Secondary 3 Physics Practice Paper 5

TuitionGoWhere Practice Paper Answers - Physics Secondary 3

Version: 5 of 5 | Mechanics

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Section A: Multiple Choice (Questions 1–10)

Each question: 2 marks. Section Total: 20 marks

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1. B — $2.0 \text{ mm}$

Teaching note: A typical coin thickness is about 2 mm (or 0.2 cm). The micrometer screw gauge measures small dimensions precisely. Option A ($0.20 \text{ cm} = 2.0 \text{ mm}$) has the correct value but micrometer readings are typically given in mm. However, $2.0 \text{ mm}$ is the standard form. Option D is 20 mm (2 cm) — too thick for one coin. Option C is 20 mm — far too large. The most reasonable measurement is 2.0 mm.

Common mistake: Confusing mm and cm units; forgetting that $1 \text{ cm} = 10 \text{ mm}$.

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2. D — Weight

Teaching note: A vector quantity has both magnitude and direction.

• Mass (A): scalar — amount of matter, no direction
• Time (B): scalar — no direction
• Speed (C): scalar — magnitude of velocity only
• Weight (D): vector — force due to gravity, acts downward toward Earth's centre

Weight = $mg$, where $g$ is acceleration due to gravity (vector). Weight is a force, and all forces are vectors.

Common mistake: Confusing mass and weight — mass is scalar, weight is vector.

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3. B — $2.0 \text{ m/s}^2$

Teaching note: Acceleration = gradient of velocity-time graph.

From the graph description: velocity increases from $0$ to $20 \text{ m/s}$ in $10 \text{ s}$.

$a = \frac{\Delta v}{\Delta t} = \frac{20 - 0}{10 - 0} = \frac{20}{10} = 2.0 \text{ m/s}^2$

Step-by-step:

• Identify initial velocity: $u = 0 \text{ m/s}$
• Identify final velocity: $v = 20 \text{ m/s}$
• Identify time interval: $\Delta t = 10 \text{ s}$
• Apply formula: $a = \frac{v-u}{t} = \frac{20}{10} = 2.0 \text{ m/s}^2$

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4. C — The velocity is zero but the acceleration is $10 \text{ m/s}^2$ downward.

Teaching note: At the highest point of vertical motion:

• Velocity is momentarily zero — the ball stops before changing direction
• Acceleration is NOT zero — gravity still acts, so $a = g = 10 \text{ m/s}^2$ downward
• The ball is still under the influence of Earth's gravitational field

This is a common exam trap. Never confuse "momentarily at rest" with "no forces acting." Newton's Second Law ($F = ma$) tells us that if $F_{grav} = mg$ acts, then $a = g$.

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5. B — $20 \text{ N}$

Teaching note: Constant velocity means zero acceleration, so net force = zero (Newton's First Law).

For horizontal motion: $\sum F = ma = 0$

Therefore: Applied force − Friction = 0

$F_{applied} = F_{friction} = 20 \text{ N}$

Marking point: Recognition that constant velocity $\Rightarrow$ balanced forces [1 mark]; correct answer [1 mark].

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6. B — $13 \text{ N}$

Teaching note: Forces at right angles → use Pythagoras' theorem.

$F_{resultant} = \sqrt{F_1^2 + F_2^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ N}$

Step-by-step:

• Draw vector diagram: 12 N east, 5 N north, resultant as hypotenuse
• Apply Pythagoras: $c^2 = a^2 + b^2$
• Calculate: $\sqrt{169} = 13 \text{ N}$
• Direction: $\tan^{-1}(5/12) = 22.6°$ north of east (not asked, but useful)

Common mistake: Simply adding (17 N) or subtracting (7 N) — only valid for collinear forces.

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7. C — $16 \text{ kW}$

Teaching note: Average power = work done / time = change in kinetic energy / time (or $P = Fv$ average).

