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A Level H2 Physics Practice Paper 1

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TuitionGoWhere Practice Paper - Physics H2 A-Level

TuitionGoWhere Secondary School (AI)

Field	Details
Subject:	Physics (H2)
Level:	A-Level
Paper:	Practice Paper — Mechanics
Version:	1 of 5
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 [ ].
You are advised to show all working clearly, including formulas and substitutions.
Assume $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.
Unless otherwise specified, air resistance may be neglected.
The total mark for this paper is 60.

Section A: Short Answer Questions [20 marks]

Answer all questions 1–10. Each question carries 2 marks.

1. State the principle of conservation of linear momentum.

[2]

2. A car accelerates uniformly from rest to a speed of $24 \text{ m s}^{-1}$ in $8.0 \text{ s}$ . Calculate the acceleration of the car.

[2]

3. Define work done by a force and state its SI unit.

[2]

4. A ball is thrown vertically upwards with an initial speed of $15 \text{ m s}^{-1}$ . Calculate the maximum height reached by the ball.

[2]

5. State Newton's first law of motion.

[2]

6. A force of $12 \text{ N}$ acts on an object of mass $4.0 \text{ kg}$ on a frictionless horizontal surface. Calculate the acceleration of the object.

[2]

7. Distinguish between scalar and vector quantities, giving one example of each.

[2]

8. A car of mass $1200 \text{ kg}$ travels at a constant speed of $18 \text{ m s}^{-1}$ . Calculate its kinetic energy.

[2]

9. State the condition for an object to be in translational equilibrium.

[2]

10. A stone is dropped from the top of a building. It takes $3.0 \text{ s}$ to reach the ground. Calculate the height of the building.

[2]

Section B: Structured Questions [25 marks]

Answer all questions 11–15.

11. A $0.50 \text{ kg}$ trolley A moves at $3.0 \text{ m s}^{-1}$ on a frictionless horizontal track and collides with a stationary trolley B of mass $1.5 \text{ kg}$ . After the collision, the two trolleys stick together and move as one.

<image_placeholder> id: Q11-fig1 type: diagram linked_question: Q11 description: A horizontal frictionless track showing two trolleys before and after collision. Trolley A (mass 0.50 kg) moving right at 3.0 m/s towards stationary trolley B (mass 1.5 kg). After collision, combined trolleys move together to the right. labels: Trolley A (0.50 kg), Trolley B (1.5 kg), v_A = 3.0 m/s (before), v_B = 0 (before), v' = ? (after, combined) values: m_A = 0.50 kg, m_B = 1.5 kg, u_A = 3.0 m/s, u_B = 0 m/s must_show: Both trolleys clearly labelled with masses, velocity arrows before and after collision, direction of motion indicated </image_placeholder>

(a) Calculate the velocity of the combined trolleys immediately after the collision.

(3 marks)

(b) Determine whether kinetic energy is conserved in this collision. Show your working clearly and state the type of collision.

(3 marks)

(c) Explain, using Newton's third law, why trolley B exerts a force on trolley A during the collision.

(2 marks)

12. A small ball is projected horizontally at $8.0 \text{ m s}^{-1}$ from the top of a cliff $45 \text{ m}$ high.

<image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: A cliff of height 45 m with a ball projected horizontally from the top at 8.0 m/s. The ball follows a parabolic trajectory to the ground. Horizontal distance from base of cliff to landing point labelled as R. labels: Cliff height h = 45 m, initial horizontal velocity u_x = 8.0 m/s, initial vertical velocity u_y = 0, parabolic trajectory, range R, g = 9.81 m/s² downward values: h = 45 m, u_x = 8.0 m/s, u_y = 0, g = 9.81 m/s² must_show: Cliff with height marked, horizontal velocity arrow at top, parabolic path, ground level, range R indicated </image_placeholder>

(a) Calculate the time taken for the ball to reach the ground.

(2 marks)

(b) Calculate the horizontal distance $R$ from the base of the cliff to the point where the ball lands.

(2 marks)

(c) Calculate the speed of the ball just before it hits the ground.

(3 marks)

13. A $60 \text{ kg}$ student stands in a lift.

(a) The lift accelerates upwards at $1.5 \text{ m s}^{-2}$ . Calculate the reaction force exerted by the lift floor on the student.

(3 marks)

(b) The lift then moves at constant velocity. State the reaction force and explain your answer using Newton's laws.

