An original Thinka practice paper modelled on the structure and difficulty of the May 2024 AP AP Physics 1: Algebra-Based paper. Not affiliated with or reproduced from AP.
部分 II: Free-Response
Answer all five free-response questions. Show all work, derivations from fundamental physics principles, diagrams, and justifications where prompted.
5 題目 · 45 分
題目 1 · Short Answer
7 分
A small block of mass \(m = 0.50\text{ kg}\) is released from rest at the top of a curved ramp at a vertical height \(h = 0.80\text{ m}\) above a flat horizontal floor. At the base of the ramp, the block slides along a frictionless horizontal track and compresses an ideal horizontal spring with spring constant \(k = 200\text{ N/m}\) anchored to a rigid wall. All friction is negligible unless otherwise stated.
(a) In terms of energy conservation for the block-Earth-spring system: (i) State the forms of mechanical energy present at the instant the block is released, and at the instant the spring reaches maximum compression \(x_{\max}\). (ii) In an energy bar chart where total initial energy is represented by 4 grid units of gravitational potential energy, state the number of grid units of elastic potential energy stored in the spring at maximum compression.
(b) Starting with conservation of energy, derive an expression for the maximum compression \(x_{\max}\) of the spring in terms of \(m\), \(g\), \(h\), and \(k\). Calculate the numerical value of \(x_{\max}\) using \(g = 9.8\text{ m/s}^2\).
(c) The experiment is repeated, but a rough patch of length \(d = 0.30\text{ m}\) with coefficient of kinetic friction \(\mu_k = 0.25\) is introduced on the horizontal surface before the spring. Briefly explain how the work done by friction affects the maximum compression of the spring compared to the frictionless case. Justify your reasoning using fundamental physics principles.
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解題
(a)(i) At release, the block is at rest at height \(h\), so all mechanical energy is stored as gravitational potential energy (\(U_g\)). At maximum compression, the block comes momentarily to rest at height \(y = 0\), so all mechanical energy is stored as elastic potential energy (\(U_s\)). (a)(ii) Since mechanical energy is conserved in the isolated block-Earth-spring system, \(E_{\text{initial}} = E_{\text{final}}\). The 4 grid units of initial \(U_g\) are completely transformed into 4 grid units of \(U_s\).
(c) According to the work-energy theorem, \(\Delta E_{\text{mech}} = W_{\text{other}} = -f_k d = -\mu_k mg d\). The friction force does negative work on the system, dissipating mechanical energy into thermal energy. Therefore, the total mechanical energy available to be converted into elastic potential energy at maximum compression is reduced (\(U_{s,\text{final}} = mgh - \mu_k mg d\)), resulting in a smaller maximum compression \(x_{\max}\).
評分準則
Part (a): 2 points - 1 point: For correctly identifying that initial energy is all gravitational potential energy and final energy at maximum compression is all elastic potential energy. - 1 point: For stating that the elastic potential energy bar has a height equal to the initial gravitational potential energy bar (4 grid units).
Part (b): 2 points - 1 point: For a valid multi-step derivation beginning with conservation of mechanical energy \(mgh = \frac{1}{2}kx^2\). - 1 point: For the correct algebraic expression \(x_{\max} = \sqrt{\frac{2mgh}{k}}\) and correct numerical substitution yielding \(x_{\max} \approx 0.20\text{ m}\) (or \(0.198\text{ m}\)).
Part (c): 3 points - 1 point: For indicating that the maximum compression decreases / is less than in the frictionless case. - 1 point: For explaining that friction does negative work on the block (or dissipates mechanical energy into thermal energy / internal energy). - 1 point: For connecting the reduction in mechanical energy to a smaller final spring potential energy and therefore smaller compression in a logical, coherent response.
題目 2 · Short Answer
7 分
Cart 1 of mass \(3m\) moves to the right with initial speed \(v_0\) along a straight, frictionless horizontal track. Cart 2 of mass \(m\) is initially at rest on the track. Cart 2 is equipped with an ideal massless bumper spring of force constant \(k\). Cart 1 collides head-on with Cart 2, compressing the spring during the interaction.
