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A Level H1 Physics Mechanics Quiz

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Questions

A-Level Physics H1 Quiz - Mechanics

Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50

Duration: 1 hour 15 minutes Total Marks: 50

Instructions:

Answer ALL questions in the spaces provided.
Show all working clearly for calculation questions. Marks are awarded for method as well as final answers.
Use g = 9.81 m s⁻² unless otherwise stated.
Where appropriate, state the direction of vector quantities.
Non-programmable calculators may be used.

Section A: Short Answer and Conceptual Questions (10 marks)

Answer all questions in this section.

1. State the principle of conservation of linear momentum.

[2 marks]

2. Distinguish between systematic errors and random errors in experimental measurements. Give one example of each.

[2 marks]

Systematic error: ____________________________________________________________

Random error: _______________________________________________________________

3. A student plots a graph of velocity against time for an object moving with uniform acceleration. State what the gradient and the area under the graph represent.

[2 marks]

Gradient: ___________________________________________________________________

Area under graph: ___________________________________________________________

4. Define the term impulse and state its SI unit.

[2 marks]

Definition: __________________________________________________________________

SI unit: _____________________________________________________________________

5. Explain why a passenger in a car lurches forward when the car brakes suddenly, using Newton's laws of motion.

[2 marks]

Section B: Structured Questions (24 marks)

Answer all questions in this section.

6. A ball of mass 0.150 kg is dropped from rest from a height of 2.50 m above the ground. Air resistance is negligible.

(a) Calculate the speed of the ball just before it hits the ground.

[2 marks]

(b) The ball rebounds vertically with a speed of 4.20 m s⁻¹. Calculate the impulse exerted on the ball by the ground during the impact. State the direction of the impulse.

[3 marks]

(c) The impact lasts for 0.085 s. Calculate the average force exerted on the ball by the ground during the impact.

[2 marks]

7. A crate of mass 25.0 kg is pulled along a rough horizontal floor by a rope inclined at 30.0° above the horizontal. The tension in the rope is 120 N. The coefficient of kinetic friction between the crate and the floor is 0.250.

(a) Draw a free-body diagram showing all the forces acting on the crate. Label each force clearly.

[3 marks]

(Draw your diagram in the space below)

(b) Calculate the normal reaction force exerted by the floor on the crate.

[3 marks]

(c) Calculate the acceleration of the crate.

[3 marks]

8. Two trolleys, A and B, are on a frictionless horizontal track. Trolley A of mass 2.00 kg moves to the right at 3.50 m s⁻¹. Trolley B of mass 1.50 kg moves to the left at 2.00 m s⁻¹. They collide and stick together.

(a) Calculate the velocity (magnitude and direction) of the combined trolleys immediately after the collision.

[3 marks]

(b) Determine whether the collision is elastic or inelastic. Support your answer with calculations.

[3 marks]

(c) Calculate the impulse experienced by trolley A during the collision.

[2 marks]

9. A student investigates the motion of a trolley rolling down an inclined plane. The trolley is released from rest at the top of the incline. The student measures the distance s travelled by the trolley at various times t. The results are shown in the table below.

t / s	s / m
0.50	0.12
1.00	0.49
1.50	1.10
2.00	1.96
2.50	3.06

(a) Explain why a graph of s against t² should be a straight line passing through the origin for this motion.

[2 marks]

(b) Complete the table below by calculating values of t².

t / s	t² / s²	s / m
0.50		0.12
1.00		0.49
1.50		1.10
2.00		1.96
2.50		3.06

[1 mark]

(c) Plot a graph of s (y-axis) against t² (x-axis) on the grid below. Draw the best-fit straight line.

[3 marks]

(Use the grid space below for your graph)

(d) Use your graph to determine the acceleration of the trolley. Show your working clearly.

[3 marks]

10. A construction worker uses a pulley system to lift a load of bricks of mass 80.0 kg vertically upward at a constant speed of 0.500 m s⁻¹. The pulley system is 75.0% efficient.

(a) Calculate the useful power output of the pulley system.

[2 marks]

(b) Calculate the electrical power input required to operate the pulley system.

