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A Level H1 Physics Practice Paper 4

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Questions

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A-Level Physics H1 Quiz - Mechanics

Name: ________________________
Class: ________________________
Date: ________________________
Score: _____ / 80
Duration: 90 minutes

Instructions

  • Answer ALL questions in the spaces provided.
  • The number of marks for each question or part question is shown in brackets [ ].
  • You are advised to show all working clearly, including formulas and substitutions.
  • Where appropriate, take g=9.81 m s−2g = 9.81 \text{ m s}^{-2}.
  • Unless otherwise stated, assume air resistance is negligible.

Section A: Multiple Choice [15 marks]

Answer ALL questions.

1. Which of the following is a vector quantity?

A. Speed
B. Distance
C. Energy
D. Displacement

[1]

2. A ball is thrown vertically upward with an initial speed of 20 m s−120 \text{ m s}^{-1}. What is its acceleration at the highest point of its trajectory? (Take upward as positive.)

A. 0 m s−20 \text{ m s}^{-2}
B. −9.81 m s−2-9.81 \text{ m s}^{-2}
C. +9.81 m s−2+9.81 \text{ m s}^{-2}
D. −20 m s−2-20 \text{ m s}^{-2}

[1]

3. A car accelerates uniformly from rest to 30 m s−130 \text{ m s}^{-1} in 6.0 s6.0 \text{ s}. What is the distance travelled during this time?

A. 60 m60 \text{ m}
B. 90 m90 \text{ m}
C. 120 m120 \text{ m}
D. 180 m180 \text{ m}

[1]

4. A 2.0 kg2.0 \text{ kg} object moving at 4.0 m s−14.0 \text{ m s}^{-1} collides with a stationary 3.0 kg3.0 \text{ kg} object. After the collision, the two objects stick together and move as one. What is their common velocity after the collision?

A. 1.0 m s−11.0 \text{ m s}^{-1}
B. 1.6 m s−11.6 \text{ m s}^{-1}
C. 2.0 m s−12.0 \text{ m s}^{-1}
D. 2.4 m s−12.4 \text{ m s}^{-1}

[1]

5. A force of 12 N12 \text{ N} acts on a body of mass 4.0 kg4.0 \text{ kg} on a frictionless horizontal surface. What is the acceleration of the body?

A. 0.33 m s−20.33 \text{ m s}^{-2}
B. 3.0 m s−23.0 \text{ m s}^{-2}
C. 8.0 m s−28.0 \text{ m s}^{-2}
D. 48 m s−248 \text{ m s}^{-2}

[1]

6. Which of the following statements about Newton's Third Law is correct?

A. Action and reaction forces act on the same body.
B. Action and reaction forces are equal in magnitude and opposite in direction.
C. Action and reaction forces cancel each other out.
D. Action and reaction forces must be contact forces.

[1]

7. A block of mass 5.0 kg5.0 \text{ kg} rests on a horizontal table. What is the normal contact force exerted by the table on the block?

A. 0 N0 \text{ N}
B. 5.0 N5.0 \text{ N}
C. 49 N49 \text{ N}
D. 98 N98 \text{ N}

[1]

8. A projectile is launched at an angle of 30∘30^\circ above the horizontal with an initial speed of 40 m s−140 \text{ m s}^{-1}. What is the horizontal component of its initial velocity?

A. 20 m s−120 \text{ m s}^{-1}
B. 203 m s−120\sqrt{3} \text{ m s}^{-1}
C. 34.6 m s−134.6 \text{ m s}^{-1}
D. 40 m s−140 \text{ m s}^{-1}

[1]

9. A 1.5 kg1.5 \text{ kg} ball moving at 6.0 m s−16.0 \text{ m s}^{-1} strikes a wall and rebounds at 4.0 m s−14.0 \text{ m s}^{-1} in the opposite direction. What is the magnitude of the change in momentum of the ball?