Method 1 — Using energy:

• $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 800 \times 20^2 = \frac{1}{2} \times 800 \times 400 = 160\,000 \text{ J}$
• $P = \frac{KE}{t} = \frac{160\,000}{10} = 16\,000 \text{ W} = 16 \text{ kW}$

Method 2 — Using force and velocity:

• $a = \frac{v-u}{t} = \frac{20-0}{10} = 2 \text{ m/s}^2$
• $F = ma = 800 \times 2 = 1600 \text{ N}$
• Average $v = 10 \text{ m/s}$; $P = F \times v_{avg} = 1600 \times 10 = 16\,000 \text{ W} = 16 \text{ kW}$

Or using $P = Fv$ at final instant: $1600 \times 20 = 32 \text{ kW}$, then average = $16 \text{ kW}$.

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8. B — $20 \text{ m/s}$

Teaching note: Use conservation of energy or suvat equations.

Method — Energy:

• $PE_{lost} = KE_{gained}$
• $mgh = \frac{1}{2}mv^2$
• $gh = \frac{1}{2}v^2$
• $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20 \text{ m/s}$

Method — suvat:

• $s = 20 \text{ m}$, $u = 0$, $a = 10 \text{ m/s}^2$, find $v$
• $v^2 = u^2 + 2as = 0 + 2 \times 10 \times 20 = 400$
• $v = 20 \text{ m/s}$

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9. B — $20 \text{ N}$

Teaching note: Principle of moments: clockwise moment = anticlockwise moment.

Taking moments about the fulcrum:

• Anticlockwise moment (load side) = $40 \times 0.3 = 12 \text{ Nm}$
• Clockwise moment (effort side) = $E \times 0.6$

For equilibrium: $E \times 0.6 = 40 \times 0.3$

$E = \frac{40 \times 0.3}{0.6} = \frac{12}{0.6} = 20 \text{ N}$

Marking: Correct moment equation [1 mark]; correct substitution and answer [1 mark].

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10. B — Parabola opening upward

Teaching note: $KE = \frac{1}{2}mv^2$

For constant mass: $KE \propto v^2$

This is a quadratic relationship — graph of $y \propto x^2$ is a parabola opening upward from the origin.

$v$	$KE$
0	0
2	$2m$
4	$8m$
6	$18m$

The values increase faster than linearly — characteristic of $v^2$ dependence.

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Section B: Structured Questions (Questions 11–17)

Section Total: 40 marks

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11. (a) Acceleration during Stage 1 [2 marks]

$a = \frac{v-u}{t} = \frac{6-0}{4} = 1.5 \text{ m/s}^2$

Marking: Formula [1 mark]; correct substitution and answer with unit [1 mark].

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(b) Distance during Stage 2 [2 marks]

At constant velocity: $s = vt = 6 \times 10 = 60 \text{ m}$

Marking: Formula or recognition of constant velocity [1 mark]; correct answer with unit [1 mark].

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(c) Velocity-time graph [3 marks]

Image pending generation: graph for Q11c.

Marking: Correct shape — acceleration, constant velocity, deceleration [1 mark]; correct values on axes [1 mark]; correct time and velocity values at transition points [1 mark].

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(d) Total distance [2 marks]

Distance = area under v-t graph:

• Stage 1: triangle = $\frac{1}{2} \times 4 \times 6 = 12 \text{ m}$
• Stage 2: rectangle = $6 \times 10 = 60 \text{ m}$
• Stage 3: triangle = $\frac{1}{2} \times 6 \times 6 = 18 \text{ m}$

Total = $12 + 60 + 18 = 90 \text{ m}$

Or using average velocity: Stage 1 avg = 3 m/s, distance = 12 m; Stage 2 = 60 m; Stage 3 avg = 3 m/s, distance = 18 m.

Marking: Correct method (areas or kinematic equations) [1 mark]; correct final answer with unit [1 mark].