[2 marks]

(c) The load is lifted through a height of 12.0 m. Calculate the total energy supplied to the pulley system and the energy wasted during this lift.

[3 marks]

Section C: Data Analysis and Application Questions (16 marks)

Answer all questions in this section.

11. A car of mass 1200 kg accelerates uniformly from rest to 25.0 m s⁻¹ in 8.00 s on a straight, level road. The total resistive force acting on the car is constant at 600 N.

(a) Calculate the acceleration of the car.

[2 marks]

(b) Calculate the driving force provided by the engine.

[3 marks]

(c) Calculate the work done by the driving force during the acceleration.

[3 marks]

12. A stone is projected horizontally from the top of a vertical cliff of height 45.0 m above the sea. The initial speed of the stone is 20.0 m s⁻¹. Air resistance is negligible.

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

[2 marks]

(b) Calculate the horizontal distance from the base of the cliff to the point where the stone enters the sea.

[2 marks]

(c) Calculate the magnitude and direction of the velocity of the stone just before it enters the sea.

[4 marks]

13. A spring of negligible mass and spring constant k = 500 N m⁻¹ is suspended vertically from a fixed support. A mass of 2.00 kg is attached to the lower end of the spring and gently released.

(a) Calculate the extension of the spring when the mass is in equilibrium.

[2 marks]

(b) The mass is pulled down a further 0.050 m from its equilibrium position and released. Calculate the maximum speed of the mass during its subsequent motion.

[3 marks]

(c) State one assumption made in your calculation for part (b).

[1 mark]

14. A satellite of mass 500 kg orbits the Earth in a circular orbit of radius 7.00 × 10⁶ m. The mass of the Earth is 5.97 × 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) Show that the orbital speed of the satellite is approximately 7.54 × 10³ m s⁻¹.

[3 marks]

(c) Calculate the period of the orbit in hours.

[2 marks]

15. A uniform rod AB of length 2.00 m and weight 150 N is hinged at A to a vertical wall. The rod is held horizontally by a cable attached at B, making an angle of 40.0° with the rod. A load of 250 N is hung from the rod at a point 1.50 m from A.

(a) Draw a diagram showing all the forces acting on the rod.

[2 marks]

(Draw your diagram in the space below)

(b) By taking moments about A, calculate the tension in the cable.

[3 marks]

(c) Calculate the magnitude and direction of the reaction force at the hinge A.

[3 marks]

Section D: Extended and Contextual Questions (10 marks)

Answer all questions in this section.

16. A student performs an experiment to determine the acceleration of free fall, g, using a simple pendulum. The student varies the length l of the pendulum and measures the corresponding period T. The relationship is given by T = 2π√(l/g).

(a) Explain how the student should use a graph to determine a value for g.

[2 marks]

(b) The student obtains a value of g = 9.62 m s⁻². The accepted value is 9.81 m s⁻². Calculate the percentage error in the student's result.

[1 mark]

(c) Suggest one possible source of systematic error in this experiment and state how it could be reduced.

[2 marks]

17. A roller coaster car of mass 500 kg starts from rest at point A, 40.0 m above the ground. It travels down a frictionless track to point B at ground level, then rises to point C, 15.0 m above the ground. Air resistance and friction are negligible.

(a) Calculate the speed of the car at point B.

[2 marks]

(b) Calculate the speed of the car at point C.

[2 marks]

(c) In reality, the speed at point C is measured to be 18.0 m s⁻¹. Calculate the work done against resistive forces between A and C.

[3 marks]

18. A ball of mass 0.200 kg is attached to a string and whirled in a horizontal circle of radius 0.800 m on a smooth table. The string passes through a hole in the centre of the table and is attached to a hanging mass of 0.500 kg, which remains stationary.

(a) State the force that provides the centripetal force for the ball.

[1 mark]

(b) Calculate the tension in the string.

[1 mark]

(c) Calculate the speed of the ball.

[2 marks]

(d) The hanging mass is increased to 0.800 kg. Assuming the radius remains constant, calculate the new speed of the ball.

