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A Level H1 Physics Practice Paper 2

Free A Level H1 Physics Practice Paper 2, LongCat Exam version, with questions, answers, and A Level-style practice for Singapore students.

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A Level H1 Physics From Real Exams Generated by LongCat 2.0 LLM Updated 2026-07-10

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Secondary School (AI)

Field	Details
Subject:	Physics
Level:	A-Level H1
Paper:	Practice Paper — Mechanics & Associated Topics
Version:	2 of 5
Duration:	1 hour 30 minutes
Total Marks:	60

Name: ___________________________ Class: ___________ Date: _______________

Instructions to Candidates

Write your name, class, and date in the spaces provided above.
Answer ALL questions in the spaces provided.
Write in dark blue or black pen.
You may use a pencil for any diagrams, graphs, or working.
Non-programmable calculators are permitted.
The total mark for this paper is 60.
The number of marks for each question or part question is shown in brackets [ ].
You are reminded of the need for clear presentation in your answers.

Section A: Multiple Choice [10 marks]

Questions 1–10: Each question is worth 1 mark. Choose the one best answer.

1. A ball is thrown vertically upwards. At the highest point of its trajectory, which of the following is correct?

A. Velocity is zero and acceleration is zero. B. Velocity is zero and acceleration is $9.8 \text{ m s}^{-2}$ downwards. C. Velocity is maximum and acceleration is $9.8 \text{ m s}^{-2}$ upwards. D. Velocity is maximum and acceleration is zero.

2. A car accelerates uniformly from rest to $24 \text{ m s}^{-1}$ in $8.0 \text{ s}$ . What distance does it travel in this time?

A. $48 \text{ m}$ B. $96 \text{ m}$ C. $144 \text{ m}$ D. $192 \text{ m}$

3. Which of the following is a vector quantity?

A. Energy B. Power C. Speed D. Momentum

4. A $2.0 \text{ kg}$ object moving at $3.0 \text{ m s}^{-1}$ collides with a stationary $4.0 \text{ kg}$ object. After the collision, the two objects stick together. What is their common velocity?

A. $0.5 \text{ m s}^{-1}$ B. $1.0 \text{ m s}^{-1}$ C. $1.5 \text{ m s}^{-1}$ D. $2.0 \text{ m s}^{-1}$

5. A force of $12 \text{ N}$ acts on an object and displaces it by $5.0 \text{ m}$ in the direction of the force. The work done by the force is:

A. $2.4 \text{ J}$ B. $17 \text{ J}$ C. $60 \text{ J}$ D. $120 \text{ J}$

6. A projectile is launched horizontally from a cliff at $15 \text{ m s}^{-1}$ . Ignoring air resistance, which statement about its motion is correct?

A. Its horizontal velocity increases. B. Its horizontal velocity decreases. C. Its horizontal velocity remains constant. D. Its vertical acceleration increases.

7. A uniform beam of weight $40 \text{ N}$ is supported at both ends. A $20 \text{ N}$ load is placed at the centre of the beam. What is the reaction force at each support?

A. $10 \text{ N}$ B. $20 \text{ N}$ C. $30 \text{ N}$ D. $40 \text{ N}$

8. A car of mass $1000 \text{ kg}$ travelling at $20 \text{ m s}^{-1}$ is brought to rest by a constant braking force over a distance of $50 \text{ m}$ . The magnitude of the braking force is:

A. $2000 \text{ N}$ B. $4000 \text{ N}$ C. $6000 \text{ N}$ D. $8000 \text{ N}$

9. An object of mass $5.0 \text{ kg}$ is raised vertically through a height of $10 \text{ m}$ . The gain in gravitational potential energy is: (Take $g = 9.8 \text{ m s}^{-2}$ )

A. $49 \text{ J}$ B. $98 \text{ J}$ C. $490 \text{ J}$ D. $980 \text{ J}$

10. A $0.50 \text{ kg}$ ball moving at $6.0 \text{ m s}^{-1}$ strikes a wall and rebounds at $4.0 \text{ m s}^{-1}$ . The magnitude of the change in momentum of the ball is:

A. $1.0 \text{ kg m s}^{-1}$ B. $3.0 \text{ kg m s}^{-1}$ C. $5.0 \text{ kg m s}^{-1}$ D. $7.0 \text{ kg m s}^{-1}$

Section B: Structured Questions [30 marks]

Answer ALL questions. Show your working clearly.