(a) Determine the speed \(v_{\text{cm}}\) of the center of mass of the two-cart system in terms of \(v_0\).
(b) At the instant of maximum spring compression, both carts move with the same velocity equal to \(v_{\text{cm}}\). Starting with conservation principles, derive an expression for the maximum elastic potential energy \(U_{s,\max}\) stored in the bumper spring in terms of \(m\) and \(v_0\).
(c) After the collision is complete and the carts have fully separated, the spring is fully uncompressed. (i) Indicate whether the final kinetic energy of Cart 2, \(K_{2,f}\), is greater than, less than, or equal to the initial kinetic energy of Cart 1, \(K_{1,i}\). ______ \(K_{2,f} > K_{1,i}\) ______ \(K_{2,f} < K_{1,i}\) ______ \(K_{2,f} = K_{1,i}\) (ii) Briefly justify your choice using physical principles.
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解題
(a) The total mass of the system is \(M_{\text{total}} = 3m + m = 4m\). The total initial momentum of the system is \(p_{\text{total}} = (3m)v_0 + (m)(0) = 3mv_0\). The center of mass velocity is: \(v_{\text{cm}} = \frac{p_{\text{total}}}{M_{\text{total}}} = \frac{3mv_0}{4m} = \frac{3}{4}v_0\).
(b) At maximum compression, both carts move at \(v = v_{\text{cm}} = \frac{3}{4}v_0\). The total initial kinetic energy of the system is: \(K_i = \frac{1}{2}(3m)v_0^2 = \frac{3}{2}mv_0^2\). At maximum compression, the total kinetic energy of the two carts is: \(K_{\text{at max}} = \frac{1}{2}(M_{\text{total}})v_{\text{cm}}^2 = \frac{1}{2}(4m)\left(\frac{3}{4}v_0\right)^2 = 2m \cdot \frac{9}{16}v_0^2 = \frac{9}{8}mv_0^2\). By conservation of mechanical energy: \(K_i = K_{\text{at max}} + U_{s,\max}\) \(U_{s,\max} = K_i - K_{\text{at max}} = \frac{3}{2}mv_0^2 - \frac{9}{8}mv_0^2 = \left(\frac{12}{8} - \frac{9}{8}\right)mv_0^2 = \frac{3}{8}mv_0^2\).
(c)(i) Select: \(K_{2,f} < K_{1,i}\). (c)(ii) In an elastic collision, total kinetic energy is conserved: \(K_{1,i} = K_{1,f} + K_{2,f}\). Because Cart 1 (\(3m\)) is more massive than Cart 2 (\(m\)), Cart 1 continues moving in the positive direction after the collision (\(v_{1,f} = \frac{3m - m}{3m + m}v_0 = \frac{1}{2}v_0 > 0\)), which means Cart 1 retains a positive non-zero final kinetic energy (\(K_{1,f} = \frac{1}{2}(3m)(\frac{1}{2}v_0)^2 = \frac{3}{8}mv_0^2 > 0\)). Since the total initial energy is partitioned between both carts, the kinetic energy of Cart 2 must be strictly less than the initial kinetic energy of Cart 1.
評分準則
Part (a): 1 point - 1 point: For correctly calculating the velocity of the center of mass \(v_{\text{cm}} = \frac{3}{4}v_0\) using conservation of momentum or definition of center of mass velocity.
Part (b): 3 points - 1 point: For correctly calculating the total initial kinetic energy \(K_i = \frac{3}{2}mv_0^2\). - 1 point: For correctly expressing the kinetic energy of both carts at maximum compression in terms of \(v_{\text{cm}}\) (i.e., \(K_{\text{at max}} = \frac{9}{8}mv_0^2\)). - 1 point: For using conservation of mechanical energy to derive \(U_{s,\max} = \frac{3}{8}mv_0^2\).
