O Level Physics Mechanics Quiz

O-Level Physics Quiz - Mechanics — Answer Key and Marking Scheme

Total Marks: 50

Section A: Short Answer (10 marks)

1. State the principle of moments.

• Answer: For a body in rotational equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments about the same pivot. [Accept: Σ clockwise moments = Σ anticlockwise moments]
• Marks: 2 marks (1 for "sum of clockwise moments", 1 for "equals sum of anticlockwise moments" or equivalent clear statement)

2. Define acceleration and state its SI unit.

• Answer: Acceleration is the rate of change of velocity. SI unit: m/s² (metre per second squared).
• Marks: 2 marks (1 for definition, 1 for unit)

3. A force of 15 N acts on a body of mass 3.0 kg. Calculate the acceleration produced.

• Answer: a = F/m = 15 / 3.0 = 5.0 m/s²
• Marks: 2 marks (1 for correct formula, 1 for correct answer with unit)

4. State the principle of conservation of energy.

• Answer: Energy cannot be created or destroyed; it can only be transformed from one form to another. The total energy of an isolated system remains constant.
• Marks: 2 marks (1 for "cannot be created or destroyed", 1 for "transformed/transferred" or "total energy constant")

5. A student pushes a box with a force of 50 N over a distance of 4.0 m. Calculate the work done by the student.

• Answer: W = F × d = 50 × 4.0 = 200 J
• Marks: 2 marks (1 for formula, 1 for correct answer with unit)

Section B: Structured Questions (20 marks)

6. Car acceleration problem.

(a) Calculate the acceleration of the car.

• Answer: a = (v - u) / t = (20 - 0) / 8.0 = 2.5 m/s²
• Marks: 2 marks (1 for formula/substitution, 1 for correct answer with unit)

(b) Calculate the resultant force acting on the car.

• Answer: F = ma = 1200 × 2.5 = 3000 N (or 3.0 × 10³ N)
• Marks: 2 marks (1 for formula, 1 for correct answer with unit; allow ecf from part (a))

7. Explain, in terms of forces, why the car maintains this constant speed.

• Answer: When the car travels at constant speed, the resultant force is zero. The driving force is balanced by the frictional forces (air resistance and road friction). According to Newton's first law, an object continues at constant velocity when the net force acting on it is zero.
• Marks: 2 marks (1 for "resultant/net force is zero" or "forces are balanced", 1 for identifying driving force balanced by friction/resistive forces)

8. Metre rule moments problem.

(a) Calculate the clockwise moment about the pivot due to the 3.0 N weight.

• Answer: Distance from pivot = 80 - 50 = 30 cm = 0.30 m. Moment = F × d = 3.0 × 0.30 = 0.90 N m
• Marks: 2 marks (1 for correct distance, 1 for correct moment with unit)

(b) Calculate the anticlockwise moment about the pivot due to the 2.0 N weight.

• Answer: Distance from pivot = 50 - 20 = 30 cm = 0.30 m. Moment = F × d = 2.0 × 0.30 = 0.60 N m
• Marks: 2 marks (1 for correct distance, 1 for correct moment with unit)

(c) Determine whether the metre rule is in equilibrium. Explain your answer.

• Answer: The rule is NOT in equilibrium. The clockwise moment (0.90 N m) is greater than the anticlockwise moment (0.60 N m). For equilibrium, the clockwise and anticlockwise moments must be equal.
• Marks: 2 marks (1 for stating "not in equilibrium", 1 for explanation comparing moments)

9. State where an additional 1.0 N weight should be placed to balance the rule.

• Answer: Additional anticlockwise moment needed = 0.90 - 0.60 = 0.30 N m. Distance from pivot = 0.30 / 1.0 = 0.30 m = 30 cm. The weight should be placed at the 20 cm mark (50 - 30 = 20 cm) on the same side as the 2.0 N weight. [Accept: 20 cm mark or 30 cm to the left of the pivot]
• Marks: 2 marks (1 for calculating required moment, 1 for correct position)

10. Hydraulic jack problem.

(a) Calculate the area of the small piston.

• Answer: Radius = 1.0 cm = 0.010 m. Area = πr² = 3.14 × (0.010)² = 3.14 × 10⁻⁴ m²
• Marks: 2 marks (1 for correct radius conversion, 1 for correct area with unit)

(b) Using Pascal's principle, calculate the force exerted by the large piston.

• Answer: Area of large piston: radius = 5.0 cm = 0.050 m. A₂ = 3.14 × (0.050)² = 7.85 × 10⁻³ m². P₁ = P₂ → F₁/A₁ = F₂/A₂ → F₂ = F₁ × (A₂/A₁) = 100 × (7.85 × 10⁻³ / 3.14 × 10⁻⁴) = 100 × 25 = 2500 N
• Marks: 2 marks (1 for correct area ratio or pressure equality, 1 for correct force with unit)

(c) Explain why the hydraulic jack is able to lift a much heavier load than the applied force.

