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A Level H2 Physics Practice Paper 1

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H2
Level: A-Level
Paper: Practice Paper — Mechanics
Duration: 1 hour 30 minutes
Total Marks: 60
Name: ___________________________
Class: ___________________________
Date: ___________________________

Instructions

Answer all questions in the spaces provided.
The number of marks for each question is shown in brackets [ ].
Unless otherwise stated, numerical answers should be given to 2 or 3 significant figures.
The use of an approved scientific calculator is expected where appropriate.
You may lose marks if you do not show your working, if you do not use appropriate units, or if you do not give your final answer to a suitable degree of precision.
Take the gravitational field strength $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A: Multiple Choice [15 marks]

Questions 1–15: Each question is worth 1 mark. Choose the one best answer.

1. A ball is thrown vertically upwards with an initial speed of $20 \text{ m s}^{-1}$ . Ignoring air resistance, what is the maximum height reached by the ball?

A. $10.2 \text{ m}$
B. $20.4 \text{ m}$
C. $40.8 \text{ m}$
D. $20.0 \text{ m}$

2. Which of the following is a vector quantity?

A. Kinetic energy
B. Power
C. Momentum
D. Work done

3. A car accelerates uniformly from rest to $30 \text{ m s}^{-1}$ in $6.0 \text{ s}$ . What is the distance travelled by the car during this time?

A. $90 \text{ m}$
B. $180 \text{ m}$
C. $60 \text{ m}$
D. $45 \text{ m}$

4. A force of $12 \text{ N}$ acts on an object of mass $3.0 \text{ kg}$ on a frictionless surface. What is the acceleration of the object?

A. $0.25 \text{ m s}^{-2}$
B. $4.0 \text{ m s}^{-2}$
C. $36 \text{ m s}^{-2}$
D. $15 \text{ m s}^{-2}$

5. An object moves in a horizontal circle of radius $2.0 \text{ m}$ at a constant speed of $4.0 \text{ m s}^{-1}$ . What is the centripetal acceleration of the object?

A. $2.0 \text{ m s}^{-2}$
B. $4.0 \text{ m s}^{-2}$
C. $8.0 \text{ m s}^{-2}$
D. $16.0 \text{ m s}^{-2}$

6. A stone is released from rest and falls freely under gravity. What is the ratio of the distance fallen in the first second to the distance fallen in the third second?

A. $1:3$
B. $1:5$
C. $1:9$
D. $1:7$

7. A $2.0 \text{ kg}$ object moving at $6.0 \text{ m s}^{-1}$ collides with a stationary $4.0 \text{ kg}$ object. After the collision, the two objects stick together. What is their common velocity after the collision?

A. $1.0 \text{ m s}^{-1}$
B. $2.0 \text{ m s}^{-1}$
C. $3.0 \text{ m s}^{-1}$
D. $4.0 \text{ m s}^{-1}$

8. A uniform plank of length $4.0 \text{ m}$ and weight $200 \text{ N}$ is supported at both ends. A child of weight $400 \text{ N}$ stands $1.0 \text{ m}$ from the left end. What is the reaction force at the right support?

A. $150 \text{ N}$
B. $200 \text{ N}$
C. $250 \text{ N}$
D. $300 \text{ N}$

9. A satellite orbits the Earth at a height where the gravitational field strength is $2.45 \text{ N kg}^{-1}$ . If the radius of the Earth is $6.37 \times 10^6 \text{ m}$ , what is the orbital radius of the satellite?

A. $1.27 \times 10^7 \text{ m}$
B. $2.55 \times 10^7 \text{ m}$
C. $6.37 \times 10^6 \text{ m}$
D. $1.91 \times 10^7 \text{ m}$

10. A projectile is launched at an angle of $30°$ above the horizontal with an initial speed of $40 \text{ m s}^{-1}$ . What is the horizontal range of the projectile? (Ignore air resistance.)

A. $80.0 \text{ m}$
B. $141 \text{ m}$
C. $70.7 \text{ m}$
D. $122 \text{ m}$

11. A spring of spring constant $50 \text{ N m}^{-1}$ is compressed by $0.10 \text{ m}$ . What is the elastic potential energy stored in the spring?

