An original Thinka practice paper modelled on the structure and difficulty of the May 2023 AP AP Physics 1: Algebra-Based paper. Not affiliated with or reproduced from AP.
部分 II: Free-Response
Answer all 5 questions. Show all work, derivations, and reasoning. Calculator and AP Physics 1 equation sheet permitted.
5 題目 · 45 分
題目 1 · Short Answer
7 分
A small sphere of mass \(m_0\) is attached to a light, nonstretching string of length \(L\) hung from the ceiling to form a simple pendulum. The sphere is pulled back until the string makes a small angle \(\theta_0\) with the vertical and is released from rest at time \(t = 0\). Air resistance and the mass of the string are negligible.
(a) The period of oscillation of this pendulum is \(T_1\). The original sphere is replaced with a sphere of mass \(4m_0\), and the string is replaced with one of length \(4L\). Calculate the new period of oscillation \(T_2\) in terms of \(T_1\).
(b) As the sphere swings from its release point at \(\theta_0\) to the lowest point of its path (\(\theta = 0\)), describe how the magnitude of the net restoring force on the sphere changes. Justify your answer using physics principles.
(c) i. Derive an expression for the maximum speed \(v_{\text{max}}\) of the sphere of mass \(m_0\) as it passes through the lowest point in terms of \(g\), \(L\), and \(\theta_0\). ii. If the initial release angle is doubled from \(\theta_0\) to \(2\theta_0\) (where the angle still remains small enough for the small-angle approximation \(\cos\theta \approx 1 - \frac{\theta^2}{2}\) to hold), by what factor does the total mechanical energy of the sphere-Earth system change? Briefly explain your reasoning.
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解題
(a) The period of a simple pendulum executing small-angle oscillations is given by \(T = 2\pi \sqrt{\frac{L}{g}}\). The period depends only on the length \(L\) and gravitational acceleration \(g\), and is independent of mass \(m\). When the length becomes \(4L\): \[ T_2 = 2\pi \sqrt{\frac{4L}{g}} = 2\left(2\pi \sqrt{\frac{L}{g}}\right) = 2T_1 \]
(b) The restoring force acting on the pendulum bob is the tangential component of the gravitational force, given by \(F_{\text{restoring}} = m_0 g \sin\theta\). As the sphere moves from \(\theta = \theta_0\) to \(\theta = 0\), the angle \(\theta\) continuously decreases toward zero. Because \(\sin\theta\) decreases as \(\theta\) decreases, the restoring force decreases continuously throughout this portion of the swing, reaching zero at the lowest point (equilibrium position).
(c) i. Taking the lowest point as the reference level for gravitational potential energy (\(U_g = 0\)), the initial height relative to the lowest point is \(h = L - L\cos\theta_0 = L(1 - \cos\theta_0)\). Applying conservation of mechanical energy to the sphere-Earth system: \[ E_i = E_f \] \[ m_0 g h = \frac{1}{2} m_0 v_{\text{max}}^2 \] \[ g L (1 - \cos\theta_0) = \frac{1}{2} v_{\text{max}}^2 \] \[ v_{\text{max}} = \sqrt{2gL(1 - \cos\theta_0)} \]
ii. Using the small-angle approximation \(1 - \cos\theta \approx \frac{\theta^2}{2}\), the total mechanical energy is: \[ E = m_0 g L(1 - \cos\theta) \approx \frac{1}{2} m_0 g L \theta^2 \] Because energy is proportional to the square of the angular amplitude (\(E \propto \theta^2\)), doubling the amplitude from \(\theta_0\) to \(2\theta_0\) increases the total mechanical energy by a factor of \(2^2 = 4\).
評分準則
(a) [2 points total] • 1 point: For correctly identifying or applying the relationship \(T = 2\pi\sqrt{L/g}\) and recognizing that the period is independent of mass. • 1 point: For the correct calculation showing \(T_2 = 2T_1\).
