Secondary 3 Combined Science Semestral Assessment 2 (End of Year) Paper 2

TuitionGoWhere Exam Practice (AI) — SA2 Practice Paper Answer Key

Subject: Combined Science
Level: Secondary 3
Paper: SA2 Physical Sciences Practice (Version 2 of 5)
Total Marks: 65

Section A (22 marks)

1. [2 marks]
Energy cannot be created or destroyed, only converted from one form to another (or transferred). Total energy in a closed system is constant.
Teaching note: Must include both “not created/destroyed” and “converted/transferred” for full marks. Common mistake: stating only “energy is conserved” without explanation.

2. [2 marks]
$GPE = mgh = 4.0 \times 10 \times 2.0 = 80\ \text{J}$
Working: $m=4.0\ \text{kg}$, $g=10\ \text{N/kg}$, $h=2.0\ \text{m}$. Substitute into $mgh$.
Marking: 1 mark for correct formula and substitution, 1 mark for final answer with unit.

3. [2 marks]
$v = \frac{d}{t} = \frac{100}{5.0} = 20\ \text{m/s}$
Working: speed = distance ÷ time.
Marking: 1 for method, 1 for answer with unit.

4. [1 mark]
Thermal energy (heat) / sound energy.
Note: Accept “heat” as correct.

5. [1 mark]
The battery provides electrical energy / pushes charges around the circuit (source of emf).
Based on image: Battery shown as two lines (long/short) in series loop with bulb.

6. [1 mark]
Conduction.

7. [2 marks]
$W = Fd = 3.0 \times 0.50 = 1.5\ \text{J}$
Marking: 1 for formula/substitution, 1 for answer.

8. [3 marks]
Scalar: quantity with magnitude only, e.g. mass / speed / time. [1]
Vector: quantity with magnitude and direction, e.g. force / velocity / weight. [1]
Difference: vector has direction, scalar does not. [1]
Common mistake: giving distance as vector or displacement as scalar.

Section B (20 marks)

9.
(a) [2] $a = \frac{\Delta v}{\Delta t} = \frac{8-0}{4-0} = 2.0\ \text{m/s}^2$. 1 mark method, 1 answer.
(b) [3] Distance = area under graph = triangle + rectangle + triangle
= $\frac{1}{2}(4)(8) + (6)(8) + \frac{1}{2}(2)(8) = 16 + 48 + 8 = 72\ \text{m}$.
Marking: 1 for identifying areas, 1 for partial calc, 1 for final.

10. [2 marks]
$I = \frac{V}{R} = \frac{12}{4.0} = 3.0\ \text{A}$. 1 method, 1 answer.

11. [3 marks]
Clockwise moment = anticlockwise moment about pivot.
$2.0 \times 30 = W \times 30 \Rightarrow W = 2.0\ \text{N}$.
Marking: 1 for principle of moments, 1 for distances (30 cm each), 1 for W.
Image: pivot at 50 cm, load at 20 cm (30 cm left), W at 80 cm (30 cm right).

12. [1 mark]
Convection (if heater below) or conduction; for immersion heater, conduction/convection. In typical beaker, thermal energy transferred by convection in water and conduction from heater. Accept “convection”.

13. [3 marks]
Loss in GPE = gain in KE.
$KE = mgh = 0.20 \times 10 \times 5.0 = 10\ \text{J}$.
Marking: 1 for concept, 1 substitution, 1 answer.
(If using $v^2=2gh$ then KE = $\frac{1}{2}mv^2$ same result.)

14. [3 marks]
Snell’s law: $n_1 \sin i = n_2 \sin r$. Air $n_1=1.0$, glass $n_2=1.5$.
$\sin r = \frac{\sin 40^\circ}{1.5} = \frac{0.643}{1.5} = 0.429 \Rightarrow r = 25.4^\circ \approx 25^\circ$.
Marking: 1 formula, 1 calc, 1 answer.
Image: incident 40° to normal, refracted closer to normal.

Section C (23 marks)

15. [4 marks]
At highest point: max gravitational PE, zero KE. [1]
As it falls, PE converts to KE. [1]
At lowest point: max KE, min PE. [1]
Total mechanical energy constant (ignoring air resistance). [1]
Teaching: Energy changes form, total conserved.

16.
(a) [2] $W = mgh = 500 \times 10 \times 10 = 50\,000\ \text{J} = 50\ \text{kJ}$. 1 method, 1 answer.
(b) [2] $P = \frac{W}{t} = \frac{50\,000}{20} = 2500\ \text{W}$. 1 method, 1 answer.
(c) [1] Constant speed / no friction / g=10.

17. [3 marks]
Spoon is metal, good conductor. [1]
Thermal energy conducted from hot soup through spoon material. [1]
Particles vibrate more, pass energy to neighbouring particles to handle. [1]
Note: Process is conduction.

18.
(a) [3] $\frac{1}{R_T} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1+2}{6} = \frac{3}{6} = 0.5 \Rightarrow R_T = 2.0\ \Omega$. 1 formula, 1 calc, 1 answer.
(b) [2] $I = \frac{V}{R_T} = \frac{9.0}{2.0} = 4.5\ \text{A}$. 1 method, 1 answer.

19.
(a) [1] Transverse wave.
(b) [2] $v = f\lambda = 2.0 \times 0.20 = 0.40\ \text{m/s}$. (λ=20 cm=0.20 m) 1 method, 1 answer.
(c) [1] Transverse: oscillations perpendicular to direction of travel; longitudinal: parallel.

20.
(a) [2] $KE = \frac{1}{2}mv^2 = 0.5 \times 800 \times 20^2 = 160\,000\ \text{J}$. 1 method, 1 answer.
(b) [3] $a = \frac{0-20}{4} = -5\ \text{m/s}^2$; $F = ma = 800 \times 5 = 4000\ \text{N}$.
Or using work: $F = \frac{KE}{d}$, $d = \frac{v}{2}t = 40\ \text{m}$, $F=160000/40=4000\ \text{N}$. 1 method, 1 calc, 1 answer.
(c) [1] Converted to heat/sound by brakes.

Total marks check: Section A 22 + B 20 + C 23 = 65 ✓