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A Level H2 Physics Practice Paper 5
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Questions
TuitionGoWhere Practice Paper - Physics H2 A-Level
TuitionGoWhere Exam Practice (AI)
Subject: Physics H2 (9478) Level: A-Level Paper: Practice Paper 5 (Mechanics) Duration: 1 hour 30 minutes Total Marks: 60
Name: _________________________ Class: _________________________ Date: _________________________
Instructions to Candidates
- This paper consists of 20 questions on the topic of Mechanics.
- Answer all questions in the spaces provided.
- The number of marks is given in brackets [ ] at the end of each question or part question.
- You are reminded of the need for clear presentation in your answers.
- Where calculations are required, show all working.
- Take g = 9.81 m s⁻² unless otherwise stated.
Section A: Principles and Definitions (Questions 1–5)
Answer all questions in this section.
1. State the principle of conservation of linear momentum.
[2 marks]
2. Define the term impulse and state its SI unit.
[2 marks]
3. State Newton's second law of motion in terms of momentum.
[2 marks]
4. Define work as used in mechanics and state the condition under which no work is done by a force acting on a moving object.
[2 marks]
5. State the principle of conservation of energy.
[2 marks]
Section B: Calculations and Applications (Questions 6–15)
Answer all questions in this section. Show all working clearly.
6. A ball of mass 0.45 kg is dropped from rest and falls vertically through a height of 2.5 m before hitting the ground. It rebounds to a height of 1.8 m.
(a) Calculate the speed of the ball just before it hits the ground.
[2 marks]
(b) Calculate the speed of the ball just after it leaves the ground.
[2 marks]
(c) Calculate the change in momentum of the ball during the impact with the ground. State the direction of this change.
[3 marks]
7. A car of mass 1200 kg accelerates uniformly from rest to a speed of 25 m s⁻¹ in 8.0 s along a straight horizontal road. The total resistive force acting on the car is 600 N.
(a) Calculate the acceleration of the car.
[2 marks]
(b) Calculate the net force acting on the car.
[2 marks]
(c) Calculate the driving force produced by the engine.
[2 marks]
8. A particle of mass 0.20 kg performs simple harmonic motion with amplitude 0.080 m and period 2.5 s.
(a) Calculate the angular frequency of the motion.
[2 marks]
(b) Calculate the maximum acceleration of the particle.
[2 marks]
(c) Calculate the maximum kinetic energy of the particle.
[2 marks]
9. A block of mass 3.0 kg slides down a smooth plane inclined at 30° to the horizontal. The block starts from rest and travels 4.0 m along the incline.
(a) Calculate the component of the weight of the block acting down the plane.
[2 marks]
(b) Calculate the acceleration of the block down the plane.
[2 marks]
(c) Calculate the speed of the block after travelling 4.0 m.
[2 marks]
10. Two trolleys, A and B, are on a frictionless horizontal track. Trolley A of mass 2.0 kg moves with velocity 3.0 m s⁻¹ towards stationary trolley B of mass 1.0 kg. After the collision, trolley A moves with velocity 1.0 m s⁻¹ in the same direction.
(a) Calculate the velocity of trolley B after the collision.
[3 marks]
(b) Determine whether the collision is elastic or inelastic. Support your answer with calculations.
[3 marks]
11. A stone of mass 0.50 kg is tied to a string of length 1.2 m and whirled in a vertical circle at constant speed. At the top of the circle, the tension in the string is 2.0 N.
(a) Calculate the centripetal force required at the top of the circle.
[2 marks]
(b) Calculate the speed of the stone.
[2 marks]
(c) Calculate the tension in the string at the bottom of the circle.
[2 marks]
12. A satellite of mass 500 kg orbits the Earth in a circular orbit of radius 7.0 × 10⁶ m. The mass of the Earth is 6.0 × 10²⁴ kg and the gravitational constant G = 6.67 × 10⁻¹¹ N m² kg⁻².
(a) Calculate the gravitational force acting on the satellite.
[2 marks]
(b) Calculate the orbital speed of the satellite.
[2 marks]
(c) Calculate the period of the orbit.
[2 marks]
13. A spring of spring constant 200 N m⁻¹ is compressed by 0.15 m. A block of mass 0.50 kg is placed against the compressed spring on a horizontal frictionless surface. The spring is released.
(a) Calculate the elastic potential energy stored in the compressed spring.
[2 marks]
(b) Calculate the speed of the block when it leaves the spring.
[2 marks]
(c) The block then encounters a rough horizontal surface with coefficient of kinetic friction 0.40. Calculate the distance the block travels on the rough surface before coming to rest.