(b) [2 points total] • 1 point: For stating that the restoring force decreases (or reaches zero at the bottom). • 1 point: For a valid justification referencing the tangential component of gravity (\(F_{\text{net}} = mg\sin\theta\)) decreasing as angle \(\theta\) decreases.
(c)(i) [2 points total] • 1 point: For a valid statement of conservation of energy with correct expression for the vertical displacement \(h = L(1-\cos\theta_0)\). • 1 point: For correctly solving for \(v_{\text{max}} = \sqrt{2gL(1 - \cos\theta_0)}\).
(c)(ii) [1 point total] • 1 point: For stating that the energy increases by a factor of 4 with valid reasoning connecting energy to the square of the amplitude (\(E \propto \theta_0^2\)).
題目 2 · Short Answer
7 分
A uniform thin rod of mass \(M\) and length \(D\) is free to rotate in a horizontal plane about a frictionless vertical pivot fixed at one of its ends. The rotational inertia of the rod about this end pivot is \(I_{\text{rod}} = \frac{1}{3}MD^2\). The rod is initially at rest on a frictionless horizontal table. A small piece of sticky clay of mass \(m\) slides along the table with speed \(v_0\) in a direction perpendicular to the rod. The clay strikes the free end of the rod (a distance \(D\) from the pivot) and sticks to it immediately.
(a) Write an expression for the magnitude of the angular momentum \(L_i\) of the clay-rod system about the pivot immediately before the collision in terms of \(m\), \(v_0\), and \(D\).
(b) Derive an expression for the angular speed \(\omega\) of the clay-rod system immediately after the collision in terms of \(m\), \(M\), \(v_0\), and \(D\).
(c) Is mechanical energy conserved in the collision between the clay and the rod? Briefly justify your answer using physics principles.
(d) Suppose the experiment is repeated under identical conditions, except the clay ball strikes and sticks to the rod at its midpoint (a distance \(D/2\) from the pivot) with the same initial speed \(v_0\). Is the resulting angular speed immediately after the collision greater than, less than, or equal to the angular speed obtained in part (b)? Justify your answer without purely relying on algebraic re-derivation.
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解題
(a) Before the collision, the rod is stationary, so its angular momentum is zero. The clay behaves as a point mass moving at distance \(D\) perpendicular to the pivot. Therefore, the angular momentum of the system about the pivot is: \[ L_i = m v_0 D \]
(b) During the collision, the pivot exerts an external horizontal force on the rod, but this force produces zero torque about the pivot axis because its lever arm is zero. Therefore, angular momentum about the pivot is conserved: \[ L_i = L_f \] \[ m v_0 D = I_{\text{total}} \omega \] The total rotational inertia after the collision is the sum of the rotational inertia of the rod and that of the stuck clay (treated as a point particle at distance \(D\)): \[ I_{\text{total}} = I_{\text{rod}} + I_{\text{clay}} = \frac{1}{3}MD^2 + m D^2 = \left(\frac{1}{3}M + m\right)D^2 \] Substituting into the conservation equation: \[ m v_0 D = \left(\frac{1}{3}M + m\right)D^2 \omega \] \[ \omega = \frac{m v_0 D}{\left(\frac{1}{3}M + m\right)D^2} = \frac{m v_0}{\left(\frac{1}{3}M + m\right)D} \]
(c) No, mechanical (kinetic) energy is not conserved. The collision is completely inelastic because the clay sticks to the rod. Non-conservative deformation forces do work during the collision, converting mechanical kinetic energy into thermal energy and internal strain energy.
(d) Selection: Greater than. Justification: When the clay hits at \(D/2\), the initial angular momentum is halved (\(L_i' = m v_0 (D/2) = \frac{1}{2} L_i\)). However, the rotational inertia contributed by the clay depends on the square of the distance from the pivot, decreasing by a factor of 4 (from \(mD^2\) to \(m(D/2)^2 = \frac{1}{4}mD^2\)). Consequently, the denominator \(I_{\text{total}}\) decreases more substantially relative to the numerator in \(\omega = \frac{L}{I}\), resulting in a larger angular speed.