11. (a) State Newton's first law of motion. [2]

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

(b) A book of mass $1.5 \text{ kg}$ rests on a horizontal table. Draw a free-body diagram showing all the forces acting on the book. Label each force clearly. [2]

(c) Explain, using Newton's first law, why the book remains at rest. [1]

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

[Total: 5 marks]

12. A car starts from rest and accelerates uniformly at $2.5 \text{ m s}^{-2}$ for $10 \text{ s}$ .

(a) Calculate the final velocity of the car. [2]

(b) Calculate the distance travelled by the car during this time. [2]

(c) The car then travels at constant velocity for a further $20 \text{ s}$ . Calculate the total distance travelled over the entire $30 \text{ s}$ period. [2]

[Total: 6 marks]

13. (a) State the principle of conservation of linear momentum. [2]

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

(b) A $0.060 \text{ kg}$ tennis ball moving horizontally at $25 \text{ m s}^{-1}$ is struck by a racket. The ball moves off in the opposite direction at $30 \text{ m s}^{-1}$ . The contact time between the ball and racket is $0.0050 \text{ s}$ .

(i) Calculate the change in momentum of the ball. [2]

(ii) Calculate the average force exerted by the racket on the ball. [2]

[Total: 6 marks]

14. A ball is projected from ground level at an angle of $35°$ above the horizontal with an initial speed of $20 \text{ m s}^{-1}$ . Air resistance is negligible. Take $g = 9.8 \text{ m s}^{-2}$ .

(a) Show that the horizontal component of the initial velocity is $16.4 \text{ m s}^{-1}$ . [1]

(b) Calculate the vertical component of the initial velocity. [1]

(c) Calculate the maximum height reached by the ball. [3]

(d) Calculate the horizontal range of the projectile. [3]

[Total: 8 marks]

15. A crane lifts a load of mass $500 \text{ kg}$ vertically upwards at constant speed through a height of $12 \text{ m}$ in $15 \text{ s}$ . Take $g = 9.8 \text{ m s}^{-2}$ .

(a) Calculate the weight of the load. [1]

(b) Calculate the work done by the crane in lifting the load. [2]

(c) Calculate the power output of the crane. [2]

(d) State the gain in gravitational potential energy of the load. Explain why this is equal to the work done by the crane. [2]

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

[Total: 7 marks]

Section C: Data Interpretation & Extended Response [20 marks]

Answer ALL questions.

16. The velocity–time graph below shows the motion of a car along a straight road.

Graph for Q16 (ALEVEL Physics H1)

Generated graph for Q16.

(a) Describe the motion of the car during the first $10 \text{ s}$ . [1]

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

(b) Calculate the acceleration of the car during the first $10 \text{ s}$ . [2]

(c) Calculate the total distance travelled by the car in the $30 \text{ s}$ period. [3]

(d) Sketch an acceleration–time graph for the entire $30 \text{ s}$ motion. Show numerical values on both axes. [3]

Graph for Q16d (ALEVEL Physics H1)

Generated graph for Q16d.

[Total: 9 marks]

17. A student investigates the motion of a trolley along a friction-compensated slope. The student releases the trolley from rest and uses a motion sensor to record its velocity at various times. The results are shown in the table below.

Time $t$ / s	0.0	0.5	1.0	1.5	2.0	2.5	3.0
Velocity $v$ / $\text{m s}^{-1}$	0.00	0.40	0.80	1.20	1.60	2.00	2.40

(a) State what type of motion the data suggests. Justify your answer. [2]

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

(b) Plot a graph of velocity (vertical axis) against time (horizontal axis) on the grid provided. [3]

Graph for Q17b (ALEVEL Physics H1)

Generated graph for Q17b.

(c) Determine the acceleration of the trolley from your graph. Show clearly how you obtained your answer. [2]

(d) The student now doubles the mass of the trolley and repeats the experiment, keeping the slope angle the same. State and explain whether the acceleration will change. [2]

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

[Total: 9 marks]

18. (a) Define the term work done by a force. [1]

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

(b) A box of mass $8.0 \text{ kg}$ is pulled along a rough horizontal floor by a force of $50 \text{ N}$ applied at an angle of $25°$ above the horizontal. The box moves a distance of $6.0 \text{ m}$ . The frictional force acting on the box is $18 \text{ N}$ .