A. $0.25 \text{ J}$
B. $0.50 \text{ J}$
C. $2.5 \text{ J}$
D. $5.0 \text{ J}$

12. An object of mass $5.0 \text{ kg}$ is lifted vertically at constant speed through a height of $3.0 \text{ m}$ . What is the work done against gravity?

A. $15 \text{ J}$
B. $49 \text{ J}$
C. $147 \text{ J}$
D. $441 \text{ J}$

13. A car of mass $1000 \text{ kg}$ travels around a flat circular bend of radius $50 \text{ m}$ at a constant speed of $20 \text{ m s}^{-1}$ . What is the minimum coefficient of static friction between the tyres and the road?

A. $0.41$
B. $0.82$
C. $0.20$
D. $1.63$

14. A ball is dropped from a height of $20 \text{ m}$ . Just before it hits the ground, what fraction of its total energy is kinetic? (Ignore air resistance.)

A. $25\%$
B. $50\%$
C. $75\%$
D. $100\%$

15. Two forces of magnitudes $8.0 \text{ N}$ and $6.0 \text{ N}$ act at right angles to each other. What is the magnitude of the resultant force?

A. $2.0 \text{ N}$
B. $10.0 \text{ N}$
C. $14.0 \text{ N}$
D. $7.0 \text{ N}$

Section B: Structured Questions [25 marks]

Answer all questions. Show your working clearly.

16. [4 marks]

A car of mass $1200 \text{ kg}$ is travelling along a straight horizontal road. The engine exerts a driving force of $3600 \text{ N}$ and the total resistive force acting on the car is $1200 \text{ N}$ .

(a) Calculate the acceleration of the car. [2 marks]

..................................................................................................................................

(b) The car starts from rest. Calculate the speed of the car after $8.0 \text{ s}$ . [2 marks]

..................................................................................................................................

17. [5 marks]

A small ball of mass $0.20 \text{ kg}$ is projected horizontally from the top of a cliff with a speed of $15 \text{ m s}^{-1}$ . The ball hits the ground $3.0 \text{ s}$ later.

(a) Calculate the horizontal distance from the base of the cliff where the ball lands. [2 marks]

..................................................................................................................................

(b) Calculate the vertical component of the velocity of the ball just before it hits the ground. [2 marks]

..................................................................................................................................

18. [6 marks]

A trolley of mass $0.80 \text{ kg}$ moving at $2.5 \text{ m s}^{-1}$ on a frictionless horizontal track collides with a stationary trolley of mass $1.2 \text{ kg}$ . After the collision, the two trolleys stick together and move as one.

(a) State the principle of conservation of linear momentum. [1 mark]

..................................................................................................................................

(b) Calculate the velocity of the combined trolleys immediately after the collision. [3 marks]

..................................................................................................................................

(c) Determine whether kinetic energy is conserved in this collision. Show your working and state the type of collision. [2 marks]

..................................................................................................................................

19. [5 marks]

A uniform beam of weight $400 \text{ N}$ and length $5.0 \text{ m}$ is hinged at a wall and supported by a cable attached to the far end of the beam, as shown in the diagram. The cable makes an angle of $30°$ with the horizontal. The beam is horizontal and in equilibrium.

<image_placeholder> id: Q19-fig1 type: diagram linked_question: Q19 description: A horizontal beam of length 5.0 m hinged at the left end to a vertical wall. A cable is attached to the right end of the beam, making an angle of 30° above the horizontal, connecting to a point on the wall above the hinge. The beam has its weight (400 N) acting at its midpoint (2.5 m from hinge). A load of 200 N hangs from the right end of the beam. labels: Wall (left), hinge, beam (5.0 m, horizontal), cable at 30° above horizontal, weight 400 N at midpoint, load 200 N at right end values: Beam length = 5.0 m, beam weight = 400 N at 2.5 m from hinge, load = 200 N at 5.0 m from hinge, cable angle = 30° above horizontal must_show: Hinge at left wall, horizontal beam, cable from right end to wall above at 30°, weight arrow at midpoint labelled 400 N, load arrow at right end labelled 200 N, angle 30° clearly marked between cable and horizontal </image_placeholder>

(a) By taking moments about the hinge, calculate the tension $T$ in the cable. [3 marks]

..................................................................................................................................