評分準則
(a) [1 point total] • 1 point: For the correct expression \(L_i = mv_0 D\).
(b) [3 points total] • 1 point: For using conservation of angular momentum: \(L_i = L_f\) or \(mv_0 D = I_{\text{total}}\omega\). • 1 point: For correctly finding \(I_{\text{total}} = \frac{1}{3}MD^2 + mD^2\). • 1 point: For a complete and correct algebraic derivation yielding \(\omega = \frac{mv_0}{(\frac{1}{3}M + m)D}\).
(c) [1 point total] • 1 point: For stating that mechanical energy is not conserved because the inelastic collision converts mechanical energy into internal/thermal energy.
(d) [2 points total] • 1 point: For selecting 'Greater than'. • 1 point: For a valid justification explaining that although initial angular momentum decreases by a factor of 2, the clay's contribution to rotational inertia decreases by a factor of 4 (\(r^2\) dependence), which leads to an overall increase in angular velocity \(\omega = L/I\).
題目 3 · long-free-response
12 分
Students conduct an experiment to determine the magnitude of the constant acceleration \(a\) of a block as it slides to rest across a horizontal tabletop due to kinetic friction. A motion sensor positioned at the release point measures the initial launch speed \(v_0\) of the block at time \(t = 0\). The \(+x\)-axis is defined along the track in the direction of motion with its origin at the release point. The block slides along the track until it comes to rest after traveling a stopping distance \(\Delta x\). The students collect the data shown in the table below.
(a) i. Indicate which quantities could be graphed to yield a straight line whose slope could be used to determine the acceleration \(a\) of the block. You may use the remaining columns in the table, as needed, to record any quantities (including units) that are not already in the table.
ii. Plot the appropriate quantities to create a graph that can be used to determine the magnitude of the acceleration \(a\) of the block. Clearly scale and label all axes (including units), as appropriate. Draw a straight line that best represents the data.
iii. Using the best-fit line you drew in part (a)(ii), calculate an experimental value for the magnitude of the acceleration \(a\) of the block.
(b) The students are asked to determine an experimental value for the coefficient of kinetic friction \(\mu_{\text{exp}}\) between the block and the tabletop. i. What additional physical quantity do the students need to know or measure in order to calculate \(\mu_{\text{exp}}\) from \(a\)?
ii. Write an expression for \(\mu_{\text{exp}}\) in terms of \(a\) and any necessary fundamental physical constants.
(c) The students calculate the value of \(\mu_{\text{exp}}\) to be significantly greater than the manufacturer's listed coefficient of friction for the smooth surface. i. What is a physical reason, other than air resistance or human error, that could lead to this difference in the experimentally determined value of \(\mu_{\text{exp}}\)?
ii. Briefly explain how the physical reason you identified in part (c)(i) would lead to an increase in the experimentally determined value of \(\mu_{\text{exp}}\).
(d) On the axes below, sketch the position \(x\) and velocity \(v\) of the block as functions of time \(t\) from the moment of release (\(t = 0\)) until the block comes to rest at time \(t_{\text{stop}}\).
