2024 AP AP Physics C: Mechanics Practice Paper with Answers

Part (a)(i):
1. Conservation of mechanical energy for the launch from the spring (from \(t = 0\) to \(t = t_1\)):
\[\frac{1}{2} k x_0^2 = \frac{1}{2} m v_1^2 \implies v_1 = x_0\sqrt{\frac{k}{m}}\]
2. Work done by kinetic friction along the rough patch of length \(L\):
\[W_{\text{f}} = -F_{\text{f}} L = -\mu m g L\]
3. Applying the work-energy theorem between \(t_1\) and \(t_3\):
\[\frac{1}{2} m v_3^2 - \frac{1}{2} m v_1^2 = -\mu m g L\]
\[\frac{1}{2} m v_3^2 = \frac{1}{2} k x_0^2 - \mu m g L\]
\[v_3^2 = \frac{k}{m} x_0^2 - 2\mu g L\]
\[v_3 = \sqrt{\frac{k}{m} x_0^2 - 2\mu g L}\]

Part (a)(ii):
1. Conservation of linear momentum during the inelastic collision at \(t = t_4\):
\[m v_3 = (m + 2m) v_{\text{sys}} \implies v_{\text{sys}} = \frac{1}{3} v_3\]
2. Conservation of mechanical energy during the pendulum swing (from immediately after collision to maximum height \(h_{\text{max}}\)):
\[\frac{1}{2}(3m)v_{\text{sys}}^2 = (3m)g h_{\text{max}}\]
\[h_{\text{max}} = \frac{v_{\text{sys}}^2}{2g} = \frac{\left(\frac{1}{3} v_3\right)^2}{2g} = \frac{v_3^2}{18g}\]
3. Relating height to angle \(\theta_{\text{max}}\):
\[h_{\text{max}} = \ell(1 - \cos\theta_{\text{max}})\]
\[\ell(1 - \cos\theta_{\text{max}}) = \frac{\frac{k}{m} x_0^2 - 2\mu g L}{18g}\]
\[1 - \cos\theta_{\text{max}} = \frac{k x_0^2 - 2\mu m g L}{18 m g \ell}\]
\[\cos\theta_{\text{max}} = 1 - \frac{k x_0^2 - 2\mu m g L}{18 m g \ell}\]

Part (b)(i):
- \(0 \le t \le t_1\): Starts at \(p = 0\) and increases nonlinearly (sinusoidally/concave down) up to \(p_1 = m v_1\).
- \(t_1 \le t \le t_2\): Constant horizontal line at value \(p_1\).
- \(t_2 \le t \le t_3\): Decreases linearly from \(p_1\) to \(p_3 = m v_3\).
- \(t_3 \le t \le t_4\): Constant horizontal line at value \(p_3\).
- At \(t_4\): Instantaneous drop in momentum of block \(m\) from \(p_3\) to \(p_{\text{after}} = m v_{\text{sys}} = \frac{1}{3} p_3\).
- \(t_4 \le t \le t_5\): Continuous decrease from \(\frac{1}{3} p_3\) to \(0\) at \(t_5\).

Part (b)(ii):
During \(t_2 \le t \le t_3\), the only horizontal force acting on block \(m\) is the constant force of kinetic friction \(F_{\text{f}} = \mu m g\) opposing motion. By Newton's second law / impulse-momentum theorem:
\[\frac{dp}{dt} = \Sigma F = -\mu m g = \text{constant}\]
Because the net force is constant and negative, the slope \(\frac{dp}{dt}\) of the momentum-time graph is a constant negative value, producing a downward-sloping straight line.

Part (c):
Select: \(T_{\text{new}} = T_0\).
Justification: For small angular amplitudes, the period of a simple pendulum is given by \(T = 2\pi \sqrt{\frac{\ell}{g}}\), which depends only on the length of the string \(\ell\) and the acceleration due to gravity \(g\). It is independent of the initial speed, energy, mass, and amplitude of oscillation.

(a)(i) (3 points total)
- 1 point: For applying conservation of mechanical energy for the spring decompression or equating initial elastic energy to kinetic energy.
- 1 point: For applying the work-energy theorem with the correct negative work done by friction (\(W_{\text{f}} = -\mu m g L\)).
- 1 point: For a correct expression for \(v_3\).

