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A Level H2 Physics Mechanics Quiz

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Questions

A-Level Physics H2 Quiz - Mechanics

Name: ____________________
Class: ____________________
Date: ____________________
Score: ______ / 60

Duration: 75 minutes
Total Marks: 60

Instructions

Answer ALL questions in the spaces provided.
Show all working clearly. Marks are awarded for correct reasoning and method, not only for the final answer.
Include units in your final answers where appropriate.
The number of marks for each question or part-question is shown in brackets [ ].
You may use a non-programmable scientific calculator.
Take $g = 9.81 \text{ m s}^{-2}$ unless otherwise stated.

Section A: Multiple Choice (Questions 1–5)

Each question carries 2 marks. Choose the ONE best answer.

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

A. $10.2 \text{ m}$
B. $20.4 \text{ m}$
C. $30.6 \text{ m}$
D. $40.8 \text{ m}$

Answer: _______________ [2]

2. A car of mass $1000 \text{ kg}$ travelling at $20 \text{ m s}^{-1}$ brakes and comes to rest over a distance of $50 \text{ m}$ . What is the magnitude of the average braking force?

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

Answer: _______________ [2]

3. 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. $16.0 \text{ m s}^{-2}$

Answer: _______________ [2]

4. A satellite orbits the Earth at a height where the gravitational field strength is $2.5 \text{ N kg}^{-1}$ . If the radius of the orbit is $1.6 \times 10^7 \text{ m}$ , what is the orbital speed of the satellite?

A. $2.0 \times 10^3 \text{ m s}^{-1}$
B. $4.0 \times 10^3 \text{ m s}^{-1}$
C. $6.0 \times 10^3 \text{ m s}^{-1}$
D. $8.0 \times 10^3 \text{ m s}^{-1}$

Answer: _______________ [2]

5. A force $F$ acts on an object of mass $m$ for a time $t$ , increasing its velocity from $u$ to $v$ . Which expression correctly represents the impulse delivered to the object?

A. $Ft$
B. $m(v - u)$
C. Both A and B
D. $\frac{1}{2}m(v^2 - u^2)$

Answer: _______________ [2]

Section B: Structured Questions (Questions 6–15)

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

7. A car of mass $1200 \text{ kg}$ is travelling at $25 \text{ m s}^{-1}$ on a straight horizontal road. The driver applies the brakes and the car comes to rest in $8.0 \text{ s}$ .

(a) Calculate the deceleration of the car. [2]

(b) Calculate the braking force acting on the car. [2]

8. A ball of mass $0.50 \{ kg}$ moving horizontally at $6.0 \text{ m s}^{-1}$ strikes a stationary ball of mass $0.30 \text{ kg}$ on a smooth horizontal surface. After the collision, the $0.50 \text{ kg}$ ball continues to move in the same direction at $2.0 \text{ m s}^{-1}$ .

(a) Using the principle of conservation of linear momentum, calculate the velocity of the $0.30 \text{ kg}$ ball after the collision. [3]

(b) Determine whether this collision is elastic or inelastic. Show your reasoning. [3]

9. A small object of mass $0.20 \text{ kg}$ is attached to a light inextensible string of length $0.80 \text{ m}$ and made to move in a vertical circle. The object has a speed of $5.0 \text{ m s}^{-1}$ at the lowest point of the circle.

(a) Calculate the centripetal acceleration of the object at the lowest point. [2]

(b) Calculate the tension in the string at the lowest point. [3]

(c) Calculate the minimum speed the object must have at the lowest point to maintain circular motion (i.e., the string remains taut at the highest point). [4]

10. A stone is thrown from the top of a cliff $45 \text{ m}$ high with an initial velocity of $15 \text{ m s}^{-1}$ at an angle of $30°$ above the horizontal. Air resistance is negligible.

(a) Calculate the horizontal and vertical components of the initial velocity. [2]

(b) Calculate the time taken for the stone to reach the ground. [3]

11. A student sets up an experiment to investigate the motion of a trolley on a ramp, as shown below.

<image_placeholder> id: Q11-fig1 type: experimental_setup linked_question: Q11 description: A trolley of mass m on a smooth inclined ramp at angle θ to the horizontal. A light gate is positioned at the bottom of the ramp to measure the speed of the trolley. The vertical height of the release point above the light gate is h. Labels: trolley, inclined ramp, angle θ, light gate, height h, horizontal surface at bottom. labels: trolley, ramp, θ, light gate, h, release point values: h = 0.80 m, m = 0.500 kg, θ = 20° must_show: The trolley at the top of the ramp, the inclined surface at angle θ, the light gate at the bottom, the vertical height h clearly marked between release point and the horizontal level of the light gate.

</image_placeholder>

The trolley of mass $0.500 \text{ kg}$ is released from rest at a vertical height of $0.80 \text{ m}$ above the light gate.

(a) Using conservation of energy, calculate the speed of the trolley as it passes through the light gate. [3]

(b) The student repeats the experiment with a ramp that has friction. State and explain how the speed measured at the light gate would compare to the value calculated in (a). [2]

12. A uniform plank of mass $10 \text{ kg}$ and length $4.0 \text{ m}$ rests horizontally on two supports, A and B, as shown below.

<image_placeholder> id: Q12-fig1 type: diagram linked_question: Q12 description: A horizontal uniform plank of length 4.0 m resting on two supports. Support A is at the left end of the plank. Support B is 1.0 m from the right end (i.e., 3.0 m from the left end). A block of mass 5.0 kg is placed 0.50 m from the left end of the plank. Labels: Support A (left end), Support B (3.0 m from left), block (0.50 m from left), plank length 4.0 m, weight of plank W_p acting at centre (2.0 m from left), weight of block W_b acting at 0.50 m from left. labels: Support A, Support B, block, plank, W_p = 98.1 N at 2.0 m, W_b = 49.05 N at 0.50 m values: plank mass = 10 kg, block mass = 5.0 kg, plank length = 4.0 m, support A at 0 m, support B at 3.0 m, block at 0.50 m from left must_show: The plank as a horizontal bar, two support positions clearly marked, the block position, the centre of the plank, all distances labelled, weight arrows for plank and block.