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解題
Part (a)(i) From the kinematic equation \(v^2 = v_0^2 + 2a_x \Delta x\), since the final velocity is \(v = 0\) and the acceleration is directed opposite to the motion (\(a_x = -a\)): \[0 = v_0^2 - 2a\Delta x \implies v_0^2 = 2a\Delta x\] Plotting \(v_0^2\) on the vertical axis versus \(\Delta x\) on the horizontal axis yields a straight line with slope equal to \(2a\). *(Alternatively, plotting \(\Delta x\) on the vertical axis versus \(v_0^2\) on the horizontal axis yields a straight line with slope equal to \(\frac{1}{2a}\).)*
Calculated values of \(v_0^2\): - For \(v_0 = 1.20\text{ m/s}\), \(v_0^2 = 1.44\text{ m}^2/\text{s}^2\) - For \(v_0 = 1.80\text{ m/s}\), \(v_0^2 = 3.24\text{ m}^2/\text{s}^2\) - For \(v_0 = 2.20\text{ m/s}\), \(v_0^2 = 4.84\text{ m}^2/\text{s}^2\) - For \(v_0 = 2.60\text{ m/s}\), \(v_0^2 = 6.76\text{ m}^2/\text{s}^2\) - For \(v_0 = 3.00\text{ m/s}\), \(v_0^2 = 9.00\text{ m}^2/\text{s}^2\)
Part (a)(ii) - Horizontal axis: \(\Delta x\text{ (m)}\), scaled from \(0\) to \(2.0\text{ m}\). - Vertical axis: \(v_0^2\text{ (m}^2/\text{s}^2\text{)}\), scaled from \(0\) to \(10.0\text{ m}^2/\text{s}^2\). - Data points are plotted accurately, and a straight line of best fit is drawn through the data.
Part (a)(iii) Choosing two points on the best-fit line: \[\text{Slope} = \frac{\Delta(v_0^2)}{\Delta(\Delta x)} = \frac{9.00 - 1.44}{1.85 - 0.30} = \frac{7.56}{1.55} \approx 4.88\text{ m/s}^2\] Since \(\text{Slope} = 2a\): \[a = \frac{\text{Slope}}{2} = \frac{4.88}{2} = 2.44\text{ m/s}^2\]
Part (b)(i) The acceleration due to gravity, \(g\) (or \(9.8\text{ m/s}^2\)).
Part (b)(ii) Using Newton's second law along the horizontal direction: \[\Sigma F_x = f_k = m a \implies \mu_{\text{exp}} m g = m a\] \[\mu_{\text{exp}} = \frac{a}{g}\]
Part (c)(i) The table was slightly tilted upward (incline in the direction of motion), so gravity has a component pointing down the ramp opposing the motion.
Part (c)(ii) If the table is tilted upward at an angle \(\theta\), the net deceleration is \(a = \mu g \cos\theta + g \sin\theta\). The measured deceleration \(a\) is greater than \(\mu_k g\) alone. When computing \(\mu_{\text{exp}} = \frac{a}{g}\), this additional deceleration from gravity is attributed entirely to friction, resulting in an experimentally determined coefficient of friction that is larger than the true value.
Part (d) - \(x(t)\) graph: Starts at \(x = 0\) with a positive initial slope, curves downward (concave down) due to negative acceleration, and reaches a horizontal slope at \(t = t_{\text{stop}}\). - \(v(t)\) graph: Starts at \(v = +v_0\) at \(t = 0\), decreases linearly with a constant negative slope, and reaches \(v = 0\) at \(t = t_{\text{stop}}\).
評分準則
Part (a)(i): (1 point) - 1 point for indicating two quantities that, when graphed together, produce a straight line whose slope can be used to determine the acceleration (e.g., $v_0^2$ vs. $\Delta x$ or $\Delta x$ vs. $v_0^2$).
Part (a)(ii): (3 points) - 1 point for having linear, appropriately scaled axes with labels and units such that the plotted points occupy more than half of the grid in both dimensions. - 1 point for plotting at least 4 of the data pairs correctly according to the chosen variables. - 1 point for drawing a single best-fit line that represents the trend of the plotted data.
Part (a)(iii): (2 points) - 1 point for correctly calculating the slope of the best-fit line using two points on the line (not necessarily original data points). - 1 point for using a correct kinematic relationship relating the calculated slope to the magnitude of acceleration $a$ (e.g., $a = \text{slope}/2$ if plotting $v_0^2$ vs. $\Delta x$).
Part (b)(i): (1 point) - 1 point for identifying the acceleration due to gravity $g$ (or Earth's gravitational field strength) as the necessary quantity.
Part (b)(ii): (1 point) - 1 point for a correct algebraic expression relating $\mu_{\text{exp}}$ to $a$ and $g$: $\mu_{\text{exp}} = \frac{a}{g}$.
