Secondary 3 Combined Science Practice Paper 5

TuitionGoWhere Practice Paper Answer Key - Combined Science Secondary 3

Version 5 of 5

SECTION A

1. (a) Reading = 12.5 + 0.28 = 12.78 mm [1]

(b) The micrometer can measure to 0.01 mm precision [1], while a metre rule only measures to 1 mm precision [1]. The micrometer's screw mechanism and smaller division scale allow finer discrimination between close measurements.

2. (a) Acceleration = gradient = $\frac{8 - 0}{10 - 0} = \frac{8}{10}$ = 0.80 m/s² [1]

(b) Distance = area under graph [1]

• Triangle: $\frac{1}{2} \times 10 \times 8 = 40$ m
• Rectangle: $10 \times 8 = 80$ m
• Triangle: $\frac{1}{2} \times 10 \times 8 = 40$ m
• Total = $40 + 80 + 40 =$ 160 m [1]

3. (a) Resultant force = $ma = 2.0 \times 4.0 =$ 8.0 N [1]

(b) Resultant force = Applied force − Frictional force [1]

• $8.0 = 12 - f$, so $f = 12 - 8 =$ 4.0 N [1]

4. (a) GPE = $mgh = 0.20 \times 10 \times 0.15 =$ 0.30 J [2]

• (1 mark for correct formula, 1 mark for final answer with units)

(b) By conservation of energy, KE at lowest point = GPE at highest point = 0.30 J [1]

5. (a) Pascal's principle: pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel [1]

(b) $\frac{F_1}{A_1} = \frac{F_2}{A_2}$ [1]

• $\frac{400}{2.0} = \frac{F_2}{8.0}$
• $F_2 = 400 \times \frac{8.0}{2.0} =$ 1600 N [1]

6. (a) $E = mc\Delta T = 0.50 \times 4200 \times (80-20) = 0.50 \times 4200 \times 60$ [1]

• $=$ 126 000 J (or 126 kJ) [1]

(b) $t = \frac{E}{P} = \frac{126000}{210} =$ 600 s (or 10 minutes) [1]

7. (a) R₂ and R₃ in parallel: $\frac{1}{R_{23}} = \frac{1}{6.0} + \frac{1}{12} = \frac{2+1}{12} = \frac{3}{12}$ [1]

• $R_{23} = 4.0$ Ω
• Total R = $4.0 + 4.0 =$ 8.0 Ω [1]

(b) $I = \frac{V}{R} = \frac{12}{8.0} =$ 1.5 A [1]

8. (a) $v = f\lambda$, so $f = \frac{v}{\lambda} = \frac{330}{0.66} =$ 500 Hz [2]

• (1 mark for formula, 1 mark for answer)

(b) Speed increases in water (sound travels faster in denser media) [0.5]. Wavelength increases since $v = f\lambda$ and frequency stays constant [0.5] = 1 mark

9. (a) $n = \frac{\sin i}{\sin r}$ [1]

(b) $\sin r = \frac{\sin 45°}{1.5} = \frac{0.707}{1.5} = 0.471$ [1]

• $r = \sin^{-1}(0.471) =$ 28.1° (accept 28°) [1]

10. (a) $\frac{V_s}{V_p} = \frac{N_s}{N_p}$, so $V_s = 240 \times \frac{80}{400} =$ 48 V [2]

• (1 mark for correct ratio, 1 mark for answer)

(b) Assumption: No energy losses (100% efficient transformer) / OR No flux leakage / OR Perfect magnetic coupling [1]

SECTION B

11. (a) Principle of conservation of energy: energy cannot be created or destroyed, only converted from one form to another [1]; the total energy in a closed system remains constant [1] = 2 marks

(b) GPE = $mgh = 800 \times 10 \times 25$ [1] $=$ 200 000 J (or 200 kJ) [1] = 2 marks

(c) GPE at A = KE at B (conservation of energy) [1]

• $200000 = \frac{1}{2} \times 800 \times v^2$ [1]
• $v^2 = \frac{400000}{800} = 500$
• $v = \sqrt{500} =$ 22.4 m/s [1] = 3 marks
• (Accept 22 m/s; allow ecf from part (b))

(d) Loss of GPE from A to C = $mg(h_A - h_C) = 800 \times 10 \times (25-15) = 800 \times 10 \times 10 = 80000$ J [1]

• This becomes KE at C: $\frac{1}{2} \times 800 \times v^2 = 80000$
• $v = \sqrt{200} =$ 14.1 m/s [1] = 2 marks
• (Alternative: GPE at C = 120 000 J, so KE at C = 200 000 − 120 000 = 80 000 J)

(e) Friction does work against the motion / friction dissipates energy as heat [1]; so actual speed will be less than calculated value [1] = 2 marks

12. (a) The electromagnet holds the steel ball until the timer is started [1]; when the current is switched off, the ball falls and the electromagnet releases, simultaneously starting the timer; the ball hits the trapdoor, breaking the circuit and stopping the timer [1] = 2 marks

(b) $s = \frac{1}{2}gt^2$, so $g = \frac{2s}{t^2} = \frac{2 \times 1.80}{(0.60)^2}$ [1]

• $= \frac{3.60}{0.36} =$ 10 m/s² [1] = 2 marks

(c) Any TWO from:

• Air resistance acts on the falling ball, slowing it slightly
• Reaction time in operating/start/stop mechanisms
• Ball not released from rest (initial push or remaining magnetization)
• Timer delay in starting or stopping
• Measured distance not exact (ball height, trapdoor thickness) [2 marks, 1 each]

(d) Random errors (unpredictable variations) average out when taking the mean [1]; repeating reduces uncertainty / gives a more reliable estimate of the true value [1] = 2 marks

13. (a) $a = \frac{v-u}{t} = \frac{0-20}{8.0} = \frac{-20}{8.0}$ [1] $=$ −2.5 m/s² (deceleration = 2.5 m/s²) [1] = 2 marks

(b) $F = ma = 1200 \times 2.5$ [1] $=$ 3000 N [1] = 2 marks

• (Accept ecf from (a); positive value for braking force)

(c) $W = F \times d = 3000 \times 80$ [1] $=$ 240 000 J (or 240 kJ) [1] = 2 marks

• (Or using work-energy: $W = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 20^2 = 240 000$ J)

(d) Kinetic energy is converted to heat / thermal energy (and sound) in the brake pads, discs, and surroundings [1]

(e) Friction between brake pads and discs does work against the motion [1]; this work done converts kinetic energy to thermal energy, raising the temperature of the brake pads [1] = 2 marks

14. (a) Boyle's law: For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume [2]

• (2 marks for complete statement with all conditions; 1 mark if one element missing)

(b) $p_1V_1 = p_2V_2$ [1]

• $150 \times 200 = p_2 \times 100$ [1]
• $p_2 = \frac{150 \times 200}{100} =$ 300 kPa [1] = 3 marks

(c) When volume decreases, molecules have less distance to travel between walls [1]; molecules hit the walls more frequently [1]; each collision still transfers the same momentum, so greater force per unit area (pressure) results [1] = 3 marks

• (Alternative: same number of molecules in smaller space → higher number density → more collisions per second)

(d) Pressure would increase (at constant volume, higher temperature means faster molecules hitting walls more frequently and with greater force) [1]

15. (a) Circuit symbol: rectangular box with arrow diagonally across it, or rheostat symbol [1]

(b) Ammeter: must be in series so all current flows through it to measure the total circuit current [1]; Voltmeter: must be in parallel so it measures potential difference across the lamp only, without changing the circuit current [1] = 2 marks

(c) $R = \frac{V}{I} = \frac{2.4}{0.30}$ [1] $=$ 8.0 Ω [1] = 2 marks

(d) $R = \frac{4.0}{0.50} =$ 8.0 Ω [1]; Resistance has stayed the same / no change [1] = 2 marks

• (Note: If resistance appears unchanged in these values, accept; typical filament lamp shows increase, but with these specific values the resistance is constant. Students should note this is unusual for a lamp and might indicate limited data range or approximations. Alternative valid answer: if student observes slight increase to 8.0 from 8.0, state "stayed approximately constant")

(e) As current increases, the filament gets hotter [1]; metal resistance increases with temperature [1]; so the lamp's resistance rises as current increases (non-ohmic behaviour) [1] = 3 marks

• (For these specific values: if resistance stayed same, explain that at these operating points the temperature may be similar, or the values are simplified. In reality, a lamp's resistance increases significantly from cold.)

16. (a) Focal length: distance from lens centre to F [1] (marking of f on diagram required)

(b) Image properties: real, magnified, inverted [2]

• (2 correct = 2 marks; 1 correct = 1 mark; 0 correct = 0 marks)

(c) $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ [1]

• $\frac{1}{10} = \frac{1}{15} + \frac{1}{v}$
• $\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{3-2}{30} = \frac{1}{30}$ [1]
• $v =$ 30 cm [1] = 3 marks

(d) $m = \frac{v}{u} = \frac{30}{15}$ [1] $=$ 2 [1] = 2 marks

• (Or $m = \frac{h_i}{h_o}$, so image height = 4.0 cm)

(e) Projector / camera / magnifying glass (when object between F and lens) / eye [1]

• (Note: for this configuration with real image, projector or photocopier is most appropriate)

17. (a) Nuclear fission [1]

(b) Ratio = $\frac{8 \times 10^{13}}{3 \times 10^{7}} = \frac{8}{3} \times 10^{6}$ [1] $=$ 2.7 × 10⁶ (approximately 2.7 million times) [1] = 2 marks

• (Accept $2.67 \times 10^6$)

(c) Energy = Power × time = $2.4 \times 10^9 \times (24 \times 3600)$ [1]

• $= 2.4 \times 10^9 \times 86400 =$ 2.07 × 10¹⁴ J (or $2.1 \times 10^{14}$ J) [1] = 2 marks

(d) Radioactive waste remains hazardous for thousands of years [1]; requires secure, long-term storage that won't leak into environment / groundwater / must withstand geological changes and human interference [1] = 2 marks

(e) Any TWO from:

• Very high energy density / small fuel mass needed
• No greenhouse gas emissions during operation / low carbon footprint
• Reliable baseload power (not intermittent like solar/wind)
• Fuel supply can be reliable / not weather-dependent
• Reduces dependence on fossil fuels [2 marks, 1 each]

(f) Fusion: light nuclei combine (fuse) to form heavier nuclei releasing energy [1]; Fission: heavy nucleus splits into smaller fragments [1] = 2 marks

• (Or: fusion requires extremely high temperatures/pressures; fusion produces less radioactive waste; fusion uses hydrogen isotopes not uranium)

END OF ANSWER KEY