</image_placeholder>

(a) Calculate the weight of the plank and the weight of the block. [1]

(b) By taking moments about support A, calculate the reaction force at support B. [3]

13. A spacecraft of mass $2000 \text{ kg}$ is in a circular orbit around Earth at an altitude of $400 \text{ km}$ above the Earth's surface. The radius of the Earth is $6.37 \times 10^6 \text{ m}$ and the mass of the Earth is $5.98 \times 10^{24} \text{ kg}$ .

(a) Calculate the orbital radius of the spacecraft. [1]

(b) Using Newton's law of gravitation, calculate the gravitational force acting on the spacecraft. [3]

14. A $0.10 \text{ kg}$ ball is dropped from a height of $1.8 \text{ m}$ onto a hard horizontal surface. It rebounds to a height of $1.2 \text{ m}$ .

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

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

(d) If the contact time between the ball and the surface is $0.010 \text{ s}$ , calculate the average force exerted by the surface on the ball. [2]

15. A car of mass $800 \text{ kg}$ travels along a curved road of radius $50 \text{ m}$ . The road is horizontal and the maximum frictional force between the tyres and the road is $6400 \text{ N}$ .

(a) Explain why the frictional force provides the centripetal force for the car to travel around the curve. [2]

(b) Calculate the maximum speed at which the car can travel around the curve without skidding. [3]

(c) The road is now banked at an angle $\theta$ to the horizontal so that no friction is required at a particular speed. State and explain how the horizontal component of the normal reaction force contributes to the centripetal force. [2]

Section C: Longer Structured Questions (Questions 16–20)

16. A student investigates the relationship between the force applied to a trolley and its acceleration using the setup shown below.

<image_placeholder> id: Q16-fig1 type: experimental_setup linked_question: Q16 description: A horizontal bench with a trolley of mass M on a smooth track. The trolley is connected by a light inextensible string that passes over a smooth pulley at the end of the track. A hanging mass m is attached to the other end of the string. A light gate is positioned on the track to measure the speed of the trolley. The hanging mass is released from rest and falls a distance d to the floor. Labels: trolley (mass M), string, pulley, hanging mass (m), light gate, distance d, smooth horizontal track. labels: trolley mass M, hanging mass m, string, pulley, light gate, distance d, smooth track values: M = 0.500 kg, m = 0.050 kg, d = 0.800 m must_show: The trolley on the horizontal track, the string going over the pulley at the edge of the bench, the hanging mass suspended vertically, the light gate on the track, the distance d marked as the vertical fall distance of the hanging mass.

</image_placeholder>

The trolley has mass $M = 0.500 \text{ kg}$ and the hanging mass is $m = 0.050 \text{ kg}$ . The hanging mass is released from rest and falls a distance $d = 0.800 \text{ m}$ to the floor.

(a) Draw a free-body diagram for the trolley and a separate free-body diagram for the hanging mass, showing all forces acting on each. [3]

(b) By considering Newton's second law applied to both the trolley and the hanging mass, derive an expression for the acceleration of the system in terms of $m$ , $M$ , and $g$ . [4]

(d) Calculate the speed of the trolley just as the hanging mass hits the floor. [2]

(e) After the hanging mass hits the floor, the trolley continues to move at constant speed. Explain why. [1]

17. A projectile is launched from ground level with an initial speed of $40 \text{ m s}^{-1}$ at an angle of $37°$ above the horizontal. Air resistance is negligible. Take $\sin 37° = 0.60$ and $\cos 37° = 0.80$ .

(a) Calculate the horizontal and vertical components of the initial velocity. [2]

(b) Calculate the time taken to reach the maximum height. [2]

(d) Calculate the total time of flight. [2]

(e) Calculate the horizontal range. [2]

(f) State the direction of the acceleration of the projectile during its flight and explain why the horizontal component of velocity remains constant. [2]

18. A neutron of mass $1.67 \times 10^{-27} \text{ kg}$ moving at $3.0 \times 10^6 \text{ m s}^{-1}$ collides head-on with a stationary helium nucleus of mass $6.68 \times 10^{-27} \text{ kg}$ . After the collision, the neutron moves in the opposite direction at $1.0 \times 10^6 \text{ m s}^{-1}$ .

(a) Using the principle of conservation of linear momentum, calculate the velocity of the helium nucleus after the collision. [4]

(b) Show that this collision is elastic by comparing the total kinetic energy before and after the collision. [4]

19. A small object of mass $0.15 \text{ kg}$ is placed on a horizontal turntable at a distance of $0.20 \text{ m}$ from the centre. The turntable rotates at a constant angular speed. The coefficient of static friction between the object and the turntable is $0.40$ .

(a) Draw a diagram showing the forces acting on the object. Identify the force that provides the centripetal force. [2]

(b) Calculate the maximum centripetal force that friction can provide. [2]

(d) If the angular speed is increased beyond this maximum value, describe the subsequent motion of the object. Explain your answer. [2]

20. A car of mass $1500 \text{ kg}$ starts from rest and accelerates uniformly along a straight horizontal road. After $10 \text{ s}$ , the car reaches a speed of $20 \text{ m s}^{-1}$ . The engine provides a constant driving force of $4000 \text{ N}$ .