A. 3.0 kg m s−13.0 \text{ kg m s}^{-1}
B. 9.0 kg m s−19.0 \text{ kg m s}^{-1}
C. 15.0 kg m s−115.0 \text{ kg m s}^{-1}
D. 21.0 kg m s−121.0 \text{ kg m s}^{-1}

[1]

10. Which pair of quantities has the same unit?

A. Work and force
B. Power and momentum
C. Energy and work
D. Impulse and energy

[1]

11. A car of mass 1200 kg1200 \text{ kg} travels around a horizontal circular track of radius 50 m50 \text{ m} at a constant speed of 20 m s−120 \text{ m s}^{-1}. What is the centripetal force acting on the car?

A. 480 N480 \text{ N}
B. 9600 N9600 \text{ N}
C. 24000 N24000 \text{ N}
D. 48000 N48000 \text{ N}

[1]

12. A spring is compressed by 0.10 m0.10 \text{ m} and stores 2.0 J2.0 \text{ J} of elastic potential energy. What is the spring constant of the spring?

A. 20 N m−120 \text{ N m}^{-1}
B. 40 N m−140 \text{ N m}^{-1}
C. 200 N m−1200 \text{ N m}^{-1}
D. 400 N m−1400 \text{ N m}^{-1}

[1]

13. A 0.50 kg0.50 \text{ kg} ball is dropped from a height of 20 m20 \text{ m}. What is its kinetic energy just before it hits the ground? (Take g=9.81 m s−2g = 9.81 \text{ m s}^{-2}.)

A. 10 J10 \text{ J}
B. 98 J98 \text{ J}
C. 196 J196 \text{ J}
D. 392 J392 \text{ J}

[1]

14. An object moves with constant velocity. Which of the following statements is true?

A. The net force on the object is zero.
B. The object has no forces acting on it.
C. The object must be at rest.
D. The acceleration of the object is increasing.

[1]

15. A 3.0 kg3.0 \text{ kg} object is lifted vertically at constant speed through a height of 5.0 m5.0 \text{ m}. How much work is done against gravity?

A. 15 J15 \text{ J}
B. 30 J30 \text{ J}
C. 147 J147 \text{ J}
D. 294 J294 \text{ J}

[1]

Section B: Structured Questions [35 marks]

Answer ALL questions.

16. (a) State the principle of conservation of linear momentum.

[2]

(b) A 0.060 kg0.060 \text{ kg} tennis ball strikes a wall horizontally at 25 m s−125 \text{ m s}^{-1} and rebounds at 18 m s−118 \text{ m s}^{-1} in the opposite direction. The contact time with the wall is 0.020 s0.020 \text{ s}.

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

[2]

    (ii) Calculate the average force exerted by the wall on the ball.

[2]

    (iii) State the direction of the force exerted by the wall on the ball.

[1]

17. A car of mass 1500 kg1500 \text{ kg} starts from rest and accelerates uniformly along a straight horizontal road. After travelling 200 m200 \text{ m}, the car reaches a speed of 30 m s−130 \text{ m s}^{-1}.

(a) Calculate the acceleration of the car.

[2]

(b) Calculate the time taken for the car to travel the 200 m200 \text{ m}.

[2]

(c) The engine of the car provides a constant driving force. Calculate the work done by the driving force over the 200 m200 \text{ m}.

[2]

(d) Calculate the average power developed by the engine during this time.

[2]

18. A ball is projected from ground level at an angle of 45∘45^\circ above the horizontal with an initial speed of 28 m s−128 \text{ m s}^{-1}. Take g=9.81 m s−2g = 9.81 \text{ m s}^{-2}.

(a) Calculate the horizontal component of the initial velocity.

[1]

(b) Calculate the vertical component of the initial velocity.

[1]

(c) Calculate the maximum height reached by the ball.

[2]

(d) Calculate the horizontal range of the ball.

[2]

(e) State, with a reason, whether air resistance would increase or decrease the range.

[2]

19. A block of mass 4.0 kg4.0 \text{ kg} is placed on a rough inclined plane that makes an angle of 30∘30^\circ with the horizontal. The coefficient of friction between the block and the plane is 0.350.35. The block is released from rest.