(i) Calculate the work done by the applied force. [2]

(ii) Calculate the work done against friction. [1]

(iii) Using the work-energy principle, calculate the final speed of the box if it started from rest. [3]

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

[Total: 7 marks]

END OF PAPER

Total: 60 marks

Answers

TuitionGoWhere Practice Paper — Physics H1 A-Level

Answer Key — Version 2 of 5

Section A: Multiple Choice [10 marks]

1. Answer: B

At the highest point of a vertical trajectory, the ball momentarily stops (velocity = 0), but gravity still acts on it, giving a downward acceleration of $9.8 \text{ m s}^{-2}$ . Acceleration due to gravity is constant throughout the motion (ignoring air resistance).

Common mistake: Choosing A — students often assume that if velocity is zero, acceleration must also be zero. Acceleration is the rate of change of velocity, not dependent on the instantaneous velocity itself.

[1 mark]

2. Answer: B

Using $s = \frac{1}{2}(u + v)t$ : $s = \frac{1}{2}(0 + 24)(8.0) = \frac{1}{2}(24)(8.0) = 96 \text{ m}$

Alternatively, using $a = \frac{v - u}{t} = \frac{24}{8.0} = 3.0 \text{ m s}^{-2}$ , then $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3.0)(64) = 96 \text{ m}$ .

[1 mark]

3. Answer: D

Momentum ( $p = mv$ ) is a vector quantity because it has both magnitude and direction (the direction of velocity). Energy, power, and speed are all scalar quantities — they have magnitude only.

[1 mark]

4. Answer: B

Using conservation of linear momentum (perfectly inelastic collision): $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ $(2.0)(3.0) + (4.0)(0) = (2.0 + 4.0)v$ $6.0 = 6.0v$ $v = 1.0 \text{ m s}^{-1}$

[1 mark]

5. Answer: C

Work done: $W = F \times d = 12 \times 5.0 = 60 \text{ J}$

Work done is the product of force and displacement in the direction of the force.

[1 mark]

6. Answer: C

In projectile motion (ignoring air resistance), there is no horizontal force, so the horizontal velocity remains constant. The vertical acceleration is always $g = 9.8 \text{ m s}^{-2}$ downwards and does not change.

[1 mark]

7. Answer: C

Total downward force = weight of beam + load = $40 + 20 = 60 \text{ N}$

By symmetry (uniform beam, load at centre), each support carries half: $\text{Reaction at each support} = \frac{60}{2} = 30 \text{ N}$

[1 mark]

8. Answer: B

First, find the deceleration using $v^2 = u^2 + 2as$ : $0 = (20)^2 + 2a(50)$ $0 = 400 + 100a$ $a = -4.0 \text{ m s}^{-2}$

Then, using $F = ma$ : $F = 1000 \times 4.0 = 4000 \text{ N}$

[1 mark]

9. Answer: C

Gain in gravitational potential energy: $\Delta E_p = mgh = 5.0 \times 9.8 \times 10 = 490 \text{ J}$

[1 mark]

10. Answer: C

Taking the initial direction as positive: $\text{Initial momentum} = 0.50 \times 6.0 = 3.0 \text{ kg m s}^{-1}$ $\text{Final momentum} = 0.50 \times (-4.0) = -2.0 \text{ kg m s}^{-1}$

Change in momentum: $\Delta p = p_{\text{final}} - p_{\text{initial}} = (-2.0) - (3.0) = -5.0 \text{ kg m s}^{-1}$

Magnitude of change in momentum = $5.0 \text{ kg m s}^{-1}$

Common mistake: Students may subtract speeds directly ( $6.0 - 4.0 = 2.0$ ) and multiply by mass to get $1.0$ , which is incorrect. Momentum is a vector; direction matters.

[1 mark]

Section B: Structured Questions [30 marks]

11. (a) [B1] Newton's first law: An object remains at rest or continues to move at constant velocity [B1] unless acted upon by a resultant (net) external force.

