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A Level H1 Physics Practice Paper 3

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H1 Level: A-Level Paper: Practice Paper 3 (Mechanics) Duration: 1 hour 30 minutes Total Marks: 60

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of 20 questions in three sections.
Answer all questions.
Write your answers in the spaces provided.
Show all working clearly for calculation questions. Marks are awarded for method as well as final answers.
Take g = 9.81 m s⁻² unless otherwise stated.
The number of marks is given in brackets [ ] at the end of each question or part question.
You are advised to spend about 1 hour 15 minutes on the questions, leaving 15 minutes for checking.

Section A: Kinematics and Dynamics (Questions 1–7)

[Total: 20 marks]

1. A car accelerates uniformly from rest to 25.0 m s⁻¹ in 8.0 s.

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

(b) Calculate the distance travelled by the car during this 8.0 s period. [2]

(c) The car then maintains a constant velocity of 25.0 m s⁻¹ for a further 12.0 s. Calculate the total distance travelled from the start. [2]

2. A ball is thrown vertically upwards with an initial speed of 18.0 m s⁻¹ from the edge of a cliff 45.0 m above the sea.

(a) Calculate the maximum height reached by the ball above the point of release. [2]

(b) Calculate the total time taken for the ball to reach the sea below the cliff. [3]

3. A stone is projected horizontally from the top of a vertical cliff 60.0 m high. The stone lands in the sea at a horizontal distance of 80.0 m from the base of the cliff.

(a) Calculate the time taken for the stone to reach the sea. [2]

(b) Calculate the initial horizontal speed of the stone. [1]

4. A block of mass 5.0 kg is pulled along a rough horizontal surface by a force of 30.0 N applied at an angle of 25° above the horizontal. The coefficient of kinetic friction between the block and the surface is 0.35.

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

(b) Calculate the normal reaction force exerted by the surface on the block. [2]

5. State Newton's three laws of motion. [3]

6. A car of mass 1200 kg travelling at 20.0 m s⁻¹ collides with a stationary car of mass 800 kg. After the collision, the two cars move together.

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

(b) Calculate the common velocity of the cars immediately after the collision. [2]

7. A tennis ball of mass 0.058 kg strikes a wall at 25.0 m s⁻¹ perpendicular to the wall and rebounds at 18.0 m s⁻¹.

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

(b) If the ball is in contact with the wall for 0.015 s, calculate the average force exerted by the wall on the ball. [2]

Section B: Forces, Work, Energy and Power (Questions 8–14)

[Total: 22 marks]

8. A uniform plank AB of length 4.0 m and weight 200 N rests on two supports at points P and Q. P is 0.50 m from end A, and Q is 1.0 m from end B. A person of weight 600 N stands on the plank at a distance x from end A.

(a) Draw a diagram showing all the forces acting on the plank. Label each force clearly. [2]

(b) Write an expression for the reaction force at Q in terms of x. [2]

9. A crate of mass 50.0 kg is pushed up a rough incline of 30° to the horizontal by a force of 400 N parallel to the incline. The coefficient of kinetic friction between the crate and the incline is 0.25.

(a) Calculate the component of the weight of the crate acting down the incline. [1]

(b) Calculate the friction force acting on the crate. [2]

10. A spring has a spring constant of 250 N m⁻¹.

(a) Calculate the work done in stretching the spring from its natural length by 0.080 m. [2]

(b) The stretched spring is used to launch a ball of mass 0.050 kg horizontally from rest. Assuming all the elastic potential energy is converted to kinetic energy, calculate the speed of the ball immediately after launch. [2]

11. A pump lifts 500 kg of water per minute through a vertical height of 12.0 m.

(a) Calculate the work done by the pump per minute. [2]

(b) Calculate the minimum power output required by the pump. [1]

12. A cyclist of total mass 85.0 kg (including bicycle) travels at a constant speed of 8.0 m s⁻¹ along a horizontal road. The total resistive force acting on the cyclist is 45.0 N.