[3 marks]
14. A projectile is launched from ground level with an initial speed of 30 m s⁻¹ at an angle of 40° above the horizontal. Air resistance is negligible.
(a) Calculate the horizontal and vertical components of the initial velocity.
[2 marks]
(b) Calculate the time taken for the projectile to reach its maximum height.
[2 marks]
(c) Calculate the horizontal range of the projectile.
[2 marks]
15. A uniform rod AB of length 2.0 m and weight 50 N is pivoted at point A. A vertical force F is applied at point B to keep the rod horizontal.
(a) Draw a diagram showing all the forces acting on the rod.
[2 marks]
(b) By taking moments about point A, calculate the magnitude of force F.
[2 marks]
(c) Calculate the magnitude and direction of the reaction force at the pivot.
[2 marks]
Section C: Data Analysis and Experimental Skills (Questions 16–20)
Answer all questions in this section.
16. An experiment is conducted to investigate the relationship between the period T of a simple pendulum and its length L. The following data is obtained:
| Length L / m | 0.40 | 0.60 | 0.80 | 1.00 | 1.20 |
|---|---|---|---|---|---|
| Period T / s | 1.27 | 1.55 | 1.79 | 2.00 | 2.19 |
The relationship is expected to be of the form T = kLⁿ, where k and n are constants.
(a) State what quantities should be plotted on the x-axis and y-axis to obtain a straight-line graph.
[2 marks]
(b) Using the data for L = 0.40 m and L = 1.20 m, calculate the value of n.
[3 marks]
(c) Hence, calculate the value of k. State the units of k.
[2 marks]
17. A student performs an experiment to determine the acceleration of free fall g using a falling ball and a trapdoor. The ball is released from rest at a measured height h above the trapdoor, and the time t for the ball to reach the trapdoor is recorded. The student plots a graph of h against t².
(a) Explain why the graph of h against t² should be a straight line passing through the origin.
[2 marks]
(b) The gradient of the graph is found to be 4.7 m s⁻². Determine the value of g from this result.
[2 marks]
(c) Suggest two sources of systematic error in this experiment and state how each could be reduced.
[4 marks]
18. A dynamics cart of mass M is placed on a horizontal track and connected by a light inextensible string passing over a smooth pulley to a hanging mass m. The system is released from rest and the acceleration a is measured for different values of m while M is kept constant.
The equation relating the quantities is: a = (mg) / (M + m)
(a) Rearrange this equation into a form suitable for a straight-line graph with 1/a on the y-axis and 1/m on the x-axis.
[2 marks]
(b) State what the gradient and y-intercept of this graph represent in terms of M and g.
[2 marks]
(c) The gradient is found to be 0.51 s² m⁻¹ and the y-intercept is 0.10 s² m⁻¹. Calculate M and g.
[3 marks]
19. A student investigates the motion of a mass oscillating on a spring. The mass is displaced and released, and the student records the displacement x at various times t. The student suspects the motion follows x = x₀ cos(ωt).
(a) Explain how the student could use a graphical method to verify that the motion is simple harmonic.
[3 marks]
(b) State what additional measurements would be needed to determine the spring constant k.
[2 marks]
(c) The student finds that the amplitude of oscillation gradually decreases with time. Explain why this happens and state one precaution that could be taken to minimise this effect in a laboratory experiment to determine k.
[3 marks]
20. In an experiment to verify the principle of conservation of momentum, two gliders on a linear air track are used. Glider A of mass 0.30 kg moves with velocity 0.50 m s⁻¹ towards stationary glider B of mass 0.20 kg. After the collision, the gliders stick together.
(a) Calculate the expected velocity of the combined gliders after the collision.
[2 marks]
(b) State any precautions that would be taken to improve the accuracy and safety of this experiment.
[4 marks]
(c) The measured velocity after collision is found to be 0.28 m s⁻¹. Calculate the percentage difference between the theoretical and experimental values. Suggest a reason for any discrepancy.
[2 marks]
END OF PAPER
Answers
TuitionGoWhere Practice Paper - Physics H2 A-Level (Mechanics)
Answer Key and Marking Scheme
Total Marks: 60
Section A: Principles and Definitions (Questions 1–5)
1. State the principle of conservation of linear momentum. [2 marks]
Answer: The total momentum of a system remains constant (1) provided no net external force acts on the system (1).
Alternative acceptable answer: In a closed/isolated system, the total momentum before an interaction equals the total momentum after the interaction (1), provided no external forces act (1).