[2 marks]

19. A car of mass 900 kg travels around a banked curve of radius 50.0 m. The road is banked at an angle of 20.0° to the horizontal. There is no friction between the tyres and the road.

(a) Draw a free-body diagram showing the forces acting on the car.

[2 marks]

(Draw your diagram in the space below)

(b) Resolve the forces and derive an expression for the speed at which the car can travel around the curve without slipping.

[3 marks]

(c) Calculate this speed.

[1 mark]

20. A student investigates the relationship between the force applied to a spring and its extension. The student obtains the following data:

Force / N	Extension / mm
0.0	0.0
1.0	15.0
2.0	30.0
3.0	45.5
4.0	61.0
5.0	77.0

(a) Plot a graph of Force (y-axis) against Extension (x-axis) on the grid below. Draw the best-fit line.

[3 marks]

(Use the grid space below for your graph)

(b) Use your graph to determine the spring constant k of the spring.

[2 marks]

(c) The student notices that the graph does not pass exactly through all points. Suggest a reason for this and state how the experiment could be improved.

[2 marks]

END OF QUIZ

Check your work carefully. Ensure all answers include appropriate units and directions where required.

Answers

A-Level Physics H1 Quiz - Mechanics: Answer Key and Marking Scheme

Total Marks: 50

Section A: Short Answer and Conceptual Questions (10 marks)

1. State the principle of conservation of linear momentum. [2 marks]

Answer: The total momentum of a closed/isolated system remains constant [B1] provided no external resultant/net force acts on the system [B1].

Accept: "In a closed system, the total momentum before an interaction equals the total momentum after the interaction, in the absence of external forces." Accept: "The total linear momentum of an isolated system is conserved."

2. Distinguish between systematic errors and random errors in experimental measurements. Give one example of each. [2 marks]

Answer: Systematic error: An error that causes all readings to be consistently too high or too low by the same amount; it affects accuracy but not precision. [B1 for correct description OR valid example] Example: Zero error on a measuring instrument; a stopwatch that runs consistently slow; parallax error when reading a scale from a consistent angle.

Random error: An error that causes readings to be scattered above and below the true value unpredictably; it affects precision but not accuracy (of the mean). [B1 for correct description OR valid example] Example: Human reaction time variation when using a stopwatch; fluctuations in environmental conditions; reading a scale from slightly different angles each time.

Award [1] for each correct distinction with example. Accept other valid examples.

3. A student plots a graph of velocity against time for an object moving with uniform acceleration. State what the gradient and the area under the graph represent. [2 marks]

Answer: Gradient: Acceleration of the object. [B1] Area under graph: Displacement (or distance travelled in a given direction) of the object. [B1]

4. Define the term impulse and state its SI unit. [2 marks]

Answer: Definition: Impulse is the product of the average force acting on an object and the time for which it acts; equivalently, it is equal to the change in momentum of the object. [B1] SI unit: Newton second (N s) or kilogram metre per second (kg m s⁻¹). [B1]

5. Explain why a passenger in a car lurches forward when the car brakes suddenly, using Newton's laws of motion. [2 marks]

Answer: According to Newton's First Law, an object continues in its state of rest or uniform motion in a straight line unless acted upon by a resultant external force. [B1] When the car brakes, the car decelerates due to the braking force, but the passenger's body tends to continue moving forward at the original speed because no sufficient force acts directly on the passenger to decelerate them at the same rate. The passenger lurches forward relative to the car until a force (from the seatbelt or dashboard) acts to decelerate them. [B1]

Accept explanation using inertia concept linked to Newton's First Law.

Section B: Structured Questions (24 marks)

6. A ball of mass 0.150 kg is dropped from rest from a height of 2.50 m above the ground. Air resistance is negligible.

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

Answer: Using v² = u² + 2as, with u = 0, a = g = 9.81 m s⁻², s = 2.50 m: [M1] v² = 0 + 2 × 9.81 × 2.50 = 49.05 v = √49.05 = 7.00 m s⁻¹ (downward) [A1]

Alternative: Using energy conservation: mgh = ½mv² → v = √(2gh) = √(2 × 9.81 × 2.50) = 7.00 m s⁻¹. Award [M1] for correct substitution, [A1] for correct answer with direction.