(a) Calculate the power output of the cyclist. [2]

(b) The cyclist now travels up a slope inclined at 5.0° to the horizontal at the same speed. The resistive force remains 45.0 N. Calculate the new power output required. [2]

13. A ball of mass 0.20 kg is dropped from a height of 15.0 m above the ground. Air resistance is negligible.

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

(b) The ball penetrates 0.040 m into soft ground before coming to rest. Calculate the average resistive force exerted by the ground on the ball. [2]

14. Explain, using the principle of conservation of energy, why the speed of a pendulum bob is maximum at its lowest point and zero at its highest points. [2]

Section C: Integrated Mechanics Problems (Questions 15–20)

[Total: 18 marks]

15. A block of mass 2.0 kg slides down a frictionless curved track from rest at point A, which is 5.0 m above the ground. At point B, the block leaves the track horizontally at a height of 2.0 m above the ground.

(a) Calculate the speed of the block at point B. [2]

(b) Calculate the horizontal distance from point B where the block lands on the ground. [2]

16. A bullet of mass 0.015 kg is fired horizontally into a wooden block of mass 2.0 kg suspended by a light string. The bullet becomes embedded in the block, and the block swings to a maximum height of 0.12 m above its initial position.

(a) Calculate the speed of the block and bullet immediately after the collision. [2]

(b) Calculate the initial speed of the bullet. [2]

17. A conveyor belt moves at a constant speed of 2.5 m s⁻¹. Sand is dropped vertically onto the belt at a rate of 40.0 kg s⁻¹.

(a) Calculate the rate of change of momentum of the sand in the horizontal direction. [2]

(b) Hence, calculate the horizontal force that must be applied to the belt to maintain its constant speed. [1]

18. Two blocks, A of mass 4.0 kg and B of mass 6.0 kg, are connected by a light inextensible string passing over a smooth pulley. Block A rests on a smooth horizontal table, and block B hangs freely.

(a) Draw free-body diagrams for both blocks. [2]

(b) Calculate the acceleration of the system. [2]

19. A rocket of initial mass 5000 kg ejects exhaust gases at a constant speed of 2000 m s⁻¹ relative to the rocket. The rate of ejection of gases is 50.0 kg s⁻¹.

(a) Calculate the thrust produced by the rocket engine. [2]

(b) Calculate the initial acceleration of the rocket, assuming it is launched vertically from the ground. [2]

20. A ball of mass 0.15 kg is attached to a string of length 0.80 m and whirled in a horizontal circle at a constant speed. The string makes an angle of 30° with the vertical.

(a) Draw a diagram showing the forces acting on the ball. [1]

(b) Calculate the tension in the string. [2]

END OF PAPER

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Answers

TuitionGoWhere Practice Paper - Physics H1 A-Level

Answer Key and Marking Scheme (Version 3)

Section A: Kinematics and Dynamics (Questions 1–7)

1. Car acceleration problem

(a) a = (v − u) / t = (25.0 − 0) / 8.0 = 3.125 m s⁻² ≈ 3.13 m s⁻² [M1, A1]

(b) s = ut + ½at² = 0 + ½ × 3.125 × (8.0)² = 100 m [M1, A1] Alternative: s = ½(u + v)t = ½(0 + 25.0) × 8.0 = 100 m

2. Vertical throw from cliff

(a) v² = u² + 2as → 0 = (18.0)² + 2(−9.81)s → s = (18.0)² / (2 × 9.81) = 16.5 m [M1, A1]

(b) Consider motion from release to sea: s = −45.0 m, u = +18.0 m s⁻¹, a = −9.81 m s⁻² s = ut + ½at² → −45.0 = 18.0t − 4.905t² [M1] 4.905t² − 18.0t − 45.0 = 0 [M1] t = [18.0 ± √(324 + 882.9)] / 9.81 = [18.0 ± √1206.9] / 9.81 = [18.0 ± 34.74] / 9.81 t = 5.38 s (positive root) [A1]

3. Horizontal projection from cliff

(a) Vertical motion: s = ut + ½at² → 60.0 = 0 + ½(9.81)t² → t = √(120.0/9.81) = 3.50 s [M1, A1]

(b) Horizontal: s = ut → 80.0 = u × 3.50 → u = 22.9 m s⁻¹ [A1]

(c) v_y = u_y + at = 0 + 9.81 × 3.50 = 34.3 m s⁻¹ [M1] v = √(22.9² + 34.3²) = √(524.4 + 1176.5) = √1700.9 = 41.2 m s⁻¹ [A1]

4. Block pulled on rough surface

(a) Free-body diagram: [B2]

Weight mg downward (49.05 N)
Normal reaction N upward
Applied force F = 30.0 N at 25° above horizontal
Friction f opposing motion (to the left if motion is to the right) Deduct 1 mark for each missing or incorrect force.

