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A Level H1 Physics Practice Paper 2

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A Level H1 Physics From Real Exams Generated by Gemma 4 31B Updated 2026-06-03
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Questions

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A-Level Physics H1 Quiz - Mechanics

Name: ____________________
Class: ____________________
Date: ____________________
Score: ________ / 55

Duration: 60 Minutes
Total Marks: 55 Marks

Instructions:

  • Answer all questions.
  • Use g=9.81 m s−2g = 9.81\text{ m s}^{-2} unless otherwise stated.
  • Show all working clearly for calculation questions.
  • Use a scientific calculator where necessary.

Section A: Fundamentals and Kinematics (Questions 1–7)

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

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  2. Write down the expressions for the momentum pp and kinetic energy KK of a particle of mass mm moving with velocity vv. [2] (a) p=p = ____________________ (b) K=K = ____________________

  3. A small metal sphere has a horizontal momentum of 0.45 N s0.45\text{ N s} and a kinetic energy of 0.12 J0.12\text{ J}. Calculate the mass and velocity of the sphere. [3]


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  4. A ball is dropped from a height of 20 m20\text{ m}. Sketch the graph of vertical speed vv against time tt for the ball's motion, taking into account the effect of air resistance. [2]


    (Space for graph)

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  5. Explain the shape of the graph you sketched in Question 4, specifically referring to the concept of terminal velocity. [2]

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  6. A projectile is launched with an initial velocity uu at an angle θ\theta to the horizontal. State the acceleration of the projectile in the horizontal direction, assuming no air resistance. [1]
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  7. A car accelerates uniformly from 10 m s−110\text{ m s}^{-1} to 25 m s−125\text{ m s}^{-1} over a distance of 100 m100\text{ m}. Calculate the acceleration of the car. [3]

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Section B: Dynamics and Forces (Questions 8–14)

  1. Define the term impulse and state its SI unit. [2]
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  2. A 0.5 kg0.5\text{ kg} block is pushed across a rough horizontal surface with a constant force of 10 N10\text{ N}. If the coefficient of kinetic friction is 0.30.3, calculate the acceleration of the block. [3]

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  3. A uniform plank AB of length 4.0 m4.0\text{ m} and weight 120 N120\text{ N} is placed across two supports. A person of weight 600 N600\text{ N} stands at a distance xx from end A. Draw a free-body diagram of the plank, labeling all forces acting on it. [3]

    (Space for diagram)

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  4. Using the scenario in Question 10, if the plank is in equilibrium and the supports are at the ends A and B, calculate the reaction force at support B when the person is 1.0 m1.0\text{ m} from end A. [4]

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  5. Two trolleys, A (1.0 kg1.0\text{ kg}) and B (2.0 kg2.0\text{ kg}), move toward each other with speeds 3.0 m s−13.0\text{ m s}^{-1} and 2.0 m s−12.0\text{ m s}^{-1} respectively. They collide and stick together. Calculate the final velocity of the combined mass. [3]

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  6. Distinguish between an elastic collision and an inelastic collision in terms of kinetic energy. [2]
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  7. A 2.0 kg2.0\text{ kg} object is acted upon by two perpendicular forces: F1=6.0 NF_1 = 6.0\text{ N} and F2=8.0 NF_2 = 8.0\text{ N}. Calculate the magnitude of the resultant acceleration. [3]

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Section C: Work, Energy, and Power (Questions 15–20)

  1. Define work done by a force and state the condition under which no work is done even if a force is applied. [2]
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  2. A 0.2 kg0.2\text{ kg} ball is thrown vertically upwards with an initial speed of 15 m s−115\text{ m s}^{-1}. Calculate the maximum height reached by the ball, ignoring air resistance. [3]

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  3. A motor lifts a 50 kg50\text{ kg} load through a height of 10 m10\text{ m} in 5.0 s5.0\text{ s}. Calculate the average power output of the motor. [3]

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  4. A car of mass 1200 kg1200\text{ kg} travels at a constant speed of 30 m s−130\text{ m s}^{-1}. If the total resistive force is 600 N600\text{ N}, calculate the power required to maintain this speed. [3]

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  5. An object of mass mm is compressed against a spring of spring constant kk by a distance xx. If the object is released, derive an expression for the maximum speed vv of the object. [3]

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  6. A 0.1 kg0.1\text{ kg} block slides down a rough inclined plane at 30∘30^\circ to the horizontal. If the block starts from rest and slides 2.0 m2.0\text{ m} before stopping, calculate the work done against friction. [4]

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Answers

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Answer Key - A-Level Physics H1 Quiz (Mechanics)

1. In a closed/isolated system, the total linear momentum remains constant provided no external forces act. [B1 for constant momentum, B1 for no external forces]

2. (a) p=mvp = mv [B1] (b) K=12mv2K = \frac{1}{2}mv^2 [B1]

3.

