A Level H1 Physics Practice Paper 1

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TuitionGoWhere Exam Practice (AI) — Physics H1 A-Level

Practice Paper: Mechanics (Version 1 of 5)

School: TuitionGoWhere Exam Practice (AI)
Subject: Physics H1
Level: A-Level
Paper: Practice Paper (Mechanics Topic Quiz)
Version: 1 of 5
Duration: 75 minutes
Total Marks: 60
Name: ________________________
Class: ________________________
Date: ________________________

Instructions:

• Answer all questions in the spaces provided.
• Show all working clearly. Use SI units and appropriate notation.
• For calculation questions, unsupported final answers may receive reduced credit.
• Section A: Short structured questions (1–10). Section B: Data and graph interpretation (11–15). Section C: Extended mechanics reasoning (16–20).

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Section A: Short Structured Questions (22 marks)

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

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2. Write down, in terms of mass $m$ and velocity $v$: (a) the linear momentum $p$; [1] (b) the kinetic energy $K$. [1]

(a) ____________________
(b) ____________________

3. A drone has a horizontal momentum of $12\ \text{N·s}$ and kinetic energy of $24\ \text{J}$. Calculate the mass and speed of the drone. [3]

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4. Define the moment of a force about a point. [2]

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5. A uniform rod AB of length $4.0\ \text{m}$ and weight $80\ \text{N}$ rests on two smooth supports at A and B. A child of weight $40\ \text{N}$ stands $1.0\ \text{m}$ from A. Draw a labelled diagram showing all forces acting on the rod. [3]

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6. A car accelerates uniformly from rest to $20\ \text{m s}^{-1}$ in $10\ \text{s}$. Calculate its acceleration. [2]

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7. State Newton’s first law of motion. [2]

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8. A block of mass $2.0\ \text{kg}$ is pulled along a horizontal surface by a constant force of $10\ \text{N}$. If frictional resistance is $4.0\ \text{N}$, calculate the acceleration of the block. [2]

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9. A projectile is launched horizontally from a cliff of height $45\ \text{m}$ with speed $30\ \text{m s}^{-1}$. Calculate the time taken to reach the ground. (Use $g = 9.8\ \text{m s}^{-2}$.) [2]

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10. State the condition for a body to be in translational equilibrium. [2]

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Section B: Data and Graph Interpretation (18 marks)

11. The velocity–time graph below shows the motion of a cyclist.

Image pending generation: graph for Q11.

(a) Calculate the acceleration during the first 10 s. [2]
(b) Determine the total distance travelled in 20 s. [2]

(a) ____________________
(b) ____________________

12. A ball of mass $0.50\ \text{kg}$ moving at $6.0\ \text{m s}^{-1}$ collides head-on with a stationary ball of mass $0.30\ \text{kg}$. After collision the $0.50\ \text{kg}$ ball moves at $2.0\ \text{m s}^{-1}$ in the same direction. Calculate the velocity of the $0.30\ \text{kg}$ ball after collision. [3]

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13. The table shows force $F$ against extension $x$ for a spring.

$x$ / mm	0	10	20	30	40
$F$ / N	0	2	4	6	8

(a) State Hooke’s law as applied to this spring. [1]
(b) Determine the spring constant $k$ in $\text{N m}^{-1}$. [2]

(a) ____________________
(b) ____________________

14. A uniform plank of length $5.0\ \text{m}$ and weight $100\ \text{N}$ is supported at its ends. A box of weight $50\ \text{N}$ is placed $2.0\ \text{m}$ from the left support. Calculate the reaction force at the right support. [3]

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15. Sketch a displacement–time graph for an object moving with constant negative acceleration from an initial positive velocity. [3]

Image pending generation: graph for Q15.

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Section C: Extended Mechanics Reasoning (20 marks)

16. Explain why a person jumping from a boat causes the boat to move backwards, using the principle of conservation of momentum. [4]

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17. A car of mass $1200\ \text{kg}$ travelling at $15\ \text{m s}^{-1}$ is brought to rest by a constant braking force of $3000\ \text{N}$. (a) Calculate the deceleration. [2] (b) Calculate the stopping distance. [2] (c) State one assumption made. [1]

(a) ____________________
(b) ____________________
(c) ____________________

18. Describe how you would determine the centre of gravity of an irregular lamina using a plumb line. [4]

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19. A satellite orbits Earth at constant speed in a circular path. Explain why its velocity is changing even though its speed is constant, and name the force providing the centripetal acceleration. [4]

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20. A block of mass $m$ is released from rest at the top of a frictionless incline of height $h$. Derive an expression for its speed at the bottom using conservation of energy. [5]

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TuitionGoWhere Exam Practice (AI) — Physics H1 A-Level

Practice Paper: Mechanics (Version 1 of 5) — Answer Key

Total Marks: 60

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Section A Answers (22 marks)

Q1. [2]
Principle: In a closed (or isolated) system, the total linear momentum remains constant. [B1]
Provided no net external force acts on the system. [B1]
Teaching note: Momentum is a vector; "closed system" means no external forces. Do not confuse with energy conservation.

