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A Level H1 Physics Practice Paper 2

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A Level H1 Physics AI Generated Generated by Gemma 4 31B Updated 2026-08-17

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Questions

A-Level Physics H1 Quiz - Mechanics

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 65

Duration: 90 Minutes
Total Marks: 65
Instructions: Answer all questions. Show all necessary working. Use $g = 9.81 \text{ m s}^{-2}$ .

Section A: Kinematics and Dynamics (Questions 1-7)

State the principle of conservation of linear momentum. [2]

\
A particle has a mass of $0.40 \text{ kg}$ , a momentum of $1.2 \text{ kg m s}^{-1}$ , and a kinetic energy of $1.8 \text{ J}$ . Calculate the velocity of the particle. [3]

\
A ball is thrown from the top of a building of height $20 \text{ m}$ with an initial vertical velocity of $5.0 \text{ m s}^{-1}$ upwards. Calculate the time taken to reach the ground. [3]

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A projectile is launched at an angle of $35^\circ$ to the horizontal with a speed of $25 \text{ m s}^{-1}$ . Determine the maximum height reached by the projectile. [3]

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Sketch a graph of velocity versus time for an object falling from rest through a viscous fluid. Label the terminal velocity $v_t$ . [2]

[Graph space: 12 lines]
Explain the shape of the graph sketched in Question 5, referring to the forces acting on the object. [3]

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A $1.5 \text{ kg}$ block is pushed across a rough horizontal surface with a constant force of $10 \text{ N}$ . If the block accelerates at $2.0 \text{ m s}^{-1}$ , calculate the magnitude of the frictional force. [3]

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Section B: Momentum and Collisions (Questions 8-13)

Distinguish between an elastic collision and an inelastic collision. [2]

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A $0.20 \text{ kg}$ trolley moving at $3.0 \text{ m s}^{-1}$ collides with a stationary $0.30 \text{ kg}$ trolley. They stick together after the collision. Calculate the common velocity. [3]

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In the collision described in Question 9, calculate the loss in kinetic energy. [3]

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Two spheres of equal mass $m$ collide elastically in two dimensions. Sphere A moves along the x-axis at $u$ . After collision, Sphere A moves at angle $30^\circ$ to the x-axis with speed $v_1$ . Express the velocity of Sphere B in terms of $u$ and $v_1$ . [4]

\
A $0.05 \text{ kg}$ tennis ball hits a wall perpendicularly at $20 \text{ m s}^{-1}$ and rebounds at $15 \text{ m s}^{-1}$ . Calculate the impulse delivered to the ball. [3]

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A $1200 \text{ kg}$ car traveling at $15 \text{ m s}^{-1}$ collides with a $800 \text{ kg}$ car traveling in the opposite direction at $10 \text{ m s}^{-1}$ . If they lock together, determine the final velocity and direction. [4]

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Section C: Forces, Equilibrium and Energy (Questions 14-20)

A uniform plank of length $4.0 \text{ m}$ and mass $20 \text{ kg}$ is supported by two pivots at its ends. A $60 \text{ kg}$ person stands $1.0 \text{ m}$ from the left end. Draw a free-body diagram of the plank. [3]

[Drawing space: 10 lines]
Using the scenario in Question 14, calculate the reaction force at the left and right pivots. [4]

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Define the term "Work Done" by a force and state its SI unit. [2]

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A $500 \text{ g}$ mass is lifted vertically through a height of $2.0 \text{ m}$ in $1.5 \text{ s}$ at a constant speed. Calculate the power output of the lifting agent. [3]

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An object of mass $2 \text{ kg}$ is compressed against a spring with spring constant $k = 500 \text{ N m}^{-1}$ by a distance of $0.1 \text{ m}$ . If the spring is released, calculate the speed of the object as it leaves the spring. [4]

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A motor lifts a $100 \text{ kg}$ crate at a constant speed of $0.4 \text{ m s}^{-1}$ . If the input electrical power is $600 \text{ W}$ , calculate the efficiency of the motor. [3]

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A block slides down a rough inclined plane at a constant velocity. Explain why the acceleration is zero and identify the forces acting parallel to the plane. [4]

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Answers

Answer Key - A-Level Physics H1 Quiz: Mechanics

Section A: Kinematics and Dynamics

[2 marks]
- In a closed/isolated system, the total linear momentum remains constant [1]
- provided no external forces act on the system [1].
[3 marks]
- $p = mv \rightarrow 1.2 = 0.40 \times v \rightarrow v = 3.0 \text{ m s}^{-1}$ [2]
- Check with KE: $K = \frac{1}{2}(0.4)(3)^2 = 1.8 \text{ J}$ . Consistent. [1]
- Answer: $3.0 \text{ m s}^{-1}$
[3 marks]
- $s = ut + \frac{1}{2}at^2 \rightarrow -20 = 5t - 4.905t^2$ [1]
- $4.905t^2 - 5t - 20 = 0$
- Using quadratic formula: $t = \frac{5 \pm \sqrt{25 - 4(4.905)(-20)}}{2(4.905)} = \frac{5 \pm 20.5}{9.81}$ [1]
- $t \approx 2.55 \text{ s}$ [1]
[3 marks]
- $u_y = 25 \sin

<stage5_exam_answers_md>
# Answer Key - A-Level Physics H1 Quiz: Mechanics

### Section A: Kinematics and Dynamics
1. **[2 marks]** 
   - In a closed/isolated system, the total linear momentum remains constant [1]
   - provided no external forces act on the system [1].

