An original Thinka practice paper modelled on the structure and difficulty of the May 2025 AP AP Physics 1: Algebra-Based paper. Not affiliated with or reproduced from AP.
Section II: Free-Response Questions
Answer all four free-response questions. Show all work and derivations starting from fundamental physics principles. Clearly label graphs, axes, and units.
4 Question · 40 marks
Question 1 · free-response
10 marks
A cart of mass \(M\) holds a projectile of mass \(\frac{1}{3}M\). The cart is on a frictionless horizontal track.
• At time \(t = 0\), the cart and projectile move together to the right across the track with a constant speed \(v_0\). • At time \(t = t_1\), an internal spring mechanism in the cart fires the projectile forward in the direction of motion, launching it so that it exits the cart with a speed of \(2v_0\) relative to the track. • At time \(t = t_2\), both the cart and the projectile continue to move to the right at their respective constant speeds, with the cart moving at speed \(v_c\).
A. i. On axes of total horizontal momentum \(p_x\) versus time \(t\), sketch a graph of the magnitude of the \(x\)-component of the total momentum of the cart-projectile system as a function of time \(t\) from \(t = 0\) until \(t > t_2\).
ii. Derive an expression for the speed \(v_c\) of the cart after time \(t = t_2\) in terms of \(M\), \(v_0\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
iii. Derive an expression for the change in kinetic energy \(\Delta K\) of the cart-projectile system from \(t = 0\) to \(t = t_2\) in terms of \(M\), \(v_0\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
B. Consider a modified scenario where an external braking mechanism applies a horizontal force to the projectile as it is fired, directed toward the left during the launch time interval \(\Delta t\).
Indicate whether the \(x\)-component of the total momentum of the cart-projectile system increases, decreases, or remains constant during \(\Delta t\).
Show answer & marking schemeHide answer & marking scheme
Worked solution
### Part A (i) Because the track is frictionless and the spring force between the cart and projectile is entirely internal to the cart-projectile system, there is no net external horizontal force acting on the system (\(\Sigma F_{x,\text{ext}} = 0\)). Therefore, the total horizontal momentum \(p_x\) of the system remains constant throughout the entire motion from \(t = 0\) to \(t > t_2\).
Graph sketch: A continuous, non-zero horizontal line of constant height \(p_x = \frac{4}{3}M v_0\) from \(t = 0\) extending past \(t_2\).
---
### Part A (ii) Start from the principle of conservation of linear momentum:
\[\sum p_i = \sum p_f\]
Initial momentum of the system: \[p_i = (M + m_p) v_0 = \left(M + \frac{1}{3}M\right) v_0 = \frac{4}{3}M v_0\]
Final momentum of the system after the launch (\(t = t_2\)): \[p_f = M v_c + m_p v_p = M v_c + \left(\frac{1}{3}M\right)(2v_0) = M v_c + \frac{2}{3}M v_0\]
Equating initial and final momentum: \[\frac{4}{3}M v_0 = M v_c + \frac{2}{3}M v_0\] \[M v_c = \frac{4}{3}M v_0 - \frac{2}{3}M v_0 = \frac{2}{3}M v_0\] \[v_c = \frac{2}{3}v_0\]
---
### Part A (iii) Start from the definition of the change in kinetic energy:
\[\Delta K = K_f - K_i\] \[K = \frac{1}{2}mv^2\]
Initial kinetic energy at \(t = 0\): \[K_i = \frac{1}{2}(M + m_p)v_0^2 = \frac{1}{2}\left(\frac{4}{3}M\right)v_0^2 = \frac{2}{3}M v_0^2\]
Justification: The impulse-momentum theorem states that \(\Delta \vec{p} = \vec{F}_{\text{net, ext}} \Delta t\). Because the braking mechanism exerts an external force in the negative \(x\)-direction (opposite to the motion) on the cart-projectile system, there is a net external horizontal force on the system. This negative net external impulse causes the total \(x\)-component of momentum of the system to decrease during the time interval \(\Delta t\).
Marking scheme
### Scoring Guidelines (10 points total)
Part A (i) (2 points): - Point A1: For sketching a constant (horizontal) \(p_x\) value during either the pre-launch (\(t < t_1\)) or post-launch (\(t > t_1\)) period. - Point A2: For sketching a single, continuous, non-zero horizontal line representing that total momentum remains constant across the entire time interval from \(t = 0\) to \(t > t_2\).