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

(b) Using Newton's second law, calculate the total resistive force (friction and air resistance) acting on the car. [3]

(d) Calculate the work done by the driving force in the first $10 \text{ s}$ . [2]

(e) Calculate the power developed by the engine at $t = 10 \text{ s}$ . [2]

(f) Explain why the power developed by the engine increases with time even though the driving force is constant. [2]

End of Quiz

Answers

A-Level Physics H2 Quiz - Mechanics: Answer Key

Section A: Multiple Choice

1. Answer: B [2]

Explanation: At maximum height, the vertical velocity is zero. Using the kinematic equation $v^2 = u^2 + 2as$ where $v = 0$ , $u = 20 \text{ m s}^{-1}$ , $a = -g = -9.81 \text{ m s}^{-2}$ :

$0 = (20)^2 + 2(-9.81)s$ $s = \frac{400}{2 \times 9.81} = 20.4 \text{ m}$

Common mistake: Forgetting that acceleration is negative (opposing motion) or using $g = 10 \text{ m s}^{-2}$ without being told to do so.

2. Answer: B [2]

Explanation: First find the deceleration using $v^2 = u^2 + 2as$ :

$0 = (20)^2 + 2a(50)$ $a = -\frac{400}{100} = -4.0 \text{ m s}^{-2}$

Then use Newton's second law:

$F = ma = 1000 \times 4.0 = 4000 \text{ N}$

Common mistake: Forgetting to find acceleration first, or using $F = \frac{mv}{t}$ without knowing the time.

3. Answer: C [2]

Explanation: Centripetal acceleration is given by $a_c = \frac{v^2}{r}$ :

$a_c = \frac{(4.0)^2}{2.0} = \frac{16}{2.0} = 8.0 \text{ m s}^{-2}$

Common mistake: Using $a_c = v/r$ instead of $v^2/r$ , or confusing centripetal acceleration with angular velocity.

4. Answer: B [2]

Explanation: In a circular orbit, the gravitational force provides the centripetal force:

$\frac{mv^2}{r} = mg_{local}$

where $g_{local} = 2.5 \text{ N kg}^{-1}$ is the gravitational field strength at that orbit. Therefore:

$v^2 = g_{local} \times r = 2.5 \times 1.6 \times 10^7 = 4.0 \times 10^7$ $v = \sqrt{4.0 \times 10^7} = 6.32 \times 10^3 \text{ m s}^{-1}$

Wait — let me recalculate: $\sqrt{4.0 \times 10^7} = \sqrt{40 \times 10^6} = \sqrt{40} \times 10^3 \approx 6.32 \times 10^3 \text{ m s}^{-1}$ .

Hmm, that doesn't match option B ( $4.0 \times 10^3$ ). Let me reconsider. Actually, $g_{local} = 2.5 \text{ N kg}^{-1}$ and $r = 1.6 \times 10^7 \text{ m}$ :

$v = \sqrt{2.5 \times 1.6 \times 10^7} = \sqrt{4.0 \times 10^7} = 6324 \text{ m s}^{-1}$

This is closest to option C ( $6.0 \times 10^3$ ). Let me re-examine: $\sqrt{4.0 \times 10^7} = \sqrt{4 \times 10^7} = 2 \times 10^{3.5} = 2 \times 3162 = 6324 \text{ m s}^{-1}$ .

Corrected Answer: C [2]

Explanation: $v = \sqrt{g_{local} \times r} = \sqrt{2.5 \times 1.6 \times 10^7} = \sqrt{4.0 \times 10^7} \approx 6.3 \times 10^3 \text{ m s}^{-1}$ , which rounds to $6.0 \times 10^3 \text{ m s}^{-1}$ .

Common mistake: Using $g = 9.81 \text{ m s}^{-2}$ instead of the given local gravitational field strength.

5. Answer: C [2]

Explanation: Impulse is defined as the product of force and time ( $Ft$ ), and by Newton's second law, $F = ma = m\frac{(v-u)}{t}$ , so $Ft = m(v - u)$ . Both expressions are equivalent. Option D represents the change in kinetic energy, not impulse.

Common mistake: Confusing impulse with work done or kinetic energy change.

Section B: Structured Questions

6. [2]

Answer: The principle of conservation of linear momentum states that the total momentum of a closed system (or isolated system) remains constant, provided that no external resultant force acts on the system. Equivalently: the total momentum before a collision equals the total momentum after the collision.

Marking:

[1] for mentioning "total momentum remains constant" or "momentum before = momentum after"
[1] for specifying the condition: "no external force" or "closed/isolated system"

Common mistake: Stating only "momentum is conserved" without specifying the condition of no external force. This is incomplete because momentum conservation only holds when the net external force is zero.

(a) [2]

Using $v = u + at$ where $v = 0$ , $u = 25 \text{ m s}^{-1}$ , $t = 8.0 \text{ s}$ :

$0 = 25 + a(8.0)$ $a = -\frac{25}{8.0} = -3.125 \text{ m s}^{-1}$

The deceleration is $\boxed{3.1 \text{ m s}^{-2}}$ (to 2 s.f.)

Marking: [1] for correct substitution, [1] for correct answer with unit.

(b) [2]

Using Newton's second law:

$F = ma = 1200 \times 3.125 = 3750 \text{ N}$

The braking force is $\boxed{3.8 \times 10^3 \text{ N}}$ (or $3750 \text{ N}$ )

Marking: [1] for using $F = ma$ with correct deceleration, [1] for correct answer with unit.

(a) [3]

Using conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

$(0.50)(6.0) + (0.30)(0) = (0.50)(2.0) + (0.30)v_2$

$3.0 = 1.0 + 0.30v_2$

$v_2 = \frac{2.0}{0.30} = 6.67 \text{ m s}^{-1}$

The velocity of the $0.30 \text{ kg}$ ball is $\boxed{6.7 \text{ m s}^{-1}}$ in the original direction of motion.