(a)(ii) (3 points total)
- 1 point: For applying conservation of linear momentum during the inelastic collision to find the post-collision speed (\(v_{\text{sys}} = \frac{1}{3} v_3\)).
- 1 point: For equating post-collision kinetic energy to gravitational potential energy to find \(h_{\text{max}}\) or \(1 - \cos\theta_{\text{max}}\).
- 1 point: For a correct final expression for \(\cos\theta_{\text{max}}\).

(b)(i) (4 points total)
- 1 point: For a curve that starts at zero and increases nonlinearly over \(0 \le t \le t_1\).
- 1 point: For horizontal segments during \(t_1 \le t \le t_2\) and \(t_3 \le t \le t_4\) reflecting zero net force / constant speed.
- 1 point: For a strictly linear decrease during the friction interval \(t_2 \le t \le t_3\).
- 1 point: For a step decrease in momentum at \(t_4\) followed by a curve decreasing to zero at \(t_5\).

(b)(ii) (3 points total)
- 1 point: For stating that the net force on the block during \(t_2 \le t \le t_3\) is the force of kinetic friction.
- 1 point: For noting that kinetic friction is constant in magnitude and opposes the direction of motion.
- 1 point: For explicitly relating the constant net force to the constant negative slope of the momentum-time graph using \(\frac{dp}{dt} = \Sigma F\).

(c) (2 points total)
- 1 point: For correctly selecting \(T_{\text{new}} = T_0\).
- 1 point: For a valid justification indicating that the small-angle period of a pendulum depends only on string length \(\ell\) and \(g\) (independent of amplitude/speed/mass).

(a) Applying Newton's second law in the horizontal direction (taking the direction of initial velocity as positive):
$$\Sigma F_x = -F_R = M a$$
$$- \gamma v = M \frac{dv}{dt}$$
$$M \frac{dv}{dt} + \gamma v = 0 \quad \text{or} \quad \frac{dv}{dt} = -\frac{\gamma}{M}v$$

(b)
i. The sketch of acceleration magnitude $|a(t)|$ begins at an initial nonzero value $|a(0)| = \frac{\gamma v_0}{M}$ on the vertical axis and decays smoothly (concave up) toward zero asymptotically as $t \to \infty$.
ii. According to Newton's second law, $a = \frac{F_{\text{net}}}{M} = \frac{\gamma v}{M}$. As the resistive force slows the cart down, the speed $v$ decreases, which in turn decreases the magnitude of the resistive force $F_R$. Because the net force decreases over time, the magnitude of acceleration must also continuously decrease.

(c) Selection: Equal to
Justification: The total distance is given by $x_{\text{tot}} = \frac{M v_0}{\gamma}$, which shows a direct linear relationship with the initial speed $v_0$. Doubling the initial speed from $v_0$ to $2v_0$ results in $x_{\text{tot, new}} = \frac{M(2v_0)}{\gamma} = 2 x_{\text{tot}}$, exactly twice the original distance.

(d)
i. Plotting $x_{\text{tot}}$ on the vertical axis versus $M$ on the horizontal axis yields a linear trend. Choosing two representative points along the best-fit line, e.g., $(0.20\text{ kg}, 0.50\text{ m})$ and $(1.00\text{ kg}, 2.50\text{ m})$:
$$\text{slope} = \frac{2.50\text{ m} - 0.50\text{ m}}{1.00\text{ kg} - 0.20\text{ kg}} = \frac{2.00\text{ m}}{0.80\text{ kg}} = 2.50\text{ m/kg}$$
(Acceptable slopes typically range between $2.40\text{ m/kg}$ and $2.60\text{ m/kg}$.)

ii. From the expression $x_{\text{tot}} = \left(\frac{v_0}{\gamma}\right) M$, the slope corresponds to:
$$\text{slope} = \frac{v_0}{\gamma} \implies \gamma = \frac{v_0}{\text{slope}}$$
$$\gamma = \frac{2.0\text{ m/s}}{2.50\text{ m/kg}} = 0.80\text{ kg/s} \text{ (or N}\cdot\text{s/m)}$$

(e)
i. Vertical axis: $\gamma$ (or $1/x_{\text{tot}}$); Horizontal axis: $N$ (number of magnets).
ii. If the damping constant $\gamma$ is directly proportional to $N$, the graph of $\gamma$ versus $N$ should yield a straight best-fit line that passes through the origin.