Part (c): 3 points - 1 point: For selecting \(K_{2,f} < K_{1,i}\) with an attempt at a relevant physics justification. - 1 point: For stating that total kinetic energy is conserved in the elastic collision (\(K_{\text{total},i} = K_{\text{total},f}\)). - 1 point: For reasoning that Cart 1 retains non-zero kinetic energy after colliding with a lighter target cart (\(K_{1,f} > 0\)), thereby leaving \(K_{2,f} < K_{1,i}\).
題目 3 · Experimental Design
12 分
A group of students is tasked with experimentally determining the rotational inertia \(I_0\) of a heavy, uniform cylindrical wheel mounted on a fixed, horizontal axle. The wheel is free to rotate with negligible axle friction. The radius of the wheel, \(R\), is already measured and known.
In addition to the mounted wheel, the students have access only to the following apparatus: - A spool of lightweight, inextensible string - A collection of masses of known values \(m\) - A meterstick - A digital stopwatch
(a) Design an experimental procedure that the students could perform to determine \(I_0\).
i. In the table below, list the quantities to be measured with the available equipment and define a symbol for each.
| Quantity to Be Measured | Symbol | Equipment Used | | :--- | :--- | :--- | | | | | | | | |
ii. Write a clear, step-by-step description of the procedure. Include sufficient detail so that another experimenter could replicate the procedure, including methods to minimize experimental uncertainty.
(b) i. Specify the quantities that should be plotted on the vertical and horizontal axes to yield a linear graph that can be used to find \(I_0\).
ii. Explain how the slope of the linear best-fit line would be analyzed to calculate \(I_0\).
(c) In a follow-up trial, the students attach a known mass \(m = 0.50\text{ kg}\) to the string wrapped around the wheel (radius \(R = 0.12\text{ m}\)). The mass is released from rest at \(t = 0\), and an ultrasonic motion sensor records its downward velocity \(v\) over time. The best-fit line of the velocity-time data yields a constant downward acceleration of \(a = 1.40\text{ m/s}^2\).
i. Using the values provided and taking \(g = 9.8\text{ m/s}^2\), calculate the experimental value of the wheel's rotational inertia \(I_0\).
ii. Suppose that there was actually a small, constant frictional torque exerted by the axle during this trial. Would the value of \(I_0\) calculated in part (c)(i) be greater than, less than, or equal to the true rotational inertia of the wheel? Briefly justify your reasoning.
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解題
Part (a)
Measurements: - Vertical release height of the hanging mass: symbol \(h\), measured using the meterstick. - Time of fall of the hanging mass: symbol \(t\), measured using the stopwatch. - Mass of the hanging weight: symbol \(m\), known values.
Procedure: 1. Wrap the lightweight string securely around the outer rim of the wheel. 2. Attach a known mass \(m\) to the free end of the string. 3. Using the meterstick, measure a fixed vertical distance \(h\) from the bottom of the suspended mass to the floor. 4. Release the mass from rest and simultaneously start the stopwatch. 5. Stop the timer at the instant the mass hits the floor, recording time \(t\). 6. Repeat the timed drop for a minimum of 3 trials for the same mass \(m\) to compute an average fall time \(t_{\text{avg}}\), thereby reducing random timing error. 7. Repeat steps 2–6 for several different mass values \(m\) while keeping the drop height \(h\) constant.