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12. (a) Weight of parcel [1 mark]

$W = mg = 2.5 \times 10 = 25 \text{ N}$

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(b) Reading when accelerating upward [3 marks]

When accelerating upward: $T - mg = ma$ (Newton's Second Law, upward positive)

$T = m(g + a) = 2.5 \times (10 + 2) = 2.5 \times 12 = 30 \text{ N}$

Step-by-step:

• Draw free body diagram: tension T upward, weight $mg$ downward [implicit]
• Net force upward: $T - mg = ma$ [1 mark]
• Substitution: $T - 25 = 2.5 \times 2 = 5$ [1 mark]
• Solve: $T = 30 \text{ N}$ [1 mark]

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(c) Explanation of increased reading [2 marks]

The spring balance measures tension in the spring. When the lift accelerates upward:

• The parcel must also accelerate upward
• By Newton's Second Law, net force must be upward
• Therefore tension > weight ($T = m(g+a) > mg$)
• The balance reads the tension, so it shows more than the actual weight

Marking: Identifies that spring balance reads tension [1 mark]; explains need for net upward force/acceleration [1 mark].

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13. (a) Component of weight parallel to plane [2 marks]

$W_{\parallel} = W \sin \theta = 80 \times \sin 30° = 80 \times 0.5 = 40 \text{ N}$

Marking: Correct formula $W \sin \theta$ or $mg \sin \theta$ [1 mark]; correct substitution and answer [1 mark].

Common mistake: Using $\cos 30°$ instead of $\sin 30°$ — always check which component you need. The parallel component uses $\sin \theta$; perpendicular uses $\cos \theta$.

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(b) Force F for constant speed [2 marks]

At constant speed: net force = 0 (equilibrium)

Forces up slope = Forces down slope: $F = W_{\parallel} + f = 40 + 15 = 55 \text{ N}$

Marking: Recognition of equilibrium/constant speed [1 mark]; correct addition of components [1 mark].

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(c) Reducing friction [2 marks]

Method: Lubricate the surface / use rollers / use smoother materials / reduce normal force

Explanation: Friction $f = \mu R$ (where $\mu$ is coefficient, $R$ is normal reaction). By reducing $\mu$ (lubrication, smoother surface) or using rolling friction instead of sliding friction, the frictional force decreases, so less applied force F is needed.

Marking: Valid method [1 mark]; clear explanation of mechanism [1 mark].

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14. (a) Kinetic energy before braking [2 marks]

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 15^2 = 600 \times 225 = 135\,000 \text{ J} = 135 \text{ kJ}$

Marking: Formula [1 mark]; correct substitution and answer with unit [1 mark].

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(b) Average braking force [3 marks]

Work done by brakes = KE lost = $135\,000 \text{ J}$

$W = F \times d \Rightarrow F = \frac{W}{d} = \frac{135\,000}{45} = 3000 \text{ N}$

Alternative — kinematics first:

• $v^2 = u^2 + 2as \Rightarrow 0 = 225 + 2a(45) \Rightarrow a = -2.5 \text{ m/s}^2$
• $F = ma = 1200 \times 2.5 = 3000 \text{ N}$

Marking: Correct energy/work method or kinematic method [1 mark]; correct substitution [1 mark]; final answer with unit [1 mark].

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(c) Stopping distance at double speed [3 marks]

Answer: More than double.

Reasoning:

• $KE \propto v^2$, so at $30 \text{ m/s}$, $KE = \frac{1}{2} \times 1200 \times 900 = 540\,000 \text{ J}$ — this is 4 times the original KE
• Work to stop: $W = F \times d = KE_{lost}$
• With same $F$: $d = \frac{KE}{F} \propto KE \propto v^2$
• New distance = $4 \times 45 = 180 \text{ m}$

This is 4 times, not 2 times, the original.

