2023 AP AP Physics C: Mechanics 模拟试题及答案详解

(a) i.
The displacement is the integral of the velocity function with respect to time:
\[ \Delta x = \int_{0}^{2.0} v(t)\,dt = \int_{0}^{2.0} (30 - 15t^2 + 5t^3)\,dt \]
\[ \Delta x = \left[ 30t - 5t^3 + \frac{5}{4}t^4 \right]_0^2 = 30(2.0) - 5(2.0)^3 + 1.25(2.0)^4 \]
\[ \Delta x = 60 - 40 + 20 = 40\text{ m} \]

(a) ii.
The acceleration is the derivative of the velocity:
\[ a(t) = \frac{dv}{dt} = -30t + 15t^2 \]
To find the maximum magnitude of acceleration, take the time derivative of \( a(t) \) and set it to zero:
\[ \frac{da}{dt} = -30 + 30t = 0 \implies t = 1.0\text{ s} \]
Evaluate the acceleration at \( t = 1.0\text{ s} \):
\[ |a_{\text{max}}| = |-30(1.0) + 15(1.0)^2| = |-15\text{ m/s}^2| = 15\text{ m/s}^2 \]
Using Newton's second law:
\[ F_{\text{max}} = M |a_{\text{max}}| = (800\text{ kg})(15\text{ m/s}^2) = 12{,}000\text{ N} \]

(b) i.
At \( t = 2.0\text{ s} \), the speed of the sled just before collision is:
\[ v_1 = 30 - 15(2.0)^2 + 5(2.0)^3 = 30 - 60 + 40 = 10\text{ m/s} \]
Applying conservation of linear momentum for the perfectly inelastic collision:
\[ M v_1 = (M + m_T) v_f \]
\[ (800\text{ kg})(10\text{ m/s}) = (800\text{ kg} + 1200\text{ kg}) v_f \]
\[ 8000 = 2000 v_f \implies v_f = 4.0\text{ m/s} \]

(b) ii.
Initial kinetic energy before collision:
\[ K_i = \frac{1}{2} M v_1^2 = \frac{1}{2}(800\text{ kg})(10\text{ m/s})^2 = 40{,}000\text{ J} \]
Final kinetic energy after collision:
\[ K_f = \frac{1}{2}(M + m_T) v_f^2 = \frac{1}{2}(2000\text{ kg})(4.0\text{ m/s})^2 = 16{,}000\text{ J} \]
Energy dissipated:
\[ E_{\text{diss}} = K_i - K_f = 40{,}000\text{ J} - 16{,}000\text{ J} = 24{,}000\text{ J} = 2.4 \times 10^4\text{ J} \]

(c) i.
Potential energy stored is the work required to compress the spring:
\[ U_s(x) = \int_0^x F_s(x')\,dx' = \int_0^x \beta x'^3\,dx' = \frac{1}{4}\beta x^4 \]

(c) ii.
By conservation of mechanical energy, all the kinetic energy of the coupled system is converted into spring potential energy at maximum compression:
\[ K_f = U_s(x_{\text{max}}) \]
\[ 16{,}000 = \frac{1}{4}(2.5 \times 10^4) x_{\text{max}}^4 = 6250 x_{\text{max}}^4 \]
\[ x_{\text{max}}^4 = \frac{16{,}000}{6250} = 2.56 \]
\[ x_{\text{max}} = (2.56)^{1/4} = \sqrt{1.6} \approx 1.26\text{ m} \]

(d)
Correct selection: Equal to.
Justification: By the work-energy theorem (\( W_{\text{net}} = \Delta K \)), the work done by the spring on the coupled system is equal to the change in the system's kinetic energy (\( W = 0 - K_f = -16{,}000\text{ J} \)). Since both springs bring the same system with the same initial kinetic energy to rest, the total work done by each spring must be identical regardless of the spring force equation.

Part (a)(i): 2 points
- 1 point: For recognizing that displacement is the integral of velocity over time and applying the integral to the polynomial function.
- 1 point: For the correct numerical answer with correct units (\(40\text{ m}\)).

Part (a)(ii): 3 points
- 1 point: For taking the derivative of the velocity function to find the acceleration function \( a(t) = -30t + 15t^2 \).
- 1 point: For setting the derivative of acceleration to zero (or using vertex/critical point analysis) to determine the time of maximum acceleration (\(t = 1.0\text{ s}\)) and evaluating \(|a_{\text{max}}| = 15\text{ m/s}^2\).
- 1 point: For multiplying \( M \) by \(|a_{\text{max}}|\) to calculate the maximum force with units (\(12{,}000\text{ N}\)).

Part (b)(i): 2 points
- 1 point: For correctly applying conservation of linear momentum with correct initial and final mass terms: \( M v_1 = (M + m_T) v_f \).
- 1 point: For correctly calculating \( v_f = 4.0\text{ m/s} \) using \( v_1 = 10\text{ m/s} \).