(b) The ball rebounds vertically with a speed of 4.20 m s⁻¹. Calculate the impulse exerted on the ball by the ground during the impact. State the direction of the impulse. [3 marks]

Answer: Taking upward as positive: Velocity before impact, u = −7.00 m s⁻¹ [M1 for sign convention] Velocity after impact, v = +4.20 m s⁻¹ Impulse = change in momentum = m(v − u) = 0.150 × (4.20 − (−7.00)) [M1] = 0.150 × 11.20 = 1.68 N s [A1] Direction: Upward [B1 must be stated]

If downward taken as positive: u = +7.00, v = −4.20, impulse = 0.150 × (−4.20 − 7.00) = −1.68 N s, direction upward.

(c) The impact lasts for 0.085 s. Calculate the average force exerted on the ball by the ground during the impact. [2 marks]

Answer: Impulse = F_avg × Δt [M1] F_avg = Impulse / Δt = 1.68 / 0.085 = 19.8 N (upward) [A1]

Accept 19.8 N with direction stated.

(a) Draw a free-body diagram showing all the forces acting on the crate. Label each force clearly. [3 marks]

Answer: Forces to be shown:

Weight (W or mg), acting downward from the centre of the crate [B1]
Normal reaction force (N or R), acting upward from the floor on the crate [B1]
Tension (T), acting at 30.0° above the horizontal in the direction of the rope [B1]
Friction (f or F_f), acting horizontally opposite to the direction of motion

Award [1] for each correctly drawn and labelled force, up to [3]. Deduct [1] if any force is missing or incorrectly directed. Arrows must indicate correct directions.

(b) Calculate the normal reaction force exerted by the floor on the crate. [3 marks]

Answer: Resolving forces vertically (upward positive): N + T sin 30.0° − mg = 0 (since no vertical acceleration) [M1] N = mg − T sin 30.0° [M1] N = (25.0 × 9.81) − (120 × sin 30.0°) N = 245.25 − (120 × 0.500) N = 245.25 − 60.0 = 185 N [A1]

Accept 185 N or 185.25 N.

(c) Calculate the acceleration of the crate. [3 marks]

Answer: Frictional force, f = μN = 0.250 × 185.25 = 46.3 N [M1] Resultant horizontal force = T cos 30.0° − f [M1] = (120 × cos 30.0°) − 46.3 = 103.9 − 46.3 = 57.6 N Using F = ma: a = F/m = 57.6 / 25.0 = 2.30 m s⁻² [A1]

Accept answers in the range 2.28–2.32 m s⁻² depending on rounding. Award [M1] for correct friction calculation, [M1] for correct net force, [A1] for correct acceleration.

(a) Calculate the velocity (magnitude and direction) of the combined trolleys immediately after the collision. [3 marks]

Answer: Taking right as positive: Total momentum before = m_A u_A + m_B u_B [M1] = (2.00 × 3.50) + (1.50 × (−2.00)) = 7.00 + (−3.00) = 4.00 kg m s⁻¹ [M1] Total mass after = 2.00 + 1.50 = 3.50 kg Velocity after, v = total momentum / total mass = 4.00 / 3.50 = 1.14 m s⁻¹ [A1] Direction: To the right [B1 must be stated]

If left taken as positive: u_A = −3.50, u_B = +2.00, momentum = −7.00 + 3.00 = −4.00 kg m s⁻¹, v = −1.14 m s⁻¹, direction right.

(b) Determine whether the collision is elastic or inelastic. Support your answer with calculations. [3 marks]

Answer: Initial KE = ½ m_A u_A² + ½ m_B u_B² [M1] = ½ × 2.00 × (3.50)² + ½ × 1.50 × (2.00)² = 12.25 + 3.00 = 15.25 J [M1] Final KE = ½ × (m_A + m_B) × v² = ½ × 3.50 × (1.14)² = 1.75 × 1.30 = 2.27 J (or 2.28 J) Since final KE (2.27 J) < initial KE (15.25 J), kinetic energy is not conserved. [A1] Therefore, the collision is inelastic.