Or calculate directly: $a = -2.5 \text{ m/s}^2$ (same braking force), then $v^2 = u^2 + 2as \Rightarrow 0 = 900 + 2(-2.5)s \Rightarrow s = 180 \text{ m}$

Marking: Correct identification (more than double) [1 mark]; explains $KE \propto v^2$ [1 mark]; shows calculation leading to factor of 4 [1 mark].

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15. (a) Work done [2 marks]

$W = F \times d = mg \times h = 500 \times 10 \times 8 = 40\,000 \text{ J}$

Or: $W = mgh = 500 \times 10 \times 8 = 40\,000 \text{ J}$

Marking: Formula [1 mark]; correct substitution and answer with unit [1 mark].

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(b) Power output [2 marks]

$P = \frac{W}{t} = \frac{40\,000}{20} = 2000 \text{ W} = 2 \text{ kW}$

Marking: Formula [1 mark]; correct answer with unit [1 mark].

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(c) Why actual power > calculated [2 marks]

The calculated value is the useful power output for lifting the load. The motor's actual power must be greater because:

• Energy is lost to heat due to friction in moving parts
• Energy is used to lift the cable/rope itself (if massive)
• Energy is lost to air resistance
• The motor is not 100% efficient

Marking: Identifies energy losses [1 mark]; names specific loss mechanisms [1 mark].

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16. (a) Velocity ratio [2 marks]

Velocity ratio = 4

Explanation: Count the number of supporting strands — rope segments that directly support the load. In the diagram, 4 strands support the movable pulley block, so the load is shared among 4 strands.

Marking: Correct value [1 mark]; correct explanation (supporting strands) [1 mark].

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(b) Effort required [3 marks]

For ideal machine: $MA = VR = 4$, so ideal effort = $\frac{600}{4} = 150 \text{ N}$

With efficiency $\eta = 75\% = 0.75$:

$\eta = \frac{MA}{VR} = \frac{\frac{L}{E}}{VR}$

$0.75 = \frac{600/E}{4} = \frac{600}{4E}$

$E = \frac{600}{4 \times 0.75} = \frac{600}{3} = 200 \text{ N}$

Or: Useful work out = $600 \times h$; work in = $E \times 4h$ (effort moves 4× distance)

$\eta = \frac{600h}{E \times 4h} = \frac{600}{4E} = 0.75$

Same result: $E = 200 \text{ N}$

Marking: Correct efficiency relationship [1 mark]; correct substitution [1 mark]; final answer [1 mark].

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(c) Why actual effort > ideal effort [2 marks]

In an ideal system: no friction, no mass of moving parts.

Actual system requires more effort because:

• Friction in pulley bearings opposes motion
• The pulleys themselves (especially movable ones) have mass that must also be lifted
• Energy is lost to these factors, so more work input is needed for the same work output

Marking: Identifies friction [1 mark]; identifies mass of pulley system [1 mark].

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17. (a) Time to reach sea [2 marks]

Vertical motion only (horizontal does not affect vertical fall time):

$u_y = 0$ (thrown horizontally), $s = 45 \text{ m}$, $a = g = 10 \text{ m/s}^2$

$s = ut + \frac{1}{2}at^2 \Rightarrow 45 = 0 + \frac{1}{2} \times 10 \times t^2$

$t^2 = 9 \Rightarrow t = 3 \text{ s}$

Marking: Correct vertical equation/values [1 mark]; correct answer [1 mark].

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(b) Horizontal distance [2 marks]

Horizontal: constant velocity $u_x = 8 \text{ m/s}$, time = 3 s

$s_x = u_x \times t = 8 \times 3 = 24 \text{ m}$

Marking: Correct formula/recognition of constant horizontal velocity [1 mark]; correct answer with unit [1 mark].

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(c) Path sketch [3 marks]

Image pending generation: sketch for Q17c.

Marking: Correct parabolic shape [1 mark]; horizontal component constant (same length arrows) [1 mark]; vertical component increasing (longer arrow lower down) [1 mark].