<image_placeholder> id: Q19-fig1 type: diagram linked_question: Q19 description: A block of mass 4.0 kg on an inclined plane at 30 degrees to the horizontal. The block is shown on the slope with forces labelled: weight mg acting vertically downward, normal reaction R perpendicular to the surface, and friction f acting up the slope (opposing motion). The angle of inclination 30° is marked between the plane and the horizontal. labels: mass = 4.0 kg, angle = 30°, coefficient of friction μ = 0.35, weight mg, normal reaction R, friction f, angle 30° marked values: m = 4.0 kg, θ = 30°, μ = 0.35, g = 9.81 m s⁻² must_show: inclined plane at 30°, block on plane, weight vector vertically down, normal reaction perpendicular to surface, friction up the slope, angle label 30°, horizontal reference line

(a) Draw a free-body diagram showing all the forces acting on the block. Label each force clearly.
[2]

(b) Calculate the component of the weight acting down the slope.

[1]

(c) Calculate the normal reaction force.

[1]

(d) Calculate the frictional force acting on the block.

[1]

(e) Determine whether the block will slide down the slope. Show your reasoning.

[2]

Section C: Free Response [30 marks]

Answer ALL questions.

20. A small sphere of mass 0.20 kg0.20 \text{ kg} is attached to a light inextensible string of length 1.5 m1.5 \text{ m}. The sphere is swung in a vertical circle. The string remains taut throughout the motion. Take g=9.81 m s−2g = 9.81 \text{ m s}^{-2}.

(a) Explain what is meant by centripetal acceleration.

[2]

(b) The sphere passes through the lowest point of the circle with a speed of 6.0 m s−16.0 \text{ m s}^{-1}.

    (i) Calculate the centripetal acceleration of the sphere at this point.

[2]

    (ii) Calculate the tension in the string at this point.

[2]

(c) The sphere is now given a different speed at the lowest point such that the tension in the string at the highest point is exactly zero.

    (i) Calculate the minimum speed the sphere must have at the highest point for the string to remain taut.

[2]

    (ii) Using the principle of conservation of energy, calculate the speed of the sphere at the lowest point.

[3]

(d) Explain why the tension in the string varies as the sphere moves around the circle.

[2]

End of Quiz

Total: 80 marks

Answers

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A-Level Physics H1 Quiz - Mechanics

Answer Key

Section A: Multiple Choice [15 marks]

1. D. Displacement
Displacement is a vector quantity because it has both magnitude and direction. Speed, distance, and energy are scalar quantities — they have magnitude only.
Common mistake: Students often confuse speed (scalar) with velocity (vector), or distance (scalar) with displacement (vector).

2. B. −9.81 m s−2-9.81 \text{ m s}^{-2}
At the highest point, the ball's velocity is momentarily zero, but its acceleration is still due to gravity throughout the flight. Taking upward as positive, gravitational acceleration is −9.81 m s−2-9.81 \text{ m s}^{-2}.
Common mistake: Students incorrectly assume acceleration is zero when velocity is zero.

3. B. 90 m90 \text{ m}
Using v=u+atv = u + at: 30=0+a(6.0)30 = 0 + a(6.0), so a=5.0 m s−2a = 5.0 \text{ m s}^{-2}.
Using s=ut+12at2s = ut + \frac{1}{2}at^2: s=0+12(5.0)(6.0)2=90 ms = 0 + \frac{1}{2}(5.0)(6.0)^2 = 90 \text{ m}.
Alternatively, average speed = 0+302=15 m s−1\frac{0+30}{2} = 15 \text{ m s}^{-1}, distance = 15×6.0=90 m15 \times 6.0 = 90 \text{ m}.

4. B. 1.6 m s−11.6 \text{ m s}^{-1}
By conservation of momentum: m1u1+m2u2=(m1+m2)vm_1 u_1 + m_2 u_2 = (m_1 + m_2)v
(2.0)(4.0)+(3.0)(0)=(2.0+3.0)v(2.0)(4.0) + (3.0)(0) = (2.0 + 3.0)v
8.0=5.0v8.0 = 5.0v
v=1.6 m s−1v = 1.6 \text{ m s}^{-1}
This is a perfectly inelastic collision — the objects stick together.