(2 marks)

(c) Finally, the lift decelerates (while still moving upwards) at $2.0 \text{ m s}^{-2}$ . Calculate the new reaction force.

(2 marks)

14. A $2.0 \text{ kg}$ block is pulled along a rough horizontal surface by a force of $20 \text{ N}$ acting at an angle of $30°$ above the horizontal. The coefficient of kinetic friction between the block and the surface is $0.25$ .

<image_placeholder> id: Q14-fig1 type: diagram linked_question: Q14 description: A block of mass 2.0 kg on a rough horizontal surface. A force of 20 N is applied at 30° above the horizontal to the right. Weight mg acts downward, normal reaction R acts upward, and friction f acts to the left. labels: Block (2.0 kg), F = 20 N at 30° above horizontal, weight W = mg downward, normal reaction R upward, friction f leftward, horizontal surface values: m = 2.0 kg, F = 20 N, θ = 30°, μ_k = 0.25, g = 9.81 m/s² must_show: Block on surface, force at angle, all four forces labelled with directions, angle marked </image_placeholder>

(a) Calculate the horizontal component of the applied force.

(1 mark)

(b) Calculate the normal reaction force exerted by the surface on the block.

(2 marks)

(c) Calculate the acceleration of the block.

(3 marks)

15. A ball of mass $0.20 \text{ kg}$ strikes a vertical wall horizontally at $10 \text{ m s}^{-1}$ and rebounds horizontally at $6.0 \text{ m s}^{-1}$ .

(a) Calculate the change in momentum of the ball.

(3 marks)

(b) If the contact time between the ball and the wall is $0.040 \text{ s}$ , calculate the average force exerted by the wall on the ball.

(2 marks)

Section C: Extended Response [15 marks]

Answer any two questions from 16–18. Each question carries 7 or 8 marks.

16. A $1500 \text{ kg}$ car travels along a straight horizontal road. The engine provides a constant driving force of $3000 \text{ N}$ . The total resistive force acting on the car is given by $F_r = 500 + 0.8v^2$ where $v$ is the speed of the car in $\text{m s}^{-1}$ and $F_r$ is in newtons.

(a) Calculate the initial acceleration of the car from rest.

(2 marks)

(b) Explain, using Newton's laws, why the acceleration of the car decreases as its speed increases.

(3 marks)

(c) Calculate the maximum speed of the car.

(3 marks)

17. A small object of mass $0.40 \text{ kg}$ is attached to one end of a light inextensible string of length $0.80 \text{ m}$ . The other end of the string is fixed. The object is made to move in a vertical circle.

<image_placeholder> id: Q17-fig1 type: diagram linked_question: Q17 description: A vertical circle of radius 0.80 m with a mass 0.40 kg attached to a string. Three positions shown: top of circle, bottom of circle, and a general position. At the top, tension T_top and weight mg both point downward toward centre. At the bottom, tension T_bottom points upward toward centre, weight mg points downward away from centre. labels: mass m = 0.40 kg, string length r = 0.80 m, top position (T_top + mg toward centre), bottom position (T_bottom toward centre, mg away from centre), vertical circle values: m = 0.40 kg, r = 0.80 m, g = 9.81 m/s² must_show: Complete vertical circle, mass at top and bottom positions, tension and weight vectors clearly shown at both positions, centre of circle indicated </image_placeholder>

(a) Calculate the minimum speed the object must have at the top of the circle for it to just maintain circular motion.

(3 marks)

(b) If the speed of the object at the bottom of the circle is $6.0 \text{ m s}^{-1}$ , calculate the tension in the string at the bottom.

(3 marks)

(c) Explain why the string is most likely to break when the object is at the bottom of the circle.

(2 marks)

18. A $5.0 \text{ kg}$ block slides from rest down a rough inclined plane that makes an angle of $25°$ with the horizontal. The length of the plane along the slope is $4.0 \text{ m}$ . The coefficient of kinetic friction between the block and the plane is $0.15$ .