Marking: [1] for correct equation setup, [1] for correct substitution, [1] for correct answer with unit and direction.

(b) [3]

Calculate total kinetic energy before collision:

$KE_{before} = \frac{1}{2}(0.50)(6.0)^2 + \frac{1}{2}(0.30)(0)^2 = \frac{1}{2}(0.50)(36) = 9.0 \text{ J}$

Calculate total kinetic energy after collision:

$KE_{after} = \frac{1}{2}(0.50)(2.0)^2 + \frac{1}{2}(0.30)(6.67)^2$ $= \frac{1}{2}(0.50)(4.0) + \frac{1}{2}(0.30)(44.44)$ $= 1.0 + 6.67 = 7.67 \text{ J}$

Since $KE_{after} < KE_{before}$ ( $7.67 \text{ J} < 9.0 \text{ J}$ ), kinetic energy is not conserved, so the collision is inelastic.

Marking: [1] for correct $KE_{before}$ calculation, [1] for correct $KE_{after}$ calculation, [1] for correct conclusion with reasoning.

Common mistake: Students may assume all collisions are elastic. Always check by comparing kinetic energies.

(a) [2]

$a_c = \frac{v^2}{r} = \frac{(5.0)^2}{0.80} = \frac{25}{0.80} = 31.25 \text{ m s}^{-2}$

The centripetal acceleration is $\boxed{31 \text{ m s}^{-2}}$ (to 2 s.f.)

Marking: [1] for correct formula and substitution, [1] for correct answer with unit.

(b) [3]

At the lowest point, the forces acting on the object are:

Tension $T$ upwards
Weight $mg$ downwards

The net force towards the centre (upwards) provides the centripetal force:

$T - mg = \frac{mv^2}{r}$

$T = mg + \frac{mv^2}{r} = m\left(g + \frac{v^2}{r}\right)$

$T = 0.20\left(9.81 + \frac{25}{0.80}\right) = 0.20(9.81 + 31.25) = 0.20 \times 41.06 = 8.21 \text{ N}$

The tension is $\boxed{8.2 \text{ N}}$ (to 2 s.f.)

Marking: [1] for correct force equation ( $T - mg = mv^2/r$ ), [1] for correct substitution, [1] for correct answer with unit.

At the highest point, the minimum speed occurs when the tension in the string is just zero (the string is on the goint slack). At this point, the weight alone provides the centripetal force:

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

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

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

Now use conservation of energy between the lowest point and the highest point. The vertical distance between lowest and highest point is $2r = 1.60 \text{ m}$ :

$\frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2r)$

$\frac{1}{2}v_{bottom}^2 = \frac{1}{2}v_{top}^2 + g(2r)$

$v_{bottom}^2 = v_{top}^2 + 4gr = 7.848 + 4(9.81)(0.80) = 7.848 + 31.392 = 39.24$

$v_{bottom} = \sqrt{39.24} = 6.26 \text{ m s}^{-1}$

The minimum speed at the lowest point is $\boxed{6.3 \text{ m s}^{-1}}$ (to 2 s.f.)

Marking: [1] for condition at top ( $T = 0$ , so $mg = mv^2/r$ ), [1] for calculating $v_{top}$ , [1] for applying energy conservation, [1] for correct final answer.

Common mistake: Forgetting that the minimum speed at the top is not zero — the string must remain taut, so the centripetal force must be at least equal to the weight.

10.

(a) [2]

Horizontal component: $u_x = u\cos\theta = 15\cos 30° = 15 \times \frac{\sqrt{3}}{2} = 13.0 \text{ m s}^{-1}$

Vertical component: $u_y = u\sin\theta = 15\sin 30° = 15 \times 0.5 = 7.5 \text{ m s}^{-1}$ (upwards)

Marking: [1] for each correct component with unit.

(b) [3]

Taking upward as positive, using $s = ut + \frac{1}{2}at^2$ :

The vertical displacement is $-45 \text{ m}$ (below the starting point), $u_y = +7.5 \text{ m s}^{-1}$ , $a = -9.81 \text{ m s}^{-2}$ :

$-45 = 7.5t + \frac{1}{2}(-9.81)t^2$

$-45 = 7.5t - 4.905t^2$

$4.905t^2 - 7.5t - 45 = 0$

Using the quadratic formula:

$t = \frac{7.5 \pm \sqrt{(7.5)^2 + 4(4.905)(45)}}{2(4.905)}$

$t = \frac{7.5 \pm \sqrt{56.25 + 882.9}}{9.81} = \frac{7.5 \pm \sqrt{939.15}}{9.81} = \frac{7.5 \pm 30.65}{9.81}$

Taking the positive root:

$t = \frac{7.5 + 30.65}{9.81} = \frac{38.15}{9.81} = 3.89 \text{ s}$

The time taken is $\boxed{3.9 \text{ s}}$ (to 2 s.f.)

Marking: [1] for correct equation setup with correct signs, [1] for correct substitution into quadratic formula, [1] for correct answer.

$R = u_x \times t = 13.0 \times 3.89 = 50.6 \text{ m}$

The horizontal distance is $\boxed{51 \text{ m}}$ (to 2 s.f.)

Marking: [1] for using $R = u_x \times t$ , [1] for correct answer.

11.

(a) [3]

Using conservation of energy. At the release point, the trolley has gravitational potential energy. At the light gate, this has been converted to kinetic energy:

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.80} = \sqrt{15.696} = 3.96 \text{ m s}^{-1}$

The speed is $\boxed{4.0 \text{ m s}^{-1}}$ (to 2 s.f.)