Q2. [2 total: (a) 1, (b) 1]
(a) $p = mv$ [1]
(b) $K = \frac{1}{2}mv^2$ [1]
Common trap: Omitting $\frac{1}{2}$ in kinetic energy.

Q3. [3]
Given $p = 12\ \text{N·s}$, $K = 24\ \text{J}$.
Use $K = \frac{p^2}{2m} \Rightarrow m = \frac{p^2}{2K} = \frac{12^2}{2\times24} = \frac{144}{48} = 3.0\ \text{kg}$ [M1+ A1]
Then $v = \frac{p}{m} = \frac{12}{3.0} = 4.0\ \text{m s}^{-1}$ [M1+ A1]
Marking: 1 for correct formula rearrangement, 1 for mass, 1 for velocity.

Q4. [2]
The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. [B1]
Unit: N·m. [B1]
Note: Must mention perpendicular distance.

Q5. [3]
Forces: weight of rod $80\ \text{N}$ downward at centre (2.0 m from A); weight of child $40\ \text{N}$ downward at 1.0 m from A; reaction $R_A$ upward at A; reaction $R_B$ upward at B. [3]
Marking: 1 for rod weight at centre, 1 for child weight position, 1 for two reactions.

Q6. [2]
$a = \frac{v-u}{t} = \frac{20-0}{10} = 2.0\ \text{m s}^{-2}$ [2]
Working: substitution and answer.

Q7. [2]
A body remains at rest or in uniform motion in a straight line unless acted upon by a net external force. [2]

Q8. [2]
Net force = $10 - 4 = 6\ \text{N}$.
$a = \frac{F}{m} = \frac{6}{2.0} = 3.0\ \text{m s}^{-2}$ [2]

Q9. [2]
Vertical motion: $s = \frac{1}{2}gt^2 \Rightarrow 45 = \frac{1}{2}(9.8)t^2 \Rightarrow t^2 = \frac{90}{9.8} = 9.18 \Rightarrow t = 3.0\ \text{s}$ [2]

Q10. [2]
The resultant force on the body is zero (vector sum of all forces = 0). [2]

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Section B Answers (18 marks)

Q11. [4 total: (a) 2, (b) 2]
(a) $a = \frac{\Delta v}{\Delta t} = \frac{8-0}{10} = 0.8\ \text{m s}^{-2}$ [2]
(b) Distance = area under graph = triangle + rectangle = $\frac{1}{2}(10)(8) + (10)(8) = 40 + 80 = 120\ \text{m}$ [2]
Image needed: graph with axes as described; area calculation verified visually.

Q12. [3]
Conservation of momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
$0.50(6.0) + 0 = 0.50(2.0) + 0.30v_2$
$3.0 = 1.0 + 0.30v_2 \Rightarrow 0.30v_2 = 2.0 \Rightarrow v_2 = 6.67\ \text{m s}^{-1}$ [3]
Marking: 1 equation, 1 substitution, 1 answer.

Q13. [3 total: (a) 1, (b) 2]
(a) Extension proportional to force applied (within limit). [1]
(b) $k = \frac{F}{x} = \frac{2\ \text{N}}{0.010\ \text{m}} = 200\ \text{N m}^{-1}$ (using any point). [2]

Q14. [3]
Take moments about left support: $R_R \times 5.0 = 100\times2.5 + 50\times2.0$
$R_R \times 5 = 250 + 100 = 350 \Rightarrow R_R = 70\ \text{N}$ [3]
Marking: 1 moments eqn, 1 calc, 1 answer.

Q15. [3]
Sketch: downward-opening parabola, starts at origin with positive slope, reaches max then decreases. Axes labelled $t$ and $s$. [3]
Image: see placeholder; shape must show decreasing gradient to zero then negative.

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Section C Answers (20 marks)

Q16. [4]
Initially total momentum = 0. [1]
Person pushes boat backward as they jump forward. [1]
Momentum before = momentum after: $m_p v_p + m_b v_b = 0$. [1]
Hence boat gains opposite momentum. [1]

Q17. [5 total: (a) 2, (b) 2, (c) 1]
(a) $a = \frac{F}{m} = \frac{-3000}{1200} = -2.5\ \text{m s}^{-2}$ (deceleration $2.5$). [2]
(b) $v^2 = u^2 + 2as \Rightarrow 0 = 15^2 + 2(-2.5)s \Rightarrow s = 45\ \text{m}$. [2]
(c) e.g. constant braking force, level road. [1]

Q18. [4]
Suspend lamina from a point, mark vertical line via plumb line. [1]
Repeat from another point. [1]
Intersection of lines is centre of gravity. [1]
State rationale: body balances at CoG. [1]

Q19. [4]
Velocity changes because direction changes continuously. [2]
Force: gravitational (or gravity). [2]

Q20. [5]
PE at top = $mgh$. [1]
KE at bottom = $\frac{1}{2}mv^2$. [1]
Conservation: $mgh = \frac{1}{2}mv^2$. [1]
Cancel $m$: $gh = \frac{1}{2}v^2$. [1]
$v = \sqrt{2gh}$. [1]