Part (a): 3 points
- 1 point: For applying Newton's second law of motion to the horizontal forces acting on the cart.
- 1 point: For indicating that the net horizontal force is equal to the negative resistive force ($- \gamma v$).
- 1 point: For a correctly substituted differential equation relating $\frac{dv}{dt}$ and $v$ in terms of the given parameters ($M, \gamma$).

Part (b): 3 points
- 1 point: For drawing an acceleration vs. time graph that starts at a positive nonzero value and decays asymptotically to zero.
- 1 point: For drawing a concave-up decaying curve.
- 1 point: For using Newton's second law ($F = Ma$) and the velocity-dependence of drag ($F_R \propto v$) to justify why acceleration decreases as speed decreases.

Part (c): 2 points
- 1 point: For selecting 'Equal to' with an attempt at a relevant justification.
- 1 point: For a correct justification showing that $x_{\text{tot}}$ is directly proportional to initial velocity $v_0$.

Part (d): 4 points
- 1 point: For correctly calculating the slope of the line of best fit using two points on the line (not necessarily raw data points unless they lie on the line).
- 1 point: For correctly relating the slope of the best-fit line to the physical expression ($\text{slope} = v_0 / \gamma$).
- 1 point: For substituting the given value of $v_0$ into the slope relationship to solve for $\gamma$.
- 1 point: For a calculated value of $\gamma$ in the range $0.75\text{ kg/s} \le \gamma \le 0.85\text{ kg/s}$ with appropriate units ($\text{kg/s}$ or $\text{N}\cdot\text{s/m}$).

Part (e): 3 points
- 1 point: For identifying appropriate quantities to plot on the vertical and horizontal axes (e.g., $\gamma$ vs. $N$).
- 1 point: For indicating that a direct proportionality requires a linear relationship / constant slope.
- 1 point: For specifying that the linear fit must pass through (or have a y-intercept consistent with) the origin.

(a) Force Diagram:
- Gravitational force of the boom \( F_{g,\text{boom}} = m_b g \) directed straight downward at the midpoint (distance \( L/2 \) from the hinge).
- Force from the hanging payload \( F_{\text{payload}} = m_p g \) directed straight downward at the end (distance \( L \) from the hinge).
- Tension force \( F_T \) directed horizontally to the left at distance \( \frac{2}{3}L \).
- Hinge force \( F_{\text{hinge}} \) exerted at the base, pointing upward and to the right to balance the downward gravitational forces and the leftward horizontal tension.

(b) Choosing the hinge as the rotation axis and applying rotational equilibrium (\( \sum \tau = 0 \)):
\( \tau_{\text{clockwise}} = \tau_{\text{counterclockwise}} \)
\( \left(\frac{L}{2}\right) m_b g \cos\beta + L m_p g \cos\beta = \left(\frac{2}{3}L\right) F_T \sin\beta \)
Dividing through by \( L \):
\( \frac{1}{2} m_b g \cos\beta + m_p g \cos\beta = \frac{2}{3} F_T \sin\beta \)
\( \left(\frac{1}{2} m_b + m_p\right) g \cos\beta = \frac{2}{3} F_T \sin\beta \)
Solving for \( F_T \):
\( F_T = \frac{3}{2} \left(\frac{1}{2} m_b + m_p\right) g \frac{\cos\beta}{\sin\beta} = \left(\frac{3}{4} m_b + \frac{3}{2} m_p\right) g \cot\beta \)

(c) Less than.
Justification: The required counter-torque provided by the cable to support the weight of the boom and payload remains unchanged because the positions and weights of the boom and payload are identical. When the cable is angled upward, the angle between the cable and the boom becomes closer to \( 90^\circ \), thereby increasing the perpendicular component of the tension force (or the effective lever arm). Consequently, a smaller tension \( F_{T,\text{new}} \) is needed to produce the same supporting torque.

