A Level H1 Physics Practice Paper 3

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

Practice Paper: Mechanics (Version 3 of 5)

School: TuitionGoWhere Exam Practice (AI)
Subject: Physics H1
Level: A-Level
Paper: Practice Paper (Mechanics Topic Quiz)
Version: 3
Duration: 75 minutes
Total Marks: 60
Name: ___________________________
Class: ___________________________
Date: ___________________________

Instructions:

• This practice paper contains 20 questions on Mechanics only.
• Section A (Q1–5): Short structured recall and definition (12 marks)
• Section B (Q6–13): Calculation and proof (24 marks)
• Section C (Q14–20): Data interpretation and extended response (24 marks)
• Show all working clearly. Use SI units.
• Write answers in the spaces provided.

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Section A (12 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 $E_k$. [1]

(a) ___________________
(b) ___________________

3. A body moves with constant velocity. State the resultant force acting on it. [1]

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

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

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Section B (24 marks)

6. 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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7. A car of mass $950\ \text{kg}$ accelerates from rest to $25\ \text{m s}^{-1}$ in $10\ \text{s}$. Calculate the average resultant force on the car. [3]

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8. A ball is thrown vertically upward with initial speed $18\ \text{m s}^{-1}$. Using $g = 9.8\ \text{m s}^{-2}$, calculate the maximum height reached. [3]

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9. A uniform plank AB of length $4.0\ \text{m}$ and weight $200\ \text{N}$ rests on two supports at A and B. A child of weight $300\ \text{N}$ stands $1.0\ \text{m}$ from A.
(a) Draw a labelled diagram showing all forces on the plank. [2]
(b) Calculate the reaction force at support B. [3]

(a)

Image pending generation: diagram for Q9.

(b) ___________________________________________________________

10. A projectile is fired at $30^\circ$ above horizontal with speed $40\ \text{m s}^{-1}$. Calculate the horizontal range ignoring air resistance. Use $g = 9.8\ \text{m s}^{-2}$. [4]

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11. A $0.15\ \text{kg}$ ball moving at $8.0\ \text{m s}^{-1}$ collides head-on with a stationary $0.10\ \text{kg}$ ball. After collision the $0.15\ \text{kg}$ ball moves at $2.0\ \text{m s}^{-1}$ in same direction. Calculate the velocity of the second ball. [3]

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12. A mass of $2.0\ \text{kg}$ is whirled in a horizontal circle of radius $1.5\ \text{m}$ at constant speed $6.0\ \text{m s}^{-1}$. Calculate the centripetal force. [3]

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13. A crane lifts a $500\ \text{kg}$ load vertically at constant speed $0.20\ \text{m s}^{-1}$. Calculate the useful power output. Use $g = 9.8\ \text{m s}^{-2}$. [3]

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Section C (24 marks)

14. The velocity–time graph below shows a cyclist’s motion.

Image pending generation: graph for Q14.

(a) Calculate the total displacement. [2]
(b) State the acceleration during 0–5 s. [1]

(a) ___________________
(b) ___________________

15. Explain why a falling object reaches terminal velocity. [3]

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16. A satellite orbits Earth at radius $r$ with speed $v$. Show that $v = \sqrt{\frac{GM}{r}}$. [3]

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17. A $2.0\ \text{kg}$ block is pushed $3.0\ \text{m}$ up a smooth incline of angle $30^\circ$ at constant speed. Calculate work done against gravity. Use $g = 9.8\ \text{m s}^{-2}$. [3]

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18. Two wires X and Y are parallel and carry currents in the same direction.

Image pending generation: diagram for Q18.

Draw arrows on each wire to show the magnetic force due to the other. [2]

19. A student claims: “In an inelastic collision, both momentum and kinetic energy are conserved.” Discuss this statement. [4]

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20. A ball of mass $0.20\ \text{kg}$ is dropped from height $5.0\ \text{m}$ and rebounds to $3.2\ \text{m}$. Calculate the energy lost and the impulse of the ground on the ball. Use $g = 9.8\ \text{m s}^{-2}$. [5]

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

Practice Paper: Mechanics (Version 3) — Answer Key

Total Marks: 60
Topic: Mechanics

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Section A

1. [2 marks]
State the principle of conservation of linear momentum.
Answer: In a closed (or isolated) system, the total linear momentum remains constant provided no net external force acts.
Teaching note: Momentum before = momentum after. Must include “closed/isolated system” and “no external force” for both marks. Common trap: confusing with energy conservation.

2. [2 marks]
(a) $p = mv$ [1]
(b) $E_k = \frac{1}{2}mv^2$ [1]
Teaching note: Kinetic energy has the $\frac{1}{2}$ factor; omitting it is a common error.

3. [1 mark]
Resultant force = $0\ \text{N}$ (or zero).
Teaching note: Constant velocity ⇒ acceleration zero ⇒ $F_{\text{net}} = ma = 0$.

4. [2 marks]
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. ($M = F \times d_\perp$)
Teaching note: Award [B1] for product of force and perpendicular distance, [B1] for correct definition wording.

5. [2 marks]
A body remains at rest or in uniform motion in a straight line unless acted upon by a net external force.
Teaching note: This is Newton’s first law; mention both rest and uniform motion.