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(d) Kinetic energy change and impact speed [4 marks]

KE change:

• Horizontal $v_x$ stays constant at $8 \text{ m/s}$ (no horizontal acceleration)
• Vertical $v_y$ increases: $v_y = gt = 10 \times 3 = 30 \text{ m/s}$ at impact (or $v_y^2 = 2gs = 900$, so $v_y = 30 \text{ m/s}$)
• Speed at impact: $v = \sqrt{v_x^2 + v_y^2} = \sqrt{64 + 900} = \sqrt{964} \approx 31.0 \text{ m/s}$

KE increases because:

• Gravitational PE converts to KE as height decreases
• Total mechanical energy conserved (no air resistance)
• Speed increases, so $KE = \frac{1}{2}mv^2$ increases

Calculation: $v = \sqrt{8^2 + 30^2} = \sqrt{64 + 900} = \sqrt{964} = 31.0 \text{ m/s}$

Or by energy: $mgh = \frac{1}{2}mv^2 - \frac{1}{2}mu_x^2$

$10 \times 45 = \frac{1}{2}(v^2 - 64)$ $900 = v^2 - 64$ $v^2 = 964$ $v = 31.0 \text{ m/s}$

Marking: Explains KE increases due to PE conversion [1 mark]; states horizontal velocity constant [1 mark]; calculates vertical component or uses energy method [1 mark]; final speed with method [1 mark].

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Section C: Data Analysis and Application (Questions 18–20)

Section Total: 15 marks

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18. (a) Graph plotting [3 marks]

Image pending generation: graph for Q18a.

Marking: Correct axes and scale [1 mark]; all points correct (allow ±0.2 cm) [1 mark]; line of best fit showing Hooke's Law region and deviation [1 mark].

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(b) Range of Hooke's Law [2 marks]

Answer: $0$ to $8 \text{ N}$ (approximately, up to between 8 and 10 N)

Reasoning: Hooke's Law states that extension is directly proportional to force ($F = kx$ or $x \propto F$). This produces a straight line through the origin. The graph is linear up to $F = 8 \text{ N}$ (check: $x/F = 1.2/2 = 0.6$, $2.4/4 = 0.6$, $3.6/6 = 0.6$, $4.8/8 = 0.6$ — constant ratio). At 10 N and 12 N, the ratio increases ($6.2/10 = 0.62$, $8.5/12 = 0.71$), showing deviation from proportionality.

Marking: Correct range approximately 0–8 N [1 mark]; evidence of constant ratio/straight line check [1 mark].

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(c) Spring constant [3 marks]

From linear region: $k = \frac{F}{x}$ (using $F = kx$ in form $k = F/x$)

Using $F = 8 \text{ N}$, $x = 4.8 \text{ cm} = 0.048 \text{ m}$:

$k = \frac{8}{0.048} = 166.7 \text{ N/m}$

Or using gradient: $\frac{8 - 0}{0.048 - 0} = \frac{8}{0.048} = 166.7 \text{ N/m}$

Accept 167 N/m or 170 N/m.

Step-by-step:

• Identify Hooke's Law region [implicit in choice of point]
• Convert cm to m: $4.8 \text{ cm} = 0.048 \text{ m}$ [crucial mark]
• Calculate $k = F/x$ [1 mark]
• Final answer with unit N/m [1 mark]

Marking: Correct formula [1 mark]; unit conversion from cm to m [1 mark]; correct final value [1 mark].

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(d) Why Hooke's Law ceases [2 marks]

At higher forces:

• The elastic limit of the material is exceeded
• Permanent deformation occurs — the spring does not return to original length
• The atomic/molecular bonds in the spring material are being stretched beyond their elastic capability
• The spring may be approaching its limit of proportionality

Marking: States elastic limit/limit of proportionality exceeded [1 mark]; explains permanent deformation or atomic level change [1 mark].