(b) Vertical equilibrium: N + F sin 25° = mg [M1] N = (5.0 × 9.81) − 30.0 sin 25° = 49.05 − 12.68 = 36.4 N [A1]

(c) Friction: f = μN = 0.35 × 36.37 = 12.7 N [M1] Horizontal net force: F cos 25° − f = ma [M1] 30.0 cos 25° − 12.73 = 5.0a → 27.19 − 12.73 = 5.0a → a = 2.89 m s⁻² [A1]

5. Newton's three laws

First Law: A body remains at rest or in uniform motion in a straight line unless acted upon by a net external force. [B1]
Second Law: The rate of change of momentum of a body is directly proportional to the net force acting on it and takes place in the direction of the force. (F = ma for constant mass) [B1]
Third Law: For every action, there is an equal and opposite reaction. (If body A exerts a force on body B, body B exerts an equal and opposite force on body A.) [B1]

6. Car collision

(a) In a closed system (no external forces), the total linear momentum before collision equals the total linear momentum after collision. [B1]

(b) m₁u₁ + m₂u₂ = (m₁ + m₂)v [M1] (1200 × 20.0) + (800 × 0) = (1200 + 800)v → 24000 = 2000v → v = 12.0 m s⁻¹ [A1]

(c) Initial KE = ½(1200)(20.0)² + 0 = 240,000 J [M1] Final KE = ½(2000)(12.0)² = 144,000 J [M1] KE is not conserved (240,000 ≠ 144,000), therefore the collision is inelastic. [A1]

7. Tennis ball and wall

(a) Taking direction towards wall as positive: Initial momentum = 0.058 × (+25.0) = +1.45 kg m s⁻¹ Final momentum = 0.058 × (−18.0) = −1.044 kg m s⁻¹ [M1] Change in momentum = −1.044 − (+1.45) = −2.49 kg m s⁻¹ Magnitude of change = 2.49 kg m s⁻¹ (away from wall) [A1]

(b) F = Δp / Δt = 2.494 / 0.015 [M1] F = 166 N (away from wall) [A1]

Section B: Forces, Work, Energy and Power (Questions 8–14)

8. Plank equilibrium

(a) Diagram: [B2]

Weight of plank 200 N downward at centre (2.0 m from A)
Weight of person 600 N downward at distance x from A
Reaction R_P upward at P (0.50 m from A)
Reaction R_Q upward at Q (3.0 m from A, since Q is 1.0 m from B and AB = 4.0 m)

(b) Taking moments about P: [M1] Clockwise moments = Anticlockwise moments 200(2.0 − 0.50) + 600(x − 0.50) = R_Q(3.0 − 0.50) 200(1.5) + 600(x − 0.50) = R_Q(2.5) 300 + 600x − 300 = 2.5R_Q R_Q = 240x [A1]

(c) At point of tipping about P, R_Q = 0 [M1] From vertical equilibrium: R_P = 200 + 600 = 800 N Taking moments about P: 200(1.5) = 600(x − 0.50) 300 = 600x − 300 → 600x = 600 → x = 1.0 m [A1]

9. Crate on incline

(a) Component down incline = mg sin 30° = 50.0 × 9.81 × 0.5 = 245 N [A1]

(b) Normal reaction: N = mg cos 30° = 50.0 × 9.81 × 0.866 = 425 N [M1] Friction: f = μN = 0.25 × 424.8 = 106 N [A1]

10. Spring and projectile

(a) W = ½kx² = ½ × 250 × (0.080)² [M1] W = 125 × 0.0064 = 0.80 J [A1]

(b) ½mv² = 0.80 → v = √(2 × 0.80 / 0.050) [M1] v = √32 = 5.66 m s⁻¹ [A1]

11. Water pump

(a) Work per minute = mgh = 500 × 9.81 × 12.0 = 58,860 J [M1, A1]

(b) P = W/t = 58,860 / 60 = 981 W [A1]

(c) Electrical power input = Useful power / efficiency = 981 / 0.75 = 1308 W [M1] P = VI → I = P/V = 1308 / 240 = 5.45 A [A1]