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Section B

6. [3 marks]
Given $p = 12\ \text{N·s}$, $E_k = 24\ \text{J}$.
$p = mv \Rightarrow v = p/m$
$E_k = \frac{1}{2}mv^2 = \frac{p^2}{2m} \Rightarrow m = \frac{p^2}{2E_k} = \frac{12^2}{2 \times 24} = \frac{144}{48} = 3.0\ \text{kg}$ [M1+ A1]
$v = p/m = 12 / 3.0 = 4.0\ \text{m s}^{-1}$ [M1+ A1]
Answer: mass = $3.0\ \text{kg}$, speed = $4.0\ \text{m s}^{-1}$.

7. [3 marks]
$a = (v-u)/t = (25-0)/10 = 2.5\ \text{m s}^{-2}$ [M1]
$F = ma = 950 \times 2.5 = 2375\ \text{N}$ [M1]
Answer: $2.38 \times 10^3\ \text{N}$ (or $2375\ \text{N}$) [A1]

8. [3 marks]
At max height, $v=0$. $v^2 = u^2 - 2gh \Rightarrow h = u^2/(2g)$ [M1]
$h = 18^2 / (2 \times 9.8) = 324 / 19.6 = 16.5\ \text{m}$ [M1+A1]
Answer: $16.5\ \text{m}$

9. [5 marks]
(a) [2] Diagram: plank horizontal, $R_A$ up at A, $R_B$ up at B, $200\ \text{N}$ down at centre (2.0 m), $300\ \text{N}$ down at 1.0 m from A.
(b) [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]
Answer: $175\ \text{N}$

10. [4 marks]
$u_x = 40\cos30^\circ = 34.64\ \text{m s}^{-1}$, $u_y = 40\sin30^\circ = 20\ \text{m s}^{-1}$ [M1]
Time of flight $T = 2u_y/g = 40/9.8 = 4.08\ \text{s}$ [M1]
Range $R = u_x T = 34.64 \times 4.08 = 141\ \text{m}$ [M1+A1]
Answer: $141\ \text{m}$

11. [3 marks]
Conservation of momentum: $0.15\times8 + 0 = 0.15\times2 + 0.10\times v$ [M1]
$1.2 = 0.30 + 0.10v \Rightarrow 0.10v = 0.90$ [M1]
$v = 9.0\ \text{m s}^{-1}$ [A1]
Answer: $9.0\ \text{m s}^{-1}$ same direction.

12. [3 marks]
$F_c = mv^2/r = 2.0 \times 6.0^2 / 1.5 = 2.0 \times 36 / 1.5 = 48\ \text{N}$ [M2+A1]
Answer: $48\ \text{N}$

13. [3 marks]
Force = weight = $mg = 500 \times 9.8 = 4900\ \text{N}$ [M1]
Power $P = Fv = 4900 \times 0.20 = 980\ \text{W}$ [M1+A1]
Answer: $980\ \text{W}$

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Section C

14. [3 marks]
(a) Area = triangle(0–5) + rectangle(5–15) + triangle(15–20)
$= \frac{1}{2}\times5\times10 + 10\times10 + \frac{1}{2}\times5\times10 = 25+100+25 = 150\ \text{m}$ [M1+A1]
(b) $a = (10-0)/5 = 2.0\ \text{m s}^{-2}$ [A1]
Answer: (a) $150\ \text{m}$ (b) $2.0\ \text{m s}^{-2}$

15. [3 marks]
As object falls, air resistance increases with speed. When air resistance equals weight, net force is zero, acceleration zero, speed constant = terminal velocity. [B1 for increasing resistance, B1 for equality of forces, B1 for constant speed]

16. [3 marks]
Gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$ [M1]
Cancel $m$ and one $r$: $\frac{GM}{r} = v^2$ [M1]
$v = \sqrt{\frac{GM}{r}}$ [A1]
Answer: shown.

17. [3 marks]
Vertical height gain $h = 3.0\sin30^\circ = 1.5\ \text{m}$ [M1]
Work = $mgh = 2.0 \times 9.8 \times 1.5 = 29.4\ \text{J}$ [M1+A1]
Answer: $29.4\ \text{J}$

18. [2 marks]
Parallel currents attract: arrow on X toward Y, arrow on Y toward X (horizontal). [B1 each]

19. [4 marks]
Momentum conserved in all collisions (closed system) [B1]. Kinetic energy NOT conserved in inelastic collisions; some converted to heat/sound/deformation [B2]. Statement false because KE not conserved [B1].

20. [5 marks]
Initial PE = $mgh_1 = 0.20\times9.8\times5.0 = 9.8\ \text{J}$ [M1]
Rebound PE = $0.20\times9.8\times3.2 = 6.272\ \text{J}$ [M1]
Energy lost = $9.8 - 6.272 = 3.53\ \text{J}$ [A1]
Speed before impact $u = \sqrt{2gh_1} = \sqrt{98} = 9.90\ \text{m s}^{-1}$ down [M1]
Speed after $v = \sqrt{2gh_2} = \sqrt{62.72} = 7.92\ \text{m s}^{-1}$ up [M1]
Impulse = $\Delta p = m(v - (-u)) = 0.20(7.92 + 9.90) = 3.56\ \text{N·s}$ up [A1]
Answer: Energy lost $3.53\ \text{J}$, impulse $3.56\ \text{N·s}$ upward.