Secondary 4 Pure Physics Mechanics Quiz

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Secondary 4 Pure Physics Quiz - Mechanics

Name: ___________________________
Class: ______________
Date: ______________
Score: ______________ / 40

Duration: 60 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Show your working clearly where calculations are required.
• Use $g = 10\ \text{m/s}^2$ unless stated otherwise.
• Section A: Multiple-choice style short responses (1 mark each).
• Section B: Structured short answers (2 marks each).
• Section C: Extended calculations and explanations (3 marks each).

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Section A (Questions 1–5, 1 mark each)

1. A scalar quantity has only magnitude. Which of the following is a scalar quantity?
______ (write one: velocity / speed / force / displacement)

2. A car travels 100 m in 20 s at constant speed. What is its average speed?
______ m/s

3. State the SI unit for force.

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4. What is the moment of a force about a pivot?
______ (write: turning effect / linear push / energy / mass)

5. Pressure is defined as force per unit ______.
______ (write: area / volume / time / mass)

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Section B (Questions 6–10, 2 marks each)

6. A boy of mass 50 kg stands on a weighing scale. Calculate his weight.
Weight = ______ N
[Show space: _______________]

7. A object accelerates from $4\ \text{m/s}$ to $10\ \text{m/s}$ in 3 s. Calculate its acceleration.
Acceleration = ______ $\text{m/s}^2$
[Show space: _______________]

8. State the principle of moments for a body in equilibrium.

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9. A hydraulic press has input area $0.02\ \text{m}^2$ and output area $0.20\ \text{m}^2$. A force of 50 N is applied at the input. Calculate the output force.
Output force = ______ N
[Show space: _______________]

10. A block is at rest on a table. Draw a free-body diagram description: name the two forces acting and their direction.

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Section C (Questions 11–20, 3 marks each)

11. A car accelerates uniformly from rest to $20\ \text{m/s}$ in 10 s.
(a) Calculate its acceleration.
(b) Calculate the distance travelled.
[Show space: _______________]

12.

Image pending generation: graph for Q12.

Using the graph, calculate the total displacement of the cyclist from $t=0$ to $t=12$ s.

13. A 2 kg object is acted on by a net force of 6 N. Calculate its acceleration and explain how Newton’s second law applies.
[Show space: _______________]

14. A uniform rod of length 2.0 m and weight 20 N is pivoted at its centre. A 10 N force is applied at one end downwards. Calculate the force needed at the other end to balance it.
[Show space: _______________]

15. Explain why a skydiver reaches terminal velocity during free fall. Include forces in your answer.

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16. A liquid column of height 5 m has density $800\ \text{kg/m}^3$. Calculate the pressure at the bottom due to the liquid. ($g=10\ \text{m/s}^2$)
[Show space: _______________]

17. A ball of mass 0.5 kg is dropped from height 20 m. Calculate its gravitational potential energy at the top and its speed just before hitting the ground (ignore air resistance).
[Show space: _______________]

18. A man pushes a box with 40 N across a rough floor at constant speed. State the frictional force and explain using Newton’s first law.

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19.

Image pending generation: experimental_setup for Q19.

Calculate the component of weight parallel to the slope. Hence state what minimum force up the slope is needed to hold it at rest if friction is ignored.

20. A crane lifts a 200 kg load by 15 m in 10 s. Calculate the work done and the power output.
[Show space: _______________]

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Secondary 4 Pure Physics Quiz - Mechanics: Answer Key

Total Marks: 40
Topic: Mechanics
Level: Secondary 4 Pure Physics

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Section A (1 mark each)

1. speed
Teaching note: Scalar quantities have magnitude only. Speed = distance/time, no direction. Velocity, force, displacement are vectors.

2. 5
Working: average speed = distance / time = $100 / 20 = 5\ \text{m/s}$.

3. newton (N)
Teaching note: Force is measured in newtons, symbol N.

4. turning effect
Teaching note: Moment = turning effect of a force about a pivot.

5. area
Teaching note: Pressure $P = F / A$, so force per unit area.