Part A (ii) (2 points): - Point A3: For writing a valid conservation of linear momentum equation (e.g., \(p_i = p_f\) or \((M + m_p)v_0 = M v_c + m_p v_p\)). - Point A4: For substituting correct values for masses and speeds to obtain the isolated final expression \(v_c = \frac{2}{3}v_0\).
Part A (iii) (3 points): - Point A5: For a multistep derivation that begins with the correct definition for kinetic energy or change in kinetic energy (\(K = \frac{1}{2}mv^2\) and \(\Delta K = K_f - K_i\)). - Point A6: For correctly substituting the initial mass (\(\frac{4}{3}M\)) and the post-launch speeds/masses of both the cart and projectile into the kinetic energy expressions. - Point A7: For obtaining a correct, isolated final expression for \(\Delta K\) consistent with part A (ii) (e.g., \(\Delta K = \frac{2}{9}M v_0^2\)).
Part B (3 points): - Point B1: For selecting "Decreases". (Scored independently of the explanation) - Point B2: For identifying that the braking force is an external force acting on the cart-projectile system in the negative direction. - Point B3: For connecting the net external force/impulse to the change in system momentum (i.e., \(\vec{J}_{\text{ext}} = \Delta \vec{p}\), indicating that a net external force causes a decrease in total system momentum).
Question 2 · free-response
12 marks
A cart of mass \( M \) is released from rest at position \( A \) at the top of a curved, frictionless track at a vertical height \( H \) above the horizontal ground. The lower portion of the track is horizontal and level with the ground. On the horizontal section, an ideal spring with spring constant \( k \) is fixed at one end to a rigid wall.
The cart moves down the track and reaches the horizontal section. At position \( B \), the cart encounters the free end of the uncompressed spring. The cart compresses the spring by a maximum distance \( x_{\text{max}} \) until it momentarily comes to rest at position \( C \).
Let the gravitational potential energy \( U_g \) of the cart-spring-Earth system be defined to be zero on the horizontal ground (\( y = 0 \)).
A. An energy bar chart is provided that represents the kinetic energy \( K \) of the cart, the gravitational potential energy \( U_g \) of the cart-spring-Earth system, and the elastic potential energy \( U_s \) of the spring at the instant the spring is compressed by a distance \( x = \frac{1}{2}x_{\text{max}} \). At this instant, the chart shows \( K = 6E_0 \), \( U_g = 0 \), and \( U_s = 2E_0 \).
Draw shaded bars that represent \( K \), \( U_g \), and \( U_s \) to complete the energy bar charts for: 1. The instant the cart is released from rest at position \( A \). 2. The instant the cart first makes contact with the uncompressed spring at position \( B \).
Guidelines: • Shaded bars must start at the dashed line representing zero energy. • Represent any energy value equal to zero with a distinct line on the zero-energy line. • The relative heights of the shaded bars must reflect the magnitudes consistent with the scale where total mechanical energy is \( 8E_0 \).
B. Starting with conservation of energy, derive an expression for the spring constant \( k \). Express your answer in terms of \( M \), \( H \), \( x_{\text{max}} \), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
C. On a graph of Energy as a function of the spring compression \( x \) from \( x = 0 \) to \( x = x_{\text{max}} \): (i) Sketch and label a line or curve that represents the total mechanical energy \( E \) of the cart-spring-Earth system. (ii) Sketch and label a line or curve that represents the kinetic energy \( K \) of the cart as a function of compression \( x \).
D. Indicate whether the magnitude of the acceleration of the cart \( a_{1/2} \) at compression \( x = \frac{1}{2}x_{\text{max}} \) is greater than, less than, or equal to the magnitude of the acceleration \( a_{\text{max}} \) at maximum compression \( x = x_{\text{max}} \).