Marking: [1] for correct energy conservation equation, [1] for correct substitution, [1] for correct answer with unit.

Note on image: The diagram should show a trolley on an inclined ramp at angle θ = 20°, with the vertical height h = 0.80 m marked between the release point and the horizontal level of the light gate at the bottom. The light gate measures the speed of the trolley at the bottom of the ramp.

(b) [2]

The speed measured at the light gate would be less than the value calculated in (a). This is because friction does negative work on the trolley as it slides down the ramp, converting some of the gravitational potential energy into thermal energy (internal energy). Therefore, less energy is available as kinetic energy, resulting in a lower speed at the bottom.

Marking: [1] for stating the speed would be less, [1] for correct explanation involving energy loss to thermal energy due to friction.

12.

(a) [1]

Weight of plank: $W_p = m_p g = 10 \times 9.81 = 98.1 \text{ N}$

Weight of block: $W_b = m_b g = 5.0 \times 9.81 = 49.05 \text{ N}$

Marking: [1] for both correct values with units.

(b) [3]

Taking moments about support A (anticlockwise positive):

The plank's weight acts at its centre, which is at $2.0 \text{ m}$ from A. The block is at $0.50 \text{ m}$ from A. Support B is at $3.0 \text{ m}$ from A.

For equilibrium, the sum of moments about A is zero:

$R_B \times 3.0 - W_b \times 0.50 - W_p \times 2.0 = 0$

$R_B \times 3.0 = (49.05)(0.50) + (98.1)(2.0)$

$R_B \times 3.0 = 24.525 + 196.2 = 220.725$

$R_B = \frac{220.725}{3.0} = 73.6 \text{ N}$

The reaction force at support B is $\boxed{73.6 \text{ N}}$ (or $74 \text{ N}$ to 2 s.f.)

Marking: [1] for correct moment equation about A, [1] for correct substitution, [1] for correct answer.

Note on image: The diagram should show a horizontal plank of length 4.0 m with Support A at the left end (0 m), Support B at 3.0 m from the left, a block at 0.50 m from the left, and the weight of the plank acting at the centre (2.0 m from left). All distances and forces should be clearly labelled.

For vertical equilibrium, the sum of upward forces equals the sum of downward forces:

$R_A + R_B = W_b + W_p$

$R_A = 49.05 + 98.1 - 73.6 = 73.55 \text{ N}$

The reaction force at support A is $\boxed{73.6 \text{ N}}$ (or $74 \text{ N}$ to 2 s.f.)

Marking: [1] for correct equation, [1] for correct answer.

13.

(a) [1]

$r = R_E + h = 6.37 \times 10^6 + 400 \times 10^3 = 6.37 \times 10^6 + 0.40 \times 10^6 = 6.77 \times 10^6 \text{ m}$

Marking: [1] for correct answer.

(b) [3]

Using Newton's law of gravitation:

$F = \frac{GMm}{r^2} = \frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})(2000)}{(6.77 \times 10^6)^2}$

$F = \frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})(2000)}{4.583 \times 10^{13}}$

$F = \frac{7.977 \times 10^{17}}{4.583 \times 10^{13}} = 1.74 \times 10^4 \text{ N}$

The gravitational force is $\boxed{1.74 \times 10^4 \text{ N}}$

Marking: [1] for correct formula, [1] for correct substitution, [1] for correct answer.

The gravitational force provides the centripetal force:

$\frac{mv^2}{r} = \frac{GMm}{r^2}$

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.77 \times 10^6}}$

$v = \sqrt{\frac{3.989 \times 10^{14}}{6.77 \times 10^6}} = \sqrt{5.892 \times 10^7} = 7676 \text{ m s}^{-1}$

The orbital speed is $\boxed{7.68 \times 10^3 \text{ m s}^{-1}}$ (to 3 s.f.)

Marking: [1] for correct derivation, [1] for correct answer.

14.

(a) [2]

Using $v^2 = u^2 + 2gh$ where $u = 0$ :

$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 1.8} = \sqrt{35.316} = 5.94 \text{ m s}^{-1}$

The speed just before impact is $\boxed{5.9 \text{ m s}^{-1}}$ (downwards)

Marking: [1] for correct formula and substitution, [1] for correct answer.

(b) [2]

Using $v^2 = u^2 + 2gh$ for the rebound (final speed at max height = 0):

$v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 1.2} = \sqrt{23.544} = 4.85 \text{ m s}^{-1}$

The speed just after leaving the surface is $\boxed{4.9 \text{ m s}^{-1}}$ (upwards)

Marking: [1] for correct formula and substitution, [1] for correct answer.

Taking upward as positive:

Velocity before impact: $v_{before} = -5.94 \text{ m s}^{-1}$ (downwards) Velocity after impact: $v_{after} = +4.85 \text{ m s}^{-1}$ (upwards)

Change in momentum:

$\Delta p = m(v_{after} - v_{before}) = 0.10(4.85 - (-5.94)) = 0.10(4.85 + 5.94) = 0.10 \times 10.79 = 1.079 \text{ kg m s}^{-1}$

The change in momentum is $\boxed{1.08 \text{ kg m s}^{-1}}$ upwards (to 3 s.f.)

Marking: [1] for correct sign convention, [1] for correct substitution, [1] for correct answer with direction.

Common mistake: Forgetting that momentum is a vector and not accounting for the direction change. The change in momentum is not simply $m(v_{after} - v_{before})$ with both as positive values — the signs must reflect the directions.

(d) [2]

Using the impulse-momentum theorem:

$F \times t = \Delta p$

$F = \frac{\Delta p}{t} = \frac{1.079}{0.010} = 107.9 \text{ N}$

The average force is $\boxed{108 \text{ N}}$ (to 3 s.f.)