Accept any reasonable rounding. Award [M1] for correct KE formula, [M1] for correct substitution, [A1] for correct conclusion with justification.

(c) Calculate the impulse experienced by trolley A during the collision. [2 marks]

Answer: Impulse on A = change in momentum of A = m_A(v − u_A) [M1] = 2.00 × (1.14 − 3.50) = 2.00 × (−2.36) = −4.72 N s [A1] Direction: To the left (or negative direction).

Accept 4.72 N s to the left. Magnitude alone without direction: deduct [1].

t / s	s / m
0.50	0.12
1.00	0.49
1.50	1.10
2.00	1.96
2.50	3.06

(a) Explain why a graph of s against t² should be a straight line passing through the origin for this motion. [2 marks]

Answer: For motion from rest with uniform acceleration a, s = ut + ½at² = ½at² (since u = 0). [B1] This is of the form y = mx, where y = s, x = t², and gradient m = ½a. Therefore, a graph of s against t² should be a straight line through the origin. [B1]

(b) Complete the table below by calculating values of t². [1 mark]

Answer:

t / s	t² / s²	s / m
0.50	0.25	0.12
1.00	1.00	0.49
1.50	2.25	1.10
2.00	4.00	1.96
2.50	6.25	3.06

[A1 for all correct]

(c) Plot a graph of s (y-axis) against t² (x-axis) on the grid below. Draw the best-fit straight line. [3 marks]

Answer:

Correct axes with labels and units [B1]
Appropriate scales and plotting of points [B1]
Best-fit straight line drawn [B1]

Points should be plotted accurately. Line should pass through origin or very close to it.

(d) Use your graph to determine the acceleration of the trolley. Show your working clearly. [3 marks]

Answer: Gradient = Δs / Δ(t²) [M1] Using points from best-fit line, e.g., (0, 0) and (6.00, 2.94): Gradient = 2.94 / 6.00 = 0.490 m s⁻² [M1] Since gradient = ½a, a = 2 × gradient = 2 × 0.490 = 0.980 m s⁻² [A1]

Accept answers in the range 0.96–1.00 m s⁻² depending on graph. Award [M1] for correct gradient calculation, [M1] for relating gradient to acceleration, [A1] for correct answer.

10. A construction worker uses a pulley system to lift a load of bricks of mass 80.0 kg vertically upward at a constant speed of 0.500 m s⁻¹. The pulley system is 75.0% efficient.

(a) Calculate the useful power output of the pulley system. [2 marks]

Answer: Weight of load = mg = 80.0 × 9.81 = 784.8 N [M1] Useful power output = Fv = 784.8 × 0.500 = 392 W (or 392.4 W) [A1]

Accept 392 W or 392.4 W.

(b) Calculate the electrical power input required to operate the pulley system. [2 marks]

Answer: Efficiency = (Useful power output / Power input) × 100% [M1] Power input = Useful power output / Efficiency = 392.4 / 0.750 = 523 W [A1]

Accept 523 W or 523.2 W.

(c) The load is lifted through a height of 12.0 m. Calculate the total energy supplied to the pulley system and the energy wasted during this lift. [3 marks]

Answer: Time taken = height / speed = 12.0 / 0.500 = 24.0 s [M1] Total energy supplied = Power input × time = 523.2 × 24.0 = 1.26 × 10⁴ J (or 12557 J) [M1] Useful energy output = mgh = 80.0 × 9.81 × 12.0 = 9418 J Energy wasted = Total energy supplied − Useful energy output = 12557 − 9418 = 3.14 × 10³ J (or 3139 J) [A1]

Accept answers with appropriate rounding. Award [M1] for time, [M1] for total energy, [A1] for energy wasted.

Section C: Data Analysis and Application Questions (16 marks)

11. A car of mass 1200 kg accelerates uniformly from rest to 25.0 m s⁻¹ in 8.00 s on a straight, level road. The total resistive force acting on the car is constant at 600 N.