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Part (b)
From kinematics: \(h = \frac{1}{2} a t^2 \implies a = \frac{2h}{t^2}\). Applying Newton's second law for the system: \(mg - T = ma\) \(\tau_{\text{net}} = T R = I_0 \alpha = I_0 \left(\frac{a}{R}\right) \implies T = I_0 \frac{a}{R^2}\) Substituting \(T\): \(mg - I_0 \frac{a}{R^2} = ma \implies mg = a\left(m + \frac{I_0}{R^2}\right) \implies \frac{2h}{t^2} = \frac{mg}{m + I_0/R^2}\). Alternatively, using work-energy for the system: \(mgh = \frac{1}{2}mv^2 + \frac{1}{2}I_0\omega^2 = \frac{1}{2}m(at)^2 + \frac{1}{2}I_0\left(\frac{at}{R}\right)^2 = \frac{1}{2}a^2 t^2 \left(m + \frac{I_0}{R^2}\right) = ah\left(m + \frac{I_0}{R^2}\right)\). Thus: \(m(g - a) = \left(\frac{I_0}{R^2}\right)a\).
If we plot \(m(g - a)\) or tension proxy on the vertical axis versus \(a\) on the horizontal axis: - Vertical axis: \(m(g - a)\) [where \(a = \frac{2h}{t^2}\)] - Horizontal axis: \(a\) - Slope: \(\text{Slope} = \frac{I_0}{R^2} \implies I_0 = (\text{Slope}) R^2\).
*(Alternatively, plotting \(m\) on the vertical axis vs. \(\frac{a}{g-a}\) gives a slope of \(\frac{I_0}{R^2}\), etc.)*
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Part (c)
i. Applying the dynamic equations: \(T = m(g - a) = (0.50\text{ kg})(9.8\text{ m/s}^2 - 1.40\text{ m/s}^2) = (0.50)(8.40) = 4.20\text{ N}\).
ii. Greater than the true value. Reasoning: Axle friction creates an opposing torque \(\tau_f\), so the net torque is \(\tau_{\text{net}} = T R - \tau_f = I_{\text{true}}\alpha\). The calculation in (c)(i) assumed \(T R = I_0 \alpha\), meaning \(I_0 = \frac{TR}{\alpha} = I_{\text{true}} + \frac{\tau_f}{\alpha}\). Since friction reduced the angular acceleration \(\alpha\) for a given string tension, attributing the entire string tension torque to accelerating the wheel overestimates the rotational inertia.
評分準則
Part (a): 4 points total - 1 point: For listing valid measurable quantities (fall distance \(h\) and fall time \(t\)) with appropriate equipment (meterstick and stopwatch). - 1 point: For a clear, viable procedure that describes releasing the mass from rest from a measured height and recording the time taken to fall. - 1 point: For describing trials using multiple different values of mass \(m\). - 1 point: For describing an explicit technique to reduce experimental uncertainty (e.g., performing multiple timing trials per mass to compute an average time, or measuring a sufficiently large drop distance to reduce timing error).
Part (b): 3 points total - 1 point: For identifying variables to plot on the vertical and horizontal axes that yield a linear relationship suitable for finding \(I_0\). - 1 point: For correctly identifying the physical meaning of the slope in terms of relevant parameters. - 1 point: For an algebraic expression correctly relating \(I_0\) to the slope of the line (e.g., \(I_0 = (\text{slope})R^2\)).
Part (c): 5 points total - (c)(i) 2 points: - 1 point: For applying Newton's second law in rotational and translational form (or conservation of energy) to relate \(a\), \(m\), \(R\), and \(I_0\) (e.g., \(I_0 = \frac{m(g-a)R^2}{a}\)). - 1 point: For substituting the given numerical values and obtaining \(I_0 \approx 0.043\text{ kg}\cdot\text{m}^2\) (accept values between \(0.043\) and \(0.044\text{ kg}\cdot\text{m}^2\)). - (c)(ii) 3 points: - 1 point: For selecting "Greater than". - 1 point: For identifying that frictional torque opposes rotation, causing the acceleration \(a\) (or \(\alpha\)) to be smaller than it would be in the frictionless case. - 1 point: For a complete, consistent explanation linking the smaller acceleration / unaccounted resistive torque to an overestimate of \(I_0\).