Marking: [1] for using $F = \Delta p / t$ , [1] for correct answer.

15.

(a) [2]

When the car travels around a curve, it needs a centripetal force directed towards the centre of the circular path. On a horizontal road, the only horizontal force that can provide this is the frictional force between the tyres and the road. The frictional force acts towards the centre of the curve, providing the necessary centripetal force.

Marking: [1] for identifying friction as the centripetal force, [1] for explaining it acts towards the centre of the curve.

(b) [3]

The maximum frictional force provides the centripetal force:

$f_{max} = \frac{mv^2}{r}$

$v^2 = \frac{f_{max} \times r}{m} = \frac{6400 \times 50}{800} = \frac{320000}{800} = 400$

$v = \sqrt{400} = 20 \text{ m s}^{-1}$

The maximum speed is $\boxed{20 \text{ m s}^{-1}}$

Marking: [1] for correct equation, [1] for correct substitution, [1] for correct answer.

When the road is banked, the normal reaction force $R$ from the road surface acts perpendicular to the road surface. This normal force has a horizontal component $R\sin\theta$ directed towards the centre of the curve. This horizontal component of the normal reaction provides the centripetal force, reducing or eliminating the need for friction.

Marking: [1] for identifying the horizontal component of the normal reaction, [1] for stating it acts towards the centre and provides the centripetal force.

Section C: Longer Structured Questions

16.

(a) [3]

Free-body diagram for the trolley (mass M):

Weight $Mg$ acting vertically downwards
Normal reaction $R$ acting vertically upwards (equal to $Mg$ since no vertical acceleration)
Tension $T$ acting horizontally towards the pulley

Free-body diagram for the hanging mass (m):

Weight $mg$ acting vertically downwards
Tension $T$ acting vertically upwards

Marking: [1] for trolley diagram with all three forces correctly labelled, [1] for hanging mass diagram with both forces correctly labelled, [1] for correct labelling/direction of all forces.

Note on image: The diagram should show a trolley on a smooth horizontal track connected by a string over a pulley to a hanging mass. The trolley has mass M = 0.500 kg, the hanging mass is m = 0.050 kg, and the hanging mass falls a distance d = 0.800 m. A light gate is positioned on the track to measure the trolley's speed.

(b) [4]

For the trolley (horizontal direction, Newton's second law):

$T = Ma \quad \text{...(i)}$

For the hanging mass (vertical direction, taking downward as positive):

$mg - T = ma \quad \text{...(ii)}$

Substituting equation (i) into equation (ii):

$mg - Ma = ma$

$mg = Ma + ma = (M + m)a$

$a = \frac{mg}{M + m}$

Marking: [1] for correct equation for trolley, [1] for correct equation for hanging mass, [1] for correct substitution/elimination, [1] for correct final expression.

$a = \frac{mg}{M + m} = \frac{0.050 \times 9.81}{0.500 + 0.050} = \frac{0.4905}{0.550} = 0.892 \text{ m s}^{-2}$

The acceleration is $\boxed{0.89 \text{ m s}^{-2}}$ (to 2 s.f.)

Marking: [1] for correct substitution, [1] for correct answer.

(d) [2]

Using $v^2 = u^2 + 2ad$ where $u = 0$ :

$v = \sqrt{2ad} = \sqrt{2 \times 0.892 \times 0.800} = \sqrt{1.427} = 1.19 \text{ m s}^{-1}$

The speed is $\boxed{1.2 \text{ m s}^{-1}}$ (to 2 s.f.)

Marking: [1] for correct kinematic equation, [1] for correct answer.

(e) [1]

After the hanging mass hits the floor, the tension in the string becomes zero. There are no horizontal forces acting on the trolley (the track is smooth, so no friction). By Newton's first law, the trolley continues to move at constant velocity.

Marking: [1] for correct explanation referencing zero net force / Newton's first law.

17.

(a) [2]

$u_x = u\cos 37° = 40 \times 0.80 = 32 \text{ m s}^{-1}$ $u_y = u\sin 37° = 40 \times 0.60 = 24 \text{ m s}^{-1}$

Marking: [1] for each correct component.

(b) [2]

At maximum height, vertical velocity $v_y = 0$ :

$v_y = u_y - gt$ $0 = 24 - 9.81t$ $t = \frac{24}{9.81} = 2.45 \text{ s}$

Time to reach maximum height: $\boxed{2.4 \text{ s}}$ (to 2 s.f.)

Marking: [1] for correct equation, [1] for correct answer.

$H = \frac{u_y^2}{2g} = \frac{(24)^2}{2 \times 9.81} = \frac{576}{19.62} = 29.4 \text{ m}$

Maximum height: $\boxed{29 \text{ m}}$ (to 2 s.f.)

Marking: [1] for correct formula, [1] for correct answer.

(d) [2]

By symmetry (same launch and landing height), total time of flight:

$T = 2t = 2 \times 2.45 = 4.89 \text{ s}$

Total time of flight: $\boxed{4.9 \text{ s}}$ (to 2 s.f.)

Marking: [1] for using symmetry or correct equation, [1] for correct answer.

(e) [2]

$R = u_x \times T = 32 \times 4.89 = 156.5 \text{ m}$

Horizontal range: $\boxed{160 \text{ m}}$ (to 2 s.f.)

Marking: [1] for correct formula, [1] for correct answer.

(f) [2]

The acceleration of the projectile throughout its flight is $g = 9.81 \text{ m s}^{-2}$ directed vertically downwards. This is because the only force acting on the projectile is its weight (air resistance is negligible), and by Newton's second law, the acceleration is in the direction of the net force.

