A Level H1 Physics Practice Paper 5

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Exam Practice (AI)

Subject: Physics H1
Level: A-Level
Paper: Practice Paper (Version 5 of 5)
Duration: 1 hour 30 minutes
Total Marks: 60
Name: ___________________________
Class: ___________________________
Date: ___________________________

Instructions:

• Answer all questions in Sections A, B, and C.
• Show all working clearly where calculation is required.
• Use SI units unless otherwise stated.
• An approved calculator may be used.

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Section A: Foundations of Mechanics (Questions 1–8) [24 marks]

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

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

(a) _________________________
(b) _________________________

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

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

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5. A uniform plank AB has length $4.0\ \text{m}$ and weight $200\ \text{N}$. It rests on two supports at A and B. A child of weight $300\ \text{N}$ stands $1.0\ \text{m}$ from A.
Draw a labelled diagram showing all forces acting on the plank. [3]

Image pending generation: diagram for Q5.

6. For the plank in Q5, calculate the reaction force at support B. [3]

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

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8. Sketch a velocity–time graph for an object moving with constant deceleration from $15\ \text{m s}^{-1}$ to rest in $5\ \text{s}$. Label axes with units. [2]

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Section B: Motion and Forces (Questions 9–14) [20 marks]

9. A ball is thrown vertically upward with initial speed $20\ \text{m s}^{-1}$. Using $g = 9.8\ \text{m s}^{-2}$, calculate the maximum height reached. [3]

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10. Describe how the gradient of a displacement–time graph is related to the motion of an object. [2]

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11. A block of mass $5.0\ \text{kg}$ is pulled along a horizontal surface by a force of $30\ \text{N}$ and experiences friction of $5\ \text{N}$. Calculate the acceleration of the block. [3]

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12.

Image pending generation: graph for Q12.

Using the graph in Q12-fig1, determine:
(a) the acceleration during the first $4\ \text{s}$, [2]
(b) the total distance travelled. [2]

(a) _________________________
(b) _________________________

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

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14. 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. ($g = 9.8\ \text{m s}^{-2}$) [3]

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Section C: Energy, Collisions and Circular Motion (Questions 15–20) [16 marks]

15. A $2.0\ \text{kg}$ object moving at $4.0\ \text{m s}^{-1}$ collides with a stationary $3.0\ \text{kg}$ object. They stick together. Calculate their common final speed. [3]

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16. Explain the difference between an elastic and an inelastic collision in terms of kinetic energy. [2]

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17. A mass of $0.50\ \text{kg}$ is whirled in a horizontal circle of radius $1.2\ \text{m}$ at $3.0\ \text{rev s}^{-1}$. Calculate the centripetal force. [3]

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18. A crane lifts a $500\ \text{kg}$ load through $20\ \text{m}$ in $10\ \text{s}$. Calculate the useful power output. ($g = 9.8\ \text{m s}^{-2}$) [3]

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19. State what is meant by gravitational field strength at a point. [2]

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20. A satellite orbits Earth at constant speed in a circular path. Explain why it is accelerating even though its speed is constant. [3]

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End of Paper

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TuitionGoWhere Practice Paper - Physics H1 A-Level (Version 5) Answer Key

Total Marks: 60

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Section A

Q1. [2]
Principle: In a closed (or isolated) system, the total linear momentum remains constant provided no net external force acts.
Marking: B1 for "total momentum constant / before = after"; B1 for "closed system / no external force".
Teaching: Momentum is conserved vectorially; external impulses change total momentum.

Q2. [2]
(a) $p = mv$ [1]
(b) $E_k = \frac{1}{2}mv^2$ [1]
Teaching: Momentum is vector, KE is scalar; do not omit ½.

Q3. [3]
Given $p = 12\ \text{N·s}$, $E_k = 24\ \text{J}$.
Use $E_k = \frac{p^2}{2m} \Rightarrow m = \frac{p^2}{2E_k} = \frac{12^2}{2 \times 24} = \frac{144}{48} = 3.0\ \text{kg}$ [M1+A1]
$v = p/m = 12 / 3.0 = 4.0\ \text{m s}^{-1}$ [M1]
Answer: mass $3.0\ \text{kg}$, speed $4.0\ \text{m s}^{-1}$.

Q4. [2]
Moment = force × perpendicular distance from point to line of action.
B1 definition, B1 perpendicular distance.

Q5. [3]
Forces: $R_A$ up at A, $R_B$ up at B, $200\ \text{N}$ down at 2.0 m from A, $300\ \text{N}$ down at 1.0 m from A.
Marking: 1 each for correct four forces with positions; deduct if weight at centre omitted.

Q6. [3]
Take moments about A:
$R_B \times 4.0 = 200 \times 2.0 + 300 \times 1.0$ [M1]
$R_B \times 4 = 400 + 300 = 700$ [M1]
$R_B = 175\ \text{N}$ [A1]
Teaching: Plank uniform → weight at midpoint.

Q7. [2]
$a = \frac{\Delta v}{t} = \frac{20 - 0}{10} = 2.0\ \text{m s}^{-2}$ [2]

Q8. [2]
Straight line from (0,15) to (5,0); axes labelled v (m s⁻¹) and t (s). [2]

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Section B

Q9. [3]
$v^2 = u^2 + 2as$, at top $v=0$, $a = -9.8$:
$0 = 20^2 - 2(9.8)h \Rightarrow h = 400 / 19.6 = 20.4\ \text{m}$ [3]

Q10. [2]
Gradient = velocity (rate of change of displacement with time). [2]

Q11. [3]
Net force = $30 - 5 = 25\ \text{N}$ [M1]
$a = F/m = 25 / 5.0 = 5.0\ \text{m s}^{-2}$ [M1+A1]

Q12. [4]
(a) $a = (12-0)/4 = 3.0\ \text{m s}^{-2}$ [2]
(b) Area = triangle + rectangle + triangle = ½×4×12 + 6×12 + ½×4×12 = 24+72+24 = 120 m [2]

Q13. [2]
Body remains at rest or uniform motion unless acted by net external force. [2]

Q14. [3]
Vertical: $s = ½gt^2 \Rightarrow 45 = ½(9.8)t^2$ [M1]
$t^2 = 90/9.8 = 9.18 \Rightarrow t = 3.03\ \text{s}$ [M1+A1]

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Section C

Q15. [3]
Momentum before = $2.0×4.0 = 8.0\ \text{kg m s}^{-1}$ [M1]
After: $(2.0+3.0)v = 8.0 \Rightarrow v = 1.6\ \text{m s}^{-1}$ [M1+A1]

Q16. [2]
Elastic: total KE conserved. Inelastic: KE not conserved (some lost). [2]

Q17. [3]
$\omega = 2\pi f = 2\pi×3.0 = 18.85\ \text{rad s}^{-1}$ [M1]
$F = m\omega^2 r = 0.50 × (18.85)^2 × 1.2 = 213\ \text{N}$ [M1+A1]

Q18. [3]
Work = $mgh = 500×9.8×20 = 98000\ \text{J}$ [M1]
Power = $98000/10 = 9800\ \text{W}$ [M1+A1]

Q19. [2]
Gravitational force per unit mass at point. [2]

Q20. [3]
Direction of velocity changes continuously; acceleration is towards centre (centripetal). [3]