5. B. 3.0 m s−23.0 \text{ m s}^{-2}
Using Newton's Second Law: F=maF = ma
12=4.0×a12 = 4.0 \times a
a=3.0 m s−2a = 3.0 \text{ m s}^{-2}

6. B. Action and reaction forces are equal in magnitude and opposite in direction.
Newton's Third Law states that for every action, there is an equal and opposite reaction. These forces act on different bodies, so they do not cancel each other out. They can be contact or non-contact (e.g., gravitational) forces.
Common mistake: Students think action-reaction pairs act on the same body or cancel out.

7. C. 49 N49 \text{ N}
The normal contact force balances the weight of the block: R=mg=5.0×9.81=49.05 N≈49 NR = mg = 5.0 \times 9.81 = 49.05 \text{ N} \approx 49 \text{ N}.
Common mistake: Students forget to multiply by gg and simply write the mass value.

8. B. 203 m s−120\sqrt{3} \text{ m s}^{-1} (or 34.6 m s−134.6 \text{ m s}^{-1})
Horizontal component: vx=vcos⁡θ=40cos⁡30∘=40×32=203≈34.6 m s−1v_x = v \cos\theta = 40 \cos 30^\circ = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.6 \text{ m s}^{-1}.
Note: Both B and C are numerically equivalent, but B is the exact form. In an exam, either would be accepted.

9. C. 15.0 kg m s−115.0 \text{ kg m s}^{-1}
Taking the initial direction as positive:
Initial momentum = 1.5×6.0=9.0 kg m s−11.5 \times 6.0 = 9.0 \text{ kg m s}^{-1}
Final momentum = 1.5×(−4.0)=−6.0 kg m s−11.5 \times (-4.0) = -6.0 \text{ kg m s}^{-1}
Change in momentum = final − initial = −6.0−9.0=−15.0 kg m s−1-6.0 - 9.0 = -15.0 \text{ kg m s}^{-1}
Magnitude = 15.0 kg m s−115.0 \text{ kg m s}^{-1}
Common mistake: Students subtract speeds instead of momenta, or forget to account for the direction change.

10. C. Energy and work
Both energy and work have the unit joule (J). Work is the transfer of energy.

  • Work and force: work is in joules (J), force is in newtons (N) — different.
  • Power and momentum: power is in watts (W), momentum is in kg m s⁻¹ — different.
  • Impulse and energy: impulse is in N s (or kg m s⁻¹), energy is in joules (J) — different.

11. B. 9600 N9600 \text{ N}
Centripetal force: F=mv2r=1200×(20)250=1200×40050=48000050=9600 NF = \frac{mv^2}{r} = \frac{1200 \times (20)^2}{50} = \frac{1200 \times 400}{50} = \frac{480000}{50} = 9600 \text{ N}

12. D. 400 N m−1400 \text{ N m}^{-1}
Elastic potential energy: E=12kx2E = \frac{1}{2}kx^2
2.0=12×k×(0.10)22.0 = \frac{1}{2} \times k \times (0.10)^2
2.0=12×k×0.012.0 = \frac{1}{2} \times k \times 0.01
k=2.0×20.01=400 N m−1k = \frac{2.0 \times 2}{0.01} = 400 \text{ N m}^{-1}

13. B. 98 J98 \text{ J}
By conservation of energy: loss in gravitational potential energy = gain in kinetic energy
KE=mgh=0.50×9.81×20=98.1 J≈98 JKE = mgh = 0.50 \times 9.81 \times 20 = 98.1 \text{ J} \approx 98 \text{ J}
Common mistake: Students use v2=2ghv^2 = 2gh and then 12mv2\frac{1}{2}mv^2 but make arithmetic errors.

14. A. The net force on the object is zero.
By Newton's First Law, an object moving with constant velocity (which includes being at rest) has zero net force acting on it. This does not mean no forces act — it means the forces are balanced.
Common mistake: Students confuse "no net force" with "no forces at all."