(b) Calculate the magnitude of the reaction force at the hinge. [2 marks]

..................................................................................................................................

20. [5 marks]

A small object of mass $0.50 \text{ kg}$ is attached to one end of a string of length $1.2 \text{ m}$ . The other end of the string is fixed, and the object moves in a horizontal circle at constant speed, so that the string makes an angle of $25°$ with the vertical.

<image_placeholder> id: Q20-fig1 type: diagram linked_question: Q20 description: A conical pendulum: a string of length 1.2 m fixed at the upper end, with a mass of 0.50 kg at the lower end. The mass moves in a horizontal circle. The string makes an angle of 25° with the vertical. The radius of the circular path is shown as r = L sin(25°). labels: Fixed point at top, string length L = 1.2 m, angle θ = 25° from vertical, mass m = 0.50 kg, radius r of circular path, tension T along string, weight mg downward values: m = 0.50 kg, L = 1.2 m, θ = 25°, g = 9.81 m s⁻² must_show: Fixed point, string at 25° to vertical, mass at bottom, horizontal circular path indicated, angle 25° clearly marked, length 1.2 m labelled, weight mg and tension T arrows shown </image_placeholder>

(a) Calculate the tension in the string. [2 marks]

..................................................................................................................................

(b) Calculate the speed of the object. [3 marks]

..................................................................................................................................

Section C: Long Structured Questions [20 marks]

Answer all questions. Show your working clearly.

21. [10 marks]

A ball of mass $0.40 \text{ kg}$ is dropped from a height of $5.0 \text{ m}$ above the ground. After hitting the ground, the ball rebounds to a height of $3.2 \text{ m}$ . Assume air resistance is negligible throughout.

(a) Calculate the speed of the ball just before it hits the ground. [2 marks]

..................................................................................................................................

(b) Calculate the speed of the ball just after it leaves the ground. [2 marks]

..................................................................................................................................

(d) The ball is in contact with the ground for $0.020 \text{ s}$ . Calculate the average force exerted by the ground on the ball. [2 marks]

..................................................................................................................................

(e) State whether the collision is elastic or inelastic. Justify your answer. [1 mark]

..................................................................................................................................

22. [10 marks]

A spacecraft of mass $2.0 \times 10^4 \text{ kg}$ is in a circular orbit around the Earth at an altitude of $400 \text{ km}$ above the Earth's surface.

Data:

Mass of Earth $M = 5.97 \times 10^{24} \text{ kg}$
Radius of Earth $R = 6.37 \times 10^6 \text{ m}$
Gravitational constant $G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$

(a) Show that the orbital speed of the spacecraft is approximately $7.7 \times 10^3 \text{ m s}^{-1}$ . [3 marks]

..................................................................................................................................

(b) Calculate the kinetic energy of the spacecraft. [2 marks]

..................................................................................................................................

(d) The spacecraft fires its engines to move to a higher orbit. State and explain what happens to the total mechanical energy of the spacecraft. [2 marks]

..................................................................................................................................

Answers

TuitionGoWhere Practice Paper — Physics H2 A-Level

Answer Key — Mechanics Practice Paper

Section A: Multiple Choice [15 marks]

1. B — $20.4 \text{ m}$

Method: Using $v^2 = u^2 + 2as$ with $v = 0$ , $u = 20 \text{ m s}^{-1}$ , $a = -9.81 \text{ m s}^{-2}$ : $0 = (20)^2 + 2(-9.81)s$ $s = \frac{400}{19.62} = 20.4 \text{ m}$

2. C — Momentum

Explanation: Momentum ( $p = mv$ ) has both magnitude and direction, making it a vector. Kinetic energy, power, and work done are all scalar quantities — they have magnitude only.