2. Define the term impulse and state its SI unit. [2 marks]
Answer: Impulse is the product of the average force acting on an object and the time for which it acts (1). SI unit: N s (or kg m s⁻¹) (1).
Accept: Impulse = change in momentum (1) with correct unit (1).
3. State Newton's second law of motion in terms of momentum. [2 marks]
Answer: The rate of change of momentum of an object is directly proportional to the net force acting on it (1) and takes place in the direction of the force (1).
Accept: F = Δp/Δt or F = dp/dt with explanation (1 mark each for rate of change and direction/proportionality).
4. Define work as used in mechanics and state the condition under which no work is done by a force acting on a moving object. [2 marks]
Answer: Work is the product of the force and the displacement in the direction of the force (1). No work is done when the force is perpendicular to the displacement (1).
Accept: W = Fs cos θ with explanation (1 mark for definition, 1 mark for perpendicular condition).
5. State the principle of conservation of energy. [2 marks]
Answer: Energy cannot be created or destroyed (1); it can only be transferred or transformed from one form to another (1). The total energy of an isolated system remains constant (1).
Award 2 marks for any two of the above points.
Section B: Calculations and Applications (Questions 6–15)
6. Ball dropped and rebounding.
(a) Speed just before hitting ground. [2 marks]
v² = u² + 2as = 0 + 2(9.81)(2.5) = 49.05 (1) v = √49.05 = 7.00 m s⁻¹ (1)
(b) Speed just after leaving ground. [2 marks]
v² = u² + 2as → 0 = u² + 2(-9.81)(1.8) (1) u² = 35.316 → u = 5.94 m s⁻¹ (1)
(c) Change in momentum. [3 marks]
Taking downward as positive: Initial momentum = 0.45 × 7.00 = 3.15 kg m s⁻¹ downward (1) Final momentum = 0.45 × (-5.94) = -2.67 kg m s⁻¹ (i.e., 2.67 kg m s⁻¹ upward) (1) Change in momentum = -2.67 - 3.15 = -5.82 kg m s⁻¹ Magnitude = 5.82 kg m s⁻¹, direction: upward (1)
7. Car acceleration.
(a) Acceleration. [2 marks]
a = (v - u)/t = (25 - 0)/8.0 (1) = 3.125 m s⁻² ≈ 3.13 m s⁻² (1)
(b) Net force. [2 marks]
F_net = ma = 1200 × 3.125 (1) = 3750 N ≈ 3.75 × 10³ N (1)
(c) Driving force. [2 marks]
F_driving - F_resistive = F_net (1) F_driving = 3750 + 600 = 4350 N ≈ 4.35 × 10³ N (1)
8. Simple harmonic motion.
(a) Angular frequency. [2 marks]
ω = 2π/T = 2π/2.5 (1) = 2.51 rad s⁻¹ (1)
(b) Maximum acceleration. [2 marks]
a_max = ω²x₀ = (2.51)² × 0.080 (1) = 0.504 m s⁻² (1)
(c) Maximum kinetic energy. [2 marks]
v_max = ωx₀ = 2.51 × 0.080 = 0.201 m s⁻¹ (1) KE_max = ½mv_max² = ½ × 0.20 × (0.201)² = 4.04 × 10⁻³ J (1)
Alternative: KE_max = ½mω²x₀² = ½ × 0.20 × (2.51)² × (0.080)² = 4.03 × 10⁻³ J
9. Block on incline.
(a) Component of weight down plane. [2 marks]
mg sin θ = 3.0 × 9.81 × sin 30° (1) = 3.0 × 9.81 × 0.5 = 14.7 N (1)
(b) Acceleration down plane. [2 marks]
F = ma → a = F/m = 14.7/3.0 (1) = 4.90 m s⁻² (1)
(c) Speed after 4.0 m. [2 marks]
v² = u² + 2as = 0 + 2(4.90)(4.0) = 39.2 (1) v = √39.2 = 6.26 m s⁻¹ (1)
10. Trolley collision.
(a) Velocity of trolley B after collision. [3 marks]
By conservation of momentum: m_A u_A + m_B u_B = m_A v_A + m_B v_B (1) (2.0)(3.0) + (1.0)(0) = (2.0)(1.0) + (1.0)(v_B) (1) 6.0 = 2.0 + v_B → v_B = 4.0 m s⁻¹ (1)
(b) Elastic or inelastic? [3 marks]
Initial KE = ½(2.0)(3.0)² + 0 = 9.0 J (1) Final KE = ½(2.0)(1.0)² + ½(1.0)(4.0)² = 1.0 + 8.0 = 9.0 J (1) Since initial KE = final KE, the collision is elastic (1).