Justify your answer by referencing the net force exerted on the cart or the energy relationships described in parts A–C.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Part A: From the given chart at \( x = \frac{1}{2}x_{\text{max}} \), total mechanical energy is \( E_{\text{total}} = K + U_g + U_s = 6E_0 + 0 + 2E_0 = 8E_0 \). Since the track is frictionless and the system is isolated, total mechanical energy is conserved and equals \( 8E_0 \) at all positions. 1. At position \( A \), the cart is at rest (\( K = 0 \)) at height \( H \), and the spring is uncompressed (\( U_s = 0 \)). Thus, \( U_g = 8E_0 \). The bar chart has \( U_g = 8E_0 \), \( K = 0 \), and \( U_s = 0 \). 2. At position \( B \), the cart is at \( y = 0 \) (\( U_g = 0 \)) and the spring has not yet compressed (\( U_s = 0 \)). Thus, all energy is kinetic: \( K = 8E_0 \). The bar chart has \( K = 8E_0 \), \( U_g = 0 \), and \( U_s = 0 \).
Part B: Starting from conservation of mechanical energy for the closed system: \[ E_i = E_f \] \[ U_{g,i} + K_i + U_{s,i} = U_{g,f} + K_f + U_{s,f} \] Taking initial state at position \( A \) and final state at position \( C \): \[ M g H + 0 + 0 = 0 + 0 + \frac{1}{2} k x_{\text{max}}^2 \] \[ M g H = \frac{1}{2} k x_{\text{max}}^2 \] Solving for \( k \): \[ k = \frac{2 M g H}{x_{\text{max}}^2} \]
Part C: (i) Total mechanical energy is constant at all compression values: sketch a horizontal line at \( E = 8E_0 \) extending from \( x = 0 \) to \( x = x_{\text{max}} \). (ii) Since \( U_s(x) = \frac{1}{2}kx^2 \) and \( U_g = 0 \) along the horizontal track, the kinetic energy is: \[ K(x) = E - U_s(x) = 8E_0 - \frac{1}{2}kx^2 \] This is a concave-down (inverted) parabola starting at \( (0, 8E_0) \), passing through \( (\frac{1}{2}x_{\text{max}}, 6E_0) \), and ending at \( (x_{\text{max}}, 0) \).
Part D: Select: \( a_{1/2} < a_{\text{max}} \) Justification: On the horizontal surface, the only horizontal force acting on the cart is the restoring force of the spring, \( |F_{\text{net}}| = |F_s| = kx \). By Newton's second law, the magnitude of the acceleration is \( a = \frac{kx}{M} \). Since acceleration is directly proportional to spring compression \( x \), at \( x = \frac{1}{2}x_{\text{max}} \) the force and acceleration are half of their maximum values at \( x = x_{\text{max}} \). Therefore, \( a_{1/2} < a_{\text{max}} \).
Marking scheme
Part A (3 points): • Point A1: For drawing a single bar in the Position A chart representing only gravitational potential energy \( U_g \). • Point A2: For drawing a single bar in the Position B chart representing only kinetic energy \( K \). • Point A3: For drawing bars in both charts whose heights each correspond to a total energy of \( 8E_0 \).
Part B (4 points): • Point B1: For stating a fundamental conservation of energy equation (e.g., \( E_i = E_f \) or \( \Delta E = 0 \)). • Point B2: For equating the initial gravitational potential energy to the maximum elastic potential energy (\( U_g = U_s \)). • Point B3: For correctly substituting \( M g H \) for \( U_g \) and \( \frac{1}{2}kx_{\text{max}}^2 \) for \( U_s \). • Point B4: For obtaining the correct expression \( k = \frac{2 M g H}{x_{\text{max}}^2} \).
Part C (3 points): • Point C1: For drawing a horizontal line at \( E = 8E_0 \) labeled \( E \) continuous from \( x = 0 \) to \( x = x_{\text{max}} \). • Point C2: For drawing a curve for \( K \) that starts at \( (0, 8E_0) \) and decreases to \( 0 \) at \( x = x_{\text{max}} \). • Point C3: For drawing the curve for \( K \) as concave downward (inverted quadratic dependence on \( x \)).
Part D (2 points): • Point D1: For correctly selecting \( a_{1/2} < a_{\text{max}} \). • Point D2: For a correct justification connecting the spring force \( F_s = kx \) (or the magnitude of the slope of the \( K(x) \) curve) to Newton's second law \( a = \frac{F_{\text{net}}}{M} \).
Question 3 · free-response
10 marks
Students are investigating simple harmonic motion using an oscillating system. The students have a horizontal track with negligible friction, an ideal spring of known spring constant \(k_0\) fixed to a rigid end-stop, and an object of unknown mass \(m_0\) that can be connected to the free end of the spring. The students have access to a standard laboratory timer and a meterstick, but they do not have a balance or scale to measure mass directly.