15. C. 147 J147 \text{ J}
Work done against gravity: W=mgh=3.0×9.81×5.0=147.15 J≈147 JW = mgh = 3.0 \times 9.81 \times 5.0 = 147.15 \text{ J} \approx 147 \text{ J}
Common mistake: Students forget to include gg in the calculation.

Section B: Structured Questions [35 marks]

16. (a) [2 marks]
The principle of conservation of linear momentum states that the total momentum of a closed/isolated system remains constant, provided that no external forces act on the system.
Equivalently: the total momentum before an interaction equals the total momentum after the interaction.
Marking: [B1] for "total momentum is constant/unchanged"; [B1] for "no external forces" or "closed/isolated system" qualifier.

(b) (i) [2 marks]
Taking the initial direction of motion as positive:
Initial momentum = 0.060×25=1.5 kg m s−10.060 \times 25 = 1.5 \text{ kg m s}^{-1}
Final momentum = 0.060×(−18)=−1.08 kg m s−10.060 \times (-18) = -1.08 \text{ kg m s}^{-1}
Change in momentum = final − initial = −1.08−1.5=−2.58 kg m s−1-1.08 - 1.5 = -2.58 \text{ kg m s}^{-1}
Magnitude of change in momentum = 2.58 kg m s−12.58 \text{ kg m s}^{-1} (or 2.6 kg m s−12.6 \text{ kg m s}^{-1} to 2 s.f.)
Marking: [M1] for correct substitution into momentum change formula; [A1] for correct answer with unit.

(b) (ii) [2 marks]
Using the impulse-momentum theorem: FΔt=ΔpF \Delta t = \Delta p
F=ΔpΔt=2.580.020=129 N≈130 NF = \frac{\Delta p}{\Delta t} = \frac{2.58}{0.020} = 129 \text{ N} \approx 130 \text{ N}
Marking: [M1] for using F=Δp/ΔtF = \Delta p / \Delta t; [A1] for correct answer.

(b) (iii) [1 mark]
The force exerted by the wall on the ball is in the opposite direction to the initial motion of the ball (i.e., away from the wall).
This is because the wall pushes the ball back, causing it to rebound.
Marking: [B1] for correct direction stated.

17. (a) [2 marks]
Using v2=u2+2asv^2 = u^2 + 2as:
(30)2=0+2×a×200(30)^2 = 0 + 2 \times a \times 200
900=400a900 = 400a
a=2.25 m s−2a = 2.25 \text{ m s}^{-2}
Marking: [M1] for correct substitution; [A1] for correct answer with unit.

(b) [2 marks]
Using v=u+atv = u + at:
30=0+2.25×t30 = 0 + 2.25 \times t
t=302.25=13.3 st = \frac{30}{2.25} = 13.3 \text{ s}
Marking: [M1] for correct method; [A1] for correct answer.

(c) [2 marks]
Work done = force × distance = ma×sma \times s
F=1500×2.25=3375 NF = 1500 \times 2.25 = 3375 \text{ N}
W=3375×200=675000 J=675 kJW = 3375 \times 200 = 675000 \text{ J} = 675 \text{ kJ}
Alternatively, work done = change in kinetic energy = 12mv2−0=12(1500)(30)2=675000 J\frac{1}{2}mv^2 - 0 = \frac{1}{2}(1500)(30)^2 = 675000 \text{ J}
Marking: [M1] for correct method; [A1] for correct answer.

(d) [2 marks]
Average power = Work donetime=67500013.3=50752 W≈50.8 kW\frac{\text{Work done}}{\text{time}} = \frac{675000}{13.3} = 50752 \text{ W} \approx 50.8 \text{ kW}
Alternatively, P=Fvaverage=3375×15=50625 W≈50.6 kWP = Fv_{\text{average}} = 3375 \times 15 = 50625 \text{ W} \approx 50.6 \text{ kW}
Marking: [M1] for correct method; [A1] for correct answer.

18. (a) [1 mark]
vx=vcos⁡θ=28cos⁡45∘=28×22=19.8 m s−1v_x = v \cos\theta = 28 \cos 45^\circ = 28 \times \frac{\sqrt{2}}{2} = 19.8 \text{ m s}^{-1}
Marking: [B1] for correct answer.

