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O Level Physics Mechanics Quiz
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Questions
O-Level Physics Quiz - Mechanics
Name: _________________________ Class: _________________________ Date: _________________________ Score: ______ / 50
Duration: 45 minutes Total Marks: 50
Instructions:
- Answer ALL questions in the spaces provided.
- Show all working clearly for calculation questions.
- State units in all final numerical answers.
- Take g = 10 m/s² unless otherwise stated.
- The number of marks is given in brackets [ ] at the end of each question or part question.
Section A: Kinematics and Dynamics (Questions 1–5)
[Total: 12 marks]
1. A car accelerates uniformly from rest to 25 m/s in 10 seconds.
(a) Calculate the acceleration of the car. [2]
(b) Calculate the distance travelled by the car during these 10 seconds. [2]
2. A stone is dropped from the top of a cliff and takes 3.0 seconds to reach the ground. Air resistance is negligible.
(a) State the velocity of the stone when it reaches the ground. [1]
(b) Calculate the height of the cliff. [2]
3. A box of mass 15 kg is pushed across a rough horizontal floor with a constant force of 60 N. The box moves with a constant velocity of 2.0 m/s.
(a) Explain why the box moves with constant velocity even though a force is applied. [2]
(b) Calculate the magnitude of the frictional force acting on the box. [1]
4. A student plots a velocity-time graph for a cyclist travelling along a straight road. The graph shows three sections:
- Section P: A straight line sloping upwards from (0, 0) to (4, 8).
- Section Q: A horizontal straight line from (4, 8) to (10, 8).
- Section R: A straight line sloping downwards from (10, 8) to (14, 0).
(a) Describe the motion of the cyclist in section Q. [1]
(b) Calculate the total distance travelled by the cyclist. [2]
5. A ball is thrown vertically upwards with an initial speed of 15 m/s. Air resistance is negligible.
(a) Calculate the initial kinetic energy of the ball if its mass is 0.40 kg. [1]
(b) State the velocity of the ball at its maximum height. [1]
Section B: Forces, Moments, and Pressure (Questions 6–10)
[Total: 13 marks]
6. A uniform plank of weight 200 N and length 4.0 m rests on a pivot at its centre. A boy of weight 450 N stands 1.2 m from the left end of the plank.
(a) Calculate the clockwise moment about the pivot due to the boy's weight. [2]
(b) State the principle of moments. [1]
(c) A girl stands on the right side of the plank to balance it horizontally. Calculate the distance from the pivot at which the girl, of weight 540 N, must stand. [2]
7. A rectangular block of metal measures 0.20 m × 0.15 m × 0.10 m and has a mass of 24 kg. The block rests on a horizontal surface with its largest face in contact with the surface.
(a) Calculate the pressure exerted by the block on the surface. [3]
(b) Explain how the pressure would change if the block were placed on its smallest face instead. [1]
8. A hydraulic press has a small piston of area 0.005 m² and a large piston of area 0.25 m². A force of 40 N is applied to the small piston.
(a) State the principle that allows the hydraulic press to multiply force. [1]
(b) Calculate the force exerted by the large piston. [2]
(c) State one assumption made in your calculation. [1]
9. A uniform beam of length 3.0 m and weight 150 N is supported at its two ends. A load of 200 N is placed 1.0 m from the left end.
(a) Draw a diagram showing all the forces acting on the beam. [1]
(b) By taking moments about the left support, calculate the upward force exerted by the right support. [2]
10. A student investigates the motion of a trolley on a friction-compensated runway. The trolley of mass 0.80 kg is pulled by a falling mass of 0.20 kg attached by a string passing over a smooth pulley.