Teaching note: This is also called the law of inertia. The key idea is that a resultant force is needed to change motion, not to maintain it.

[2 marks]

(b) Free-body diagram should show:

Weight ( $W$ ) acting vertically downward from the centre of the book, labelled $W = mg$ or "weight"
Normal reaction force ( $N$ or $R$ ) acting vertically upward from the bottom surface of the book, labelled "normal reaction" or " $N$ "

[B1] for correct downward force (weight), correctly labelled [B1] for correct upward force (normal reaction), correctly labelled, approximately equal in length to weight arrow

Note: The two arrows should be approximately equal in length since the book is in equilibrium.

[2 marks]

(c) The book remains at rest because the resultant force on it is zero — the weight is balanced by the normal reaction. By Newton's first law, since there is no resultant force, the book remains in its state of rest.

[1 mark]

12. (a) Using $v = u + at$ : $v = 0 + (2.5)(10) = 25 \text{ m s}^{-1}$

[B1] for correct equation or method [B1] for correct answer with unit: $25 \text{ m s}^{-1}$

[2 marks]

(b) Using $s = ut + \frac{1}{2}at^2$ : $s = 0 + \frac{1}{2}(2.5)(10)^2 = \frac{1}{2}(2.5)(100) = 125 \text{ m}$

Alternatively, $s = \frac{1}{2}(u + v)t = \frac{1}{2}(0 + 25)(10) = 125 \text{ m}$

[B1] for correct equation or method [B1] for correct answer with unit: $125 \text{ m}$

[2 marks]

(c) Distance during constant velocity phase: $s_2 = v \times t = 25 \times 20 = 500 \text{ m}$

Total distance: $s_{\text{total}} = 125 + 500 = 625 \text{ m}$

[B1] for calculating distance during constant velocity phase [B1] for correct total distance with unit: $625 \text{ m}$

[2 marks]

13. (a) [B1] In a closed/isolated system, the total momentum remains constant [B1] provided no external forces act on the system.

Teaching note: A "closed system" means no mass enters or leaves, and "no external forces" means the net external force is zero. Both conditions are needed for momentum conservation.

[2 marks]

(b) Taking the initial direction of the ball as positive:

(i) Initial momentum: $p_i = 0.060 \times 25 = 1.50 \text{ kg m s}^{-1}$

Final momentum: $p_f = 0.060 \times (-30) = -1.80 \text{ kg m s}^{-1}$

Change in momentum: $\Delta p = p_f - p_i = (-1.80) - (1.50) = -3.30 \text{ kg m s}^{-1}$

Magnitude = $3.30 \text{ kg m s}^{-1}$

[B1] for correct substitution into momentum change formula [B1] for correct answer: $3.30 \text{ kg m s}^{-1}$ (or $3.3 \text{ kg m s}^{-1}$ )

[2 marks]

(ii) Using the impulse-momentum theorem: $F \Delta t = \Delta p$ $F = \frac{|\Delta p|}{\Delta t} = \frac{3.30}{0.0050} = 660 \text{ N}$

[B1] for correct use of $F = \frac{\Delta p}{\Delta t}$ [B1] for correct answer with unit: $660 \text{ N}$

[2 marks]

14. (a) Horizontal component: $v_x = v \cos\theta = 20 \cos 35° = 20 \times 0.8192 = 16.4 \text{ m s}^{-1} \text{ (to 3 s.f.)}$

[B1] for correct working shown

[1 mark]

(b) Vertical component: $v_y = v \sin\theta = 20 \sin 35° = 20 \times 0.5736 = 11.5 \text{ m s}^{-1} \text{ (to 3 s.f.)}$

[B1] for correct answer: $11.5 \text{ m s}^{-1}$

[1 mark]

(c) At maximum height, vertical velocity $v_y = 0$ . Using $v^2 = u^2 + 2as$ vertically: $0 = (11.5)^2 + 2(-9.8)h$ $0 = 132.25 - 19.6h$ $h = \frac{132.25}{19.6} = 6.75 \text{ m}$

[B1] for correct equation [B1] for correct substitution [B1] for correct answer: $6.75 \text{ m}$ (or $6.7 \text{ m}$ to 2 s.f.)