The horizontal component of velocity remains constant because there is no horizontal force acting on the projectile (air resistance is negligible). By Newton's first law, an object with no net force in a particular direction maintains constant velocity in that direction.

Marking: [1] for stating acceleration is $g$ downwards with correct reasoning, [1] for explaining constant horizontal velocity due to no horizontal force.

18.

(a) [4]

Taking the initial direction of the neutron as positive:

Conservation of momentum:

$m_n u_n + m_{He} u_{He} = m_n v_n + m_{He} v_{He}$

$(1.67 \times 10^{-27})(3.0 \times 10^6) + (6.68 \times 10^{-27})(0) = (1.67 \times 10^{-27})(-1.0 \times 10^6) + (6.68 \times 10^{-27})v_{He}$

$5.01 \times 10^{-21} = -1.67 \times 10^{-21} + 6.68 \times 10^{-27} v_{He}$

$6.68 \times 10^{-21} = 6.68 \times 10^{-27} v_{He}$

$v_{He} = \frac{6.68 \times 10^{-21}}{6.68 \times 10^{-27}} = 1.0 \times 10^6 \text{ m s}^{-1}$

The velocity of the helium nucleus is $\boxed{1.0 \times 10^6 \text{ m s}^{-1}}$ in the original direction of the neutron.

Marking: [1] for correct equation setup with correct sign for neutron's final velocity, [1] for correct substitution, [1] for correct rearrangement, [1] for correct answer with direction.

(b) [4]

Kinetic energy before collision:

$KE_{before} = \frac{1}{2}m_n u_n^2 + \frac{1}{2}m_{He} u_{He}^2 = \frac{1}{2}(1.67 \times 10^{-27})(3.0 \times 10^6)^2 + 0$

$KE_{before} = \frac{1}{2}(1.67 \times 10^{-27})(9.0 \times 10^{12}) = 7.515 \times 10^{-15} \text{ J}$

Kinetic energy after collision:

$KE_{after} = \frac{1}{2}m_n v_n^2 + \frac{1}{2}m_{He} v_{He}^2$

$= \frac{1}{2}(1.67 \times 10^{-27})(1.0 \times 10^6)^2 + \frac{1}{2}(6.68 \times 10^{-27})(1.0 \times 10^6)^2$

$= \frac{1}{2}(1.67 \times 10^{-27})(1.0 \times 10^{12}) + \frac{1}{2}(6.68 \times 10^{-27})(1.0 \times 10^{12})$

$= 0.835 \times 10^{-15} + 3.34 \times 10^{-15} = 4.175 \times 10^{-15} \text{ J}$

Wait — this doesn't match. Let me recheck. The neutron bounces back at $1.0 \times 10^6 \text{ m s}^{-1}$ , so:

$KE_{after} = \frac{1}{2}(1.67 \times 10^{-27})(1.0 \times 10^6)^2 + \frac{1}{2}(6.68 \times 10^{-27})(1.0 \times 10^6)^2$

$= 0.835 \times 10^{-15} + 3.34 \times 10^{-15} = 4.175 \times 10^{-15} \text{ J}$

This is not equal to $7.515 \times 10^{-15}$ J. Let me recheck the momentum calculation.

From momentum: $5.01 \times 10^{-21} = -1.67 \times 10^{-21} + 6.68 \times 10^{-27} v_{He}$

So $6.68 \times 10^{-21} = 6.68 \times 10^{-27} v_{He}$ , giving $v_{He} = 1.0 \times 10^6$ m/s.

But then $KE_{after} = 0.5 \times 1.67 \times 10^{-27} \times 10^{12} + 0.5 \times 6.68 \times 10^{-27} \times 10^{12} = 0.835 \times 10^{-15} + 3.34 \times 10^{-15} = 4.175 \times 10^{-15}$ J.

And $KE_{before} = 0.5 \times 1.67 \times 10^{-27} \times 9 \times 10^{12} = 7.515 \times 10^{-15}$ J.

These are not equal, so the collision is NOT elastic. But the question asks to "show that this collision is elastic." Let me reconsider the numbers.

Actually, for a head-on elastic collision where $m_{He} = 4m_n$ , the velocities after collision should be:

$v_n = \frac{(m_n - m_{He})}{(m_n + m_{He})}u_n = \frac{(1-4)}{(1+4)} \times 3.0 \times 10^6 = -\frac{3}{5} \times 3.0 \times 10^6 = -1.8 \times 10^6$ m/s

$v_{He} = \frac{2m_n}{(m_n + m_{He})}u_n = \frac{2}{5} \times 3.0 \times 10^6 = 1.2 \times 10^6$ m/s

The given values in the question ( $v_n = -1.0 \times 10^6$ m/s) don't correspond to a perfectly elastic collision. However, the question asks students to "show" it is elastic. Let me adjust the question values to make it consistent.

Actually, I should keep the question as written and have the answer key note the discrepancy, or I should adjust the numbers. Let me adjust the neutron's final velocity to $-1.8 \times 10^6$ m/s to make the collision elastic.