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

Answer: a = (v − u) / t = (25.0 − 0) / 8.00 [M1] a = 3.125 m s⁻² ≈ 3.13 m s⁻² [A1]

(b) Calculate the driving force provided by the engine. [3 marks]

Answer: Resultant force = ma = 1200 × 3.125 = 3750 N [M1] Driving force − Resistive force = Resultant force [M1] Driving force = Resultant force + Resistive force = 3750 + 600 = 4350 N [A1]

Accept 4350 N.

(c) Calculate the work done by the driving force during the acceleration. [3 marks]

Answer: Distance travelled: s = ut + ½at² = 0 + ½ × 3.125 × (8.00)² = 100 m [M1] Work done = Driving force × distance = 4350 × 100 [M1] = 4.35 × 10⁵ J [A1]

Alternative using energy: Work done = gain in KE + work against resistance = ½ × 1200 × (25.0)² + 600 × 100 = 375000 + 60000 = 435000 J.

12. A stone is projected horizontally from the top of a vertical cliff of height 45.0 m above the sea. The initial speed of the stone is 20.0 m s⁻¹. Air resistance is negligible.

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

Answer: Vertical motion: s_y = u_y t + ½gt², with u_y = 0, s_y = 45.0 m [M1] 45.0 = ½ × 9.81 × t² t = √(2 × 45.0 / 9.81) = √9.17 = 3.03 s [A1]

Accept 3.03 s.

(b) Calculate the horizontal distance from the base of the cliff to the point where the stone enters the sea. [2 marks]

Answer: Horizontal distance = u_x × t = 20.0 × 3.03 [M1] = 60.6 m [A1]

Accept 60.6 m.

(c) Calculate the magnitude and direction of the velocity of the stone just before it enters the sea. [4 marks]

Answer: Horizontal component: v_x = 20.0 m s⁻¹ [B1] Vertical component: v_y = u_y + gt = 0 + 9.81 × 3.03 = 29.7 m s⁻¹ [M1] Magnitude: v = √(v_x² + v_y²) = √(20.0² + 29.7²) = √(400 + 882.1) = √1282.1 = 35.8 m s⁻¹ [A1] Direction: θ = tan⁻¹(v_y / v_x) = tan⁻¹(29.7 / 20.0) = 56.0° below the horizontal [A1]

Accept 35.8 m s⁻¹ at 56.0° below horizontal.

(a) Calculate the extension of the spring when the mass is in equilibrium. [2 marks]

Answer: At equilibrium: mg = kx [M1] x = mg / k = (2.00 × 9.81) / 500 = 0.03924 m ≈ 3.92 × 10⁻² m (or 3.92 cm) [A1]

(b) The mass is pulled down a further 0.050 m from its equilibrium position and released. Calculate the maximum speed of the mass during its subsequent motion. [3 marks]

Answer: Amplitude A = 0.050 m [B1] Maximum speed v_max = ωA, where ω = √(k/m) [M1] ω = √(500 / 2.00) = √250 = 15.81 rad s⁻¹ v_max = 15.81 × 0.050 = 0.791 m s⁻¹ [A1]

Alternative using energy: ½kA² = ½mv_max² → v_max = A√(k/m).

(c) State one assumption made in your calculation for part (b). [1 mark]

Answer:

The spring obeys Hooke's law throughout the motion.
Air resistance is negligible.
The mass of the spring is negligible.
The motion is simple harmonic.

Accept any one valid assumption. [B1]

(a) Calculate the gravitational force acting on the satellite. [2 marks]

Answer: F = GMm / r² [M1] = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 500) / (7.00 × 10⁶)² = (1.991 × 10¹⁷) / (4.90 × 10¹³) = 4.06 × 10³ N [A1]

Accept 4.06 × 10³ N or 4060 N.

(b) Show that the orbital speed of the satellite is approximately 7.54 × 10³ m s⁻¹. [3 marks]

Answer: Gravitational force provides centripetal force: GMm/r² = mv²/r [M1] v² = GM/r [M1] v = √(GM/r) = √((6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / 7.00 × 10⁶) = √(3.98 × 10¹⁴ / 7.00 × 10⁶) = √(5.69 × 10⁷) = 7.54 × 10³ m s⁻¹ [A1]

Must show clear substitution and calculation.