(b) [1 mark]
vy=vsin⁡θ=28sin⁡45∘=28×22=19.8 m s−1v_y = v \sin\theta = 28 \sin 45^\circ = 28 \times \frac{\sqrt{2}}{2} = 19.8 \text{ m s}^{-1}
Marking: [B1] for correct answer.

(c) [2 marks]
At maximum height, vertical velocity = 0.
Using vy2=uy2−2ghv_y^2 = u_y^2 - 2gh:
0=(19.8)2−2(9.81)h0 = (19.8)^2 - 2(9.81)h
h=(19.8)22×9.81=392.0419.62=19.98 m≈20.0 mh = \frac{(19.8)^2}{2 \times 9.81} = \frac{392.04}{19.62} = 19.98 \text{ m} \approx 20.0 \text{ m}
Marking: [M1] for correct method; [A1] for correct answer.

(d) [2 marks]
Time of flight: t=2uyg=2×19.89.81=4.04 st = \frac{2u_y}{g} = \frac{2 \times 19.8}{9.81} = 4.04 \text{ s}
Range: R=vx×t=19.8×4.04=79.9 m≈80.0 mR = v_x \times t = 19.8 \times 4.04 = 79.9 \text{ m} \approx 80.0 \text{ m}
Alternatively, using R=v2sin⁡2θg=(28)2sin⁡90∘9.81=7849.81=79.9 mR = \frac{v^2 \sin 2\theta}{g} = \frac{(28)^2 \sin 90^\circ}{9.81} = \frac{784}{9.81} = 79.9 \text{ m}
Marking: [M1] for correct method; [A1] for correct answer.

(e) [2 marks]
Air resistance would decrease the range.
Reason: Air resistance opposes the motion of the ball, reducing both the horizontal and vertical components of velocity throughout the flight. This reduces the horizontal speed (shortening the range) and also reduces the maximum height and time of flight.
Marking: [B1] for "decrease"; [B1] for valid reason involving opposition to motion/energy loss.

19. (a) [2 marks]
The free-body diagram should show:

  • Weight (mgmg) acting vertically downward from the centre of the block
  • Normal reaction (RR) acting perpendicular to the surface, away from the surface
  • Frictional force (ff) acting up the slope (opposing the direction of motion/tendency to slide)
    Marking: [B1] for weight and normal reaction correctly drawn; [B1] for friction correctly drawn in the correct direction.

(b) [1 mark]
Component of weight down the slope = mgsin⁡θ=4.0×9.81×sin⁡30∘=4.0×9.81×0.5=19.62 N≈19.6 Nmg\sin\theta = 4.0 \times 9.81 \times \sin 30^\circ = 4.0 \times 9.81 \times 0.5 = 19.62 \text{ N} \approx 19.6 \text{ N}
Marking: [B1] for correct answer.

(c) [1 mark]
Normal reaction = mgcos⁡θ=4.0×9.81×cos⁡30∘=4.0×9.81×0.866=33.98 N≈34.0 Nmg\cos\theta = 4.0 \times 9.81 \times \cos 30^\circ = 4.0 \times 9.81 \times 0.866 = 33.98 \text{ N} \approx 34.0 \text{ N}
Marking: [B1] for correct answer.

(d) [1 mark]
Frictional force = μR=0.35×33.98=11.89 N≈11.9 N\mu R = 0.35 \times 33.98 = 11.89 \text{ N} \approx 11.9 \text{ N}
Marking: [B1] for correct answer.

(e) [2 marks]
The component of weight down the slope (19.6 N19.6 \text{ N}) is greater than the maximum frictional force (11.9 N11.9 \text{ N}).
Therefore, there is a net force down the slope, and the block will slide down.
Net force down slope = 19.6−11.9=7.7 N19.6 - 11.9 = 7.7 \text{ N}
Marking: [M1] for comparing the two forces; [A1] for correct conclusion that the block will slide.

