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A Level H1 Physics Practice Paper 1

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TuitionGoWhere Practice Paper - Physics H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Physics H1 (8867) Level: A-Level Paper: Practice Paper 1 (Mechanics) Version: 1 of 5 Duration: 1 hour 30 minutes Total Marks: 60

Name: _________________________ Class: _________________________ Date: _________________________

Instructions to Candidates

This paper consists of two sections.
Answer all questions.
Write your answers in the spaces provided.
The number of marks is given in brackets [ ] at the end of each question or part question.
You may use a scientific calculator.
Take g = 9.81 m s⁻² unless otherwise stated.
Show all working clearly; marks are awarded for method as well as final answers.

Section A: Structured Questions (40 marks)

Answer all questions in this section.

1. A student measures the acceleration due to gravity, g, by timing a ball falling from rest through a measured vertical distance.

The student uses a metre rule to measure the distance fallen and a digital stopwatch to measure the time of fall. The following readings are obtained for a nominal drop height of 2.00 m:

Times recorded: 0.64 s, 0.62 s, 0.65 s, 0.63 s, 0.66 s

(a) Calculate the mean time of fall and state its uncertainty. [2]

(b) The distance fallen is measured as 2.00 ± 0.01 m. Using the mean time from (a), calculate the experimental value of g. [2]

(c) The accepted value of g is 9.81 m s⁻². Calculate the percentage difference between the experimental value and the accepted value. Suggest one reason for any discrepancy. [2]

[Total: 6 marks]

2. A car of mass 1200 kg accelerates uniformly from rest to 25.0 m s⁻¹ in 8.0 s along a straight, level road.

(a) Calculate the acceleration of the car. [1]

(b) Calculate the resultant force acting on the car during this acceleration. [1]

(c) The driving force provided by the engine is 5200 N. Calculate the total resistive force acting on the car. [2]

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

[Total: 6 marks]

3. A ball of mass 0.150 kg is thrown vertically upwards with an initial speed of 12.0 m s⁻¹ from a point 1.50 m above the ground. Air resistance may be neglected.

(a) Calculate the maximum height reached by the ball above the ground. [3]

(b) Calculate the speed of the ball just before it hits the ground. [2]

(c) Sketch a graph showing how the velocity of the ball varies with time from the moment it is thrown until it hits the ground. Label the axes with appropriate values. [3]

[Total: 8 marks]

4. A wooden block of mass 2.50 kg rests on a rough horizontal surface. The coefficient of static friction between the block and the surface is 0.45. The coefficient of kinetic friction is 0.35.

(a) Calculate the minimum horizontal force required to make the block just begin to move. [2]

(b) A horizontal force of 12.0 N is applied to the block. Determine whether the block moves, and if so, calculate its acceleration. [3]

(c) The block is now placed on a slope inclined at 20° to the horizontal. The coefficient of static friction remains 0.45. Determine, with calculation, whether the block will slide down the slope when released from rest. [3]

[Total: 8 marks]

5. A trolley A of mass 1.50 kg moves at 3.00 m s⁻¹ to the right on a frictionless track. It collides head-on with trolley B of mass 2.00 kg moving at 1.50 m s⁻¹ to the left. After the collision, trolley A moves at 0.80 m s⁻¹ to the left.

(a) State the principle of conservation of linear momentum. [2]

(b) Calculate the velocity of trolley B after the collision. [3]

(c) Determine, by calculation, whether the collision is elastic or inelastic. [3]

(d) Calculate the impulse experienced by trolley A during the collision. State the direction of this impulse. [2]

[Total: 10 marks]

6. A uniform plank AB of length 4.00 m and weight 300 N rests horizontally on two supports. Support P is located at end A. Support Q is located 1.00 m from end B. A painter of weight 700 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, in terms of x, for the reaction force at support Q. Take moments about a suitable point. [3]

(c) The maximum reaction force that support Q can provide without collapsing is 1200 N. Calculate the maximum distance x from end A that the painter can stand safely. [3]

[Total: 8 marks]

Section B: Free-Response Questions (20 marks)

Answer all questions in this section.

7. A student investigates the motion of a ball-bearing falling through oil in a tall measuring cylinder. The ball-bearing is released from rest just below the surface of the oil.