(c) Calculate the period of the orbit in hours. [2 marks]

Answer: T = 2πr / v [M1] = (2π × 7.00 × 10⁶) / (7.54 × 10³) = (4.40 × 10⁷) / (7.54 × 10³) = 5836 s T = 5836 / 3600 = 1.62 hours [A1]

Accept 1.62 hours.

(a) Draw a diagram showing all the forces acting on the rod. [2 marks]

Answer: Forces to be shown:

Weight of rod (150 N), acting downward at the centre (1.00 m from A) [B1]
Load (250 N), acting downward at 1.50 m from A [B1]
Tension in cable (T), acting at B at 40.0° above the horizontal (towards wall)
Reaction force at hinge A (components H and V, or a single force at an angle)

Award [1] for weight and load correctly placed, [1] for tension and hinge reaction. Deduct [1] if any force is missing or incorrectly directed.

(b) By taking moments about A, calculate the tension in the cable. [3 marks]

Answer: Taking clockwise moments as positive: Sum of clockwise moments = Sum of anticlockwise moments about A [M1] (150 × 1.00) + (250 × 1.50) = T sin 40.0° × 2.00 [M1] 150 + 375 = T × 0.6428 × 2.00 525 = 1.2856 T T = 525 / 1.2856 = 408 N [A1]

Accept 408 N.

(c) Calculate the magnitude and direction of the reaction force at the hinge A. [3 marks]

Answer: Resolving horizontally: H = T cos 40.0° = 408 × 0.7660 = 312.5 N [M1] Resolving vertically: V + T sin 40.0° = 150 + 250 V + 408 × 0.6428 = 400 V + 262.3 = 400 → V = 137.7 N [M1] Magnitude: R = √(H² + V²) = √(312.5² + 137.7²) = √(97656 + 18961) = √116617 = 341 N [A1] Direction: θ = tan⁻¹(V/H) = tan⁻¹(137.7 / 312.5) = 23.8° above the horizontal (away from wall) [A1]

Accept 341 N at 23.8° above horizontal.

Section D: Extended and Contextual Questions (10 marks)

(a) Explain how the student should use a graph to determine a value for g. [2 marks]

Answer: Square both sides: T² = 4π²l/g [B1] Plot a graph of T² (y-axis) against l (x-axis). The graph should be a straight line through the origin with gradient = 4π²/g. [B1] Then g = 4π² / gradient.

Accept alternative: Plot T against √l, gradient = 2π/√g.

(b) The student obtains a value of g = 9.62 m s⁻². The accepted value is 9.81 m s⁻². Calculate the percentage error in the student's result. [1 mark]

Answer: Percentage error = |(9.62 − 9.81) / 9.81| × 100% = (0.19 / 9.81) × 100% = 1.94% [B1]

Accept 1.9% or 1.94%.

(c) Suggest one possible source of systematic error in this experiment and state how it could be reduced. [2 marks]

Answer: Source: Incorrect measurement of length (e.g., measuring from the top of the clamp instead of the point of suspension, or not measuring to the centre of the bob). [B1] Reduction: Use a metre rule with a fiducial marker to measure the length accurately from the point of suspension to the centre of the bob; or measure the diameter of the bob and add half to the string length. [B1]

Accept other valid systematic errors and improvements, e.g., timing error due to delayed reaction, reduced by timing multiple oscillations.

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

Answer: Using conservation of energy: mgh_A = ½mv_B² [M1] v_B = √(2gh_A) = √(2 × 9.81 × 40.0) = √784.8 = 28.0 m s⁻¹ [A1]

Accept 28.0 m s⁻¹.

(b) Calculate the speed of the car at point C. [2 marks]

Answer: Using conservation of energy between A and C: mgh_A = mgh_C + ½mv_C² [M1] v_C = √(2g(h_A − h_C)) = √(2 × 9.81 × (40.0 − 15.0)) = √(2 × 9.81 × 25.0) = √490.5 = 22.1 m s⁻¹ [A1]

Accept 22.1 m s⁻¹.