3. A — $90 \text{ m}$

Method: Using $s = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 30)(6.0) = 90 \text{ m}$

4. B — $4.0 \text{ m s}^{-2}$

Method: Using Newton's second law $F = ma$ : $a = \frac{F}{m} = \frac{12}{3.0} = 4.0 \text{ m s}^{-2}$

5. C — $8.0 \text{ m s}^{-2}$

Method: Centripetal acceleration $a_c = \frac{v^2}{r} = \frac{(4.0)^2}{2.0} = \frac{16}{2.0} = 8.0 \text{ m s}^{-2}$

6. B — $1:5$

Method: Distance fallen in time $t$ from rest: $s = \frac{1}{2}gt^2$

Distance in 1st second: $s_1 = \frac{1}{2}g(1^2) = \frac{1}{2}g$
Distance in first 3 seconds: $s_3 = \frac{1}{2}g(3^2) = \frac{9}{2}g$
Distance in 3rd second only: $s_{3\text{rd}} = s_3 - s_2 = \frac{9}{2}g - \frac{4}{2}g = \frac{5}{2}g$
Ratio $s_1 : s_{3\text{rd}} = \frac{1}{2}g : \frac{5}{2}g = 1:5$

7. B — $2.0 \text{ m s}^{-1}$

Method: Conservation of momentum (perfectly inelastic collision): $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(2.0)(6.0) + (4.0)(0) = (2.0 + 4.0)v$ $12 = 6.0v \implies v = 2.0 \text{ m s}^{-1}$

8. A — $150 \text{ N}$

Method: Taking moments about the left support (anticlockwise positive): $R_R \times 4.0 = 200 \times 2.0 + 400 \times 1.0$ $4.0 R_R = 400 + 400 = 800$ $R_R = 200 \text{ N}$

Wait — let me recalculate. Taking moments about the left support:

Plank weight $200 \text{ N}$ acts at midpoint = $2.0 \text{ m}$ from left: moment = $200 \times 2.0 = 400 \text{ N m}$
Child weight $400 \text{ N}$ at $1.0 \text{ m}$ from left: moment = $400 \times 1.0 = 400 \text{ N m}$
Right support reaction $R_R$ at $4.0 \text{ m}$ from left

$R_R \times 4.0 = 400 + 400 = 800$ $R_R = 200 \text{ N}$

Then $R_L = 200 + 400 - 200 = 400 \text{ N}$ ... Hmm, that gives $R_R = 200 \text{ N}$ , which is option B.

Let me re-read the question. The child stands $1.0 \text{ m}$ from the left end. Taking moments about the left support: $R_R \times 4.0 = 200 \times 2.0 + 400 \times 1.0 = 400 + 400 = 800$ $R_R = 200 \text{ N}$

The answer is B — $200 \text{ N}$

9. A — $1.27 \times 10^7 \text{ m}$

Method: Gravitational field strength $g = \frac{GM}{r^2}$ , so: $r = \sqrt{\frac{GM}{g}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{2.45}}$ $r = \sqrt{\frac{3.983 \times 10^{14}}{2.45}} = \sqrt{1.626 \times 10^{14}} = 1.27 \times 10^7 \text{ m}$

10. B — $141 \text{ m}$

Method: Range of a projectile: $R = \frac{u^2 \sin 2\theta}{g}$ $R = \frac{(40)^2 \sin 60°}{9.81} = \frac{1600 \times 0.866}{9.81} = \frac{1385.6}{9.81} = 141 \text{ m}$

11. A — $0.25 \text{ J}$

Method: Elastic potential energy $E = \frac{1}{2}kx^2 = \frac{1}{2}(50)(0.10)^2 = \frac{1}{2}(50)(0.01) = 0.25 \text{ J}$

12. C — $147 \text{ J}$

Method: Work done against gravity $W = mgh = 5.0 \times 9.81 \times 3.0 = 147 \text{ J}$

13. B — $0.82$

Method: For circular motion on a flat road, friction provides centripetal force: $\mu mg = \frac{mv^2}{r}$ $\mu = \frac{v^2}{rg} = \frac{(20)^2}{50 \times 9.81} = \frac{400}{490.5} = 0.82$

14. D — $100\%$

Explanation: Ignoring air resistance, mechanical energy is conserved. At the point just before hitting the ground, all the gravitational potential energy has been converted to kinetic energy. Therefore, $100\%$ of the total energy is kinetic.

