship (e.g., \(m(g-a)\) vs. \(a\), or \(1/a\) vs. \(1/m\), or \(Tr\) vs. \(\alpha\)).

Part B(ii): 3 points
- 1 point for clearly labeled axes with appropriate quantities, units, and consistent linear scales.
- 1 point for correctly plotting all data points consistent with the quantities indicated in B(i).
- 1 point for drawing an appropriate best-fit straight line showing an even distribution of points.

Part C: 2 points
- 1 point for correctly relating the slope of the best-fit line to the rotational inertia \(I\) (e.g., \(\text{Slope} = I/r^2\)).
- 1 point for calculating an experimental value for \(I\) consistent with the drawn best-fit line (acceptable range: \(0.014\text{ to }0.018\text{ kg}\cdot\text{m}^2\)).

### Part A
Claim: \(\omega_D > \omega_H\)

Justification:
Both objects experience the same applied force at the same radial distance from the axle, meaning the torque exerted on each object is identical. However, the hoop has all of its mass distributed at its outer rim, whereas the disk has its mass distributed continuously from the axis to the rim. Consequently, the hoop has a greater rotational inertia than the disk. By Newton's second law for rotation, an object with smaller rotational inertia will experience a greater angular acceleration when subjected to the same torque. Because both objects start from rest and undergo constant angular acceleration for the same duration of time, the disk achieves a greater final angular speed.

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### Part B
Begin with Newton's second law for both translational and rotational motion:

For the falling block of mass \(m\):
$$\Sigma F_y = mg - T = ma$$

For the rotating cylinder with rotational inertia \(I\):
$$\Sigma \tau = TR = I\alpha$$

Since the string unwinds without slipping, the linear and angular accelerations are related by:
$$a = \alpha R \implies \alpha = \frac{a}{R}$$

Substitute \(\alpha\) into the rotational torque equation to find the string tension \(T\):
$$T R = I\left(\frac{a}{R}\right) \implies T = \frac{Ia}{R^2}$$

Substitute this expression for \(T\) into the translational equation:
$$mg - \frac{Ia}{R^2} = ma$$
$$mg = a\left(m + \frac{I}{R^2}\right)$$
$$a = \frac{mg}{m + \frac{I}{R^2}}$$

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### Part C
Claim: Less than (\(T_D < T_H\))

Justification:
From the equation of motion for the falling block, \(mg - T = ma\), the tension can be expressed as \(T = m(g - a)\). Since Disk D has a smaller rotational inertia than Hoop H (\(I_D < I_H\)), the downward acceleration of the block attached to Disk D is greater (\(a_D > a_H\)). A larger downward acceleration requires a smaller upward opposing tension force; therefore, \(T_D < T_H\).

Part A (3 points total)
- 1 point (Point A1): For indicating \(\omega_D > \omega_H\).
- 1 point (Point A2): For a correct comparison of the rotational inertias based on mass distribution (i.e., disk has lower rotational inertia than the hoop because its mass is closer to the axis).
- 1 point (Point A3): For linking the same applied torque and smaller rotational inertia to a greater angular acceleration and greater final angular speed.

Part B (3 points total)
- 1 point (Point B1): For a multi-step derivation that applies Newton's second law in both translational form (\(mg - T = ma\)) and rotational form (\(\tau = I\alpha\) or \(TR = I\alpha\)).
- 1 point (Point B2): For using the kinematic constraint \(a = \alpha R\) in an attempt to combine the translational and rotational equations.
- 1 point (Point B3): For obtaining a correct symbolic expression for \(a\) in terms of given quantities:
$$a = \frac{mg}{m + \frac{I}{R^2}}$$

Part C (2 points total)
- 1 point (Point C1): For indicating "Less than" (or \(T_D < T_H\)).
- 1 point (Point C2): For a correct justification connecting the larger downward acceleration of the block attached to the disk (or the smaller rotational inertia of the disk) to a smaller upward tension force using \(T = m(g - a)\) or \(T = \frac{Ia}{R^2}\).