<image_placeholder> id: Q18-fig1 type: diagram linked_question: Q18 description: An inclined plane at angle 25° to the horizontal. A block of mass 5.0 kg is at the top of the slope. The slope length is 4.0 m. The vertical height h is shown. Forces on block: weight mg downward, normal reaction R perpendicular to slope, friction f up the slope. labels: Block (5.0 kg), incline angle θ = 25°, slope length s = 4.0 m, height h, weight mg downward, normal reaction R perpendicular to surface, friction f up the slope values: m = 5.0 kg, θ = 25°, s = 4.0 m, μ_k = 0.15, g = 9.81 m/s² must_show: Inclined plane with angle marked, block at top, slope length and height indicated, all three forces on block labelled with directions </image_placeholder>

(a) Using Newton's second law, derive an expression for the acceleration of the block down the slope.

(3 marks)

(b) Calculate the speed of the block at the bottom of the slope.

(2 marks)

(c) Instead of using Newton's laws, use the principle of conservation of energy (accounting for work done against friction) to verify your answer to part (b).

(3 marks)

End of Paper

Total: 60 marks

Answers

TuitionGoWhere Practice Paper - Physics H2 A-Level

Answer Key — Practice Paper: Mechanics (Version 1 of 5)

Section A: Short Answer Questions [20 marks]

1. [2]

The principle of conservation of linear momentum states that:

In a closed system (or isolated system), the total momentum before an interaction equals the total momentum after the interaction, provided no external resultant force acts on the system.

Marking:

1 mark for stating that total momentum remains constant / is conserved.
1 mark for specifying the condition: no external resultant force / closed/isolated system.

Common mistakes:

Simply writing "momentum is conserved" without mentioning the condition of no external force — this only scores 1 mark.
Confusing with conservation of energy.

2. [2]

Using $a = \frac{v - u}{t}$ :

$a = \frac{24 - 0}{8.0} = 3.0 \text{ m s}^{-2}$

Marking:

1 mark for correct formula or method.
1 mark for correct answer with unit.

Answer: $3.0 \text{ m s}^{-2}$

3. [2]

Definition: Work done by a force is the product of the force and the displacement in the direction of the force.

$W = F \cdot s \cdot \cos\theta$

where $F$ is the force, $s$ is the displacement, and $\theta$ is the angle between the force and displacement.

SI unit: joule (J), where $1 \text{ J} = 1 \text{ N m}$ .

Marking:

1 mark for correct definition (force × displacement in direction of force).
1 mark for correct SI unit (joule or N m).

4. [2]

At maximum height, final velocity $v = 0$ .

Using $v^2 = u^2 + 2as$ :

$0 = (15)^2 + 2(-9.81)(h)$

$h = \frac{225}{2 \times 9.81} = \frac{225}{19.62} = 11.5 \text{ m}$

Marking:

1 mark for correct substitution into appropriate kinematic equation.
1 mark for correct answer (accept 11.4–11.5 m depending on $g$ used).

Answer: $11.5 \text{ m}$ (or $11.4 \text{ m}$ if $g = 10 \text{ m s}^{-2}$ used)

5. [2]

Newton's first law of motion states:

An object remains at rest or continues to move at a constant velocity unless acted upon by a resultant external force.

Marking:

1 mark for stating constant velocity / rest condition.
1 mark for stating the condition of no resultant external force.

Common mistakes:

Omitting "resultant" force.
Only stating "object at rest stays at rest" without mentioning constant velocity motion.

6. [2]

Using Newton's second law, $F = ma$ :

$a = \frac{F}{m} = \frac{12}{4.0} = 3.0 \text{ m s}^{-2}$

Marking:

1 mark for correct formula.
1 mark for correct answer with unit.

Answer: $3.0 \text{ m s}^{-2}$

7. [2]

	Scalar	Vector
Definition	A quantity with magnitude only	A quantity with magnitude and direction
Example	Speed, mass, energy, time	Velocity, force, momentum, displacement

Marking:

1 mark for correct distinction (magnitude only vs. magnitude and direction).
1 mark for one correct example of each.

8. [2]

$KE = \frac{1}{2}mv^2 = \frac{1}{2}(1200)(18)^2 = \frac{1}{2}(1200)(324) = 194400 \text{ J}$

$KE = 1.94 \times 10^5 \text{ J} \quad (\text{or } 194 \text{ kJ})$

Marking:

1 mark for correct formula and substitution.
1 mark for correct answer.

Answer: $1.94 \times 10^5 \text{ J}$

9. [2]

An object is in translational equilibrium when the resultant (net) force acting on it is zero.

This means: $\sum F = 0$

The object may be at rest or moving with constant velocity.