15. B — $10.0 \text{ N}$

Method: For perpendicular forces, use Pythagoras' theorem: $F = \sqrt{8.0^2 + 6.0^2} = \sqrt{64 + 36} = \sqrt{100} = 10.0 \text{ N}$

Section B: Structured Questions [25 marks]

16. [4 marks]

(a) [2 marks]

Using Newton's second law: $F_{\text{net}} = ma$ $3600 - 1200 = 1200 \times a$ $a = \frac{2400}{1200} = 2.0 \text{ m s}^{-2}$

Marking:

[1] for correct net force calculation ( $3600 - 1200 = 2400 \text{ N}$ )
[1] for correct answer $a = 2.0 \text{ m s}^{-2}$

(b) [2 marks]

Using $v = u + at$ : $v = 0 + 2.0 \times 8.0 = 16 \text{ m s}^{-1}$

Marking:

[1] for correct substitution into kinematic equation
[1] for correct answer $v = 16 \text{ m s}^{-1}$

17. [5 marks]

(a) [2 marks]

Horizontal motion (constant velocity): $x = v_x \times t = 15 \times 3.0 = 45 \text{ m}$

Marking:

[1] for using $x = v_x t$ (horizontal velocity is constant)
[1] for correct answer $45 \text{ m}$

(b) [2 marks]

Vertical motion (free fall from rest): $v_y = gt = 9.81 \times 3.0 = 29.4 \text{ m s}^{-1}$

Marking:

[1] for using $v = gt$ (initial vertical velocity is zero)
[1] for correct answer $29 \text{ m s}^{-1}$ (or $29.4 \text{ m s}^{-1}$ )

(c) [1 mark]

Speed is the magnitude of the resultant velocity: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{15^2 + 29.4^2} = \sqrt{225 + 864.4} = \sqrt{1089.4} = 33 \text{ m s}^{-1}$

Marking:

[1] for correct answer $33 \text{ m s}^{-1}$ (accept $33.0 \text{ m s}^{-1}$ )

18. [6 marks]

(a) [1 mark]

The total momentum of a system remains constant (or is conserved) provided that no net external force acts on the system.

Marking:

[1] for a complete statement including both the constancy of momentum AND the condition of no external force / closed system

Common mistake: Simply stating "momentum is conserved" without mentioning the condition of no external force — this would not receive full credit.

(b) [3 marks]

Using conservation of momentum: $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(0.80)(2.5) + (1.2)(0) = (0.80 + 1.2)v$ $2.0 = 2.0v$ $v = 1.0 \text{ m s}^{-1}$

Marking:

[1] for correct equation showing conservation of momentum
[1] for correct substitution of values
[1] for correct answer $v = 1.0 \text{ m s}^{-1}$ in the original direction

(c) [2 marks]

Initial kinetic energy: $KE_i = \frac{1}{2}(0.80)(2.5)^2 = \frac{1}{2}(0.80)(6.25) = 2.5 \text{ J}$

Final kinetic energy: $KE_f = \frac{1}{2}(2.0)(1.0)^2 = 1.0 \text{ J}$

Since $KE_f < KE_i$ (kinetic energy decreased from $2.5 \text{ J}$ to $1.0 \text{ J}$ ), kinetic energy is not conserved. This is a perfectly inelastic collision (the objects stick together).

Marking:

[1] for calculating both kinetic energies and showing $KE$ is not conserved
[1] for identifying the collision as perfectly inelastic

19. [5 marks]

Visual reference: The diagram shows a horizontal beam of length $5.0 \text{ m}$ hinged at the left wall, with a cable at the right end making $30°$ above the horizontal. The beam weight ( $400 \text{ N}$ ) acts at the midpoint ( $2.5 \text{ m}$ from hinge), and a load of $200 \text{ N}$ hangs from the right end.

(a) [3 marks]

Taking moments about the hinge (anticlockwise positive):

The tension $T$ acts at the far end ( $5.0 \text{ m}$ from hinge). The vertical component of tension is $T \sin 30°$ (perpendicular to the beam).