  • p=mv  ⟹  v=p/mp = mv \implies v = p/m
  • K=12mv2  ⟹  K=12m(p/m)2=p2/2mK = \frac{1}{2}mv^2 \implies K = \frac{1}{2}m(p/m)^2 = p^2 / 2m [M1]
  • m=p2/2K=(0.45)2/(2×0.12)=0.2025/0.24=0.844 kgm = p^2 / 2K = (0.45)^2 / (2 \times 0.12) = 0.2025 / 0.24 = 0.844\text{ kg} [A1]
  • v=0.45/0.844=0.533 m s−1v = 0.45 / 0.844 = 0.533\text{ m s}^{-1} [A1]

4. Graph should show a curve starting from origin, increasing gradient initially, then flattening off to a horizontal asymptote (terminal velocity). [B2]

5. As speed increases, the air resistance (drag) increases. [B1] The net downward force (W−DragW - \text{Drag}) decreases, causing acceleration to decrease until it becomes zero when air resistance equals weight, resulting in a constant terminal velocity. [B1]

6. 0 m s−20\text{ m s}^{-2} (Acceleration is zero in the horizontal direction). [B1]

7. v2=u2+2as  ⟹  (25)2=(10)2+2a(100)v^2 = u^2 + 2as \implies (25)^2 = (10)^2 + 2a(100) [M1]

  • 625=100+200a625 = 100 + 200a
  • 525=200a525 = 200a
  • a=2.625 m s−2a = 2.625\text{ m s}^{-2} [A2]

8. Impulse is the product of the force acting on an object and the time interval over which it acts (or the change in momentum). [B1] Unit: N s\text{N s} or kg m s−1\text{kg m s}^{-1}. [B1]

9. Fnet=Fpush−fk=10−(0.3×0.5×9.81)F_{\text{net}} = F_{\text{push}} - f_k = 10 - (0.3 \times 0.5 \times 9.81) [M1]

  • Fnet=10−1.47=8.53 NF_{\text{net}} = 10 - 1.47 = 8.53\text{ N} [M1]
  • a=Fnet/m=8.53/0.5=17.06 m s−2a = F_{\text{net}} / m = 8.53 / 0.5 = 17.06\text{ m s}^{-2} [A1]

10. Diagram must show:

  • Weight of plank (120 N120\text{ N}) acting at the center (2.0 m2.0\text{ m} from A). [B1]
  • Weight of person (600 N600\text{ N}) acting at distance xx. [B1]
  • Upward reaction forces RAR_A and RBR_B at ends A and B. [B1]

11. Take moments about A:

  • ∑Clockwise Moments=∑Anti-clockwise Moments\sum \text{Clockwise Moments} = \sum \text{Anti-clockwise Moments}
  • (600×1.0)+(120×2.0)=RB×4.0(600 \times 1.0) + (120 \times 2.0) = R_B \times 4.0 [M1]
  • 600+240=4.0RB600 + 240 = 4.0 R_B [M1]
  • 840=4.0RB840 = 4.0 R_B
  • RB=210 NR_B = 210\text{ N} [A2]

12. mAuA+mBuB=(mA+mB)vm_A u_A + m_B u_B = (m_A + m_B)v (taking direction of A as positive) [M1]

  • (1.0×3.0)+(2.0×−2.0)=(1.0+2.0)v(1.0 \times 3.0) + (2.0 \times -2.0) = (1.0 + 2.0)v
  • 3.0−4.0=3.0v3.0 - 4.0 = 3.0v [M1]
  • −1.0=3.0v  ⟹  v=−0.333 m s−1-1.0 = 3.0v \implies v = -0.333\text{ m s}^{-1} (opposite to A's initial direction) [A1]

13. In an elastic collision, total kinetic energy is conserved. [B1] In an inelastic collision, total kinetic energy is not conserved (some is converted to heat/sound). [B1]

14. Fres=62+82=36+64=10 NF_{\text{res}} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ N} [M1]

  • a=Fres/m=10/2.0=5.0 m s−2a = F_{\text{res}} / m = 10 / 2.0 = 5.0\text{ m s}^{-2} [A2]

15. Work done is the product of the force and the displacement in the direction of the force (W=Fdcos⁡θW = Fd \cos \theta). [B1] No work is done if the force is perpendicular to the displacement (θ=90∘\theta = 90^\circ). [B1]

16. mgh=12mv2  ⟹  h=v2/2gmgh = \frac{1}{2}mv^2 \implies h = v^2 / 2g [M1]

  • h=(15)2/(2×9.81)=225/19.62h = (15)^2 / (2 \times 9.81) = 225 / 19.62 [M1]
  • h=11.47 mh = 11.47\text{ m} [A1]

17. W=mgh=50×9.81×10=4905 JW = mgh = 50 \times 9.81 \times 10 = 4905\text{ J} [M1]

  • P=W/t=4905/5.0=981 WP = W / t = 4905 / 5.0 = 981\text{ W} [A2]

18. P=FvP = Fv (for constant speed) [M1]

  • P=600×30=18,000 WP = 600 \times 30 = 18,000\text{ W} or 18 kW18\text{ kW} [A2]

19. Energy conservation: 12kx2=12mv2\frac{1}{2}kx^2 = \frac{1}{2}mv^2 [M1]

  • v2=kx2/mv^2 = kx^2 / m [M1]
  • v=xk/mv = x \sqrt{k/m} [A1]

20. Change in Energy = Work done by friction

  • ΔE=mgh−0=m(ssin⁡30∘)g\Delta E = mgh - 0 = m(s \sin 30^\circ)g [M1]
  • Wfric=0.1×(2.0×0.5)×9.81W_{\text{fric}} = 0.1 \times (2.0 \times 0.5) \times 9.81 [M1]
  • Wfric=0.1×1.0×9.81=0.981 JW_{\text{fric}} = 0.1 \times 1.0 \times 9.81 = 0.981\text{ J} [A2]