• Answer: The pressure applied to the small piston is transmitted equally throughout the fluid (Pascal's principle). Since the large piston has a much larger area, the force produced (F = P × A) is proportionally larger. The force is multiplied by the ratio of the areas of the two pistons.
• Marks: 2 marks (1 for pressure transmission, 1 for force multiplication due to larger area)

Section C: Data Analysis and Application (20 marks)

11. Cyclist velocity-time graph problem.

(a) Describe the motion of the cyclist during the first 5 seconds.

• Answer: The cyclist accelerates uniformly from rest to 10 m/s. [Accept: constant/uniform acceleration]
• Marks: 2 marks (1 for "accelerates", 1 for "uniformly/constantly" or "from rest")

(b) Calculate the acceleration of the cyclist between t = 0 s and t = 5 s.

• Answer: a = (v - u) / t = (10 - 0) / 5 = 2.0 m/s²
• Marks: 2 marks (1 for formula, 1 for correct answer with unit)

12. Calculate the total distance travelled by the cyclist in the 25 seconds.

• Answer: Distance = area under v-t graph.
  • 0-5 s: area of triangle = ½ × 5 × 10 = 25 m
  • 5-15 s: area of rectangle = 10 × 10 = 100 m
  • 15-20 s: area of trapezium = ½ × (10 + 5) × 5 = 37.5 m
  • 20-25 s: area of triangle = ½ × 5 × 5 = 12.5 m
  • Total distance = 25 + 100 + 37.5 + 12.5 = 175 m
• Marks: 3 marks (1 for method using area under graph, 1 for correct areas of at least two sections, 1 for correct total)

13. Calculate the average speed of the cyclist over the entire journey.

• Answer: Average speed = total distance / total time = 175 / 25 = 7.0 m/s
• Marks: 2 marks (1 for formula, 1 for correct answer with unit; allow ecf from part (c))

14. Spring investigation problem.

(a) Plot a graph of force (y-axis) against extension (x-axis).

• Answer: Graph should show:
  • Axes labelled: Force/N on y-axis, Extension/cm on x-axis
  • Appropriate scales
  • All 6 points plotted correctly
  • Straight line through origin (best-fit line)
• Marks: 4 marks (1 for correct axes with labels and units, 1 for appropriate scales, 1 for correct plotting of points, 1 for best-fit straight line)

15. Determine the spring constant from your graph.

• Answer: Spring constant k = gradient = ΔF/Δx. Using points (0,0) and (10.0 cm, 4.0 N): k = 4.0 / 10.0 = 0.40 N/cm. [Accept: 40 N/m if converted to metres]
• Marks: 3 marks (1 for identifying gradient, 1 for correct calculation, 1 for correct unit; allow ecf from graph)

Section D: Further Applications (10 marks)

16. Calculate the elastic potential energy stored in the spring when the extension is 10.0 cm.

• Answer: EPE = ½ × F × x = ½ × 4.0 × 0.10 = 0.20 J [or EPE = ½kx² = ½ × 40 × (0.10)² = 0.20 J]
• Marks: 2 marks (1 for formula, 1 for correct answer with unit)

17. Explain why the 6.0 N weight produces an extension of 18.0 cm, which does not follow the pattern.

• Answer: The spring has exceeded its elastic limit / limit of proportionality. Beyond this point, Hooke's law no longer applies, and the extension is no longer directly proportional to the applied force. The spring may have undergone plastic deformation.
• Marks: 2 marks (1 for "exceeded elastic limit/limit of proportionality", 1 for "Hooke's law no longer applies" or "extension no longer proportional to force")

18. A ball of mass 0.5 kg is dropped from a height of 20 m. Calculate its potential energy before it is dropped.

• Answer: GPE = mgh = 0.5 × 10 × 20 = 100 J
• Marks: 2 marks (1 for formula, 1 for correct answer with unit)

19. Using the principle of conservation of energy, calculate the speed of the ball just before it hits the ground. (Neglect air resistance)

• Answer: GPE at top = KE at bottom → mgh = ½mv² → 100 = ½ × 0.5 × v² → v² = 400 → v = 20 m/s
• Marks: 2 marks (1 for equating GPE and KE, 1 for correct answer with unit)

20. Explain why, in reality, the speed of the ball just before hitting the ground would be slightly less than the calculated value.

• Answer: In reality, air resistance acts on the ball, doing work against its motion. Some of the initial GPE is converted into thermal energy (heat) due to air resistance, rather than all being converted to KE. Therefore, the final KE, and thus the speed, is less than the theoretical value.
• Marks: 2 marks (1 for identifying air resistance, 1 for explaining energy conversion to thermal energy)

END OF ANSWER KEY