Clockwise moments (due to weights): $\text{Moment}_{\text{clockwise}} = 400 \times 2.5 + 200 \times 5.0 = 1000 + 1000 = 2000 \text{ N m}$

Anticlockwise moment (due to tension): $\text{Moment}_{\text{anticlockwise}} = T \sin 30° \times 5.0$

For equilibrium, anticlockwise = clockwise: $T \sin 30° \times 5.0 = 2000$ $T \times 0.5 \times 5.0 = 2000$ $T = \frac{2000}{2.5} = 800 \text{ N}$

Marking:

[1] for identifying all forces and their perpendicular distances from the hinge
[1] for correct moment equation with $\sin 30°$ component
[1] for correct answer $T = 800 \text{ N}$

(b) [2 marks]

Resolving forces horizontally: $R_x = T \cos 30° = 800 \times \cos 30° = 800 \times 0.866 = 693 \text{ N}$

Resolving forces vertically: $R_y + T \sin 30° = 400 + 200$ $R_y + 800 \times 0.5 = 600$ $R_y = 600 - 400 = 200 \text{ N}$

Magnitude of reaction at hinge: $R = \sqrt{R_x^2 + R_y^2} = \sqrt{693^2 + 200^2} = \sqrt{480249 + 40000} = \sqrt{520249} = 721 \text{ N}$

Marking:

[1] for resolving tension into components and finding both $R_x$ and $R_y$
[1] for correct magnitude $R = 721 \text{ N}$ (accept $720 \text{ N}$ )

20. [5 marks]

Visual reference: A conical pendulum with string length $1.2 \text{ m}$ , mass $0.50 \text{ kg}$ , string at $25°$ to the vertical.

(a) [2 marks]

Resolving vertically (the mass has no vertical acceleration): $T \cos 25° = mg$ $T = \frac{mg}{\cos 25°} = \frac{0.50 \times 9.81}{\cos 25°} = \frac{4.905}{0.9063} = 5.41 \text{ N}$

Marking:

[1] for correct vertical equilibrium equation $T \cos\theta = mg$
[1] for correct answer $T = 5.4 \text{ N}$ (or $5.41 \text{ N}$ )

(b) [3 marks]

The radius of the circular path: $r = L \sin 25° = 1.2 \times \sin 25° = 1.2 \times 0.4226 = 0.507 \text{ m}$

The horizontal component of tension provides the centripetal force: $T \sin 25° = \frac{mv^2}{r}$

Substituting $T = \frac{mg}{\cos 25°}$ : $\frac{mg}{\cos 25°} \times \sin 25° = \frac{mv^2}{r}$ $g \tan 25° = \frac{v^2}{r}$ $v = \sqrt{rg \tan 25°} = \sqrt{0.507 \times 9.81 \times \tan 25°}$ $v = \sqrt{0.507 \times 9.81 \times 0.4663} = \sqrt{2.319} = 1.52 \text{ m s}^{-1}$

Marking:

[1] for correct radius calculation $r = L \sin\theta$
[1] for correct centripetal force equation $T\sin\theta = \frac{mv^2}{r}$
[1] for correct answer $v = 1.5 \text{ m s}^{-1}$ (or $1.52 \text{ m s}^{-1}$ )

Section C: Long Structured Questions [20 marks]

21. [10 marks]

(a) [2 marks]

Using $v^2 = u^2 + 2gh$ with $u = 0$ : $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 5.0} = \sqrt{98.1} = 9.90 \text{ m s}^{-1}$

Marking:

[1] for correct equation/substitution
[1] for correct answer $9.90 \text{ m s}^{-1}$ (downward)

(b) [2 marks]

Using $v^2 = u^2 + 2gh$ for the rebound (final velocity at max height $= 0$ ): $0 = u^2 - 2 \times 9.81 \times 3.2$ $u = \sqrt{2 \times 9.81 \times 3.2} = \sqrt{62.78} = 7.92 \text{ m s}^{-1}$

Marking:

[1] for correct equation/substitution
[1] for correct answer $7.92 \text{ m s}^{-1}$ (upward)

(c) [3 marks]

Taking upward as positive:

Velocity just before impact: $v_{\text{before}} = -9.90 \text{ m s}^{-1}$ (downward)

Velocity just after impact: $v_{\text{after}} = +7.92 \text{ m s}^{-1}$ (upward)

Change in momentum: $\Delta p = m(v_{\text{after}} - v_{\text{before}}) = 0.40 \times (7.92 - (-9.90))$ $\Delta p = 0.40 \times 17.82 = 7.13 \text{ kg m s}^{-1}$

The change in momentum is $7.13 \text{ kg m s}^{-1}$ upward.

