A Level H1 Physics Practice Paper 4

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI) — Version 4 of 5

Subject: Physics H1
Level: A-Level
Paper: Practice Paper (Topic: Mechanics)
Duration: 1 hour 15 minutes
Total Marks: 60
Name: ________________________
Class: ________________________
Date: ________________________

Instructions:

• This practice paper contains 20 questions on the topic of Mechanics (Syllabus 8867, Section II).
• Answer all questions in the spaces provided.
• Show all working clearly. Use SI units and appropriate notation.
• Marks for each question are shown in brackets.
• The total marks are 60 and must be completed within 1 hour 15 minutes.

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Section A: Foundations of Mechanics (Questions 1–7) — 21 marks

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

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

(a) _________________________
(b) _________________________

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

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4. [3] A uniform plank AB of length $4.0\ \text{m}$ and weight $80\ \text{N}$ rests on two supports at A and B. A child of weight $40\ \text{N}$ stands $1.0\ \text{m}$ from A. Draw and label all forces acting on the plank.

Image pending generation: diagram for Q4.

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5. [3] For the plank in Q4, calculate the reaction force at support B.

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6. [4] A block of mass $5.0\ \text{kg}$ is pulled along a horizontal surface by a force of $20\ \text{N}$ at $30^\circ$ above the horizontal. The frictional force is $4.0\ \text{N}$. Calculate the acceleration of the block.

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7. [4] A car of mass $1000\ \text{kg}$ travelling at $20\ \text{m s}^{-1}$ is brought to rest in $5.0\ \text{s}$ by a constant braking force. Calculate the magnitude of the braking force and the distance travelled during braking.

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Section B: Motion and Collisions (Questions 8–14) — 21 marks

8. [3] A ball is thrown horizontally from a cliff of height $45\ \text{m}$ with speed $15\ \text{m s}^{-1}$. Calculate the time taken to reach the ground and the horizontal distance travelled.

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9. [4] A projectile is launched at $37^\circ$ above the horizontal with initial speed $25\ \text{m s}^{-1}$. Calculate the maximum height reached. (Use $g = 9.8\ \text{m s}^{-2}$.)

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10. [4] A $2.0\ \text{kg}$ trolley moving at $3.0\ \text{m s}^{-1}$ collides with a stationary $1.5\ \text{kg}$ trolley. They stick together. Calculate their common final velocity and the kinetic energy lost.

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11. [3] Two identical carts of mass $1.0\ \text{kg}$ move towards each other at $2.0\ \text{m s}^{-1}$ each. After an elastic collision, they rebound. State the final velocity of each cart and explain why kinetic energy is conserved.

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12. [3] The graph below shows the velocity–time graph for a particle.

Image pending generation: graph for Q12.

Calculate the total displacement of the particle.

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13. [2] Define the term "impulse" and state its unit.

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14. [2] A force–time graph for a collision has a rectangular shape of height $50\ \text{N}$ and width $0.10\ \text{s}$. Calculate the impulse delivered.

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Section C: Circular Motion and Gravitation (Questions 15–20) — 18 marks

15. [3] A mass of $0.50\ \text{kg}$ is whirled in a horizontal circle of radius $0.80\ \text{m}$ at constant speed $4.0\ \text{m s}^{-1}$. Calculate the centripetal force acting on it.

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16. [3] A satellite orbits Earth at a height where gravitational field strength is $6.0\ \text{N kg}^{-1}$. If its mass is $200\ \text{kg}$, calculate the gravitational force on it.

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17. [3] Explain why a passenger in a turning car feels pushed outward, with reference to Newton's first law.

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18. [3] A geostationary satellite has an orbital period of 24 hours. State one characteristic of its orbit and explain its use in communications.

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19. [2] Write the equation for centripetal acceleration in terms of speed $v$ and radius $r$.

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20. [1] State the direction of the centripetal force on a planet in circular orbit around the Sun.

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End of Practice Paper
Total Marks: 60

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TuitionGoWhere Practice Paper — Answer Key (Version 4)

Subject: Physics H1
Level: A-Level
Topic: Mechanics
Total Marks: 60

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Section A: Foundations of Mechanics (Q1–7) — 21 marks

Q1 [2 marks]
Teaching note: Conservation of linear momentum applies to a closed system (no net external force).

• In a closed (or isolated) system, the total linear momentum remains constant. [B1]
• This holds provided no external resultant force acts on the system. [B1]
  Common trap: Omitting "closed system" or confusing with energy conservation.

Q2 [2 marks]
(a) $p = mv$ [B1]
(b) $E_k = \frac{1}{2}mv^2$ [B1]
Teaching note: Momentum is vector, kinetic energy is scalar; do not forget the $\frac{1}{2}$.

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

Q4 [3 marks]
Teaching note: For equilibrium, show all downward weights and upward reactions.
Expected diagram (see Q4-fig1):

• Weight of plank $80\ \text{N}$ downward at centre (2.0 m from A)
• Child weight $40\ \text{N}$ downward at 1.0 m from A
• Reaction $R_A$ upward at A, $R_B$ upward at B
  [M1 for plank weight at centre, M1 for child weight, M1 for two reactions]
  Marking: 3 marks for fully labelled correct forces.