Part (b)(ii): 2 points
- 1 point: For calculating initial kinetic energy (\(40{,}000\text{ J}\)) and post-collision kinetic energy (\(16{,}000\text{ J}\)).
- 1 point: For calculating the difference to find dissipated energy with units (\(24{,}000\text{ J}\) or \(2.4 \times 10^4\text{ J}\)).

Part (c)(i): 2 points
- 1 point: For using \( U_s(x) = \int F_s(x)\,dx \) with integration from \( 0 \) to \( x \).
- 1 point: For the correct expression \( U_s(x) = \frac{1}{4}\beta x^4 \).

Part (c)(ii): 2 points
- 1 point: For equating the mechanical energy after collision to the potential energy expression from part (c)(i).
- 1 point: For calculating the correct value of maximum compression with units (\(1.26\text{ m}\) or \(1.27\text{ m}\)).

Part (d): 2 points
- 1 point: For selecting 'Equal to' with an attempt at a relevant justification.
- 1 point: For a valid justification referencing the work-energy theorem or the fact that both springs remove the same amount of kinetic energy from the system.

Part (a)
The rotational inertia of the uniform rod about its end pivot is:
\[ I_{\text{rod}} = \frac{1}{3}ML_0^2 \]
The point mass collar at distance \( d \) has rotational inertia:
\[ I_{\text{collar}} = md^2 \]
Thus, the total rotational inertia is:
\[ I = \frac{1}{3}ML_0^2 + md^2 \]

Part (b)
The distance to the center of mass \( x_{\text{cm}} \) from the pivot is:
\[ x_{\text{cm}} = \frac{M\left(\frac{L_0}{2}\right) + md}{M+m} \]
The restoring torque for small angle \( \theta \) is:
\[ \tau_{\text{net}} = -(M+m)g x_{\text{cm}} \sin\theta \approx -\left(\frac{1}{2}ML_0 + md\right)g\theta \]
Using Newton's second law for rotation, \( \tau_{\text{net}} = I\alpha = I\frac{d^2\theta}{dt^2} \):
\[ I\frac{d^2\theta}{dt^2} + \left(\frac{1}{2}ML_0 + md\right)g\theta = 0 \]
The angular frequency of simple harmonic oscillation is:
\[ \omega_0 = \sqrt{\frac{\left(\frac{1}{2}ML_0 + md\right)g}{I}} \]
Thus, the period \( T = \frac{2\pi}{\omega_0} \) is:
\[ T = 2\pi \sqrt{\frac{I}{\left(\frac{1}{2}ML_0 + md\right)g}} = 2\pi \sqrt{\frac{\frac{1}{3}ML_0^2 + md^2}{\left(\frac{1}{2}ML_0 + md\right)g}} \]

Part (c)
i. Using two representative points on the linear fit, e.g., \( (0.296, 1.19) \) and \( (0.576, 2.32) \):
\[ \text{Slope} = \frac{\Delta Y}{\Delta I} = \frac{2.32 - 1.19}{0.576 - 0.296} = \frac{1.13}{0.280} \approx 4.036\text{ s}^2/\text{m} \]

ii. From the linearized theoretical equation \( Y = \left(\frac{4\pi^2}{g}\right) I \), we have:
\[ \text{Slope} = \frac{4\pi^2}{g} \implies g = \frac{4\pi^2}{\text{Slope}} = \frac{4\pi^2}{4.036} \approx 9.78\text{ m/s}^2 \]

Part (d)
Since \( g = \frac{4\pi^2}{\text{Slope}} \), an overestimated value of \( g \) corresponds to an underestimated slope \( \frac{\Delta Y}{\Delta I} \). A systematic error such as the measured period \( T \) being recorded as smaller than it actually was (e.g., human reaction time delay in stopping a timer vs starting it, or using an improperly calibrated photogate) reduces \( Y \), lowering the slope and leading to an artificially larger value of \( g \).

Part (e)
i. For the rod alone, \( I = \frac{1}{3}ML_0^2 \) and the center of mass is at \( h = \frac{L_0}{2} \). By conservation of mechanical energy:
\[ \Delta U = Mg\frac{L_0}{2}(1 - \cos\theta_{\text{max}}) \approx Mg\frac{L_0}{2}\left(\frac{\theta_{\text{max}}^2}{2}\right) = \frac{1}{4}MgL_0\theta_{\text{max}}^2 \]
Equating this to rotational kinetic energy \( K_{\text{max}} = \frac{1}{2}I\omega_{\text{max}}^2 = \frac{1}{2}\left(\frac{1}{3}ML_0^2\right)\omega_{\text{max}}^2 = \frac{1}{6}ML_0^2\omega_{\text{max}}^2 \):
\[ \frac{1}{6}ML_0^2\omega_{\text{max}}^2 = \frac{1}{4}MgL_0\theta_{\text{max}}^2 \implies \omega_{\text{max}} = \theta_{\text{max}} \sqrt{\frac{3g}{2L_0}} \]
Substituting values (\( L_0 = 1.20\text{ m}, g = 9.8\text{ m/s}^2, \theta_{\text{max}} = 0.10\text{ rad} \)):
\[ \omega_{\text{max}} = 0.10 \sqrt{\frac{3(9.8)}{2(1.20)}} = 0.10 \sqrt{12.25} = 0.10(3.5) = 0.35\text{ rad/s} \]