The students are asked to collect experimental data that will allow them to produce a linear graph whose slope can be used to determine the mass \(m_0\) of the object.
A. Describe an experimental procedure to collect data that would allow the students to determine \(m_0\). Include any specific steps necessary to reduce experimental uncertainty.
B. Describe how the data collected in part A could be graphed and explain how that graph would be analyzed to determine \(m_0\).
In a second experiment, the students suspend a metal cylinder of unknown mass \(M\) vertically from various lightweight springs, each having a different known spring constant \(k\). For each spring, the cylinder is pulled down a small distance and released from rest so that it oscillates vertically. The students measure the period \(T\) of vertical oscillation for each spring constant. Table 1 shows the measured values of \(k\) and \(T\).
The students correctly determine that the theoretical relationship between the period \(T\), the mass \(M\), and the spring constant \(k\) is given by: \[ T^2 = \frac{4\pi^2 M}{k} \]
The students create a graph with \(\frac{1}{k}\) plotted on the horizontal axis.
C. i. Indicate what measured or calculated quantity could be plotted on the vertical axis to yield a linear graph whose slope can be used to determine an experimental value for the mass \(M\) of the cylinder.
ii. Complete the data table below with the calculated values to be plotted on both axes, clearly label the vertical axis including appropriate units, and plot the data points.
iii. Draw a straight line of best fit for the plotted data points.
D. Using the best-fit line that you drew in part C (iii), calculate an experimental value for the mass \(M\) of the cylinder.
Show answer & marking schemeHide answer & marking scheme
Worked solution
### Part A 1. Attach the object of mass \(m_0\) to the spring on the horizontal track. 2. Displace the mass a small horizontal distance from its equilibrium position and release it from rest. 3. Using the timer, measure the time required for multiple complete oscillations (e.g., 10 to 20 cycles) and divide the total elapsed time by the number of cycles to find the period \(T\). 4. To reduce experimental uncertainty: - Repeat the timing measurement for at least 3 trials at the same initial displacement and calculate the average period. - Systematically attach additional known small masses \(\Delta m\) to \(m_0\) (or vary the system parameters) and measure the period \(T\) for each combination.
### Part B - The relationship between period and mass is \(T = 2\pi \sqrt{\frac{m}{k_0}} \implies T^2 = \frac{4\pi^2}{k_0} m\). - If additional known masses \(\Delta m\) are added, plot \(T^2\) on the vertical axis as a function of \(\Delta m\) on the horizontal axis. - The resulting graph will be linear with a slope equal to \(\frac{4\pi^2}{k_0}\) and a vertical-axis intercept of \(T_0^2 = \frac{4\pi^2 m_0}{k_0}\). - Thus, the unknown mass can be calculated from the intercept: \(m_0 = \frac{k_0 \cdot (\text{vertical intercept})}{4\pi^2}\). *(Alternatively, measuring \(T\) for multiple trials and using \(m_0 = \frac{k_0 T^2}{4\pi^2}\) with an average slope analysis is fully valid.)*
### Part C i. Vertical axis: \(T^2\) (in units of \(\text{s}^2\)) *(Note: Plotting \(\frac{T^2}{4\pi^2}\) is also acceptable.)*
ii. & iii. Calculated values for the table: - For \(k = 10\text{ N/m}\): \(\frac{1}{k} = 0.100\text{ m/N}\), \(T^2 = (1.00)^2 = 1.00\text{ s}^2\) - For \(k = 16\text{ N/m}\): \(\frac{1}{k} = 0.063\text{ m/N}\), \(T^2 = (0.79)^2 \approx 0.62\text{ s}^2\) - For \(k = 25\text{ N/m}\): \(\frac{1}{k} = 0.040\text{ m/N}\), \(T^2 = (0.62)^2 \approx 0.38\text{ s}^2\) - For \(k = 40\text{ N/m}\): \(\frac{1}{k} = 0.025\text{ m/N}\), \(T^2 = (0.50)^2 = 0.25\text{ s}^2\) - For \(k = 65\text{ N/m}\): \(\frac{1}{k} = 0.015\text{ m/N}\), \(T^2 = (0.39)^2 \approx 0.15\text{ s}^2\)
The vertical axis is scaled linearly from \(0\) to \(1.20\text{ s}^2\), the five points are plotted accurately, and a straight best-fit line is drawn that passes closely through the points and near the origin.