Part (c)(i): (1 point) - 1 point for identifying a plausible physical factor other than air resistance or human error (e.g., the tabletop is inclined upward in the direction of motion, or the track surface roughness increases along the path).
Part (c)(ii): (1 point) - 1 point for correctly explaining the functional dependence: explaining how the identified physical factor increases the net stopping force/acceleration, thereby yielding a calculated value of $\mu_{\text{exp}} = a/g$ that is larger than the true surface friction coefficient.
Part (d): (2 points) - 1 point for sketching a position vs. time graph that is concave down, starts at the origin with a positive slope, and levels off to zero slope at $t_{\text{stop}}$. - 1 point for sketching a velocity vs. time graph that begins at a positive value and is a straight line with a constant negative slope reaching $v = 0$ at $t_{\text{stop}}$, consistent with constant deceleration.
題目 4 · Qualitative/Quantitative Translation
12 分
Two blocks of unequal masses, Block 1 of mass \(M\) and Block 2 of mass \(m\) (where \(M > m\)), are connected by a light, inextensible string that passes over an ideal, massless, and frictionless pulley.
Scenario 1: Block 1 of mass \(M\) is placed on a smooth horizontal table where friction is negligible, and Block 2 of mass \(m\) hangs vertically over the edge of the table. The system is held at rest and then released.
Scenario 2: The blocks are swapped so that Block 2 of mass \(m\) is placed on the smooth horizontal table, and Block 1 of mass \(M\) hangs vertically over the edge of the table. The system is again released from rest.
(a) i. On the dots below representing Block 1 and Block 2 in Scenario 1, draw and label the forces (not components) exerted on each block after the system is released. Each force must be represented by a distinct arrow starting on, and pointing away from, the appropriate dot.
ii. Is the magnitude of the acceleration of the system in Scenario 1 greater than, less than, or equal to the magnitude of the acceleration of the system in Scenario 2?
Briefly justify your prediction without using equations.
(b) i. Starting from fundamental principles of physics (such as Newton's second law), derive an expression for the magnitude of the acceleration \(a_1\) of the system in Scenario 1. Express your answer in terms of \(M\), \(m\), and \(g\).
ii. Derive an algebraic expression for the tension \(T_1\) in the string for Scenario 1 in terms of \(M\), \(m\), and \(g\).
iii. Derive an algebraic expression for the tension \(T_2\) in the string for Scenario 2 in terms of \(M\), \(m\), and \(g\).
(c) Do your mathematical expressions for \(T_1\) and \(T_2\) from parts (b)(ii) and (b)(iii) support your reasoning in part (a)(iii)? Explain your reasoning.
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解題
(a) i. Free-Body Diagrams (Scenario 1): - Block 1 (on table): - Upward normal force \(F_N\) (or \(N\)). - Downward gravitational force \(F_{g1} = Mg\). - Rightward tension force \(T_1\). - Vertical forces must be equal in length (\(F_N = Mg\)). - Block 2 (hanging): - Downward gravitational force \(F_{g2} = mg\). - Upward tension force \(T_1\). - The downward force must be longer than the upward tension force (since Block 2 accelerates downward).
(a) ii. Acceleration Comparison: - Selection: \(a_1 < a_2\) - Justification: The total inertia (total mass) of the system being accelerated is identical in both scenarios (\(M + m\)). However, the net external force causing the acceleration is provided solely by the weight of the hanging block. In Scenario 2, the hanging block has a larger mass (\(M > m\)), providing a greater net external force. With a greater net pulling force acting on the same total mass, Newton's second law dictates that the acceleration in Scenario 2 is greater than in Scenario 1 (\(a_1 < a_2\)).
(a) iii. Tension Comparison: - Selection: \(T_1 = T_2\) - Justification: The tension is the only horizontal force acting on the sliding block, so it is responsible for accelerating it. In Scenario 1, the acceleration is smaller, but it acts on the larger mass \(M\). In Scenario 2, the acceleration is larger, but it acts on the smaller mass \(m\). Because the acceleration is directly proportional to the hanging mass, the product of the sliding mass and the system's acceleration is symmetric between the two cases, resulting in equal tension forces in the string.