Marking:

2 marks for stating that the resultant/net force is zero.
Accept: "vector sum of all forces is zero" or "sum of forces in any direction is zero."

10. [2]

Using $s = ut + \frac{1}{2}at^2$ where $u = 0$ :

$h = 0 + \frac{1}{2}(9.81)(3.0)^2 = \frac{1}{2}(9.81)(9.0) = 44.1 \text{ m}$

Marking:

1 mark for correct substitution.
1 mark for correct answer (accept 44.1 m or 45 m if $g = 10$ used).

Answer: $44.1 \text{ m}$

Section B: Structured Questions [25 marks]

11. (a) [3]

By conservation of linear momentum:

$m_A u_A + m_B u_B = (m_A + m_B)v'$

$(0.50)(3.0) + (1.5)(0) = (0.50 + 1.5)v'$

$1.5 = 2.0 \times v'$

$v' = \frac{1.5}{2.0} = 0.75 \text{ m s}^{-1}$

Marking:

1 mark for stating/using conservation of momentum.
1 mark for correct substitution.
1 mark for correct answer with unit.

Answer: $0.75 \text{ m s}^{-1}$ in the original direction of motion of trolley A.

(b) [3]

Kinetic energy before collision:

$KE_{\text{before}} = \frac{1}{2}(0.50)(3.0)^2 + \frac{1}{2}(1.5)(0)^2 = \frac{1}{2}(0.50)(9.0) = 2.25 \text{ J}$

Kinetic energy after collision:

$KE_{\text{after}} = \frac{1}{2}(2.0)(0.75)^2 = \frac{1}{2}(2.0)(0.5625) = 0.5625 \text{ J}$

Since $KE_{\text{after}} < KE_{\text{before}}$ , kinetic energy is not conserved.

This is a perfectly inelastic collision (the objects stick together).

Marking:

1 mark for calculating KE before.
1 mark for calculating KE after and comparing.
1 mark for stating it is not conserved and identifying the collision type.

(c) [2]

By Newton's third law, when trolley A exerts a force on trolley B during the collision, trolley B exerts an equal and opposite force on trolley A. The two forces are equal in magnitude, opposite in direction, act on different bodies, and are of the same type (contact force).

Marking:

1 mark for stating equal and opposite forces (action-reaction pair).
1 mark for noting they act on different bodies / same type of force.

12. (a) [2]

Vertical motion: $h = \frac{1}{2}gt^2$ (since $u_y = 0$ )

$45 = \frac{1}{2}(9.81)t^2$

$t^2 = \frac{90}{9.81} = 9.174$

$t = 3.03 \text{ s}$

Marking:

1 mark for correct equation.
1 mark for correct answer.

Answer: $3.03 \text{ s}$

(b) [2]

Horizontal motion (constant velocity):

$R = u_x \times t = 8.0 \times 3.03 = 24.2 \text{ m}$

Marking:

1 mark for using horizontal velocity × time.
1 mark for correct answer.

Answer: $24.2 \text{ m}$

(c) [3]

Vertical component of velocity just before hitting ground:

$v_y = gt = 9.81 \times 3.03 = 29.7 \text{ m s}^{-1}$

Horizontal component remains constant: $v_x = 8.0 \text{ m s}^{-1}$

Resultant speed:

$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(8.0)^2 + (29.7)^2} = \sqrt{64 + 882.1} = \sqrt{946.1} = 30.8 \text{ m s}^{-1}$

Marking:

1 mark for calculating $v_y$ .
1 mark for using Pythagoras' theorem to find resultant.
1 mark for correct final answer.

Answer: $30.8 \text{ m s}^{-1}$

13. (a) [3]

Taking upward as positive. Using Newton's second law:

$R - mg = ma$

$R = m(g + a) = 60(9.81 + 1.5) = 60 \times 11.31 = 678.6 \text{ N}$

$R \approx 679 \text{ N}$

Marking:

1 mark for correct free-body diagram or equation setup.
1 mark for correct substitution.
1 mark for correct answer.

Answer: $679 \text{ N}$ (upward)

(b) [2]

When the lift moves at constant velocity, acceleration $a = 0$ .

$R = mg = 60 \times 9.81 = 588.6 \text{ N} \approx 589 \text{ N}$

Explanation: By Newton's first law, when the lift moves at constant velocity, the resultant force on the student is zero. Therefore, the reaction force equals the weight.