### Part D Choose two points on the best-fit line to find the slope: \[ \text{slope} = \frac{\Delta(T^2)}{\Delta(1/k)} = \frac{1.00\text{ s}^2 - 0.10\text{ s}^2}{0.100\text{ m/N} - 0.010\text{ m/N}} = \frac{0.90}{0.090} \approx 10.0\text{ s}^2/(\text{m/N}) \]
From the theoretical equation: \[ T^2 = (4\pi^2 M) \left(\frac{1}{k}\right) \implies \text{slope} = 4\pi^2 M \] \[ M = \frac{\text{slope}}{4\pi^2} = \frac{10.0\text{ kg}\cdot\text{s}^2}{4\pi^2} \approx 0.253\text{ kg} \approx 0.25\text{ kg} \]
Marking scheme
Part A (2 points): - 1 point (A1): For describing a valid procedure that involves displacing the attached mass, releasing it into oscillation, and measuring the time for multiple oscillations to obtain the period. - 1 point (A2): For describing a valid method to reduce experimental uncertainty (e.g., timing at least 10–20 continuous oscillations per trial, repeating multiple timing trials for an average, or using multiple known mass increments).
Part B (2 points): - 1 point (B1): For identifying appropriate quantities that yield a linear graph related to period/frequency and mass (e.g., \(T^2\) vs mass, or \(T^2\) vs \(1/k\)). - 1 point (B2): For correctly explaining how the slope (or intercept) of the linear graph is algebraically manipulated using fundamental constants and given values to calculate \(m_0\).
Part C (4 points): - 1 point (C1): For correctly specifying \(T^2\) (or an algebraic equivalent such as \(T^2/4\pi^2\)) for the vertical axis. - 1 point (C2): For providing a correctly labeled vertical axis with units (e.g., \(T^2\;(\text{s}^2)\)) and a uniform linear scale. - 1 point (C3): For correctly plotting the 5 data points consistent with the calculated table values. - 1 point (C4): For drawing a single, straight best-fit line that reasonably represents the trend of the plotted data.
Part D (2 points): - 1 point (D1): For correctly relating the slope of the best-fit line to mass \(M\) using the equation \(M = \frac{\text{slope}}{4\pi^2}\) (or consistent with the axes chosen in Part C). - 1 point (D2): For calculating a final value of \(M\) within the acceptable range of \(0.23\text{ kg}\) to \(0.27\text{ kg}\) with correct units.
Question 4 · free_response
8 marks
A uniform solid disk of mass \(M\) and radius \(R\) is mounted on a fixed horizontal axle passing through its center so that it can rotate in a vertical plane with negligible friction. The rotational inertia of the disk about this axle is \(I = \frac{1}{2}MR^2\).
In Scenario 1, a string is wrapped around the outer rim of the disk at distance \(R\) from the axle. A student pulls the string with a constant force of magnitude \(F_0\) directed tangentially to the rim. The disk starts from rest and rotates through a total angular displacement \(\Delta \theta_0\), reaching a final angular speed \(\omega_1\).
In Scenario 2, a string is wrapped around a concentric inner hub of radius \(\frac{1}{2}R\) fixed rigidly to an identical disk. The student pulls the string with the same constant force of magnitude \(F_0\) directed tangentially to the hub. The disk starts from rest and rotates through the same angular displacement \(\Delta \theta_0\), reaching a final angular speed \(\omega_2\).
A. Indicate whether \(\omega_1\) is greater than, less than, or equal to \(\omega_2\).
______ \(\omega_1 > \omega_2\)
______ \(\omega_1 < \omega_2\)
______ \(\omega_1 = \omega_2\)
Justify your answer in terms of the torque and work or rotational kinematics. Use qualitative reasoning beyond referencing equations.
B. Consider the general case where a disk of mass \(M\) and outer radius \(R\) (rotational inertia \(I = \frac{1}{2}MR^2\)) has a string wrapped around a circular hub of radius \(r\) (where \(r \le R\)). A constant tangential force of magnitude \(F\) is applied to the string as the disk rotates from rest through an angular displacement \(\Delta \theta\).