ii. Since \( \omega_{\text{max}} \propto \sqrt{g} \), if \( g \) increases to \( 2g \), \( \omega_{\text{max}} \) will increase by a factor of \( \sqrt{2} \).

Part (a): 2 points
- 1 point: For correctly applying the rotational inertia of a thin rod about its end (\( \frac{1}{3}ML_0^2 \)).
- 1 point: For adding the point mass rotational inertia (\( md^2 \)) to get the total rotational inertia.

Part (b): 3 points
- 1 point: For writing a correct torque equation or applying Newton's second law for rotation: \( \tau = I\alpha \) with \( \tau = - (M x_{\text{cm, rod}} + m d) g \sin\theta \).
- 1 point: For applying the small-angle approximation \( \sin\theta \approx \theta \) to identify the simple harmonic oscillator form \( \frac{d^2\theta}{dt^2} + \omega^2 \theta = 0 \).
- 1 point: For a correctly derived expression for \( T = 2\pi/\omega \) consistent with the expressions for \( I \) and center of mass torque.

Part (c): 3 points
- 1 point: For calculating the slope using two points on the line/data table (acceptable range: \( 3.9 - 4.2\text{ s}^2/\text{m} \)).
- 1 point: For correctly relating the slope of the line to the physical constant: \( \text{Slope} = \frac{4\pi^2}{g} \).
- 1 point: For a correct numerical calculation of \( g \) with correct units (acceptable range: \( 9.4 - 10.1\text{ m/s}^2 \)).

Part (d): 2 points
- 1 point: For identifying an appropriate source of experimental error (e.g., systematic measurement error in period \( T \), underestimating the mass/length, or friction not accounted for properly changing effective period measurements).
- 1 point: For providing a valid justification connecting the proposed source of error to an underestimated slope, which results in a larger calculated value of \( g \).

Part (e): 5 points
- 1 point: For using conservation of energy (equating potential energy change to rotational kinetic energy) OR using \( \omega_{\text{max}} = \omega_0 \theta_{\text{max}} \).
- 1 point: For correctly substituting rotational inertia \( I = \frac{1}{3}ML_0^2 \) and height change.
- 1 point: For a correct numerical answer for \( \omega_{\text{max}} \) with units (\( 0.35\text{ rad/s} \)).
- 1 point: For selecting 'increase'.
- 1 point: For a valid justification indicating that maximum angular speed is directly proportional to \( \sqrt{g} \) (or that greater gravitational torque/potential energy yields greater kinetic energy).

(a) i. The torque is applied tangentially at the rim \(r = R\):
\[\tau(t) = R F(t) = F_0 R e^{-\gamma t}\]

ii. Applying Newton's second law in rotational form:
\[\tau(t) = I_0 \alpha(t)\]
\[\alpha(t) = \frac{\tau(t)}{I_0} = \frac{F_0 R e^{-\gamma t}}{\frac{1}{2} M R^2} = \frac{2 F_0}{M R} e^{-\gamma t}\]

iii. Using \(\alpha = \frac{d\omega}{dt}\) and integrating with the initial condition \(\omega(0) = 0\):
\[\omega(t) = \int_0^t \alpha(t')\,dt' = \int_0^t \frac{2F_0}{MR} e^{-\gamma t'}\,dt'\]
\[\omega(t) = \frac{2F_0}{MR} \left[ -\frac{1}{\gamma} e^{-\gamma t'} \right]_0^t = \frac{2F_0}{\gamma MR} \left(1 - e^{-\gamma t}\right)\]

