Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 1

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TuitionGoWhere Practice Paper - Physics Secondary 3

School: TuitionGoWhere Secondary School (AI)
Subject: Physics
Level: Secondary 3
Paper: SA2 Practice Paper (Version 1 of 5)
Duration: 75 minutes
Total Marks: 60
Name: ________________________
Class: ________
Date: ____________

Instructions:

• Answer all questions in the spaces provided.
• Show your working clearly where calculation is required.
• Use $g = 10 \text{ m s}^{-2}$ unless stated otherwise.
• Section A: Multiple Choice (10 questions, 1 mark each).
  Section B: Structured Questions (6 questions, 3–4 marks each).
  Section C: Free Response (4 questions, 4–5 marks each).
• Total marks = 60.

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Section A: Multiple Choice (10 marks)

1. Which of the following quantities is a vector?
A. Speed
B. Distance
C. Mass
D. Force

2. A car travels 100 m north in 5 s at constant speed. What is its velocity?
A. $20 \text{ m s}^{-1}$
B. $20 \text{ m s}^{-1}$ north
C. $500 \text{ m s}^{-1}$ north
D. $0.05 \text{ m s}^{-1}$

3. A block of mass 2 kg is pushed with a net force of 6 N. What is its acceleration?
A. $3 \text{ m s}^{-2}$
B. $12 \text{ m s}^{-2}$
C. $0.33 \text{ m s}^{-2}$
D. $8 \text{ m s}^{-2}$

4. The moment of a force about a pivot is calculated as:
A. Force ÷ distance
B. Force × perpendicular distance
C. Force + distance
D. Force × parallel distance

5. Pressure in a liquid increases with:
A. Decreasing depth
B. Decreasing density
C. Increasing depth
D. Constant temperature only

6. A boy of mass 50 kg climbs 4 m vertically. Gain in gravitational potential energy is:
A. 200 J
B. 2000 J
C. 500 J
D. 20 J

7. Two forces of 3 N and 4 N act at right angles. Resultant force is:
A. 1 N
B. 7 N
C. 5 N
D. 12 N

8. A falling object reaches terminal velocity when:
A. Air resistance is zero
B. Weight is greater than air resistance
C. Air resistance equals weight
D. Acceleration is maximum

9. The formula for Newton's law of gravitation is $F = \frac{Gm_1m_2}{r^2}$. What does $r$ represent?
A. Radius of larger mass
B. Distance between centres of masses
C. Sum of radii
D. Surface distance

10. Work done is zero when:
A. Force is large
B. Distance moved is zero
C. Object is heavy
D. Speed is constant

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Section B: Structured Questions (21 marks)

11. (a) Define scalar and vector quantities. Give one example of each. [2]
(b) A student walks 3 m east then 4 m north. Find the resultant displacement. [2]

12. A child of mass 30 kg slides down a vertical rope with acceleration $2 \text{ m s}^{-2}$. Find the frictional force between child and rope. [3]

13. A block of mass 5 kg is pulled up a rough inclined plane at constant speed by a force of 40 N. Distance moved along plane = 2 m, vertical height gain = 1 m. Calculate:
(a) Work done by applied force [1]
(b) Gain in gravitational potential energy [1]
(c) Work done against friction [1]

14.

Image pending generation: diagram for 14.

A ring of mass 4 kg is supported by two strings as shown. Calculate tensions $T_1$ and $T_2$. [4]

15. A hydraulic press has input area $0.01 \text{ m}^2$ and output area $0.1 \text{ m}^2$. A force of 20 N is applied at input. Find output force. [3]

16. A car accelerates from rest to $20 \text{ m s}^{-1}$ in 10 s.
(a) Calculate acceleration. [1]
(b) If mass is 800 kg, find net force. [2]
(c) State Newton's first law. [1]

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Section C: Free Response (29 marks)

17. A ball is thrown vertically upward with initial speed $15 \text{ m s}^{-1}$.
(a) Find maximum height reached. [2]
(b) Time to return to start. [2]
(c) Sketch velocity-time graph and label axes. [2]

Image pending generation: graph for 17.

