A Level H2 Physics Practice Paper 4

TuitionGoWhere Exam Practice (AI) - Physics H2 A-Level

Answer Key and Marking Scheme (Version 4)

Topic: Mechanics
Total Marks: 60

Section A: Structured Questions

1. State the Principle of Conservation of Linear Momentum. [2]

• Answer: In a closed system (or isolated system) [1], the total momentum before an interaction (collision/explosion) is equal to the total momentum after the interaction, provided no external resultant force acts on the system [1].
• Note: Accept "Total momentum remains constant if net external force is zero."

2. Calculate the maximum acceleration of the ball. [3]

• Formula: $a_{max} = \omega^2 x_0$ or $a_{max} = (2\pi f)^2 x_0$ [1]
• Substitution:
  • $\omega = 2\pi(2.5) = 5\pi \text{ rad s}^{-1}$
  • $x_0 = 0.04 \text{ m}$
  • $a_{max} = (5\pi)^2 \times 0.04$ [1]
• Calculation: $a_{max} \approx 9.87 \text{ m s}^{-2}$ [1]
• Accept: $9.9 \text{ m s}^{-2}$.

3. Free-fall experiment.

• (a) Precaution for accuracy: [1]
  • Use a large height $h$ to reduce percentage uncertainty in time measurement.
  • OR: Use an electronic timer triggered by the electromagnet and trapdoor to eliminate human reaction time error.
• (b) Steel vs Plastic ball: [2]
  • Steel is denser/heavier [1].
  • Air resistance has a smaller effect on the steel ball relative to its weight compared to the plastic ball, making the assumption of free fall ($a=g$) more valid [1].

4. Circular motion of car.

• (a) Centripetal Force: [2]
  • $F_c = \frac{mv^2}{r}$ [1]
  • $F_c = \frac{1200 \times 20^2}{50} = \frac{1200 \times 400}{50} = 9600 \text{ N}$ [1]
• (b) Origin of force: [1]
  • Friction between the tires and the road.

5. Inelastic Collision. [3]

• Principle: Conservation of Momentum.
• Equation: $m_A u_A + m_B u_B = (m_A + m_B) v$ [1]
• Substitution: $(2.0 \times 3.0) + (1.0 \times 0) = (2.0 + 1.0) v$ [1]
• Calculation: $6.0 = 3.0 v \Rightarrow v = 2.0 \text{ m s}^{-1}$ [1]

6. Define gravitational field strength. [1]

• Answer: The gravitational force per unit mass acting on a small test mass placed at that point. ($g = F/m$)

7. Satellite in free fall. [2]

• Answer: The only force acting on the satellite is the gravitational pull of the Earth [1]. This force provides the centripetal acceleration required to keep it in orbit, meaning it is constantly accelerating towards the Earth (falling), even though its tangential velocity keeps it at a constant altitude [1].

8. Block on inclined plane. [3]

• Forces: Component of weight down slope $= mg \sin \theta$. Friction acts down slope (opposing motion up).
• Equation: $F_{pull} = mg \sin \theta + F_{friction}$ [1]
• Substitution: $F_{pull} = 5.0 \times 9.81 \times \sin(30^\circ) + 10$ [1]
• Calculation: $F_{pull} = 24.525 + 10 = 34.5 \text{ N}$ [1]

9. Elastic Potential Energy. [2]

• Method: Area under Force-Extension graph.
• Calculation: $EPE = \frac{1}{2} F x$ or $\frac{1}{2} \times 20 \times 0.10$ [1]
• Answer: $1.0 \text{ J}$ [1]

10. Projectile Height. [2]

• Vertical Motion: $u_y = 0$, $a = g = 9.81 \text{ m s}^{-2}$, $t = 2.0 \text{ s}$.
• Equation: $h = u_y t + \frac{1}{2} g t^2$ [1]
• Calculation: $h = 0 + 0.5 \times 9.81 \times (2.0)^2 = 19.62 \text{ m}$ [1]
• Accept: $20 \text{ m}$ (if $g=10$).

Section B: Data Analysis and Application

11. Pendulum Graph.

• (a) Graph: [1]
  • Plot $T^2$ (y-axis) against $L$ (x-axis).
• (b) Determining g: [2]
  • Rearrange formula: $T^2 = \frac{4\pi^2}{g} L$.
  • Gradient $m = \frac{4\pi^2}{g}$ [1].
  • Therefore, $g = \frac{4\pi^2}{\text{gradient}}$ [1].