Marking:

[1] for correct sign convention (recognising that velocities are in opposite directions)
[1] for correct substitution into $\Delta p = m(v - u)$
[1] for correct answer with direction: $7.13 \text{ kg m s}^{-1}$ upward

Common mistake: Forgetting that momentum is a vector and simply subtracting the magnitudes without considering direction. This would give $0.40 \times (7.92 - 9.90) = -0.792 \text{ kg m s}^{-1}$ , which is incorrect.

(d) [2 marks]

Using the impulse-momentum theorem: $F \Delta t = \Delta p$ $F = \frac{\Delta p}{\Delta t} = \frac{7.13}{0.020} = 356.5 \text{ N} \approx 357 \text{ N}$

Marking:

[1] for using $F = \frac{\Delta p}{\Delta t}$
[1] for correct answer $F = 357 \text{ N}$ (or $356 \text{ N}$ )

(e) [1 mark]

The collision is inelastic because kinetic energy is not conserved (the ball does not return to its original height of $5.0 \text{ m}$ — it only returns to $3.2 \text{ m}$ , meaning some energy was lost to heat/sound/deformation during the collision).

Marking:

[1] for stating "inelastic" with a valid justification (e.g., height decreased, or kinetic energy before ≠ kinetic energy after)

22. [10 marks]

(a) [3 marks]

For a satellite in circular orbit, gravitational force provides centripetal force: $\frac{GMm}{r^2} = \frac{mv^2}{r}$

Solving for $v$ : $v = \sqrt{\frac{GM}{r}}$

where $r = R + h = 6.37 \times 10^6 + 400 \times 10^3 = 6.77 \times 10^6 \text{ m}$

$v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}}$ $v = \sqrt{\frac{3.983 \times 10^{14}}{6.77 \times 10^6}} = \sqrt{5.883 \times 10^7} = 7670 \text{ m s}^{-1} \approx 7.7 \times 10^3 \text{ m s}^{-1} \quad \checkmark$

Marking:

[1] for equating gravitational force to centripetal force
[1] for correct orbital radius $r = R + h = 6.77 \times 10^6 \text{ m}$
[1] for correct answer $v \approx 7.7 \times 10^3 \text{ m s}^{-1}$

(b) [2 marks]

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 2.0 \times 10^4 \times (7670)^2$ $KE = 1.0 \times 10^4 \times 5.883 \times 10^7 = 5.88 \times 10^{11} \text{ J}$

Marking:

[1] for correct substitution into $KE = \frac{1}{2}mv^2$
[1] for correct answer $KE = 5.88 \times 10^{11} \text{ J}$ (or $5.9 \times 10^{11} \text{ J}$ )

(c) [3 marks]

Gravitational potential energy: $U = -\frac{GMm}{r} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 2.0 \times 10^4}{6.77 \times 10^6}$ $U = -\frac{7.966 \times 10^{18}}{6.77 \times 10^6} = -1.177 \times 10^{12} \text{ J} \approx -1.18 \times 10^{12} \text{ J}$

Marking:

[1] for correct formula $U = -\frac{GMm}{r}$
[1] for correct substitution
[1] for correct answer $U = -1.18 \times 10^{12} \text{ J}$ (negative sign required)

(d) [2 marks]

When the spacecraft moves to a higher orbit:

The total mechanical energy increases (becomes less negative).
This is because the engines do positive work on the spacecraft, adding energy to the system.
At the higher orbit, the gravitational potential energy increases (becomes less negative) and the kinetic energy decreases, but the net effect is an increase in total mechanical energy.

Marking:

[1] for stating that total mechanical energy increases
[1] for correct explanation (engines do work / energy is added to the system)

Mark Summary:

Section	Marks
A: Q1–15 (MCQ)	15
B: Q16	4
B: Q17	5
B: Q18	6
B: Q19	5
B: Q20	5
C: Q21	10
C: Q22	10
Total	60