**(b) i. Derivation of \(a_1\):** Applying Newton's second law to the entire two-block system in Scenario 1: \[\Sigma F_{\text{ext}} = m_{\text{total}} a_1\] \[mg = (M + m) a_1\] \[a_1 = \frac{mg}{M + m}\]
**(b) ii. Derivation of \(T_1\):** Applying Newton's second law horizontally to Block 1 on the frictionless table: \[\Sigma F_{x,1} = M a_1\] \[T_1 = M \left(\frac{mg}{M + m}\right) = \frac{Mmg}{M + m}\]
**(b) iii. Derivation of \(T_2\):** In Scenario 2, applying Newton's second law to the system gives \(a_2 = \frac{Mg}{M + m}\). Applying Newton's second law horizontally to Block 2 on the table: \[\Sigma F_{x,2} = m a_2\] \[T_2 = m \left(\frac{Mg}{M + m}\right) = \frac{Mmg}{M + m}\]
(c) Reasoning and Consistency: - Selection: Yes. - Explanation: The derived expressions give \(T_1 = \frac{Mmg}{M + m}\) and \(T_2 = \frac{Mmg}{M + m}\). Both expressions are mathematically identical because the numerator \(Mmg\) and denominator \(M+m\) are completely symmetric with respect to interchanging \(M\) and \(m\). This directly confirms that swapping the two blocks leaves the tension in the string unchanged, in agreement with the qualitative prediction in part (a)(iii).
評分準則
Question 3 (12 points total)
(a)(i) 2 points - 1 point: For drawing correct forces on Block 1 (upward normal force, downward gravitational force of equal length, and horizontal tension to the right) with no extraneous forces. - 1 point: For drawing correct forces on Block 2 (downward gravitational force longer than the upward tension force).
(a)(ii) 2 points - 1 point: For selecting \(a_1 < a_2\) with an attempt at physical reasoning. - 1 point: For explaining that the total mass of the system is the same in both scenarios, but the net accelerating force (weight of the hanging mass) is larger in Scenario 2.
(a)(iii) 2 points - 1 point: For selecting \(T_1 = T_2\) with an attempt at physical justification (or selecting a consistent choice based on qualitative reasoning). - 1 point: For reasoning that in Scenario 1 a smaller acceleration acts on a larger mass, whereas in Scenario 2 a larger acceleration acts on a smaller mass, balancing out to give equal tensions.
(b)(i) 2 points - 1 point: For a valid application of Newton's second law to the system or individual blocks (e.g., \(\Sigma F = m_{\text{total}} a\) or setting up two coupled equations). - 1 point: For the correct algebraic solution: \(a_1 = \frac{mg}{M + m}\).
(b)(ii) 1 point - 1 point: For correctly deriving \(T_1 = \frac{Mmg}{M + m}\) using the horizontal equation of motion for Block 1 or the vertical equation for Block 2.
(b)(iii) 1 point - 1 point: For correctly deriving \(T_2 = \frac{Mmg}{M + m}\).
(c) 2 points - 1 point: For explicitly referencing the functional form/symmetry of the derived expressions \(T_1\) and \(T_2\). - 1 point: For a valid explanation showing how the mathematical equality \(T_1 = T_2 = \frac{Mmg}{M + m}\) supports the claim from part (a)(iii).
題目 5 · PAR
7 分
A uniform solid disk of mass \(M\) and radius \(R\) is mounted on a fixed, frictionless vertical axle passing through its center. The rotational inertia of the disk about this axis is \(I = \frac{1}{2}MR^2\).
(a) A constant tangential braking force of magnitude \(F_0\) is applied at the outer rim (distance \(R\) from the center) of the disk. Determine an expression for the magnitude of the angular acceleration \(\alpha\) of the disk in terms of \(M\), \(R\), \(F_0\), and physical constants as appropriate.