(a) Calculate the weight of the falling mass. [1]
(b) Calculate the acceleration of the system. [2]
Section C: Energy, Work, and Power (Questions 11–15)
[Total: 13 marks]
11. A crane lifts a load of mass 500 kg vertically upwards through a height of 12 m at a constant speed in 15 seconds.
(a) Calculate the work done by the crane in lifting the load. [2]
(b) Calculate the useful power output of the crane. [2]
(c) The crane's motor has an input power of 5000 W. Calculate the efficiency of the crane. [2]
12. A ball of mass 0.40 kg is dropped from rest and falls freely. Air resistance is negligible.
(a) Calculate the kinetic energy of the ball after it has fallen 5.0 m. [2]
(b) Calculate the velocity of the ball after it has fallen 5.0 m. [1]
13. A cyclist and bicycle have a combined mass of 80 kg. The cyclist freewheels down a slope from rest, descending a vertical height of 5.0 m. At the bottom of the slope, the speed is 8.0 m/s.
(a) Calculate the gravitational potential energy lost by the cyclist and bicycle. [2]
(b) Calculate the kinetic energy gained. [1]
(c) Explain why the kinetic energy gained is less than the gravitational potential energy lost. [1]
14. A car of mass 1200 kg travels along a horizontal road. The driving force provided by the engine is 3000 N, and the total resistive force is 600 N.
(a) Calculate the resultant force acting on the car. [1]
(b) Calculate the acceleration of the car. [1]
15. A student lifts a book of weight 15 N from the floor onto a shelf 2.0 m high.
(a) Calculate the work done by the student. [1]
(b) State the energy change that occurs as the book is lifted. [1]
Section D: Integrated Mechanics Problems (Questions 16–20)
[Total: 12 marks]
16. A car of mass 1200 kg accelerates from 10 m/s to 30 m/s.
(a) Calculate the distance travelled during this acceleration if the acceleration is 2.0 m/s². [2]
(b) Calculate the work done by the engine to increase the car's kinetic energy, assuming no resistive forces. [2]
17. A student investigates the motion of a trolley on a friction-compensated runway. The trolley of mass 0.80 kg is pulled by a falling mass of 0.20 kg attached by a string passing over a smooth pulley.
(a) Calculate the tension in the string. [2]
(b) Explain why the tension in the string is less than the weight of the falling mass. [1]
18. A uniform beam of length 3.0 m and weight 150 N is supported at its two ends. A load of 200 N is placed 1.0 m from the left end.
(a) Calculate the upward force exerted by the left support. [1]
(b) If the load is moved closer to the right support, explain how the upward force exerted by the left support changes. [1]
19. A stone is thrown vertically upwards with an initial speed of 20 m/s. Air resistance is negligible.
(a) Calculate the maximum height reached by the stone. [2]
(b) Calculate the time taken for the stone to return to its starting point. [2]
20. A box of mass 10 kg is pulled along a horizontal floor by a force of 50 N applied at an angle of 30° above the horizontal. The box moves with constant velocity.
(a) Calculate the horizontal component of the applied force. [1]
(b) Calculate the frictional force acting on the box. [1]
(c) Explain why the normal contact force on the box is less than its weight. [1]
END OF QUIZ
Answers
O-Level Physics Quiz - Mechanics: Answer Key and Marking Scheme
Total Marks: 50
Section A: Kinematics and Dynamics (Questions 1–5)
Question 1 (a) a = (v - u)/t = (25 - 0)/10 = 2.5 m/s² [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with units
(b) s = ut + ½at² = 0 + ½(2.5)(10)² = 125 m [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with units (Alternative: s = ½(u + v)t = ½(0 + 25) × 10 = 125 m)
Question 2 (a) v = u + gt = 0 + 10 × 3.0 = 30 m/s [1 mark]
- 1 mark for correct answer with units
(b) s = ut + ½gt² = 0 + ½(10)(3.0)² = 45 m [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with units