Q5 [3 marks]
Take moments about A:
$R_B \times 4.0 = 80 \times 2.0 + 40 \times 1.0$ [M1]
$R_B \times 4.0 = 160 + 40 = 200$ [M1]
$R_B = 50\ \text{N}$ [A1]
Answer: $50\ \text{N}$.

Q6 [4 marks]
Horizontal component of pull: $F_x = 20\cos30^\circ = 17.3\ \text{N}$ [M1]
Net horizontal force: $F_{\text{net}} = 17.3 - 4.0 = 13.3\ \text{N}$ [M1]
$a = F_{\text{net}}/m = 13.3 / 5.0 = 2.66\ \text{m s}^{-2}$ [M1]
Answer: $2.7\ \text{m s}^{-2}$ (2 s.f.) [A1]
Note: Vertical forces balance, not needed for acceleration.

Q7 [4 marks]
Using $v = u + at$: $0 = 20 + a(5.0) \Rightarrow a = -4.0\ \text{m s}^{-2}$ [M1]
Braking force $F = ma = 1000 \times 4.0 = 4000\ \text{N}$ [M1]
Distance $s = ut + \frac{1}{2}at^2 = 20(5.0) + \frac{1}{2}(-4.0)(5.0)^2 = 100 - 50 = 50\ \text{m}$ [M1]
Answer: Force = $4000\ \text{N}$, distance = $50\ \text{m}$ [A1]

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Section B: Motion and Collisions (Q8–14) — 21 marks

Q8 [3 marks]
Vertical: $s = \frac{1}{2}gt^2 \Rightarrow 45 = \frac{1}{2}(9.8)t^2$ [M1]
$t = \sqrt{90/9.8} = 3.03\ \text{s}$ [A1]
Horizontal: $x = vt = 15 \times 3.03 = 45.5\ \text{m}$ [M1]
Answer: time = $3.0\ \text{s}$, distance = $45\ \text{m}$ (3 s.f.)

Q9 [4 marks]
$u_y = 25\sin37^\circ = 15.0\ \text{m s}^{-1}$ [M1]
At max height $v_y = 0$: $0 = u_y^2 - 2gh$ [M1]
$h = \frac{15.0^2}{2 \times 9.8} = \frac{225}{19.6} = 11.5\ \text{m}$ [M1]
Answer: $11.5\ \text{m}$ [A1]

Q10 [4 marks]
Momentum: $2.0(3.0) + 1.5(0) = (2.0+1.5)v \Rightarrow 6.0 = 3.5v$ [M1]
$v = 1.71\ \text{m s}^{-1}$ [A1]
KE initial: $\frac{1}{2}(2.0)(3.0)^2 = 9.0\ \text{J}$ [M1]
KE final: $\frac{1}{2}(3.5)(1.71)^2 = 5.1\ \text{J}$; lost = $9.0 - 5.1 = 3.9\ \text{J}$ [A1]
Answer: $v = 1.7\ \text{m s}^{-1}$, KE lost = $3.9\ \text{J}$.

Q11 [3 marks]
Elastic collision, equal masses, opposite equal speeds → exchange velocities. [B1]
Each rebounds at $2.0\ \text{m s}^{-1}$ opposite direction. [B1]
Kinetic energy conserved because collision is elastic by statement. [B1]
Answer: velocities $-2.0$ and $+2.0\ \text{m s}^{-1}$ (relative), KE conserved.

Q12 [3 marks]
Area under graph = displacement.
Segment 1: $\frac{1}{2}(4)(8) = 16\ \text{m}$ [M1]
Segment 2: $4 \times 8 = 32\ \text{m}$ [M1]
Segment 3: $\frac{1}{2}(2)(8) = 8\ \text{m}$; total = $56\ \text{m}$ [M1]
Answer: $56\ \text{m}$.

Q13 [2 marks]
Impulse = change in momentum (= force × time). [B1]
Unit: N·s (or kg·m·s⁻¹). [B1]

Q14 [2 marks]
Impulse = area = $50 \times 0.10 = 5.0\ \text{N·s}$ [A1+A1]
Answer: $5.0\ \text{N·s}$.

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Section C: Circular Motion and Gravitation (Q15–20) — 18 marks

Q15 [3 marks]
$F_c = \frac{mv^2}{r} = \frac{0.50 \times 4.0^2}{0.80}$ [M1]
$= \frac{0.50 \times 16}{0.80} = 10\ \text{N}$ [M1+A1]
Answer: $10\ \text{N}$.

Q16 [3 marks]
$F = mg = 200 \times 6.0 = 1200\ \text{N}$ [M1+M1+A1]
Answer: $1200\ \text{N}$.

Q17 [3 marks]
Newton's first law: object continues in straight line unless acted on by force. [B1]
Car turns, passenger tends to continue straight (inertia). [B1]
Feels "pushed outward" due to absence of centripetal force on passenger's body relative to car. [B1]

Q18 [3 marks]
Orbit directly above equator, period = 24 h (fixed above point). [B1]
Used for fixed dish communications as satellite appears stationary. [B1+B1]

Q19 [2 marks]
$a_c = \frac{v^2}{r}$ [2]

Q20 [1 mark]
Towards the centre (towards Sun). [1]

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Total Marks: 60 — marking scheme complete.