A Level H2 Physics Practice Paper 1

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

Answer Key and Marking Scheme

Paper: Practice Paper 1 (Version 1 of 5)
Topic: Mechanics

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 event (collision/explosion) is equal to the total momentum after the event, provided no external forces act [1].
• Marking Notes:
  • 1 mark for "closed/isolated system" or "no external forces".
  • 1 mark for "total momentum before = total momentum after".

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

• Given: $m = 0.15 \text{ kg}$, $A = 4.0 \text{ cm} = 0.04 \text{ m}$, $f = 2.5 \text{ Hz}$.
• Formula: $a_{\max} = \omega^2 A$ and $\omega = 2\pi f$.
• Working:
  • $\omega = 2\pi(2.5) = 5\pi \approx 15.71 \text{ rad s}^{-1}$ [1]
  • $a_{\max} = (15.71)^2 \times 0.04$ [1]
  • $a_{\max} = 9.87 \text{ m s}^{-2}$ [1]
• Answer: $9.9 \text{ m s}^{-2}$ (2 s.f.)

3. Precautions for free-fall experiment. (a) Accuracy of $h$. [1]

• Answer: Use a meter rule with mm graduations and ensure eye is level with the scale to avoid parallax error. OR Measure from the bottom of the ball to the trapdoor.
• Marking Notes: Accept specific practical details. "Be careful" is not accepted.

(b) Accuracy of $t$. [1]

• Answer: Use an electronic timer triggered by the release mechanism and impact sensor to eliminate human reaction time error. OR Repeat the experiment and take the average.
• Marking Notes: Must link to reducing error.

4. Explain what is meant by the binding energy of a nucleus. [2]

• Answer: The energy required to completely separate a nucleus into its constituent protons and neutrons [1]. OR The energy released when protons and neutrons combine to form a nucleus [1]. It is equivalent to the mass defect via $E=mc^2$ [1].
• Marking Notes: 1 mark for "separate constituents", 1 mark for "energy required/released".

5. Calculate the magnitude of the centripetal force. [2]

• Given: $m = 1200 \text{ kg}$, $v = 20 \text{ m s}^{-1}$, $r = 50 \text{ m}$.
• Formula: $F = \frac{mv^2}{r}$
• Working:
  • $F = \frac{1200 \times 20^2}{50}$ [1]
  • $F = \frac{1200 \times 400}{50} = 9600 \text{ N}$ [1]
• Answer: $9600 \text{ N}$

Section B: Calculation and Application

6. Collision of trolleys. (a) Velocity after collision. [3]

• Principle: Conservation of Momentum.
• Working:
  • $m_A u_A + m_B u_B = (m_A + m_B) v$ [1]
  • $(2.0)(3.0) + (1.0)(0) = (2.0 + 1.0) v$
  • $6.0 = 3.0 v$
  • $v = 2.0 \text{ m s}^{-1}$ [1]
  • Direction: To the right [1]
• Answer: $2.0 \text{ m s}^{-1}$ to the right.

(b) Loss in kinetic energy. [3]

• Working:
  • $KE_{\text{initial}} = \frac{1}{2} m_A u_A^2 = \frac{1}{2}(2.0)(3.0)^2 = 9.0 \text{ J}$ [1]
  • $KE_{\text{final}} = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2}(3.0)(2.0)^2 = 6.0 \text{ J}$ [1]
  • Loss $= 9.0 - 6.0 = 3.0 \text{ J}$ [1]
• Answer: $3.0 \text{ J}$

7. Projectile Motion. (a) Maximum height. [3]

• Given: $u = 30 \text{ m s}^{-1}$, $\theta = 40^\circ$.
• Vertical component: $u_y = 30 \sin 40^\circ \approx 19.28 \text{ m s}^{-1}$.
• At max height: $v_y = 0$.
• Formula: $v_y^2 = u_y^2 - 2gh$
• Working:
  • $0 = (19.28)^2 - 2(9.81)h$ [1]
  • $19.62 h = 371.7$
  • $h = 18.9 \text{ m}$ [1]
  • Answer to 2 or 3 s.f. [1]
• Answer: $19 \text{ m}$ (2 s.f.)

(b) Horizontal range. [3]

• Time of flight: $v_y = u_y - gt \Rightarrow 0 = 19.28 - 9.81 t_{\text{up}} \Rightarrow t_{\text{up}} = 1.965 \text{ s}$.
  • Total time $T = 2 \times 1.965 = 3.93 \text{ s}$. [1]
• Horizontal component: $u_x = 30 \cos 40^\circ \approx 22.98 \text{ m s}^{-1}$.
• Range: $R = u_x T$
  • $R = 22.98 \times 3.93 = 90.3 \text{ m}$ [1]
  • Correct unit and s.f. [1]
• Answer: $90 \text{ m}$ (2 s.f.)