Question 3 (a) The box moves with constant velocity because the resultant force acting on it is zero. The applied force of 60 N is exactly balanced by the frictional force of 60 N acting in the opposite direction. When forces are balanced, there is no acceleration (Newton's First Law). [2 marks]
- 1 mark for stating resultant force is zero/forces are balanced
- 1 mark for linking to constant velocity/Newton's First Law
(b) Frictional force = 60 N [1 mark]
- 1 mark for correct answer with units
Question 4 (a) The cyclist is moving with constant velocity of 8 m/s (zero acceleration). [1 mark]
- 1 mark for correct description
(b) Total distance = area under velocity-time graph = ½(4)(8) + (6)(8) + ½(4)(8) = 16 + 48 + 16 = 80 m [2 marks]
- 1 mark for method (area under graph or sum of sections)
- 1 mark for correct answer with units
Question 5 (a) KE = ½mv² = ½ × 0.40 × (15)² = 45 J [1 mark]
- 1 mark for correct answer with units
(b) Velocity at maximum height = 0 m/s [1 mark]
- 1 mark for correct answer with units
Section B: Forces, Moments, and Pressure (Questions 6–10)
Question 6 (a) Distance of boy from pivot = 2.0 - 1.2 = 0.8 m (pivot at centre, 2.0 m from left end) Clockwise moment = F × d = 450 × 0.8 = 360 N m [2 marks]
- 1 mark for correct distance from pivot
- 1 mark for correct moment with units
(b) For a body in equilibrium, the sum of clockwise moments about a pivot equals the sum of anticlockwise moments about the same pivot. [1 mark]
- 1 mark for correct statement (accept equivalent wording)
(c) Anticlockwise moment needed = 360 N m Distance = moment/force = 360/540 = 0.67 m (or 0.667 m) [2 marks]
- 1 mark for equating moments
- 1 mark for correct distance with units
Question 7 (a) Area of largest face = 0.20 × 0.15 = 0.030 m² Weight = mg = 24 × 10 = 240 N Pressure = F/A = 240/0.030 = 8000 Pa (or 8.0 × 10³ Pa) [3 marks]
- 1 mark for correct area
- 1 mark for correct force (weight)
- 1 mark for correct pressure with units
(b) The pressure would increase because the same weight acts on a smaller area. [1 mark]
- 1 mark for stating pressure increases with correct reasoning
Question 8 (a) Pascal's principle / Pressure is transmitted equally throughout an enclosed liquid. [1 mark]
- 1 mark for correct statement
(b) P₁ = P₂ → F₁/A₁ = F₂/A₂ 40/0.005 = F₂/0.25 F₂ = 40 × 0.25/0.005 = 2000 N [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with units
(c) The liquid is incompressible / No friction in the pistons / Pistons are at the same height. [1 mark]
- 1 mark for any valid assumption
Question 9 (a) Diagram should show:
- Weight of beam (150 N) acting downwards at centre (1.5 m from either end)
- Load (200 N) acting downwards 1.0 m from left end
- Upward reaction force at left support (R_L)
- Upward reaction force at right support (R_R) [1 mark]
- 1 mark for correctly labelled diagram with all four forces
(b) Taking moments about left support: Clockwise moments = Anticlockwise moments (150 × 1.5) + (200 × 1.0) = R_R × 3.0 225 + 200 = 3R_R R_R = 425/3.0 = 141.7 N (or 142 N) [2 marks]
- 1 mark for correct moment equation
- 1 mark for correct answer with units
Question 10 (a) Weight of falling mass = mg = 0.20 × 10 = 2.0 N [1 mark]
- 1 mark for correct answer with units
(b) Considering the whole system: Resultant force = Weight of falling mass = 2.0 N Total mass = 0.80 + 0.20 = 1.00 kg a = F/m = 2.0/1.00 = 2.0 m/s² [2 marks]
- 1 mark for identifying resultant force and total mass
- 1 mark for correct acceleration with units
Section C: Energy, Work, and Power (Questions 11–15)
Question 11 (a) Weight = mg = 500 × 10 = 5000 N Work done = F × d = 5000 × 12 = 60 000 J (or 60 kJ) [2 marks]
- 1 mark for correct force (weight)
- 1 mark for correct work done with units
(b) Power = work done/time = 60 000/15 = 4000 W (or 4.0 kW) [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with units
(c) Efficiency = (useful power output/total power input) × 100% = (4000/5000) × 100% = 80% [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with % sign