11. Stone in vertical circle.
(a) Centripetal force at top. [2 marks]
At top: mg + T = F_c (1) F_c = (0.50 × 9.81) + 2.0 = 4.905 + 2.0 = 6.91 N (1)
(b) Speed of stone. [2 marks]
F_c = mv²/r → v² = F_c × r/m = 6.91 × 1.2/0.50 = 16.58 (1) v = √16.58 = 4.07 m s⁻¹ (1)
(c) Tension at bottom. [2 marks]
At bottom: T - mg = F_c (1) T = F_c + mg = 6.91 + 4.905 = 11.8 N (1)
12. Satellite orbit.
(a) Gravitational force. [2 marks]
F = GMm/r² = (6.67 × 10⁻¹¹)(6.0 × 10²⁴)(500)/(7.0 × 10⁶)² (1) = (2.001 × 10¹⁷)/(4.9 × 10¹³) = 4.08 × 10³ N (1)
(b) Orbital speed. [2 marks]
F = mv²/r → v² = Fr/m = (4.08 × 10³)(7.0 × 10⁶)/500 = 5.71 × 10⁷ (1) v = √(5.71 × 10⁷) = 7.56 × 10³ m s⁻¹ (1)
Alternative: v = √(GM/r) = √[(6.67 × 10⁻¹¹)(6.0 × 10²⁴)/(7.0 × 10⁶)] = 7.56 × 10³ m s⁻¹
(c) Period of orbit. [2 marks]
T = 2πr/v = 2π(7.0 × 10⁶)/(7.56 × 10³) (1) = (4.40 × 10⁷)/(7.56 × 10³) = 5.82 × 10³ s (1)
13. Spring and block.
(a) Elastic potential energy. [2 marks]
EPE = ½kx² = ½ × 200 × (0.15)² (1) = 2.25 J (1)
(b) Speed of block. [2 marks]
EPE → KE: ½mv² = 2.25 (1) v² = 2 × 2.25/0.50 = 9.0 → v = 3.0 m s⁻¹ (1)
(c) Distance on rough surface. [3 marks]
Friction force: f = μmg = 0.40 × 0.50 × 9.81 = 1.962 N (1) Work done by friction = loss in KE: f × d = ½mv² (1) 1.962 × d = 2.25 → d = 2.25/1.962 = 1.15 m (1)
14. Projectile motion.
(a) Components of initial velocity. [2 marks]
u_x = 30 cos 40° = 30 × 0.7660 = 23.0 m s⁻¹ (1) u_y = 30 sin 40° = 30 × 0.6428 = 19.3 m s⁻¹ (1)
(b) Time to maximum height. [2 marks]
At max height, v_y = 0: v_y = u_y - gt → 0 = 19.3 - 9.81t (1) t = 19.3/9.81 = 1.97 s (1)
(c) Horizontal range. [2 marks]
Total time of flight = 2 × 1.97 = 3.94 s (1) Range = u_x × t_total = 23.0 × 3.94 = 90.6 m (1)
Alternative: Range = (u² sin 2θ)/g = (30² × sin 80°)/9.81 = 90.4 m
15. Uniform rod.
(a) Diagram. [2 marks]
Diagram showing:
- Rod AB horizontal, pivoted at A (1)
- Weight 50 N acting downward at centre (1.0 m from A)
- Force F acting upward at B (2.0 m from A)
- Reaction force R at pivot A acting upward (1)
(b) Force F by moments. [2 marks]
Taking moments about A: Clockwise moment = anticlockwise moment (1) 50 × 1.0 = F × 2.0 → F = 25 N (1)
(c) Reaction at pivot. [2 marks]
Vertical equilibrium: R + F = 50 (1) R = 50 - 25 = 25 N, acting upward (1)
Section C: Data Analysis and Experimental Skills (Questions 16–20)
16. Pendulum experiment.
(a) Quantities for straight-line graph. [2 marks]
y-axis: log T (or ln T) (1) x-axis: log L (or ln L) (1)
(b) Value of n. [3 marks]
T = kLⁿ → log T = n log L + log k Using two points: n = (log T₂ - log T₁)/(log L₂ - log L₁) (1) = (log 2.19 - log 1.27)/(log 1.20 - log 0.40) (1) = (0.3404 - 0.1038)/(0.0792 - (-0.3979)) = 0.2366/0.4771 = 0.496 ≈ 0.50 (1)
(c) Value of k with units. [2 marks]
Using L = 1.00 m, T = 2.00 s: k = T/Lⁿ = 2.00/(1.00)⁰·⁵⁰ = 2.00 (1) Units: s m⁻⁰·⁵⁰ (1)
Note: The theoretical value is T = 2π√(L/g), so n = 0.5 and k = 2π/√g ≈ 2.01 s m⁻⁰·⁵⁰.