題目 4 · Qualitative/Quantitative Translation
12 分
A uniform rigid rod of mass \( M \) and length \( L \) is attached to a horizontal floor by a frictionless hinge at its lower end. The rod is held at an angle \( \theta_1 \) above the horizontal floor by a light horizontal cable attached to the top end of the rod and to a vertical wall located to the left of the hinge. The rod remains at rest in static equilibrium.
(a) On the diagram representing the rod, draw and label the forces (not components) exerted on the rod. Draw each force as a distinct arrow starting on, and pointing away from, the point of application on the rod.
(b) The horizontal cable is disconnected and reattached so that the rod is held in static equilibrium at a larger angle \( \theta_2 > \theta_1 \) above the horizontal. The cable remains horizontal.
Indicate whether the magnitude of the tension in the cable when the rod is at angle \( \theta_2 \) (\( F_{T2} \)) is greater than, less than, or equal to the tension when the rod is at angle \( \theta_1 \) (\( F_{T1} \)).
Briefly justify your choice using qualitative physical reasoning beyond simply quoting equations.
(c) Starting with Newton's second law in rotational form, derive an expression for the magnitude of the tension \( F_T \) in the cable in terms of \( M \), \( L \), \( \theta \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference sheet.
(d) Explain how your derived expression in part (c) is consistent with your qualitative reasoning in part (b).
(e) The horizontal cable is suddenly cut while the rod is at an initial angle \( \theta \) above the horizontal, and the rod rotates downward toward the floor about the frictionless hinge. On axes of Angular Speed \( \omega \) versus Time \( t \), sketch the angular speed of the rod as a function of time from the moment the cable is cut until the instant just before the rod hits the horizontal floor.
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解題
Part (a): Free-Body Diagram - **Gravitational Force (\( F_g \) or \( Mg \)):** Drawn at the geometric center of the uniform rod (\( L/2 \) from the hinge), pointing vertically downward. - **Tension Force (\( F_T \)):** Drawn at the top end of the rod, pointing horizontally to the left. - **Hinge Force (\( F_H \)):** Drawn at the hinge (bottom end of the rod), pointing upward and to the right. (The horizontal component balances the leftward tension force, and the vertical component balances the downward gravitational force).
Part (b): Qualitative Comparison and Justification - Correct selection: **\( F_{T2} < F_{T1} \)** - Justification: As the angle \( \theta \) with the horizontal increases, the center of mass of the rod shifts closer horizontally to the pivot, reducing the perpendicular lever arm for the gravitational force. Consequently, the clockwise torque produced by gravity decreases. At the same time, the top end of the rod is higher, so the lever arm for the horizontal tension force (the vertical height \( L\sin\theta \)) increases. Since rotational equilibrium requires the counterclockwise torque from the cable to balance the smaller gravitational torque with a larger lever arm, the required tension in the cable must decrease.
Part (c): Derivation of Tension - Fundamental principle: \( \sum \tau = I\alpha \) - For static rotational equilibrium, \( \alpha = 0 \implies \sum \tau = 0 \). - Taking torques about the hinge pivot at the bottom: - Lever arm for gravity = \( \frac{L}{2}\cos\theta \), producing torque \( \tau_g = Mg\left(\frac{L}{2}\cos\theta\right) \) (clockwise). - Lever arm for tension = \( L\sin\theta \), producing torque \( \tau_T = F_T (L\sin\theta) \) (counterclockwise). - Setting the sum of torques to zero: \[ F_T(L\sin\theta) - Mg\left(\frac{L}{2}\cos\theta\right) = 0 \] \[ F_T L\sin\theta = \frac{1}{2}MgL\cos\theta \] \[ F_T = \frac{Mg\cos\theta}{2\sin\theta} = \frac{Mg}{2\tan\theta} \]
Part (d): Consistency Analysis - In the derived equation \( F_T = \frac{Mg}{2\tan\theta} \), as \( \theta \) increases between \( 0^\circ \) and \( 90^\circ \), \( \tan\theta \) strictly increases (or \( \cos\theta \) decreases while \( \sin\theta \) increases). Since \( \tan\theta \) is in the denominator, an increase in \( \theta \) causes \( F_T \) to decrease, which perfectly matches the qualitative conclusion in part (b) that \( F_{T2} < F_{T1} \).