Question 12 (a) Loss in GPE = Gain in KE mgh = 0.40 × 10 × 5.0 = 20 J KE gained = 20 J [2 marks]
- 1 mark for equating GPE loss to KE gain
- 1 mark for correct answer with units
(b) KE = ½mv² → 20 = ½ × 0.40 × v² v² = 100 → v = 10 m/s [1 mark]
- 1 mark for correct answer with units
Question 13 (a) GPE lost = mgh = 80 × 10 × 5.0 = 4000 J (or 4.0 kJ) [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with units
(b) KE gained = ½mv² = ½ × 80 × (8.0)² = 2560 J [1 mark]
- 1 mark for correct answer with units
(c) Some of the gravitational potential energy is converted to thermal energy (heat) due to friction and air resistance acting on the cyclist and bicycle. [1 mark]
- 1 mark for identifying energy dissipated as heat due to resistive forces
Question 14 (a) Resultant force = Driving force - Resistive force = 3000 - 600 = 2400 N [1 mark]
- 1 mark for correct answer with units
(b) a = F/m = 2400/1200 = 2.0 m/s² [1 mark]
- 1 mark for correct answer with units
Question 15 (a) Work done = F × d = 15 × 2.0 = 30 J [1 mark]
- 1 mark for correct answer with units
(b) Chemical energy (in the student's body) is converted to gravitational potential energy (of the book). [1 mark]
- 1 mark for correct energy change description
Section D: Integrated Mechanics Problems (Questions 16–20)
Question 16 (a) v² = u² + 2as (30)² = (10)² + 2(2.0)s 900 = 100 + 4s s = 800/4 = 200 m [2 marks]
- 1 mark for correct formula/substitution
- 1 mark for correct answer with units
(b) Initial KE = ½ × 1200 × (10)² = 60 000 J Final KE = ½ × 1200 × (30)² = 540 000 J Work done = Change in KE = 540 000 - 60 000 = 480 000 J (or 480 kJ) [2 marks]
- 1 mark for correct method (change in KE)
- 1 mark for correct answer with units
Question 17 (a) Considering the trolley alone: T = ma = 0.80 × 2.0 = 1.6 N [2 marks]
- 1 mark for correct method (using trolley's mass and acceleration from Q10)
- 1 mark for correct answer with units (Alternative: Considering falling mass: 2.0 - T = 0.20 × 2.0 → T = 1.6 N)
(b) The tension is less than the weight because the falling mass is accelerating downwards. The resultant force on the falling mass is its weight minus the tension (mg - T = ma), so T = mg - ma, which is less than mg. [1 mark]
- 1 mark for correct explanation linking resultant force and acceleration
Question 18 (a) Vertically: R_L + R_R = 150 + 200 = 350 N From Q9(b), R_R = 141.7 N R_L = 350 - 141.7 = 208.3 N (or 208 N) [1 mark]
- 1 mark for correct answer with units
(b) The upward force exerted by the left support would decrease. Moving the load closer to the right support reduces its clockwise moment about the left support, so the right support takes more of the load, reducing the force needed at the left support. [1 mark]
- 1 mark for correct change and reasoning
Question 19 (a) At maximum height, v = 0 v² = u² + 2as → 0 = (20)² + 2(-10)s 0 = 400 - 20s → s = 20 m [2 marks]
- 1 mark for correct formula/substitution (with a = -10 m/s²)
- 1 mark for correct answer with units
(b) Time to reach maximum height: v = u + at → 0 = 20 + (-10)t → t = 2.0 s Total time = 2 × 2.0 = 4.0 s [2 marks]
- 1 mark for time to maximum height
- 1 mark for total time (or using s = ut + ½at² with s = 0)
Question 20 (a) Horizontal component = F cos θ = 50 × cos 30° = 50 × 0.866 = 43.3 N (or 43 N) [1 mark]
- 1 mark for correct answer with units
(b) Since the box moves with constant velocity, resultant horizontal force is zero. Frictional force = Horizontal component = 43.3 N [1 mark]
- 1 mark for correct answer with units
(c) The normal contact force is less than the weight because the applied force has an upward vertical component (50 sin 30° = 25 N). This upward component partially supports the weight of the box, reducing the force exerted by the box on the floor, and hence the normal contact force. [1 mark]
- 1 mark for correct explanation mentioning upward vertical component
END OF ANSWER KEY