(a) Describe the forces acting on the ball-bearing during its descent. Explain how these forces change as the ball-bearing accelerates. [4]

(b) Explain what is meant by terminal velocity. State the condition for terminal velocity to be reached. [2]

(c) The ball-bearing has mass 0.025 kg and reaches a terminal velocity of 0.40 m s⁻¹. Calculate the viscous drag force acting on the ball-bearing at terminal velocity. [2]

(d) Sketch a graph showing how the velocity of the ball-bearing varies with time from the moment of release until terminal velocity is reached. Explain the shape of your graph. [4]

[Total: 12 marks]

8. A pendulum consists of a small metal sphere of mass 0.200 kg attached to a light inextensible string of length 1.20 m. The sphere is pulled aside so that the string makes an angle of 35° with the vertical and is then released from rest. Air resistance may be neglected.

(a) Calculate the vertical height through which the sphere rises from its lowest point to its release point. [2]

(b) Using the principle of conservation of energy, calculate the speed of the sphere as it passes through the lowest point of its swing. [2]

(c) Calculate the tension in the string when the sphere is at the lowest point of its swing. [2]

(d) Explain why the tension in the string is greater than the weight of the sphere at the lowest point. [2]

[Total: 8 marks]

END OF PAPER

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Answers

TuitionGoWhere Practice Paper - Physics H1 A-Level (Answers)

Paper: Practice Paper 1 (Mechanics) Version: 1 of 5 Total Marks: 60

Section A: Structured Questions

(a) Mean time = (0.64 + 0.62 + 0.65 + 0.63 + 0.66) / 5 = 3.20 / 5 = 0.64 s [1] Uncertainty = (max − min) / 2 = (0.66 − 0.62) / 2 = 0.02 s [1] Time = 0.64 ± 0.02 s

(b) s = ut + ½at²; u = 0, so s = ½gt² → g = 2s / t² [1] g = 2 × 2.00 / (0.64)² = 4.00 / 0.4096 = 9.77 m s⁻² [1]

(c) Percentage difference = |9.77 − 9.81| / 9.81 × 100% = 0.41% [1] Possible reason: Human reaction time in starting/stopping the stopwatch introduces random error / air resistance reduces acceleration slightly / metre rule reading uncertainty. [1]

[Total: 6 marks]

(a) a = (v − u) / t = (25.0 − 0) / 8.0 = 3.125 ≈ 3.13 m s⁻² [1]

(b) F = ma = 1200 × 3.125 = 3750 N [1]

(c) Resultant force = Driving force − Resistive force [1] 3750 = 5200 − F_resistive → F_resistive = 5200 − 3750 = 1450 N [1]

(d) s = ut + ½at² = 0 + ½ × 3.125 × (8.0)² = 100 m [2] (Alternative: s = (u + v)t/2 = (0 + 25.0) × 8.0 / 2 = 100 m)

[Total: 6 marks]

(a) At maximum height, v = 0. Using v² = u² + 2as: 0 = (12.0)² + 2(−9.81)s [1] s = 144 / (2 × 9.81) = 7.34 m [1] Maximum height above ground = 7.34 + 1.50 = 8.84 m [1]

(b) From maximum height to ground: u = 0, s = 8.84 m v² = u² + 2as = 0 + 2 × 9.81 × 8.84 = 173.4 [1] v = 13.2 m s⁻¹ [1]

(c) Graph should show:

Initial velocity = +12.0 m s⁻¹ (upward positive) [1]
Straight line with negative gradient (−9.81 m s⁻²) [1]
Velocity zero at t = 12.0/9.81 = 1.22 s (maximum height)
Final velocity = −13.2 m s⁻¹ at t ≈ 2.57 s [1]
Axes labelled: velocity/m s⁻¹ (y-axis), time/s (x-axis)

[Total: 8 marks]

(a) Maximum static friction = μ_s × N = μ_s × mg = 0.45 × 2.50 × 9.81 [1] = 11.04 N ≈ 11.0 N [1]

(b) Applied force 12.0 N > 11.0 N, so block moves. [1] Kinetic friction = μ_k × mg = 0.35 × 2.50 × 9.81 = 8.58 N [1] Resultant force = 12.0 − 8.58 = 3.42 N a = F/m = 3.42 / 2.50 = 1.37 m s⁻² [1]

(c) Component of weight down slope = mg sin 20° = 2.50 × 9.81 × sin 20° = 8.39 N [1] Maximum static friction = μ_s × mg cos 20° = 0.45 × 2.50 × 9.81 × cos 20° = 10.37 N [1] Since 8.39 N < 10.37 N, the block will not slide. [1]

[Total: 8 marks]

(a) The principle of conservation of linear momentum states that in a closed/isolated system (where no external forces act), the total linear momentum remains constant. [1] Momentum before collision = momentum after collision. [1]