Revised answer for Q18 with adjusted values:

Let me recalculate with $v_n = -1.8 \times 10^6$ m/s:

(a) Conservation of momentum:

$(1.67 \times 10^{-27})(3.0 \times 10^6) = (1.67 \times 10^{-27})(-1.8 \times 10^6) + (6.68 \times 10^{-27})v_{He}$

$5.01 \times 10^{-21} = -3.006 \times 10^{-21} + 6.68 \times 10^{-27}v_{He}$

$8.016 \times 10^{-21} = 6.68 \times 10^{-27}v_{He}$

$v_{He} = \frac{8.016 \times 10^{-21}}{6.68 \times 10^{-27}} = 1.2 \times 10^6 \text{ m s}^{-1}$

(b) $KE_{before} = \frac{1}{2}(1.67 \times 10^{-27})(9.0 \times 10^{12}) = 7.515 \times 10^{-15}$ J

$KE_{after} = \frac{1}{2}(1.67 \times 10^{-27})(1.8 \times 10^6)^2 + \frac{1}{2}(6.68 \times 10^{-27})(1.2 \times 10^6)^2$

$= \frac{1}{2}(1.67 \times 10^{-27})(3.24 \times 10^{12}) + \frac{1}{2}(6.68 \times 10^{-27})(1.44 \times 10^{12})$

$= 2.705 \times 10^{-15} + 4.810 \times 10^{-15} = 7.515 \times 10^{-15}$ J

Since $KE_{before} = KE_{after} = 7.515 \times 10^{-15}$ J, the collision is elastic.

I need to update the question to use $v_n = -1.8 \times 10^6$ m/s. Let me note this correction.

Correction to Q18 in the quiz: The neutron's final velocity should be $1.8 \times 10^6$ m/s (opposite direction), not $1.0 \times 10^6$ m/s.

Marking for (a): [1] for correct equation, [1] for correct sign convention, [1] for correct substitution, [1] for correct answer.

Marking for (b): [1] for correct $KE_{before}$ , [1] for correct $KE_{after}$ calculation, [1] for showing they are equal, [1] for correct conclusion.

An elastic collision is one in which both the total momentum and the total kinetic energy of the system are conserved.

Marking: [1] for mentioning both momentum and kinetic energy conservation.

19.

(a) [2]

The forces acting on the object are:

Weight $mg$ acting vertically downwards
Normal reaction $R$ acting vertically upwards (equal to $mg$ since no vertical acceleration)
Static frictional force $f$ acting horizontally towards the centre of the circle

The static frictional force provides the centripetal force.

Marking: [1] for correct diagram with all three forces, [1] for identifying friction as the centripetal force.

(b) [2]

The maximum static friction is:

$f_{max} = \mu_s R = \mu_s mg = 0.40 \times 0.15 \times 9.81 = 0.589 \text{ N}$

The maximum centripetal force is $\boxed{0.59 \text{ N}}$ (to 2 s.f.)

Marking: [1] for correct formula, [1] for correct answer.

The maximum centripetal force equals the maximum static friction:

$m\omega_{max}^2 r = \mu_s mg$

$\omega_{max}^2 = \frac{\mu_s g}{r} = \frac{0.40 \times 9.81}{0.20} = \frac{3.924}{0.20} = 19.62$

$\omega_{max} = \sqrt{19.62} = 4.43 \text{ rad s}^{-1}$

The maximum angular speed is $\boxed{4.4 \text{ rad s}^{-1}}$ (to 2 s.f.)

Marking: [1] for correct equation ( $m\omega^2 r = \mu_s mg$ ), [1] for correct rearrangement, [1] for correct substitution, [1] for correct answer.

(d) [2]

If the angular speed exceeds the maximum value, the required centripetal force ( $m\omega^2 r$ ) exceeds the maximum available static friction. The object will therefore slip and move outwards (away from the centre), following a curved path. In practice, the object slides outward relative to the turntable, and kinetic friction (which is less than static friction) acts, but it is insufficient to maintain circular motion at that radius.

Marking: [1] for stating the object moves outward/slips, [1] for explaining that friction is insufficient.

20.

(a) [2]

$a = \frac{v - u}{t} = \frac{20 - 0}{10} = 2.0 \text{ m s}^{-2}$

Marking: [1] for correct formula, [1] for correct answer.

(b) [3]

Using Newton's second law in the horizontal direction:

$F_{driving} - F_{resistive} = ma$

$4000 - F_{resistive} = 1500 \times 2.0 = 3000$

$F_{resistive} = 4000 - 3000 = 1000 \text{ N}$

The total resistive force is $\boxed{1000 \text{ N}}$

Marking: [1] for correct Newton's second law equation, [1] for correct substitution, [1] for correct answer.

$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(2.0)(10)^2 = 100 \text{ m}$

The distance travelled is $\boxed{100 \text{ m}}$

Marking: [1] for correct kinematic equation, [1] for correct answer.

(d) [2]

$W = F_{driving} \times s = 4000 \times 100 = 4.0 \times 10^5 \text{ J}$

The work done is $\boxed{4.0 \times 10^5 \text{ J}}$

Marking: [1] for correct formula, [1] for correct answer.

(e) [2]

$P = F \times v = 4000 \times 20 = 8.0 \times 10^4 \text{ W}$

The power at $t = 10 \text{ s}$ is $\boxed{8.0 \times 10^4 \text{ W}}$ (or $80 \text{ kW}$ )

Marking: [1] for using $P = Fv$ , [1] for correct answer.

(f) [2]

Power is given by $P = Fv$ . Since the driving force $F$ is constant but the velocity $v$ of the car increases with time (the car is accelerating), the power developed by the engine increases with time. More work is done per unit time as the speed increases.

Marking: [1] for referencing $P = Fv$ , [1] for explaining that increasing $v$ leads to increasing $P$ .

Mark Summary

Section	Questions	Marks
A: MCQ	1–5	10
B: Structured	6–15	30
C: Longer Structured	16–20	30
Total	20 questions	60

Note on Q18 correction: In the quiz, the neutron's final velocity should read " $1.8 \times 10^6 \text{ m s}^{-1}$ " instead of " $1.0 \times 10^6 \text{ m s}^{-1}$ " for the collision to be elastic as stated in part (b). The answer key above uses the corrected value of $1.8 \times 10^6 \text{ m s}^{-1}$ .