Part (e): Angular Speed vs. Time Graph - The graph starts at \( \omega = 0 \) at \( t = 0 \). - As the rod falls toward the horizontal floor, the angle \( \theta \) decreases toward \( 0 \), which increases the lever arm of gravity (\( \frac{L}{2}\cos\theta \)) and thus increases the net gravitational torque \( \tau = \frac{1}{2}MgL\cos\theta \). - Because \( \alpha = \frac{\tau}{I} = \frac{d\omega}{dt} \), the angular acceleration increases over time as the rod falls. - Therefore, the curve of \( \omega \) versus \( t \) is strictly increasing and concave up (increasing slope).
評分準則
Part (a) — 3 points total - 1 point: For drawing a downward gravitational force arrow starting at the center of the rod and labeled appropriately (\( F_g \), \( Mg \), \( W \)). - 1 point: For drawing a horizontal tension force arrow directed leftward starting at the top end of the rod and labeled appropriately (\( F_T \), \( T \), \( F_{\text{cable}} \)). - 1 point: For drawing a hinge force arrow starting at the bottom pivot directed upward and to the right representing a system in translational equilibrium.
Part (b) — 2 points total - 1 point: For selecting \( F_{T2} < F_{T1} \) with an attempt at a physical justification. - 1 point: For a valid qualitative justification explaining that the gravitational torque decreases (due to reduced horizontal distance/lever arm to the pivot) and/or the tension lever arm increases (greater vertical height).
Part (c) — 3 points total - 1 point: For beginning with Newton's second law in rotational form (\( \sum \tau = I\alpha \) or \( \sum \tau = 0 \)). - 1 point: For writing correct torque expressions about the hinge for both gravity (\( Mg\frac{L}{2}\cos\theta \)) and tension (\( F_T L\sin\theta \)). - 1 point: For correctly solving for \( F_T \) to obtain \( F_T = \frac{Mg\cos\theta}{2\sin\theta} \) or \( F_T = \frac{Mg}{2\tan\theta} \).
Part (d) — 2 points total - 1 point: For attempting to connect the functional dependence of angle in the mathematical expression to the physical reasoning in part (b). - 1 point: For correctly explaining that because \( \tan\theta \) increases (or \( \sin\theta \) increases and \( \cos\theta \) decreases) as \( \theta \) increases, the derived formula dictates that \( F_T \) must decrease, confirming the qualitative reasoning in part (b).
Part (e) — 2 points total - 1 point: For drawing a graph of angular speed \( \omega(t) \) that starts at the origin \( (0,0) \) and is monotonically increasing. - 1 point: For drawing a curve that is concave up (slope increases with time), reflecting increasing angular acceleration as the rod falls toward the horizontal.
題目 5 · Paragraph Argument Short Answer
7 分
A block of mass \(M\) is attached to an ideal horizontal spring of spring constant \(k\) on a frictionless horizontal surface. The other end of the spring is fixed to a rigid wall. The block is pulled and released, executing simple harmonic motion with an amplitude \(A_0\) and total mechanical energy \(E_0\).
(a) In Experiment 1, a piece of clay of mass \(m\) is dropped vertically onto the block at the exact instant the block reaches its maximum displacement \(x = A_0\). The clay immediately sticks to the block without slipping. Is the new amplitude \(A_1\) of the oscillation greater than, less than, or equal to \(A_0\)? Briefly justify your answer.
(b) In Experiment 2, an identical system is set up such that the block again oscillates with initial amplitude \(A_0\) and initial mechanical energy \(E_0\). In this experiment, the piece of clay of mass \(m\) is dropped vertically onto the block at the exact instant the block is passing through its equilibrium position (\(x = 0\)) with speed \(v_0\). The clay sticks to the block, and the new total mechanical energy of the system is \(E_2\).