Starting with fundamental physics principles, derive an expression for the final angular speed \(\omega\) of the disk. Express your answer in terms of \(M\), \(R\), \(r\), \(F\), \(\Delta \theta\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.
C. Indicate whether the expression for \(\omega\) you derived in part B is or is not consistent with the claim made in part A. Briefly justify your answer by referencing your derivation in part B.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Part A
Claim: \(\omega_1 > \omega_2\)
Qualitative Justification: In both scenarios, the disks are identical and therefore have the same rotational inertia. The magnitude of torque exerted on an object is determined by the magnitude of the applied force and the perpendicular distance (lever arm) from the axis of rotation to the line of action of the force. In Scenario 1, the force is applied at a greater distance from the axle (\(R\)) than in Scenario 2 (\(R/2\)), so the torque exerted on the disk in Scenario 1 is greater. By Newton's second law for rotation, the greater torque produces a greater angular acceleration for the disk in Scenario 1. Starting from rest and rotating through the same angular displacement, the disk with the larger angular acceleration will reach a higher final angular speed. (Alternatively, the greater torque does more rotational work over the same angular displacement, giving greater rotational kinetic energy and thus a larger angular speed.)
---
Part B
Derivation:
Method 1: Using Newton's Second Law for Rotation and Kinematics Start with Newton's second law for rotation: \[\Sigma \tau = I\alpha\] The net torque exerted by the tangential force \(F\) at radius \(r\) is: \[\tau = F r\] Substituting \(\tau = Fr\) and \(I = \frac{1}{2}MR^2\): \[Fr = \left(\frac{1}{2}MR^2\right)\alpha\] \[\alpha = \frac{2Fr}{MR^2}\] Using the rotational kinematic equation for constant angular acceleration starting from rest (\(\omega_0 = 0\)): \[\omega^2 = \omega_0^2 + 2\alpha \Delta \theta\] \[\omega^2 = 2\left(\frac{2Fr}{MR^2}\right)\Delta \theta = \frac{4Fr\Delta \theta}{MR^2}\] \[\omega = \sqrt{\frac{4Fr\Delta \theta}{MR^2}} = 2\sqrt{\frac{Fr\Delta \theta}{MR^2}}\]
Consistency Evaluation: The derived expression \(\omega = \sqrt{\frac{4Fr\Delta \theta}{MR^2}}\) is consistent with the claim made in part A. In the derived equation, the final angular speed \(\omega\) is proportional to \(\sqrt{r}\) (the radius at which the force is applied), while all other variables (\(F\), \(M\), \(R\), and \(\Delta \theta\)) remain constant between the two scenarios. Since Scenario 1 has a larger radius \(r = R\) compared to Scenario 2 with \(r = R/2\), the equation predicts \(\omega_1 > \omega_2\), which matches the qualitative prediction.
Marking scheme
Part A (3 points total): - Point A1: For selecting "\(\omega_1 > \omega_2\)". - Point A2: For a justification that correctly indicates the torque exerted in Scenario 1 is greater than in Scenario 2 due to the larger lever arm (radius). - Point A3: For a justification that correctly connects the larger torque (or work done) to a greater angular acceleration (or greater rotational kinetic energy) and therefore a higher final angular speed after the same angular displacement.
Part B (3 points total): - Point B1: For beginning with a fundamental physics principle (e.g., \(\Sigma \tau = I\alpha\) or \(W = \Delta K_{\text{rot}}\)). - Point B2: For correctly substituting the expressions for torque (\(\tau = Fr\)) and rotational inertia (\(I = \frac{1}{2}MR^2\)). - Point B3: For a correct, fully simplified expression for \(\omega\) in terms of the specified variables: \(\omega = \sqrt{\frac{4Fr\Delta \theta}{MR^2}}\) (or \(2\sqrt{\frac{Fr\Delta \theta}{MR^2}}\)).
Part C (2 points total): - Point C1: For addressing the functional dependence between \(\omega\) and the radius \(r\) in the equation derived in part B (e.g., stating \(\omega\) is directly proportional to \(\sqrt{r}\) or in the numerator). - Point C2: For correctly evaluating that because \(r_1 > r_2\), the derived expression yields \(\omega_1 > \omega_2\), demonstrating consistency with the qualitative claim in part A.
Ready to test yourself?
Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.
thinka is an AI practice app for IGCSE & IB students: unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.