(b) Taking right as positive: m_A u_A + m_B u_B = m_A v_A + m_B v_B [1] (1.50)(3.00) + (2.00)(−1.50) = (1.50)(−0.80) + 2.00 v_B [1] 4.50 − 3.00 = −1.20 + 2.00 v_B 1.50 + 1.20 = 2.00 v_B → v_B = 2.70 / 2.00 = 1.35 m s⁻¹ to the right [1]

(c) Initial KE = ½(1.50)(3.00)² + ½(2.00)(1.50)² = 6.75 + 2.25 = 9.00 J [1] Final KE = ½(1.50)(0.80)² + ½(2.00)(1.35)² = 0.48 + 1.8225 = 2.30 J [1] KE is not conserved (9.00 J → 2.30 J), therefore the collision is inelastic. [1]

(d) Impulse = change in momentum = m(v − u) = 1.50(−0.80 − 3.00) [1] = 1.50 × (−3.80) = −5.70 N s Direction: to the left (negative sign indicates opposite to initial direction). [1]

[Total: 10 marks]

(a) Diagram should show:

Weight of plank (300 N) acting downwards at centre (2.00 m from A) [1]
Weight of painter (700 N) acting downwards at distance x from A
Reaction at P (R_P) acting upwards at A
Reaction at Q (R_Q) acting upwards at 3.00 m from A All forces clearly labelled with directions. [1]

(b) Taking moments about P (end A): Clockwise moments = Anticlockwise moments [1] (300 × 2.00) + (700 × x) = R_Q × 3.00 [1] R_Q = (600 + 700x) / 3.00 = 200 + 233x (where x is in metres) [1]

(c) R_Q ≤ 1200 N 200 + 233x ≤ 1200 [1] 233x ≤ 1000 → x ≤ 4.29 m [1] But plank length is 4.00 m, so painter can stand anywhere on the plank safely (x ≤ 4.00 m). [1]

[Total: 8 marks]

Section B: Free-Response Questions

(a) Forces acting on the ball-bearing:

Weight (mg) acting downwards (constant) [1]
Upthrust/buoyant force (constant, usually negligible compared to weight for a metal ball in oil) [1]
Viscous drag force acting upwards, which increases with speed [1] Initially, weight > drag, so net downward force causes acceleration. As speed increases, drag increases, reducing net force and acceleration. [1]

(b) Terminal velocity is the constant maximum velocity reached by an object falling through a fluid when the net force acting on it is zero. [1] Condition: Weight = Upthrust + Viscous drag (or simply: downward forces = upward forces). [1]

(c) At terminal velocity, net force = 0, so viscous drag = weight (assuming upthrust negligible) [1] Drag force = mg = 0.025 × 9.81 = 0.245 N [1]

(d) Graph should show:

Axes: velocity (y-axis) vs time (x-axis) [1]
Curve starting at origin (v = 0 at t = 0) [1]
Initial steep gradient (acceleration ≈ g initially when drag is small) [1]
Gradient decreasing as velocity increases (drag increases, net force decreases)
Curve approaching horizontal asymptote at v = 0.40 m s⁻¹ (terminal velocity) [1] Explanation: Initially, drag is negligible so acceleration is large. As speed increases, drag increases, reducing net force and acceleration. Eventually drag equals weight, net force is zero, and velocity becomes constant.

[Total: 12 marks]

(a) Vertical height risen: h = L − L cos θ = L(1 − cos θ) [1] h = 1.20(1 − cos 35°) = 1.20(1 − 0.8192) = 1.20 × 0.1808 = 0.217 m [1]

(b) By conservation of energy: Loss in GPE = Gain in KE mgh = ½mv² [1] v = √(2gh) = √(2 × 9.81 × 0.217) = √4.258 = 2.06 m s⁻¹ [1]

(c) At lowest point: T − mg = mv²/r (centripetal force provided by net force towards centre) [1] T = mg + mv²/r = 0.200 × 9.81 + 0.200 × (2.06)² / 1.20 = 1.962 + 0.200 × 4.244 / 1.20 = 1.962 + 0.707 = 2.67 N [1]

(d) At the lowest point, the sphere is moving in a circular arc, so there must be a resultant centripetal force directed towards the centre of the circle. [1] The tension must provide both the centripetal force (mv²/r) and balance the weight (mg). Therefore, T = mg + mv²/r, which is greater than mg alone. [1]

[Total: 8 marks]

END OF ANSWER KEY

Marking Notes:

Award marks for correct method even if final answer has arithmetic error (error carried forward where appropriate).
Deduct 1 mark for missing or incorrect units only once per question unless otherwise specified.
Accept alternative correct methods and equivalent numerical answers within reasonable rounding.
For graph questions, award marks for correct shape, key features, and labelled axes.