Secondary 1 Science Physical Sciences Quiz

Secondary 1 Science Quiz - Physical Sciences: Answer Key

Total Marks: 40

Section A: Multiple Choice

1. Answer: D – force

Explanation: A vector quantity has both magnitude and direction. Force has magnitude (measured in newtons) and direction (the direction of the push or pull). Mass, speed, and distance are scalar quantities—they have magnitude only, with no direction associated.

2. Answer: C – 30 J

Explanation: Work done = force × distance in direction of force. Here, $W = 10 \text{ N} \times 3 \text{ m} = 30 \text{ J}$. The units are correct: newton-metres equal joules.

3. Answer: B – her height above the ground increases

Explanation: Gravitational potential energy is calculated as $GPE = mgh$, where $m$ = mass, $g$ = gravitational field strength, and $h$ = height above a reference point. As the student climbs, her mass stays constant but her height $h$ increases, so GPE increases.

4. Answer: A – (useful energy output / total energy input) × 100%

Explanation: Efficiency measures how much of the input energy is converted to useful output. It is always expressed as a percentage of useful output over total input. Useful energy output can never exceed total energy input (conservation of energy), so efficiency cannot exceed 100%.

5. Answer: B – zero kinetic energy and maximum gravitational potential energy

Explanation: At the highest point, the ball momentarily stops before falling back down. Speed is zero, so kinetic energy ($KE = \frac{1}{2}mv^2$) is zero. Height is maximum, so gravitational potential energy ($GPE = mgh$) is at its maximum value for this motion.

Section B: Short Answer and Calculations

6. Chemical energy → thermal energy (heat) + kinetic energy [1 mark]

Note: Accept "chemical energy → heat energy" or with mention of movement energy. The key conversion is from stored chemical energy in food to released energy forms.

7. No work is done because there is no displacement [1]. Work done requires both a force and movement in the direction of the force ($W = F \times d$); since $d = 0$, $W = 0$ [1].

Teaching note: Students often think holding something heavy is "hard work" in everyday language, but in physics, "work" has a precise meaning. The person does expend chemical energy (muscle tension), but no mechanical work is done on the box.

8. (a) $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 15^2$ [1]

$= \frac{1}{2} \times 1200 \times 225 = 600 \times 225 = 135\,000 \text{ J} = 135 \text{ kJ}$ [1]

Marking point: Correct formula with substitution [1], correct final answer with unit [1].

(b) The kinetic energy is converted to thermal energy (heat) [1] through friction in the brakes and between tyres and road [1]. Sound energy is also produced.

Teaching note: Energy is conserved overall. The KE doesn't disappear; it transforms into less useful forms, mainly heat in the brake discs and pads.

9. (a) Weight = $mg = 250 \times 10 = 2500 \text{ N}$ [1]

(b) Work done = force × distance $= 2500 \times 8 = 20\,000 \text{ J}$ (or $2.0 \times 10^4 \text{ J}$) [2]

Marking: Use of correct force (weight) [1], correct calculation with unit [1].

Teaching note: Crane lifts at constant speed, so upward force equals weight (equilibrium). If speed were changing, Newton's Second Law would apply.

(c) Time taken = distance/speed $= 8 / 0.5 = 16 \text{ s}$ [1]

Power = work done / time $= 20\,000 / 16 = 1250 \text{ W}$ [1]

Alternative: Power = force × velocity $= 2500 \times 0.5 = 1250 \text{ W}$ [2]

10. (a) Gravitational potential energy [1]

(b) Gravitational potential energy decreases as height decreases [1]; kinetic energy increases as speed increases [1]. The sum GPE + KE remains constant (ignoring air resistance).

(c) Some energy is lost to air resistance [1], so not all initial GPE converts to KE and back to GPE; some becomes thermal energy in the air and bob [1].

Teaching note: This is why pendulums eventually stop without an external push. In a vacuum with frictionless pivot, the bob would reach exactly the same height.

11. (a) Efficiency $= \frac{400}{500} \times 100\% = 80\%$ [2]

Marking: Correct substitution [1], correct answer [1].

(b) Energy is lost to friction / thermal energy / sound / other non-useful forms [1]. By conservation of energy, useful output must be less than total input.

12. (a) Horizontal component = $60 \cos 30° = 60 \times 0.866 = 51.96 \approx 52 \text{ N}$ [2]

Marking: Correct trigonometric expression [1], correct evaluation [1].

Teaching note: Only the force component in the direction of motion does work. The vertical component ($60 \sin 30° = 30 \text{ N}$) acts perpendicular to motion.

(b) Work done = horizontal force × distance $= 51.96 \times 5 = 259.8 \approx 260 \text{ J}$ [2]

Accept use of exact $60 \cos 30°$ or rounded $52 \text{ N}$. With $F_h = 52 \text{ N}$: $W = 52 \times 5 = 260 \text{ J}$.

(c) Power = work done / time $= 259.8 / 10 = 25.98 \approx 26 \text{ W}$ [2]

Or using $P = F_h \times v$ where average velocity $= 5/10 = 0.5 \text{ m/s}$, giving $P = 51.96 \times 0.5 \approx 26 \text{ W}$.