2. **[3 marks]**
   - $p = mv \rightarrow 1.2 = 0.40 \times v \rightarrow v = 3.0 \text{ m s}^{-1}$ [2]
   - Check with KE: $K = \frac{1}{2}(0.4)(3)^2 = 1.8 \text{ J}$. Consistent. [1]
   - **Answer:** $3.0 \text{ m s}^{-1}$

3. **[3 marks]**
   - $s = ut + \frac{1}{2}at^2 \rightarrow -20 = 5t - 4.905t^2$ [1]
   - $4.905t^2 - 5t - 20 = 0$
   - Using quadratic formula: $t = \frac{5 \pm \sqrt{25 - 4(4.905)(-20)}}{2(4.905)} = \frac{5 \pm 20.5}{9.81}$ [1]
   - $t \approx 2.55 \text{ s}$ [1]

4. **[3 marks]**
   - $u_y = 25 \sin 35^\circ \approx 14.34 \text{ m s}^{-1}$ [1]
   - $v_y^2 = u_y^2 + 2as \rightarrow 0 = (14.34)^2 + 2(-9.81)h$ [1]
   - $h = \frac{205.6}{19.62} \approx 10.5 \text{ m}$ [1]

5. **[2 marks]**
   - Graph should show velocity increasing from 0 and curving asymptotically towards a horizontal line $v_t$ [2].

6. **[3 marks]**
   - Initially, only weight acts downwards, so acceleration is $g$ [1].
   - As speed increases, viscous drag increases [1].
   - When drag equals weight, net force is zero and terminal velocity is reached [1].

7. **[3 marks]**
   - $F_{net} = ma \rightarrow 10 - f = 1.5(2.0)$ [1]
   - $10 - f = 3.0$ [1]
   - $f = 7.0 \text{ N}$ [1]

### Section B: Momentum and Collisions
8. **[2 marks]**
   - Elastic: Kinetic energy is conserved [1].
   - Inelastic: Kinetic energy is not conserved (lost to heat/sound) [1].

9. **[3 marks]**
   - $m_1u_1 + m_2u_2 = (m_1+m_2)v$ [1]
   - $(0.20)(3.0) + 0 = (0.50)v$ [1]
   - $v = 1.2 \text{ m s}^{-1}$ [1]

10. **[3 marks]**
    - $KE_{initial} = \frac{1}{2}(0.20)(3)^2 = 0.9 \text{ J}$ [1]
    - $KE_{final} = \frac{1}{2}(0.50)(1.2)^2 = 0.36 \text{ J}$ [1]
    - $\Delta KE = 0.9 - 0.36 = 0.54 \text{ J}$ [1]

11. **[4 marks]**
    - X-axis: $mu = mv_1 \cos 30^\circ + mv_{Bx} \rightarrow v_{Bx} = u - v_1 \cos 30^\circ$ [2]
    - Y-axis: $0 = mv_1 \sin 30^\circ + mv_{By} \rightarrow v_{By} = -v_1 \sin 30^\circ$ [2]
    - $\vec{v}_B = (u - v_1 \cos 30^\circ)\hat{i} - (v_1 \sin 30^\circ)\hat{j}$

12. **[3 marks]**
    - $\Delta p = m(v - u) = 0.05(15 - (-20))$ [1]
    - $\Delta p = 0.05(35)$ [1]
    - $\Delta p = 1.75 \text{ N s}$ [1]

13. **[4 marks]**
    - $(1200)(15) + (800)(-10) = (1200+800)v$ [1]
    - $18000 - 8000 = 2000v$ [1]
    - $10000 = 2000v \rightarrow v = 5.0 \text{ m s}^{-1}$ [1]
    - Direction: In the direction of the $1200 \text{ kg}$ car [1].

### Section C: Forces, Equilibrium and Energy
14. **[3 marks]**
    - Diagram showing: Weight of plank at center, weight of person at $1 \text{ m}$, upward reactions $R_L$ and $R_R$ at ends [3].

15. **[4 marks]**
    - $\sum \tau_{left} = 0 \rightarrow (60 \times 9.81)(1) + (20 \times 9.81)(2) = R_R(4)$ [1]
    - $588.6 + 392.4 = 4R_R \rightarrow R_R = 245.25 \text{ N}$ [2]
    - $R_L = (60+20)9.81 - 245.25 = 784.8 - 245.25 = 539.55 \text{ N}$ [1]

16. **[2 marks]**
    - Work Done: Product of force and displacement in the direction of the force [1].
    - Unit: Joule (J) [1].

17. **[3 marks]**
    - $W = mgh = (0.5)(9.81)(2.0) = 9.81 \text{ J}$ [1]
    - $P = W/t = 9.81 / 1.5$ [1]
    - $P = 6.54 \text{ W}$ [1]

18. **[4 marks]**
    - $PE_{spring} = \frac{1}{2}kx^2 = \frac{1}{2}(500)(0.1)^2 = 2.5 \text{ J}$ [2]
    - $2.5 = \frac{1}{2}mv^2 \rightarrow 2.5 = \frac{1}{2}(2)v^2$ [1]
    - $v = \sqrt{2.5} \approx 1.58 \text{ m s}^{-1}$ [1]

19. **[3 marks]**
    - $P_{out} = Fv = (100 \times 9.81) \times 0.4 = 392.4 \text{ W}$ [1]
    - $\text{Efficiency} = \frac{392.4}{600} \times 100\%$ [1]
    - $\text{Efficiency} = 65.4\%$ [1]

20. **[4 marks]**
    - Constant velocity means $a=0$, so net force is zero [1].
    - Forces parallel to plane: Component of weight ($mg \sin \theta$) acting down [2] and Frictional force acting up [1].