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Section B (2 marks each)

6. Weight = 500 N
Working: $W = mg = 50 \times 10 = 500\ \text{N}$.
Marking: 1 mark formula, 1 mark answer with unit.

7. Acceleration = 2 m/s²
Working: $a = (v-u)/t = (10-4)/3 = 6/3 = 2\ \text{m/s}^2$.
Marking: 1 mark substitution, 1 mark answer.

8. Principle of moments: For a body in equilibrium, sum of clockwise moments about a pivot equals sum of anticlockwise moments.
Marking: 1 mark equilibrium, 1 mark equality of moments.

9. Output force = 500 N
Working: Pascal’s principle: $F_1/A_1 = F_2/A_2$ → $F_2 = F_1 \times A_2/A_1 = 50 \times 0.20/0.02 = 50 \times 10 = 500\ \text{N}$.
Marking: 1 mark method, 1 mark answer.

10. Forces: Weight (downwards), Normal reaction (upwards).
Marking: 1 mark each force named and direction correct.

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Section C (3 marks each)

11. (a) $a = 2\ \text{m/s}^2$ (b) $s = 100\ \text{m}$
Working:
(a) $a = (20-0)/10 = 2\ \text{m/s}^2$
(b) $s = ut + \frac{1}{2}at^2 = 0 + 0.5 \times 2 \times 10^2 = 100\ \text{m}$ (or area under v-t graph)
Marking: 1 mark (a), 2 marks (b) with formula and substitution.

12. Displacement = 95 m
Working: Area = triangle (0–5s): $0.5 \times 5 \times 10 = 25$; rectangle (5–10s): $5 \times 10 = 50$; triangle (10–12s): $0.5 \times 2 \times 10 = 10$; total = 85? Wait recalc: 25+50+10=85 m. Correction: total = 85 m.
(Note: graph values given: 0–5 line to 10, 5–10 flat, 10–12 down. Areas: 25+50+10=85.)
Answer: 85 m.
Marking: 1 mark reading graph, 2 marks area calculation.

13. a = 3 m/s²
Working: $F = ma$ → $a = F/m = 6/2 = 3\ \text{m/s}^2$.
Explanation: Newton’s second law states net force equals mass times acceleration; larger force or smaller mass gives greater acceleration.
Marking: 1 mark calc, 2 marks explanation.

14. 10 N
Working: Pivot at centre, each end 1.0 m from pivot. Clockwise moment = $10 \times 1 = 10\ \text{Nm}$. For balance, anticlockwise = $F \times 1 = 10$ → $F=10\ \text{N}$.
Marking: 1 mark distances, 1 mark moments eqn, 1 mark answer.

15. Terminal velocity explanation
At start, weight > air resistance, accelerates. As speed increases, drag increases until drag = weight. Net force zero, constant velocity (terminal).
Marking: 1 mark forces identified, 1 mark drag increase, 1 mark equilibrium constant v.

16. Pressure = 40 000 Pa
Working: $P = h\rho g = 5 \times 800 \times 10 = 40\,000\ \text{Pa}$.
Marking: 1 mark formula, 1 mark sub, 1 mark answer.

17. GPE = 100 J, speed = 20 m/s
Working: GPE = $mgh = 0.5 \times 10 \times 20 = 100\ \text{J}$. KE at bottom = 100 J = $\frac{1}{2}mv^2$ → $v = \sqrt{2 \times 100 / 0.5} = \sqrt{400} = 20\ \text{m/s}$.
Marking: 1 mark GPE, 2 marks speed with energy conservation.

18. Friction = 40 N
Explanation: Constant speed → net force zero (Newton’s first law). Applied force balanced by friction, so friction = 40 N opposite direction.
Marking: 1 mark value, 2 marks law explanation.

19. Parallel component = 25 N, min force = 25 N
Working: $W = mg = 5 \times 10 = 50\ \text{N}$. Parallel = $W \sin 30^\circ = 50 \times 0.5 = 25\ \text{N}$. To hold at rest, up-slope force = 25 N.
Marking: 1 mark weight, 1 mark component, 1 mark force.

20. Work = 30 000 J, Power = 3000 W
Working: $W = mgh = 200 \times 10 \times 15 = 30\,000\ \text{J}$. $P = W/t = 30\,000 / 10 = 3000\ \text{W}$.
Marking: 1 mark work, 2 marks power with unit.