Section C: Free Response [30 marks]

20. (a) [2 marks]
Centripetal acceleration is the acceleration directed towards the centre of the circular path. It is responsible for changing the direction of the velocity (not the speed) of an object moving in a circle. Its magnitude is given by a=v2ra = \frac{v^2}{r} or a=ω2ra = \omega^2 r.
Marking: [B1] for "towards the centre"; [B1] for "changes direction of velocity" or equivalent.

(b) (i) [2 marks]
ac=v2r=(6.0)21.5=361.5=24 m s−2a_c = \frac{v^2}{r} = \frac{(6.0)^2}{1.5} = \frac{36}{1.5} = 24 \text{ m s}^{-2}
Marking: [M1] for correct substitution; [A1] for correct answer with unit.

(b) (ii) [2 marks]
At the lowest point, the tension acts upward and weight acts downward. The net force towards the centre (upward) provides the centripetal force:
T−mg=mv2rT - mg = \frac{mv^2}{r}
T=mg+mv2r=(0.20×9.81)+0.20×(6.0)21.5T = mg + \frac{mv^2}{r} = (0.20 \times 9.81) + \frac{0.20 \times (6.0)^2}{1.5}
T=1.962+4.8=6.762 N≈6.8 NT = 1.962 + 4.8 = 6.762 \text{ N} \approx 6.8 \text{ N}
Marking: [M1] for correct equation T−mg=mv2/rT - mg = mv^2/r; [A1] for correct answer.

(c) (i) [2 marks]
At the highest point, for the string to remain taut with minimum tension (T = 0), the weight alone provides the centripetal force:
mg=mvtop2rmg = \frac{mv_{\text{top}}^2}{r}
vtop2=gr=9.81×1.5=14.715v_{\text{top}}^2 = gr = 9.81 \times 1.5 = 14.715
vtop=14.715=3.84 m s−1v_{\text{top}} = \sqrt{14.715} = 3.84 \text{ m s}^{-1}
Marking: [M1] for setting mg=mv2/rmg = mv^2/r; [A1] for correct answer.

(c) (ii) [3 marks]
Using conservation of energy between the lowest and highest points:
The height difference between lowest and highest point = 2r=2×1.5=3.0 m2r = 2 \times 1.5 = 3.0 \text{ m}
12mvbottom2=12mvtop2+mg(2r)\frac{1}{2}mv_{\text{bottom}}^2 = \frac{1}{2}mv_{\text{top}}^2 + mg(2r)
12vbottom2=12vtop2+g(2r)\frac{1}{2}v_{\text{bottom}}^2 = \frac{1}{2}v_{\text{top}}^2 + g(2r)
12vbottom2=12(14.715)+9.81×3.0\frac{1}{2}v_{\text{bottom}}^2 = \frac{1}{2}(14.715) + 9.81 \times 3.0
12vbottom2=7.3575+29.43=36.7875\frac{1}{2}v_{\text{bottom}}^2 = 7.3575 + 29.43 = 36.7875
vbottom2=73.575v_{\text{bottom}}^2 = 73.575
vbottom=73.575=8.58 m s−1v_{\text{bottom}} = \sqrt{73.575} = 8.58 \text{ m s}^{-1}
Marking: [M1] for correct energy conservation equation; [M1] for correct substitution of values; [A1] for correct final answer.

(d) [2 marks]
The tension varies because the weight of the sphere has a component that either adds to or subtracts from the tension depending on the position in the circle:

  • At the lowest point: tension must support the weight AND provide the centripetal force, so T=mg+mv2rT = mg + \frac{mv^2}{r} (maximum tension).
  • At the highest point: weight acts towards the centre, so T=mv2r−mgT = \frac{mv^2}{r} - mg (minimum tension).
  • At intermediate positions: the radial component of weight varies as cos⁡θ\cos\theta, causing the tension to vary continuously.
    Additionally, the speed varies with height due to energy conservation, which also affects the tension.
    Marking: [B1] for explaining that weight's radial component varies with position; [B1] for explaining the difference between top and bottom positions.

End of Answer Key

Total: 80 marks