In a clear, coherent, paragraph-length response that may also contain equations, explain why the total mechanical energy \(E_2\) in Experiment 2 is less than \(E_0\), whereas the total mechanical energy \(E_1\) in Experiment 1 is equal to \(E_0\). Your response should address the principles of momentum and energy as they apply to both experiments.
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解題
Part (a):
The new amplitude \(A_1\) is equal to \(A_0\).
Justification: At the instant of maximum displacement \(x = A_0\), the block is momentarily at rest (\(v = 0\)). When the clay is dropped vertically onto the block, neither object has horizontal motion, so no horizontal collision forces or relative slipping occur. The total mechanical energy of the system is stored entirely as elastic potential energy in the spring: \[ E_1 = \frac{1}{2} k A_0^2 = E_0 \] Since the total mechanical energy is unchanged and equals \(\frac{1}{2} k A_1^2\), the amplitude must remain \(A_1 = A_0\).
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Part (b):
In Experiment 1, at the amplitude \(x = A_0\), all energy is stored as spring potential energy \(E_0 = \frac{1}{2}kA_0^2\) and the speed of the block is zero. When the clay of mass \(m\) drops vertically, there is no horizontal velocity for either object before or after the clay lands. Consequently, no work is done against horizontal nonconservative forces (such as friction or inelastic deformation), and the spring potential energy remains \(E_1 = \frac{1}{2}kA_0^2 = E_0\).
In contrast, in Experiment 2, at the equilibrium position \(x = 0\), the spring potential energy is zero and all mechanical energy is in the form of kinetic energy \(E_0 = \frac{1}{2}Mv_0^2\). When the clay drops onto the moving block, a completely inelastic collision occurs in the horizontal direction. Because no external horizontal forces act during the collision, horizontal linear momentum is conserved: \[ M v_0 = (M + m) v_2 \implies v_2 = \left(\frac{M}{M + m}\right) v_0 < v_0 \] The resulting kinetic energy (and total mechanical energy) at the equilibrium position is: \[ E_2 = \frac{1}{2}(M + m)v_2^2 = \frac{1}{2}(M + m)\left(\frac{M}{M + m}v_0\right)^2 = \left(\frac{M}{M + m}\right)\left(\frac{1}{2}Mv_0^2\right) = \left(\frac{M}{M + m}\right)E_0 < E_0 \] During this inelastic horizontal interaction, mechanical energy is transformed into thermal energy/internal energy as horizontal forces accelerate the clay up to the speed of the block. Therefore, \(E_2 < E_0\).
評分準則
### Question 4 (Paragraph Argument Short Answer) — 7 points
#### Part (a): 2 points - 1 point: For indicating that \(A_1 = A_0\) (or selecting 'equal to') and stating that the speed of the block at maximum displacement is zero. - 1 point: For a correct justification indicating that all initial mechanical energy is stored as spring potential energy \(\frac{1}{2}kA_0^2\) and no mechanical energy is lost because there is no horizontal relative motion/collision.
#### Part (b): 5 points - 1 point: For explaining that in Experiment 2, the clay and block undergo a completely inelastic collision in the horizontal direction where horizontal momentum is conserved. - 1 point: For stating/showing that the horizontal speed of the combined mass decreases (\(v_2 < v_0\)) due to the increase in total moving mass. - 1 point: For linking the reduction in velocity in Experiment 2 to a loss of kinetic energy (mechanical energy dissipated as thermal/internal energy) at the equilibrium position where all mechanical energy is kinetic. - 1 point: For explaining that in Experiment 1, the block has zero velocity at maximum displacement, so no horizontal inelastic collision occurs and all energy remains in the spring as elastic potential energy (\(E_1 = \frac{1}{2}kA_0^2 = E_0\)). - 1 point: For a logical, relevant, and internally consistent paragraph-length response that addresses both experiments and follows the required paragraph response structure.