Marking:

1 mark for correct value.
1 mark for explanation referencing Newton's first law / zero acceleration.

(c) [2]

The lift is decelerating while moving upward, so acceleration is downward: $a = -2.0 \text{ m s}^{-2}$ .

$R = m(g - a_{\text{decel}}) = 60(9.81 - 2.0) = 60 \times 7.81 = 468.6 \text{ N}$

$R \approx 469 \text{ N}$

Marking:

1 mark for correct equation with reduced effective g.
1 mark for correct answer.

Answer: $469 \text{ N}$

14. (a) [1]

$F_x = F\cos\theta = 20\cos 30° = 20 \times 0.866 = 17.3 \text{ N}$

Answer: $17.3 \text{ N}$

(b) [2]

Resolving vertically (no vertical acceleration):

$R + F\sin\theta = mg$

$R = mg - F\sin\theta = (2.0)(9.81) - 20\sin 30°$

$R = 19.62 - 20(0.5) = 19.62 - 10.0 = 9.62 \text{ N}$

Marking:

1 mark for correct vertical force balance equation.
1 mark for correct answer.

Answer: $9.62 \text{ N}$

(c) [3]

Friction force:

$f = \mu_k R = 0.25 \times 9.62 = 2.405 \text{ N}$

Applying Newton's second law horizontally:

$F_x - f = ma$

$17.3 - 2.405 = 2.0 \times a$

$a = \frac{14.895}{2.0} = 7.45 \text{ m s}^{-2}$

Marking:

1 mark for calculating friction force.
1 mark for applying Newton's second law horizontally.
1 mark for correct answer.

Answer: $7.45 \text{ m s}^{-2}$

15. (a) [3]

Taking the initial direction of motion as positive:

Initial momentum: $p_i = 0.20 \times 10 = 2.0 \text{ kg m s}^{-1}$

Final momentum: $p_f = 0.20 \times (-6.0) = -1.2 \text{ kg m s}^{-1}$ (rebound is opposite direction)

Change in momentum:

$\Delta p = p_f - p_i = -1.2 - 2.0 = -3.2 \text{ kg m s}^{-1}$

Magnitude of change: $3.2 \text{ kg m s}^{-1}$ in the direction away from the wall.

Marking:

1 mark for correct initial and final momentum values with signs.
1 mark for correct subtraction.
1 mark for correct magnitude and direction.

Answer: $3.2 \text{ kg m s}^{-1}$ directed away from the wall.

(b) [2]

Using the impulse-momentum theorem:

$F \cdot \Delta t = \Delta p$

$F = \frac{|\Delta p|}{\Delta t} = \frac{3.2}{0.040} = 80 \text{ N}$

Marking:

1 mark for using $F = \Delta p / \Delta t$ .
1 mark for correct answer.

Answer: $80 \text{ N}$

Section C: Extended Response [15 marks]

16. (a) [2]

At rest, $v = 0$ , so $F_r = 500 + 0.8(0)^2 = 500 \text{ N}$

Resultant force: $F_{\text{net}} = 3000 - 500 = 2500 \text{ N}$

$a = \frac{F_{\text{net}}}{m} = \frac{2500}{1500} = 1.67 \text{ m s}^{-2}$

Marking:

1 mark for calculating resistive force at $v = 0$ .
1 mark for correct acceleration.

Answer: $1.67 \text{ m s}^{-2}$

(b) [3]

As the speed $v$ increases, the resistive force $F_r = 500 + 0.8v^2$ increases (since it depends on $v^2$ ).

The driving force remains constant at $3000 \text{ N}$ , so the resultant force:

$F_{\text{net}} = 3000 - F_r$

decreases as $F_r$ increases.

By Newton's second law ( $F = ma$ ), if the resultant force decreases while mass stays constant, the acceleration must decrease.

Marking:

1 mark for explaining that resistive force increases with speed.
1 mark for stating that resultant force therefore decreases.
1 mark for linking to Newton's second law to conclude acceleration decreases.

(c) [3]

At maximum speed, acceleration is zero, so resultant force is zero:

$F_{\text{driving}} = F_r$

$3000 = 500 + 0.8v_{\text{max}}^2$

$0.8v_{\text{max}}^2 = 2500$

$v_{\text{max}}^2 = 3125$

$v_{\text{max}} = 55.9 \text{ m s}^{-1}$

Marking:

1 mark for setting driving force = resistive force.
1 mark for correct algebraic manipulation.
1 mark for correct answer.