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19. (a) Acceleration of vehicle B [1 mark]

$a = \frac{v-u}{t} = \frac{30-0}{8} = 3.75 \text{ m/s}^2$

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(b) Kinetic energy of vehicle C [2 marks]

$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 2000 \times 30^2 = 1000 \times 900 = 900\,000 \text{ J} = 900 \text{ kJ}$

Marking: Formula [1 mark]; correct answer [1 mark].

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(c) Greatest average power [4 marks]

Average power = work done / time = gain in kinetic energy / time

Vehicle	KE gained (J)	Time (s)	Average Power (W)
A	$\frac{1}{2} \times 1500 \times 900 = 675\,000$	10	67,500
B	$\frac{1}{2} \times 1200 \times 900 = 540\,000$	8	67,500
C	$\frac{1}{2} \times 2000 \times 900 = 900\,000$	12	75,000
D	$\frac{1}{2} \times 1000 \times 900 = 450\,000$	6	75,000

Vehicles C and D tie at 75 kW.

Wait — rechecking:

• D: $450\,000 / 6 = 75\,000$ W
• C: $900\,000 / 12 = 75\,000$ W

Both have the same average power. If we need one answer, we can note they are equal, or D has greater power-to-mass ratio. But the question asks for greatest average power: C and D are equal at 75 kW, which is greater than A and B (67.5 kW).

Marking: Correct KE formula/application for at least 2 vehicles [1 mark]; calculates all four or identifies method [1 mark]; correct calculation for C or D [1 mark]; identifies C and D equal and highest [1 mark].

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20. (a) Crumple zones and Newton's Second Law [3 marks]

Newton's Second Law: $F = ma = m\frac{\Delta v}{\Delta t} = \frac{\Delta(mv)}{\Delta t} = \frac{\Delta p}{\Delta t}$

So $F = \frac{\Delta p}{\Delta t}$ for constant mass.

During a collision:

• Momentum change $\Delta p = mv_{final} - mv_{initial}$ is fixed (depends on initial speed, final speed = 0)
• Crumple zones increase $\Delta t$ by allowing gradual deformation
• Since $F = \frac{\Delta p}{\Delta t}$, increasing $\Delta t$ decreases $F$

Less force on passengers → less injury risk.

Marking: States $F = ma$ or $F = \Delta p/\Delta t$ [1 mark]; explains that $\Delta p$ is constant [1 mark]; links increased time to reduced force [1 mark].

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(b) Seat belts and Newton's First Law [3 marks]

Newton's First Law: An object continues in its state of rest or uniform motion unless acted upon by a resultant force. Also called inertia.

When car brakes suddenly:

• The car decelerates rapidly due to braking force
• The passenger, not wearing a seat belt, has no horizontal force acting on them (ignoring friction with seat, which is small)
• By Newton's First Law, the passenger continues moving forward at the original speed
• Relative to the car, the passenger appears to be "thrown forward"
• The seat belt provides the necessary force to decelerate the passenger with the car, preventing collision with windscreen/dashboard

Marking: States Newton's First Law / inertia [1 mark]; explains passenger continues at original velocity [1 mark]; explains seat belt provides stopping force / consequence of no seat belt [1 mark].

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(c) Additional safety feature [2 marks]

Feature: Head restraints / anti-lock braking system (ABS) / passenger cell / side impact bars / collapsible steering column

Example — Head restraints:

• During rear collision, torso pushed forward by seat
• Head tends to stay in place due to inertia (Newton's First Law), then whips backward
• Head restraint provides support, reducing neck extension and whiplash injury

Example — ABS:

• Prevents wheels locking during braking
• Allows driver to steer while braking
• Reduces stopping distance on slippery surfaces by preventing skidding

Example — Passenger safety cell with crumple zones:

• Rigid cell protects occupant space
• Crumple zones absorb energy

Marking: Valid safety feature [1 mark]; correct physics principle applied [1 mark].

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END OF ANSWER KEY

Total Marks Check:
Section A: 20 | Section B: 40 | Section C: 15
Total: 75 marks ✓