8. Block on Inclined Plane. (a) Component of weight down slope. [2]

• Formula: $W_{\parallel} = mg \sin \theta$
• Working:
  • $W_{\parallel} = 5.0 \times 9.81 \times \sin 30^\circ$ [1]
  • $W_{\parallel} = 49.05 \times 0.5 = 24.5 \text{ N}$ [1]
• Answer: $24.5 \text{ N}$

(b) Frictional force. [2]

• Reasoning: Constant speed means zero acceleration, so net force is zero.
• Equation: $F_{\text{pull}} = W_{\parallel} + F_{\text{friction}}$
• Working:
  • $40 = 24.5 + F_{\text{friction}}$ [1]
  • $F_{\text{friction}} = 40 - 24.5 = 15.5 \text{ N}$ [1]
• Answer: $15.5 \text{ N}$

9. Satellite Orbit. (a) Show $v = \sqrt{\frac{GM}{r}}$. [2]

• Working:
  • Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [1]
  • Cancel $m$ and one $r$: $\frac{GM}{r} = v^2 \Rightarrow v = \sqrt{\frac{GM}{r}}$ [1]

(b) Change in orbital speed if radius increases. [2]

• Answer: The orbital speed decreases [1].
• Explanation: Since $v \propto \frac{1}{\sqrt{r}}$, as $r$ increases, $v$ decreases [1].

Section C: Data Analysis and Reasoning

10. Simple Pendulum Experiment. (a) Plot $T^2$ against $L$. [4]

• Data Processing:
  • $L=0.20, T^2=0.81$
  • $L=0.40, T^2=1.61$
  • $L=0.60, T^2=2.40$
  • $L=0.80, T^2=3.20$
  • $L=1.00, T^2=4.04$
• Marking:
  • 1 mark for correct labels and units ($T^2 / \text{s}^2$, $L / \text{m}$).
  • 1 mark for suitable scales.
  • 1 mark for all 5 points plotted correctly.
  • 1 mark for straight line of best fit through origin.

(b) Determine the gradient. [2]

• Working:
  • Gradient $= \frac{\Delta T^2}{\Delta L}$
  • Using points $(0,0)$ and $(1.00, 4.04)$: Gradient $= \frac{4.04 - 0}{1.00 - 0} = 4.04 \text{ s}^2 \text{ m}^{-1}$ [1]
  • Accept range $3.9 - 4.1$ based on line drawn. [1]

(c) Calculate $g$. [3]

• Formula: $T = 2\pi \sqrt{\frac{L}{g}} \Rightarrow T^2 = \frac{4\pi^2}{g} L$.
  • Gradient $= \frac{4\pi^2}{g}$ [1]
  • $g = \frac{4\pi^2}{\text{Gradient}}$ [1]
  • $g = \frac{4\pi^2}{4.04} = 9.77 \text{ m s}^{-2}$ [1]
• Answer: $9.8 \text{ m s}^{-2}$ (2 s.f.)

11. Hooke's Law. (a) Spring constant $k$. [2]

• Formula: $F = kx$
• Working:
  • $2.0 = k(0.04)$
  • $k = \frac{2.0}{0.04} = 50 \text{ N m}^{-1}$ [1]
  • Unit correct [1]
• Answer: $50 \text{ N m}^{-1}$

(b) Elastic potential energy. [2]

• Formula: $E = \frac{1}{2} k x^2$ OR $E = \frac{1}{2} F x$
• Working:
  • $E = \frac{1}{2}(50)(0.04)^2$ [1]
  • $E = 0.04 \text{ J}$ [1]
• Answer: $0.04 \text{ J}$

(c) Change in energy. [2]

• Answer: Quadruples [1].
• Explanation: $E \propto x^2$ (or $E \propto F^2$). Since load doubles, extension doubles. $2^2 = 4$ times the energy [1].

12. Uniform Acceleration. (a) Acceleration. [2]

• Formula: $a = \frac{v - u}{t}$
• Working:
  • $a = \frac{25 - 0}{10} = 2.5 \text{ m s}^{-2}$ [1]
  • Unit correct [1]
• Answer: $2.5 \text{ m s}^{-2}$

(b) Distance travelled. [2]

• Formula: $s = ut + \frac{1}{2}at^2$ OR Area under graph.
• Working:
  • $s = 0 + \frac{1}{2}(2.5)(10)^2$ [1]
  • $s = 125 \text{ m}$ [1]
• Answer: $125 \text{ m}$

(c) Velocity-time graph. [2]

• Sketch:
  • Axes labeled $v / \text{m s}^{-1}$ and $t / \text{s}$ [1].
  • Straight line from $(0,0)$ to $(10, 25)$ [1].