(d) i. Mass of the rod:
\( M = \int dm = \int_0^L \lambda(x)\,dx = \int_0^{2.0} (1.5 x^2)\,dx = \left[ 0.5 x^3 \right]_0^{2.0} = 0.5 (2.0)^3 = 4.0\text{ kg} \)

ii. Rotational inertia about the pivot:
\( I = \int x^2 dm = \int_0^L x^2 (\lambda(x)\,dx) = \int_0^{2.0} x^2 (1.5 x^2)\,dx = \int_0^{2.0} 1.5 x^4\,dx = \left[ \frac{1.5 x^5}{5} \right]_0^{2.0} = \left[ 0.3 x^5 \right]_0^{2.0} = 0.3 (32) = 9.6\text{ kg}\cdot\text{m}^2 \)

iii. Initial angular acceleration:
\( \tau_{\text{net}} = I \alpha \implies \alpha = \frac{\tau_{\text{net}}}{I} \)
When horizontal, the torque due to gravity on the rod is:
\( \tau_{\text{net}} = \int_0^L x g\,dm = g \int_0^{2.0} x (1.5 x^2)\,dx = 1.5 g \int_0^{2.0} x^3\,dx = 1.5(9.8) \left[ \frac{x^4}{4} \right]_0^{2.0} = 1.5(9.8)(4.0) = 58.8\text{ N}\cdot\text{m} \)
Alternatively, \( x_{\text{cm}} = \frac{\int x dm}{M} = \frac{6.0}{4.0} = 1.5\text{ m} \), so \( \tau_{\text{net}} = M g x_{\text{cm}} = (4.0)(9.8)(1.5) = 58.8\text{ N}\cdot\text{m} \).
\( \alpha = \frac{58.8\text{ N}\cdot\text{m}}{9.6\text{ kg}\cdot\text{m}^2} = 6.125\text{ rad/s}^2 \approx 6.1\text{ rad/s}^2 \)

(a) [3 points total]:
- 1 point: For drawing separate, properly positioned downward gravitational force arrows at \( L/2 \) and \( L \).
- 1 point: For drawing a leftward horizontal tension force at \( 2L/3 \).
- 1 point: For drawing a hinge force vector directed up and to the right at the base, with no extraneous forces.

(b) [4 points total]:
- 1 point: For setting up the condition for rotational equilibrium (\( \sum \tau = 0 \)).
- 1 point: For correct expressions for the torques due to gravity on the boom and the hanging payload (\( \tau_g = \frac{1}{2} L m_b g \cos\beta + L m_p g \cos\beta \)).
- 1 point: For a correct expression for the torque due to cable tension (\( \tau_T = \frac{2}{3} L F_T \sin\beta \)).
- 1 point: For correctly solving for \( F_T = \left(\frac{3}{4} m_b + \frac{3}{2} m_p\right) g \cot\beta \).

(c) [2 points total]:
- 1 point: For selecting 'Less than' with an attempt at justification.
- 1 point: For a correct justification stating that the angle between the cable and boom increases towards \( 90^\circ \) (increasing the perpendicular component of force/lever arm), so less tension is required to supply the same torque.

(d)(i) [2 points total]:
- 1 point: For setting up the integral \( M = \int_0^L \lambda(x) dx \) with correct limits and substituting \( \lambda(x) \).
- 1 point: For the correct numerical value with units: \( M = 4.0\text{ kg} \).

(d)(ii) [2 points total]:
- 1 point: For setting up the integral \( I = \int x^2 dm = \int_0^L x^2 \lambda(x) dx \).
- 1 point: For the correct integration and numerical value with units: \( I = 9.6\text{ kg}\cdot\text{m}^2 \).

(d)(iii) [2 points total]:
- 1 point: For correctly calculating the net torque \( \tau_{\text{net}} = 58.8\text{ N}\cdot\text{m} \) (via integration or \( M g x_{\text{cm}} \)).
- 1 point: For using \( \alpha = \tau_{\text{net}} / I \) to calculate \( \alpha = 6.1\text{ rad/s}^2 \) (or \( 6.13\text{ rad/s}^2 \), with correct units).