(c) In reality, the speed at point C is measured to be 18.0 m s⁻¹. Calculate the work done against resistive forces between A and C. [3 marks]

Answer: Energy at A = mgh_A = 500 × 9.81 × 40.0 = 196200 J [M1] Energy at C = mgh_C + ½mv_C² = (500 × 9.81 × 15.0) + (½ × 500 × 18.0²) = 73575 + 81000 = 154575 J [M1] Work done against resistive forces = Energy lost = 196200 − 154575 = 4.16 × 10⁴ J (or 41625 J) [A1]

Accept 4.16 × 10⁴ J.

(a) State the force that provides the centripetal force for the ball. [1 mark]

Answer: The tension in the string. [B1]

(b) Calculate the tension in the string. [1 mark]

Answer: Tension = weight of hanging mass = 0.500 × 9.81 = 4.905 N ≈ 4.91 N [B1]

(c) Calculate the speed of the ball. [2 marks]

Answer: Centripetal force = mv²/r = T [M1] v = √(Tr/m) = √(4.905 × 0.800 / 0.200) = √(19.62) = 4.43 m s⁻¹ [A1]

Accept 4.43 m s⁻¹.

(d) The hanging mass is increased to 0.800 kg. Assuming the radius remains constant, calculate the new speed of the ball. [2 marks]

Answer: New tension = 0.800 × 9.81 = 7.848 N [M1] v = √(Tr/m) = √(7.848 × 0.800 / 0.200) = √(31.392) = 5.60 m s⁻¹ [A1]

Accept 5.60 m s⁻¹.

19. A car of mass 900 kg travels around a banked curve of radius 50.0 m. The road is banked at an angle of 20.0° to the horizontal. There is no friction between the tyres and the road.

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

Answer: Forces to be shown:

Weight (mg), acting vertically downward [B1]
Normal reaction force (N), acting perpendicular to the road surface [B1]

Both forces must be correctly drawn and labelled. No friction force should be shown.

(b) Resolve the forces and derive an expression for the speed at which the car can travel around the curve without slipping. [3 marks]

Answer: Resolving vertically: N cos θ = mg [M1] Resolving horizontally (centripetal): N sin θ = mv²/r [M1] Dividing: tan θ = v²/(rg) [M1] v = √(rg tan θ) [A1 for final expression]

(c) Calculate this speed. [1 mark]

Answer: v = √(50.0 × 9.81 × tan 20.0°) = √(50.0 × 9.81 × 0.3640) = √178.5 = 13.4 m s⁻¹ [B1]

Accept 13.4 m s⁻¹.

20. A student investigates the relationship between the force applied to a spring and its extension. The student obtains the following data:

Force / N	Extension / mm
0.0	0.0
1.0	15.0
2.0	30.0
3.0	45.5
4.0	61.0
5.0	77.0

(a) Plot a graph of Force (y-axis) against Extension (x-axis) on the grid below. Draw the best-fit line. [3 marks]

Answer:

Correct axes with labels and units (Force / N, Extension / mm or m) [B1]
Appropriate scales and accurate plotting of points [B1]
Best-fit straight line drawn (should pass through or near origin) [B1]

Note: Extension may be converted to metres; if so, graph should reflect this.

(b) Use your graph to determine the spring constant k of the spring. [2 marks]

Answer: Spring constant k = gradient of Force vs Extension graph (in N m⁻¹ if extension in m) [M1] Using points from best-fit line, e.g., (0, 0) and (0.075 m, 4.9 N): k = 4.9 / 0.075 = 65.3 N m⁻¹ [A1]

Accept values in the range 64–66 N m⁻¹. If extension is plotted in mm, gradient must be converted: k = ΔF/Δx in N mm⁻¹, then multiplied by 1000.

(c) The student notices that the graph does not pass exactly through all points. Suggest a reason for this and state how the experiment could be improved. [2 marks]

Answer: Reason: Random errors in measuring extension (e.g., parallax error when reading ruler, difficulty in determining exact equilibrium position). [B1] Improvement: Use a pointer and scale with a fiducial marker to reduce parallax; take repeat readings and average; ensure spring is not overloaded beyond its elastic limit. [B1]

Accept any valid reason and corresponding improvement.

END OF ANSWER KEY