17. Determination of g.
(a) Why h vs t² is straight line through origin. [2 marks]
h = ½gt² (since u = 0) (1) This is of the form y = mx with y = h, x = t², and gradient = ½g (1) Therefore, the graph is a straight line passing through the origin.
(b) Value of g from gradient. [2 marks]
Gradient = ½g = 4.7 m s⁻² (1) g = 2 × 4.7 = 9.4 m s⁻² (1)
(c) Sources of systematic error and reduction. [4 marks]
Award 2 marks for each valid error with reduction method (1 mark for error, 1 mark for reduction):
- Reaction time error in starting/stopping timer (1) — use an electronic timing system with automatic triggering (e.g., photogate) (1)
- Parallax error in measuring height h (1) — use a metre rule with a set square or measure at eye level (1)
- Air resistance affecting the fall (1) — use a dense, streamlined object or perform in a vacuum (1)
- Delay in trapdoor mechanism (1) — calibrate the trapdoor or use a contact sensor (1)
Accept any two valid suggestions.
18. Dynamics cart experiment.
(a) Rearrangement for straight-line graph. [2 marks]
a = mg/(M + m) 1/a = (M + m)/mg = M/(mg) + 1/g (1) 1/a = (M/g)(1/m) + 1/g (1)
(b) Gradient and y-intercept. [2 marks]
Gradient = M/g (1) y-intercept = 1/g (1)
(c) Calculate M and g. [3 marks]
y-intercept = 1/g = 0.10 → g = 1/0.10 = 10 m s⁻² (1) Gradient = M/g = 0.51 → M = 0.51 × 10 = 5.1 kg (1) Values: M = 5.1 kg, g = 10 m s⁻² (1)
Accept g = 10.0 m s⁻² with appropriate significant figures.
19. Oscillating mass on spring.
(a) Graphical method to verify SHM. [3 marks]
For SHM, a = -ω²x, so acceleration is proportional to displacement and directed towards equilibrium (1). Plot a graph of acceleration a (y-axis) against displacement x (x-axis) (1). If the graph is a straight line passing through the origin with negative gradient, the motion is SHM (1).
Alternative: Measure period T for different amplitudes. If T is constant (independent of amplitude), the motion is SHM (1). Plot x against t and verify sinusoidal form (1). Check that a ∝ -x from the data (1).
(b) Additional measurements for spring constant k. [3 marks]
Need to measure the mass m attached to the spring (1). For a mass-spring system: ω² = k/m, so k = mω² (1). ω can be determined from the period: ω = 2π/T (1).
(c) Decreasing amplitude and precaution. [3 marks]
The amplitude decreases due to energy loss from the system (1), primarily due to air resistance/damping (1). Precaution: Use a more streamlined mass or enclose the system to reduce air currents (1).
Accept: Use a stiffer spring to reduce amplitude of air displacement; lubricate any guide rods; perform in a partial vacuum.
20. Conservation of momentum experiment.
(a) Expected velocity after collision. [2 marks]
By conservation of momentum: m_A u_A + m_B u_B = (m_A + m_B)v (1) (0.30)(0.50) + 0 = (0.50)v → 0.15 = 0.50v → v = 0.30 m s⁻¹ (1)
(b) Precautions for accuracy and safety. [4 marks]
Award 1 mark for each valid precaution with explanation (up to 4 marks):
- Ensure the air track is level (1) — to eliminate the effect of gravity along the track, ensuring no external horizontal forces (1)
- Use light gates or motion sensors to measure velocities accurately (1) — reduces timing errors compared to manual stopwatch (1)
- Minimise friction by ensuring adequate air flow (1) — reduces external forces that would invalidate conservation of momentum (1)
- Ensure the gliders move along a straight line (1) — prevents oblique collisions that complicate analysis (1)
- Safety: Keep hands clear of moving gliders; secure the track to prevent tipping (1) — prevents injury from fast-moving masses (1)
Accept any four valid points with clear accuracy or safety justification.
(c) Percentage difference and reason. [2 marks]
Percentage difference = |theoretical - experimental|/theoretical × 100% = |0.30 - 0.28|/0.30 × 100% = 6.67% (1)
Reason: Small amount of friction on the air track (1) OR air resistance acting on the gliders (1) OR imperfect alignment causing some energy loss (1).
Accept any valid reason for the discrepancy.
END OF ANSWER KEY