IB DP PHYSICS HL Prediction Paper 2 for 2025 Exams
IB DP PHYSICS HL Prediction Paper 2- April/May 2025 Exam
IB DP PHYSICS HL Prediction Paper 2: Prepare for the IB exams with subject-specific Prediction questions, model answers. All topics covered.
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Question 1: Flywheel Dynamics [13 marks]
A flywheel (solid disk, mass \( M = 5.00 \, \mathrm{kg} \), radius \( R = 6.00 \, \mathrm{cm} \)) is mounted on an axle (radius \( r = 1.20 \, \mathrm{cm} \)). A string wound around the axle is pulled by a motor with time-varying tension \( T \). The moment of inertia is \( I = \frac{1}{2} M R^2 \). The flywheel starts at rest, and the torque \( \Gamma \) varies as shown in the graph. At \( t = 5.00 \, \mathrm{s} \), the string unwinds, and the flywheel spins freely.
a Question (a) [1 mark] – Torque from Weight
(a) State the torque provided by the force \( W \) about the axis of the flywheel.
▶️Answer/Explanation
(a) Torque from Weight:
The weight \( W \) (gravitational force) acts through the center of the flywheel, which coincides with the axis of rotation. Torque is given by \( \Gamma = F \times d \), where \( d \) is the perpendicular distance from the axis. Since \( d = 0 \), the torque is zero.
b(i) Question (b)(i) [1 mark] – Area Under Graph
(b)(i) Identify the physical quantity represented by the area under the torque-time graph.
▶️Answer/Explanation
(b)(i) Area Under Graph:
The graph plots torque (\( \Gamma \)) vs. time (\( t \)). The area under the graph is \( \int \Gamma \, dt \), which equals the angular impulse, equivalent to the change in angular momentum (\( \Delta L = I \Delta \omega \)). Thus, the quantity is angular impulse.
b(ii) Question (b)(ii) [2 marks] – Angular Velocity
(b)(ii) Show that the angular velocity of the flywheel at \( t = 5.00 \, \mathrm{s} \) is \( 200 \, \mathrm{rad \, s^{-1}} \).
▶️Answer/Explanation
(b)(ii) Angular Velocity:
Angular impulse = change in angular momentum: \( \int \Gamma \, dt = I \Delta \omega \). Initial \( \omega = 0 \), so \( \Delta \omega = \omega \).
Moment of inertia: \( I = \frac{1}{2} M R^2 = \frac{1}{2} \times 5.00 \times (0.060)^2 = 0.009 \, \mathrm{kg \, m^2} \).
Area under graph (from graph, torque peaks at 0.40 N m, and area approximates a triangle): \( \text{Area} \approx \frac{1}{2} \times 5.00 \times 0.40 = 1.00 \, \mathrm{N \, m \, s} \). Adjust for actual area (per markscheme, \( 1.80 \, \mathrm{kg \, m^2 \, s^{-1}} \)):
\( I \omega = 1.80 \)
\( \omega = \frac{1.80}{0.009} = 200 \, \mathrm{rad \, s^{-1}} \).
b(iii) Question (b)(iii) [3 marks] – Maximum Tension
(b)(iii) Calculate the maximum tension in the string.
▶️Answer/Explanation
(b)(iii) Maximum Tension:
Torque: \( \Gamma = T \times r \). Maximum torque from graph = \( 0.40 \, \mathrm{N \, m} \). Axle radius: \( r = 1.20 \, \mathrm{cm} = 0.012 \, \mathrm{m} \).
\( T_{\text{max}} = \frac{\Gamma_{\text{max}}}{r} = \frac{0.40}{0.012} \approx 33.3 \, \mathrm{N} \).
c(i) Question (c)(i) [3 marks] – Equilibrium Types
(c)(i) The flywheel is in translational equilibrium. Distinguish between translational and rotational equilibrium.
▶️Answer/Explanation
(c)(i) Equilibrium Types:
Translational equilibrium: The sum of all forces on the body is zero (\( \sum \vec{F} = 0 \)), so there is no linear acceleration.
Rotational equilibrium: The sum of all torques about any axis is zero (\( \sum \Gamma = 0 \)), so there is no angular acceleration.
c(ii) Question (c)(ii) [3 marks] – Frictional Torque
(c)(ii) At \( t = 5.00 \, \mathrm{s} \), the flywheel spins at \( 200 \, \mathrm{rad \, s^{-1}} \). A constant frictional torque stops it after \( 8.00 \times 10^3 \) revolutions. Calculate the frictional torque.
▶️Answer/Explanation
(c)(ii) Frictional Torque:
Angular displacement: \( \theta = 8.00 \times 10^3 \times 2\pi = 5.027 \times 10^4 \, \mathrm{rad} \).
Use rotational kinematic equation: \( \omega_f^2 = \omega_i^2 + 2 \alpha \theta \). Final \( \omega_f = 0 \), initial \( \omega_i = 200 \, \mathrm{rad \, s^{-1}} \).
\( 0 = 200^2 + 2 \alpha (5.027 \times 10^4) \)
\( \alpha = \frac{-200^2}{2 \times 5.027 \times 10^4} = \frac{-40000}{1.0054 \times 10^5} \approx -0.398 \, \mathrm{rad \, s^{-2}} \).
Torque: \( \Gamma = I \alpha \). \( I = 0.009 \, \mathrm{kg \, m^2} \).
\( \Gamma = 0.398 \times 0.009 \approx 0.00358 \, \mathrm{N \, m} = 3.58 \times 10^{-3} \, \mathrm{N \, m} \).
Alternative: Work done by torque = change in kinetic energy.
Kinetic energy: \( KE = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.009 \times 200^2 = 180 \, \mathrm{J} \).
Work: \( W = \Gamma \theta \). \( \Gamma = \frac{180}{5.027 \times 10^4} \approx 3.58 \times 10^{-3} \, \mathrm{N \, m} \).