12. Rocket Motion.

• (a) Initial Acceleration: [3]
  • Resultant Force $F_{net} = \text{Thrust} - \text{Weight}$ [1]
  • $W = mg = 5000 \times 9.81 = 49050 \text{ N}$
  • $F_{net} = 80000 - 49050 = 30950 \text{ N}$ [1]
  • $a = \frac{F_{net}}{m} = \frac{30950}{5000} = 6.19 \text{ m s}^{-2}$ [1]
• (b) Effect of Mass Decrease: [2]
  • Acceleration increases [1].
  • Since $a = \frac{F_{thrust} - mg}{m}$, as $m$ decreases, the denominator decreases and the net force increases (as weight decreases), leading to larger $a$ [1].

13. Cyclist on Circular Track.

• (a) Force: [1]
  • Static Friction.
• (b) Maximum Speed: [3]
  • Centripetal force provided by max friction: $\frac{mv^2}{r} = \mu mg$ [1]
  • $v^2 = \mu g r$
  • $v = \sqrt{0.8 \times 9.81 \times 20}$ [1]
  • $v = \sqrt{156.96} \approx 12.5 \text{ m s}^{-1}$ [1]

14. Gravitational Forces.

• (a) Comparison: [1]
  • The forces are equal in magnitude (Newton's Third Law).
• (b) Distance Change: [1]
  • Force is inversely proportional to $r^2$. If $r$ doubles, force becomes $\frac{1}{4}$ of the original value.

15. Bouncing Ball.

• (a) Elastic/Inelastic: [2]
  • Inelastic [1].
  • Kinetic energy is not conserved (height decreased, so PE and thus KE after bounce is less than before) [1].
• (b) Fraction of Energy Lost: [2]
  • $PE \propto h$.
  • $E_{initial} \propto h$, $E_{final} \propto 0.8h$.
  • Energy Lost $= h - 0.8h = 0.2h$.
  • Fraction lost $= \frac{0.2h}{h} = 0.2$ (or 20%) [2].

Section C: Long Structured Questions

16. Trolley Motion Sensor.

• (a) Sketch: [1]
  • Straight line starting from origin with positive gradient.
• (b) Acceleration: [1]
  • Acceleration is the gradient (slope) of the velocity-time graph.
• (c) Real-world Gradient: [2]
  • The gradient would decrease [1].
  • As velocity increases, air resistance increases, reducing the resultant force and thus the acceleration [1].

17. Conical Pendulum.

• (a) Free-body Diagram: [2]
  • Weight ($mg$) acting vertically downwards [1].
  • Tension ($T$) acting along the string towards the pivot [1].
• (b) Derivation: [4]
  • Vertical resolution: $T \cos \theta = mg$ (1) [1]
  • Horizontal resolution: $T \sin \theta = \frac{mv^2}{r}$ (2) [1]
  • Divide (2) by (1): $\frac{T \sin \theta}{T \cos \theta} = \frac{mv^2/r}{mg}$ [1]
  • $\tan \theta = \frac{v^2}{rg}$ [1]

18. Car Power.

• (a) Max Speed: [3]
  • At constant max speed, Driving Force $F_D = \text{Resistive Force } F_R = 1500 \text{ N}$ [1].
  • $P = F_D v$ [1]
  • $60000 = 1500 v \Rightarrow v = \frac{60000}{1500} = 40 \text{ m s}^{-1}$ [1]
• (b) Climbing Hill: [2]
  • Speed decreases [1].
  • Component of weight acts down the slope, increasing the total opposing force. Since Power is constant ($P=Fv$), an increase in required Force leads to a decrease in Velocity [1].

19. Impulse.

• (a) Definition: [1]
  • Impulse is the product of the average force and the time interval over which it acts ($I = F \Delta t$), or the change in momentum.
• (b) Calculation: [2]
  • Impulse = Area under Force-Time graph.
  • Area of triangle $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.10 \times 10$ [1]
  • $I = 0.5 \text{ N s}$ [1]
• (c) Final Velocity: [2]
  • $I = \Delta p = m(v - u)$
  • $0.5 = 0.5 (v - 0)$ [1]
  • $v = 1.0 \text{ m s}^{-1}$ [1]

20. Geostationary Satellite.

• (a) Conditions: [2]
  • Orbital period is 24 hours (same as Earth's rotation) [1].
  • Orbits in the same direction as Earth's rotation (West to East) [1].
• (b) Equator: [2]
  • The center of the orbit must be the center of the Earth [1].
  • To remain stationary above a fixed point on Earth, the orbit plane must coincide with the equatorial plane; otherwise, the satellite would oscillate North and South relative to the observer [1].
• (c) Advantage/Disadvantage: [2]
  • Advantage: Satellite appears stationary, so ground antennas do not need tracking mechanisms [1].
  • Disadvantage: High altitude leads to significant signal delay (latency) and weaker signal strength requiring high power/large dishes [1].