[3 marks]

(d) Time of flight: Using $v = u + at$ for the upward phase: $0 = 11.5 - 9.8t_{\text{up}}$ $t_{\text{up}} = \frac{11.5}{9.8} = 1.173 \text{ s}$

Total time of flight: $T = 2 \times t_{\text{up}} = 2 \times 1.173 = 2.347 \text{ s}$

Horizontal range: $R = v_x \times T = 16.4 \times 2.347 = 38.5 \text{ m}$

Alternatively, using the range formula: $R = \frac{v^2 \sin 2\theta}{g} = \frac{(20)^2 \sin 70°}{9.8} = \frac{400 \times 0.9397}{9.8} = 38.4 \text{ m}$

[B1] for finding time of flight (or using range formula) [B1] for correct substitution [B1] for correct answer: $38.4 \text{ m}$ or $38.5 \text{ m}$ (accept $38 \text{ m}$ to 2 s.f.)

[3 marks]

15. (a) Weight: $W = mg = 500 \times 9.8 = 4900 \text{ N}$

[B1] for correct answer with unit: $4900 \text{ N}$

[1 mark]

(b) Since the load moves at constant speed, the tension in the cable equals the weight. Work done: $W = F \times d = 4900 \times 12 = 58\,800 \text{ J} = 58.8 \text{ kJ}$

[B1] for correct force (equal to weight since constant speed) [B1] for correct answer: $58\,800 \text{ J}$ or $58.8 \text{ kJ}$

[2 marks]

(c) Power: $P = \frac{W}{t} = \frac{58\,800}{15} = 3920 \text{ W} = 3.92 \text{ kW}$

[B1] for correct formula [B1] for correct answer: $3920 \text{ W}$ or $3.92 \text{ kW}$

[2 marks]

(d) Gain in gravitational potential energy: $\Delta E_p = mgh = 500 \times 9.8 \times 12 = 58\,800 \text{ J}$

This is equal to the work done by the crane because the load moves at constant speed (no change in kinetic energy). By the work-energy principle, all the work done by the crane goes into increasing the gravitational potential energy of the load.

[B1] for correct value of GPE gain: $58\,800 \text{ J}$ [B1] for correct explanation linking constant speed (no $\Delta KE$ ) to work done = GPE gain

[2 marks]

Section C: Data Interpretation & Extended Response [20 marks]

16. (a) The car accelerates uniformly from rest, reaching a velocity of $20 \text{ m s}^{-1}$ in $10 \text{ s}$ .

[B1] for stating uniform/increasing velocity/acceleration from rest

[1 mark]

(b) Acceleration = gradient of velocity–time graph: $a = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{10 - 0} = \frac{20}{10} = 2.0 \text{ m s}^{-2}$

[B1] for correct method (gradient or $a = \frac{\Delta v}{\Delta t}$ ) [B1] for correct answer with unit: $2.0 \text{ m s}^{-2}$

[2 marks]

(c) Total distance = area under the velocity–time graph.

The graph forms a trapezium (or can be split into a triangle + rectangle + triangle):

Area of triangle (0–10 s): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$
Area of rectangle (10–20 s): $10 \times 20 = 200 \text{ m}$
Area of triangle (20–30 s): $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$

Total distance: $100 + 200 + 100 = 400 \text{ m}$

Alternatively, using the trapezium formula: $s = \frac{1}{2}(20 + 30) \times 20 = \frac{1}{2}(50)(20) = 500 \text{ m}$

Wait — let me recalculate. The total time is 30 s. The shape is a trapezium with parallel sides of 20 s (at $v = 20$ ) and 0 s... Actually, it's simpler to use the three regions:

Triangle: $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$
Rectangle: $10 \times 20 = 200 \text{ m}$
Triangle: $\frac{1}{2} \times 10 \times 20 = 100 \text{ m}$

Total: $100 + 200 + 100 = 400 \text{ m}$

[B1] for identifying area under graph as distance [B1] for correct calculation of at least two areas [B1] for correct total: $400 \text{ m}$

[3 marks]

(d) Acceleration–time graph:

$0$ to $10 \text{ s}$ : $a = +2.0 \text{ m s}^{-2}$ (uniform acceleration)
$10$ to $20 \text{ s}$ : $a = 0$ (constant velocity)
$20$ to $30 \text{ s}$ : $a = \frac{0 - 20}{10} = -2.0 \text{ m s}^{-2}$ (uniform deceleration)

[B1] for correct shape (three horizontal segments) [B1] for correct values ( $+2.0$ , $0$ , $-2.0 \text{ m s}^{-2}$ ) [B1] for correct time intervals ( $0$ – $10$ , $10$ – $20$ , $20$ – $30$ s)

[3 marks]

17. (a) The data suggests uniform acceleration (constant acceleration). Justification: the velocity increases by $0.40 \text{ m s}^{-1}$ in every $0.5 \text{ s}$ interval, i.e., the rate of change of velocity is constant. Alternatively, velocity is directly proportional to time ( $v \propto t$ ), which is characteristic of uniform acceleration from rest.

[B1] for identifying uniform/constant acceleration [B1] for valid justification (equal velocity increments in equal time intervals, or $v \propto t$ )

[2 marks]

(b) Graph requirements:

Axes correctly labelled with units: "Time / s" (horizontal), "Velocity / m s⁻¹" (vertical)
Suitable scales used (e.g., 1 cm = 0.5 s horizontally, 1 cm = 0.4 m s⁻¹ vertically)
All 7 points plotted correctly to within ±½ small square
Straight best-fit line drawn through the origin

[B1] for correct axes labels and units with suitable scales [B1] for correct plotting of points (allow ±½ small square) [B1] for correct best-fit straight line through origin

[3 marks]

(c) Acceleration = gradient of $v$ – $t$ graph.

Using two points on the best-fit line, e.g., $(0, 0)$ and $(3.0, 2.40)$ : $a = \frac{\Delta v}{\Delta t} = \frac{2.40 - 0}{3.0 - 0} = \frac{2.40}{3.0} = 0.80 \text{ m s}^{-2}$

[B1] for correct method (gradient calculation shown) [B1] for correct answer: $0.80 \text{ m s}^{-2}$

[2 marks]

(d) The acceleration will not change. On a friction-compensated slope, the component of gravitational force along the slope is $mg\sin\theta$ . By Newton's second law, $ma = mg\sin\theta$ , so $a = g\sin\theta$ . The acceleration is independent of mass — it depends only on $g$ and the slope angle $\theta$ . Since neither changes, the acceleration remains the same.

[B1] for stating that acceleration does not change [B1] for correct explanation ( $a = g\sin\theta$ , independent of mass)

[2 marks]

18. (a) Work done by a force is defined as the product of the force and the displacement of the object in the direction of the force. Equivalently, work done = force × displacement × $\cos\theta$ , where $\theta$ is the angle between the force and displacement.

Accept: $W = Fd\cos\theta$ with explanation of symbols.

[1 mark]

(b) (i) Horizontal component of applied force: $F_x = 50 \cos 25° = 50 \times 0.9063 = 45.3 \text{ N}$

Work done by applied force: $W = F_x \times d = 45.3 \times 6.0 = 272 \text{ J}$

Or directly: $W = Fd\cos\theta = 50 \times 6.0 \times \cos 25° = 300 \times 0.9063 = 272 \text{ J}$

[B1] for correct method ( $W = Fd\cos\theta$ or resolving force) [B1] for correct answer: $272 \text{ J}$ (or $270 \text{ J}$ to 2 s.f.)

[2 marks]

(ii) Work done against friction: $W_f = f \times d = 18 \times 6.0 = 108 \text{ J}$

[B1] for correct answer: $108 \text{ J}$

[1 mark]

(iii) By the work-energy principle, the net work done on the box equals the change in kinetic energy:

$W_{\text{net}} = \Delta KE$ $W_{\text{applied}} - W_{\text{friction}} = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Since the box starts from rest, $u = 0$ : $272 - 108 = \frac{1}{2}(8.0)v^2$ $164 = 4.0v^2$ $v^2 = 41$ $v = \sqrt{41} = 6.4 \text{ m s}^{-1}$

[B1] for correct application of work-energy principle [B1] for correct substitution [B1] for correct answer: $6.4 \text{ m s}^{-1}$

[3 marks]

END OF ANSWER KEY

Total: 60 marks

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