13. (a) Wasted energy $= 10\,000 - 7\,500 = 2\,500 \text{ J}$ [1]

(b) Any two from: thermal energy (heat), sound energy, kinetic energy in moving parts, energy to overcome friction [2]

14. (a) The cyclist is accelerating [1] (uniformly / constant acceleration from 0–5 seconds). Speed increases from 0 to 4 m/s.

(b) 6 m/s [1]

(c) Acceleration = change in velocity / time taken $= (4 - 0) / 5 = 0.8 \text{ m/s}^2$ [2]

Marking: Correct formula or method [1], correct substitution and answer with unit [1].

Section C: Structured Response

15. (a) Gravitational potential energy → kinetic energy → electrical energy [1]

Accept: GPE → KE → mechanical energy → electrical energy

(b) The water at height has gravitational potential energy ($mgh$) [1]. As it falls, this converts to kinetic energy, which turns turbines to generate electricity [1].

(c) Advantage: renewable / no greenhouse gas emissions during operation / low running costs once built [1]

Disadvantage: high initial construction cost / affects local ecosystem / requires specific geography (valley and river) / dependent on rainfall [1]

16. (a) $GPE = mgh = 800 \times 10 \times (50 - 5) = 800 \times 10 \times 45 = 360\,000 \text{ J}$ [2]

Marking: Formula and substitution [1], answer with unit [1]. Note height difference is 45 m (P relative to R).

(b) $GPE \text{ at P} = KE \text{ at R}$ (energy conservation) [1]

$360\,000 = \frac{1}{2} \times 800 \times v^2$ [1]

$v^2 = \frac{360\,000 \times 2}{800} = 900$

$v = \sqrt{900} = 30 \text{ m/s}$ [1]

Marking: Energy conservation statement or equivalent [1], correct equation [1], correct solution [1].

(c) Energy is lost to air resistance and friction between wheels and track [1]. Some GPE converts to thermal energy and sound, so less kinetic energy is gained at R, resulting in lower speed [1].

17. (a) Force applied / load / mass added [1]

(b) (i) Graph: correct axes with labels and units [1]; all points plotted correctly within ±1 mm, straight line of best fit [1]

Expected visual: Straight line through origin with positive gradient; points approximately (1, 2), (2, 4.1), (3, 5.9), (4, 8.1), (5, 10)

(ii) Reading from graph at $F = 2.5 \text{ N}$: approximately 5.0 cm (accept 4.9–5.1 cm) [1]

Method: draw vertical line from 2.5 N on x-axis to line of best fit, then horizontal to y-axis [1]

(iii) From graph, gradient = spring constant (in appropriate units)

Using $F = 2.0 \text{ N}, x = 0.041 \text{ m}$: $k = F/x = 2.0/0.041 = 48.8 \text{ N/m}$ [1]

Or using any point: $k = 1.0/0.020 = 50 \text{ N/m}$, $k = 3.0/0.059 = 50.8 \text{ N/m}$ [1]

Accept range 48–51 N/m [1]

Teaching note: Hooke's Law states $F = kx$ where $x$ must be in metres. The spring obeys Hooke's Law as the graph is a straight line through origin.

18. (a) Gravitational potential energy → kinetic energy [1]

(b) $GPE = mgh = 0.5 \times 10 \times 0.8 = 4.0 \text{ J}$ [2]

Marking: Correct formula [1], correct answer with unit [1].

(c) Energy lost to friction = 20% of 4.0 J = 0.8 J [1]

$KE \text{ at bottom} = 4.0 - 0.8 = 3.2 \text{ J}$ [1]

Alternative: $KE = 80\% \times 4.0 = 3.2 \text{ J}$ [2]

(d) $KE = \frac{1}{2}mv^2$

$3.2 = \frac{1}{2} \times 0.5 \times v^2$ [1]

$v^2 = \frac{3.2 \times 2}{0.5} = 12.8$

$v = \sqrt{12.8} = 3.58 \approx 3.6 \text{ m/s}$ [1]

19. (a) Wind causes blades to rotate [1]; kinetic energy of blades turns generator, converting kinetic energy to electrical energy [1].

(b) Efficiency $= \frac{108\,000}{432\,000} \times 100\% = 25\%$ [2]

Marking: Correct fraction [1], correct percentage [1].

(c) Any two from: kinetic energy in moving air not captured by blades; friction in turbine bearings; electrical resistance in generator; sound energy produced; turbulence and wake effects [2]

20. (a) Any two from: use same volume/mass of water; use same starting temperature; same distance from food to beaker; same apparatus setup; same room temperature/environmental conditions [2]

(b) Mass of water = $\rho \times V = 1.0 \times 50 = 50 \text{ g}$ [1]

Energy = $mc\Delta T = 50 \times 4.2 \times 24$ [1]

$= 5040 \text{ J}$ [1]

(c) Heat is lost to surroundings (air, beaker, metal stand) [1]; not all energy produced by burning food is transferred to water / incomplete combustion of food [1].

End of Answer Key