(b) i. Conservation of angular momentum about the central vertical axis before and immediately after the clay sticks:
\[L_i = L_f\]
\[I_0 \omega_1 = I_{\text{total}} \omega_f\]
where \(I_0 = \frac{1}{2}MR^2\) and \(I_{\text{total}} = I_0 + m R^2 = \frac{1}{2}MR^2 + \left(\frac{1}{4}M\right)R^2 = \frac{3}{4}MR^2\).
\[\left(\frac{1}{2}MR^2\right) \omega_1 = \left(\frac{3}{4}MR^2\right) \omega_f\]
\[\omega_f = \frac{2}{3}\omega_1 = \frac{2}{3}(4.0\text{ rad/s}) = \frac{8}{3}\text{ rad/s} \approx 2.67\text{ rad/s}\]

ii. The initial rotational kinetic energy:
\[K_i = \frac{1}{2}I_0 \omega_1^2 = \frac{1}{2}\left(\frac{1}{2}MR^2\right)\omega_1^2 = \frac{1}{4}MR^2 \omega_1^2\]
The final rotational kinetic energy:
\[K_f = \frac{1}{2}I_{\text{total}} \omega_f^2 = \frac{1}{2}\left(\frac{3}{4}MR^2\right)\left(\frac{2}{3}\omega_1\right)^2 = \frac{1}{2}\left(\frac{3}{4}MR^2\right)\left(\frac{4}{9}\omega_1^2\right) = \frac{1}{6}MR^2\omega_1^2\]
Thus, the ratio is:
\[\frac{K_f}{K_i} = \frac{\frac{1}{6}}{\frac{1}{4}} = \frac{4}{6} = \frac{2}{3} \approx 0.67\]

(c) The rotational equation of motion during braking is:
\[\tau_{\text{net}} = I_{\text{total}} \alpha = I_{\text{total}} \frac{d\omega}{dt} = -b\omega\]
Using the chain rule \(\frac{d\omega}{dt} = \frac{d\omega}{d\theta}\frac{d\theta}{dt} = \omega \frac{d\omega}{d\theta}\):
\[I_{\text{total}} \omega \frac{d\omega}{d\theta} = -b\omega\]
Dividing both sides by \(\omega\) (for \(\omega > 0\)):
\[I_{\text{total}} d\omega = -b\,d\theta\]
Integrating from initial angular speed \(\omega_f\) to final angular speed \(0\), and angular displacement \(0\) to \(\Delta\theta\):
\[I_{\text{total}} \int_{\omega_f}^0 d\omega = -b \int_0^{\Delta\theta} d\theta\]
\[I_{\text{total}} (0 - \omega_f) = -b \Delta\theta\]
\[\Delta\theta = \frac{I_{\text{total}} \omega_f}{b}\]
Substituting \(I_{\text{total}} = \frac{3}{4}MR^2\):
\[\Delta\theta = \frac{3MR^2\omega_f}{4b}\]

Part (a)(i) (1 point):
- 1 point for a correct expression for the torque on the disk as a function of time: \(\tau(t) = F_0 R e^{-\gamma t}\).

Part (a)(ii) (2 points):
- 1 point for stating or applying Newton's second law in rotational form \(\tau = I\alpha\).
- 1 point for correctly substituting \(I_0 = \frac{1}{2}MR^2\) and simplifying to get \(\alpha(t) = \frac{2F_0}{MR} e^{-\gamma t}\).

Part (a)(iii) (3 points):
- 1 point for expressing angular velocity as the integral of angular acceleration over time \(\omega(t) = \int \alpha(t)\,dt\).
- 1 point for carrying out the integration of an exponential function with correct limits or evaluating an integration constant with \(\omega(0) = 0\).
- 1 point for the correct final expression: \(\omega(t) = \frac{2F_0}{\gamma MR}(1 - e^{-\gamma t})\).

Part (b)(i) (2 points):
- 1 point for applying conservation of angular momentum about the rotation axis (\(L_i = L_f\)) and stating the correct final rotational inertia \(I_{\text{total}} = \frac{3}{4}MR^2\).
- 1 point for calculating the correct numerical value of angular speed with units (\(\omega_f = 2.67\text{ rad/s}\) or \(\frac{8}{3}\text{ rad/s}\)).

Part (b)(ii) (2 points):
- 1 point for using valid expressions for rotational kinetic energy before and after the collision.
- 1 point for the correct numerical or fractional ratio (\(\frac{2}{3}\) or \(0.67\)).

Part (c) (5 points):
- 1 point for writing Newton's second law in rotational form with the resistive torque: \(I_{\text{total}} \alpha = -b\omega\).
- 1 point for expressing \(\alpha\) in terms of derivative with respect to \(\theta\): \(\alpha = \omega \frac{d\omega}{d\theta}\) or equivalently using \(d\theta = \omega\,dt\) and integrating \(\omega(t)\).
- 1 point for separating variables and setting up correct integral limits: \(I_{\text{total}} \int_{\omega_f}^0 d\omega = -b \int_0^{\Delta\theta} d\theta\).
- 1 point for correctly performing the integration to obtain \(\Delta\theta = \frac{I_{\text{total}}\omega_f}{b}\).
- 1 point for substituting \(I_{\text{total}} = \frac{3}{4}MR^2\) to arrive at \(\Delta\theta = \frac{3MR^2\omega_f}{4b}\).