Two identical solid disks, Disk 1 and Disk 2, each having mass \(M\) and radius \(R\), are spinning freely about their central axles with the same initial angular speed \(\omega_i\). To bring each disk to rest, a braking force of constant magnitude \(F_0\) is applied tangentially to each disk until it stops: - On Disk 1, the force is applied at a distance \(r_1 = R\) from the axle. - On Disk 2, the force is applied at a distance \(r_2 = \frac{1}{2}R\) from the axle.
(b) In a clear, coherent, paragraph-length response that may also contain equations and drawings, explain why the time \(\Delta t_2\) required to bring Disk 2 to rest is twice the time \(\Delta t_1\) required to bring Disk 1 to rest, but the total arc length distance \(s_2\) along the disk surface over which the braking force is applied on Disk 2 is equal to the arc length distance \(s_1\) on Disk 1.
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解題
(a) The torque exerted on the disk by the tangential braking force applied at the rim is: \[\tau = F_0 R\] Using Newton's second law for rotation: \[\tau = I\alpha\] \[F_0 R = \left(\frac{1}{2}MR^2\right)\alpha\] Solving for the angular acceleration \(\alpha\): \[\alpha = \frac{F_0 R}{\frac{1}{2}MR^2} = \frac{2F_0}{MR}\]
(b) Angular Impulse & Stopping Time: The torque exerted on each disk is given by \(\tau = r F_0\). Since \(r_1 = R\) and \(r_2 = \frac{1}{2}R\), the torque on Disk 2 is half the torque on Disk 1 (\(\tau_2 = \frac{1}{2}\tau_1\)). Both disks are identical and have the same initial angular speed \(\omega_i\), meaning they have the same initial angular momentum \(L_i = I\omega_i\) and must undergo the same change in angular momentum \(|\Delta L| = I\omega_i\) to come to rest. By the angular impulse-momentum theorem, \(\Delta L = \tau \Delta t\). Since the torque on Disk 2 is half as large, the time required to deliver the necessary angular impulse must be twice as long: \(\Delta t_2 = 2\Delta t_1\).
Work-Energy & Stopping Distance: Both disks have the same initial rotational kinetic energy \(K_i = \frac{1}{2}I\omega_i^2\) and both finish at rest, so the net work done to stop each disk must be identical (\(W = \Delta K = K_i\)). The work done by a rotational torque over an angular displacement is \(W = \tau \Delta \theta = (F_0 r)\Delta \theta\). Because the arc length distance traveled by the point of application of the force is \(s = r\Delta \theta\), the work done by the braking force simplifies directly to \(W = F_0 s\). Since the magnitude of the applied force \(F_0\) is the same in both cases and the total work required to stop each disk is equal, the arc length distance traveled must be equal (\(s_1 = s_2\)).
評分準則
Part (a) — 1 point total - 1 point: For a correct expression for the angular acceleration of the disk: \(\alpha = \frac{2F_0}{MR}\).
Part (b) — 6 points total - 1 point: For indicating that the torque on Disk 2 is half the torque on Disk 1 because torque is proportional to the lever arm (\(\tau = r F_0\)). - 1 point: For stating or showing that both disks have the same initial angular momentum and therefore undergo the same change in angular momentum \(\Delta L\). - 1 point: For applying the angular impulse-momentum relationship (\(\Delta L = \tau \Delta t\)) to conclude that Disk 2 takes twice as long to stop as Disk 1 (\(\Delta t_2 = 2\Delta t_1\)). - 1 point: For stating or showing that both disks have the same initial rotational kinetic energy and therefore require the same work \(W = \Delta K\) to come to rest. - 1 point: For relating work to the applied force and arc length distance (\(W = \tau \Delta \theta = F_0 s\)) to show that equal work with equal force yields equal arc length distance (\(s_1 = s_2\)). - 1 point: For a logical, relevant, and internally consistent argument that follows the paragraph-length response guidelines.