Answer: $55.9 \text{ m s}^{-1}$

17. (a) [3]

At the top of the circle, both the tension and weight act toward the centre (downward). For minimum speed, tension is zero at the top, so weight alone provides the centripetal force:

$mg = \frac{mv_{\text{top}}^2}{r}$

$v_{\text{top}}^2 = gr = 9.81 \times 0.80 = 7.848$

$v_{\text{top}} = \sqrt{7.848} = 2.80 \text{ m s}^{-1}$

Marking:

1 mark for setting weight = centripetal force (T = 0 at minimum speed).
1 mark for correct substitution.
1 mark for correct answer.

Answer: $2.80 \text{ m s}^{-1}$

(b) [3]

At the bottom of the circle, tension acts toward the centre (upward) and weight acts away from the centre (downward):

$T - mg = \frac{mv_{\text{bottom}}^2}{r}$

$T = mg + \frac{mv^2}{r} = 0.40 \times 9.81 + \frac{0.40 \times (6.0)^2}{0.80}$

$T = 3.924 + \frac{0.40 \times 36}{0.80} = 3.924 + \frac{14.4}{0.80} = 3.924 + 18.0 = 21.9 \text{ N}$

Marking:

1 mark for correct equation (T − mg = mv²/r).
1 mark for correct substitution.
1 mark for correct answer.

Answer: $21.9 \text{ N}$

(c) [2]

At the bottom of the circle, the tension must support the weight of the object AND provide the centripetal force. The tension is therefore greatest at the bottom:

$T_{\text{bottom}} = mg + \frac{mv^2}{r}$

At the top, the weight contributes toward the centripetal force, so the tension is less:

$T_{\text{top}} = \frac{mv^2}{r} - mg$

Since tension is maximum at the bottom, the string is most likely to break there.

Marking:

1 mark for explaining that tension is greatest at the bottom.
1 mark for comparing with tension at the top (weight assists at top, opposes at bottom).

18. (a) [3]

Forces along the slope (taking down the slope as positive):

Component of weight down the slope: $mg\sin\theta$

Friction up the slope: $f = \mu_k R = \mu_k mg\cos\theta$

Applying Newton's second law along the slope:

$mg\sin\theta - \mu_k mg\cos\theta = ma$

$a = g(\sin\theta - \mu_k\cos\theta)$

Marking:

1 mark for correct identification of forces along the slope.
1 mark for including friction correctly.
1 mark for deriving the expression for acceleration.

(b) [2]

$a = 9.81(\sin 25° - 0.15\cos 25°)$

$a = 9.81(0.4226 - 0.15 \times 0.9063)$

$a = 9.81(0.4226 - 0.1359) = 9.81 \times 0.2867 = 2.813 \text{ m s}^{-2}$

Using $v^2 = u^2 + 2as$ with $u = 0$ :

$v = \sqrt{2 \times 2.813 \times 4.0} = \sqrt{22.50} = 4.74 \text{ m s}^{-1}$

Marking:

1 mark for correct substitution into acceleration expression.
1 mark for correct final speed.

Answer: $4.74 \text{ m s}^{-1}$

(c) [3]

Using conservation of energy:

Loss in gravitational PE = Gain in KE + Work done against friction

$mgh = \frac{1}{2}mv^2 + f \times s$

where $h = s\sin\theta = 4.0\sin 25° = 4.0 \times 0.4226 = 1.690 \text{ m}$

and $f = \mu_k mg\cos\theta = 0.15 \times 5.0 \times 9.81 \times \cos 25° = 0.15 \times 5.0 \times 9.81 \times 0.9063 = 6.667 \text{ N}$

$5.0 \times 9.81 \times 1.690 = \frac{1}{2}(5.0)v^2 + 6.667 \times 4.0$

$82.92 = 2.5v^2 + 26.67$

$2.5v^2 = 56.25$

$v^2 = 22.50$

$v = 4.74 \text{ m s}^{-1}$

This agrees with the answer from part (b), verifying the result.

Marking:

1 mark for correct energy conservation equation.
1 mark for correct calculation of height and friction work.
1 mark for correct final answer matching part (b).

Answer: $v = 4.74 \text{ m s}^{-1}$ ✓ (verified)

End of Answer Key

Total: 60 marks