18. Explain using free-body diagrams and Newton's laws how a book rests on a table without accelerating. Include force magnitudes if book mass is 2 kg. [5]

19. A wooden block of mass 10 kg is on a slope of angle 30° to horizontal. Coefficient of friction ignored for sliding component.
(a) Find component of weight down slope. [2]
(b) If friction is 30 N up slope and block accelerates down, find acceleration. [3]

20. A 60 kg athlete runs up 12 m staircase in 8 s.
(a) Calculate gain in GPE. [2]
(b) Calculate useful power. [2]
(c) If total energy used is 12000 J, find efficiency. [1]

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TuitionGoWhere Practice Paper - Physics Secondary 3 (Answers)

Version 1 of 5 — SA2 Practice

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Section A: Multiple Choice

1. D — Force is a vector (magnitude + direction). Speed, distance, mass are scalars. [1]

2. B — Velocity = displacement/time = 100 m / 5 s = 20 m s⁻¹ north (direction needed). [1]

3. A — $a = F/m = 6/2 = 3 \text{ m s}^{-2}$. [1]

4. B — Moment = Force × perpendicular distance from pivot. [1]

5. C — Liquid pressure ∝ depth ($p = h\rho g$). [1]

6. B — $\Delta PE = mgh = 50 \times 10 \times 4 = 2000 \text{ J}$. [1]

7. C — Resultant = $\sqrt{3^2+4^2} = 5 \text{ N}$ (Pythagoras). [1]

8. C — Terminal velocity: air resistance = weight, net force zero. [1]

9. B — $r$ is distance between centres of the two masses. [1]

10. B — $W = F \times d$; if $d=0$, work = 0. [1]

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Section B: Structured

11. (a) Scalar: quantity with magnitude only (e.g., speed). Vector: magnitude + direction (e.g., velocity). [2]
(b) Displacement = $\sqrt{3^2+4^2} = 5 \text{ m}$ at $\tan^{-1}(4/3) \approx 53°$ north of east. [2]

12. Weight $W = mg = 30 \times 10 = 300 \text{ N}$ down.
Net force down: $ma = 30 \times 2 = 60 \text{ N}$.
$W - f = ma \Rightarrow f = 300 - 60 = 240 \text{ N}$ up. [3]

13. (a) $W_{\text{app}} = Fd = 40 \times 2 = 80 \text{ J}$ [1]
(b) $\Delta PE = mgh = 5 \times 10 \times 1 = 50 \text{ J}$ [1]
(c) $W_{\text{fric}} = 80 - 50 = 30 \text{ J}$ [1]

14. $W = 4 \times 10 = 40 \text{ N}$.
Vertical: $T_1\sin50° + T_2\sin60° = 40$
Horizontal: $T_1\cos50° = T_2\cos60°$
From horiz: $T_1 = T_2 \cos60°/\cos50° = 0.766 T_2$
Sub: $0.766T_2(0.766) + T_2(0.866) = 40 \Rightarrow 0.587T_2+0.866T_2=40 \Rightarrow 1.453T_2=40 \Rightarrow T_2=27.5 \text{ N}$
$T_1 = 21.1 \text{ N}$. [4]

15. Pascal: $F_2/A_2 = F_1/A_1 \Rightarrow F_2 = 20 \times 0.1/0.01 = 200 \text{ N}$. [3]

16. (a) $a = (20-0)/10 = 2 \text{ m s}^{-2}$ [1]
(b) $F = ma = 800 \times 2 = 1600 \text{ N}$ [2]
(c) Object remains at rest/uniform velocity unless acted by net force. [1]

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Section C: Free Response

17. (a) $v^2 = u^2 + 2as \Rightarrow 0 = 15^2 - 2(10)h \Rightarrow h = 11.25 \text{ m}$ [2]
(b) $t_{\text{up}} = 15/10 = 1.5 \text{ s}$; total = 3.0 s [2]
(c) Graph: straight line from +15 to 0 at 1.5 s, then to -15 at 3 s. Axes labelled. [2]

18. Forces: weight $mg = 20 \text{ N}$ down, normal $N = 20 \text{ N}$ up. Net force = 0 (Newton's 1st). Balanced forces → no acceleration. Diagram shows equal opposite arrows. [5]

19. (a) $mg\sin30° = 10 \times 10 \times 0.5 = 50 \text{ N}$ down slope [2]
(b) Net = $50 - 30 = 20 \text{ N}$; $a = 20/10 = 2 \text{ m s}^{-2}$ down [3]

20. (a) $\Delta PE = 60 \times 10 \times 12 = 7200 \text{ J}$ [2]
(b) $P = 7200/8 = 900 \text{ W}$ [2]
(c) Eff = $7200/12000 \times 100 = 60\%$ [1]