12. Cyclist power

(a) At constant speed, driving force = resistive force = 45.0 N [M1] P = Fv = 45.0 × 8.0 = 360 W [A1]

(b) Additional force needed = mg sin 5.0° = 85.0 × 9.81 × sin 5.0° = 85.0 × 9.81 × 0.0872 = 72.7 N [M1] Total force = 45.0 + 72.7 = 117.7 N P = Fv = 117.7 × 8.0 = 942 W [A1]

13. Ball dropped onto soft ground

(a) By conservation of energy: KE = mgh = 0.20 × 9.81 × 15.0 [M1] KE = 29.4 J [A1]

(b) Work done by resistive force = KE lost [M1] F × 0.040 = 29.43 → F = 29.43 / 0.040 = 736 N [A1]

14. Pendulum energy explanation

At the highest points, the bob has maximum gravitational potential energy and zero kinetic energy (momentarily at rest). [B1] As the bob swings down, GPE is converted to KE. At the lowest point, all the initial GPE (relative to that point) has been converted to KE, so speed is maximum. [B1] (Energy is conserved: total mechanical energy = GPE + KE = constant, assuming no air resistance.)

Section C: Integrated Mechanics Problems (Questions 15–20)

15. Block on curved track

(a) Loss in height = 5.0 − 2.0 = 3.0 m [M1] By conservation of energy: ½mv² = mgh → v = √(2gh) = √(2 × 9.81 × 3.0) = √58.86 = 7.67 m s⁻¹ [A1]

(b) Time to fall 2.0 m: s = ½gt² → 2.0 = ½(9.81)t² → t = √(4.0/9.81) = 0.639 s [M1] Horizontal distance = v × t = 7.67 × 0.639 = 4.90 m [A1]

16. Ballistic pendulum

(a) By conservation of energy for swing: ½(m + M)v² = (m + M)gh [M1] v = √(2gh) = √(2 × 9.81 × 0.12) = √2.3544 = 1.53 m s⁻¹ [A1]

(b) By conservation of momentum: m_bullet × u = (m_bullet + M_block) × v [M1] 0.015 × u = (0.015 + 2.0) × 1.534 → u = (2.015 × 1.534) / 0.015 = 206 m s⁻¹ [A1]

17. Conveyor belt

(a) Rate of change of momentum = (mass per second) × (change in velocity) [M1] = 40.0 × (2.5 − 0) = 100 kg m s⁻² = 100 N [A1]

(b) Force required = rate of change of momentum = 100 N [A1]

18. Connected blocks

(a) Free-body diagrams: [B2] Block A (on table): Weight m_Ag downward, Normal reaction N upward, Tension T to the right. Block B (hanging): Weight m_Bg downward, Tension T upward. (1 mark each block)

(b) For block B: m_Bg − T = m_Ba → 6.0(9.81) − T = 6.0a → 58.86 − T = 6.0a [M1] For block A: T = m_Aa → T = 4.0a Substituting: 58.86 − 4.0a = 6.0a → 58.86 = 10.0a → a = 5.89 m s⁻² [A1]

19. Rocket thrust

(a) Thrust = (rate of ejection) × (exhaust speed relative to rocket) [M1] Thrust = 50.0 × 2000 = 100,000 N = 1.00 × 10⁵ N [A1]

(b) Net force = Thrust − Weight = 100,000 − (5000 × 9.81) = 100,000 − 49,050 = 50,950 N [M1] a = F/m = 50,950 / 5000 = 10.2 m s⁻² [A1]

20. Conical pendulum

(a) Diagram: [B1]

Weight mg vertically downward
Tension T along the string at 30° to vertical (Forces correctly shown and labelled)

(b) Vertical equilibrium: T cos 30° = mg [M1] T = (0.15 × 9.81) / cos 30° = 1.4715 / 0.8660 = 1.70 N [A1]

(c) Horizontal component provides centripetal force: T sin 30° = mv²/r [M1] r = L sin 30° = 0.80 × 0.5 = 0.40 m 1.699 × 0.5 = 0.15 × v² / 0.40 → 0.8495 = 0.375v² → v = √(0.8495/0.375) = 1.51 m s⁻¹ [A1]

END OF ANSWER KEY