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

TuitionGoWhere Practice Paper - Physics Secondary 3 SA2 Version 3 - Answer Key

Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

1

Answer: B (4.78 mm)

Working:

• Main scale reading = 4.5 mm
• Thimble scale reading = 28 × 0.01 mm = 0.28 mm
• Observed reading = 4.5 + 0.28 = 4.78 mm
• Zero error = +0.02 mm (positive zero error means reading is larger than actual)
• Actual diameter = Observed reading − Zero error = 4.78 − 0.02 = 4.76 mm

Wait, correction: Positive zero error means the instrument reads more than the true value. So true value = observed − zero error = 4.78 − 0.02 = 4.76 mm. That corresponds to option A.

Correct Answer: A (4.76 mm)

Marking note: Common mistake is adding the zero error instead of subtracting for positive zero error.

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2

Answer: B (1.0 m/s²)

Working:

• Acceleration = gradient of velocity-time graph
• Between t = 2 s and t = 6 s: velocity changes from 2 m/s to 6 m/s
• Time interval = 6 − 2 = 4 s
• Acceleration = (6 − 2) / 4 = 4 / 4 = 1.0 m/s²

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3

Answer: B (6.0 m/s²)

Working:

• Newton's second law: $F = ma$
• $a = F/m = 12 / 2.0 = 6.0 \text{ m/s}^2$

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4

Answer: B (20 m)

Working:

• At maximum height, final velocity $v = 0$
• Using $v^2 = u^2 + 2as$ with $a = -g = -10 \text{ m/s}^2$
• $0 = 20^2 + 2(-10)s$
• $0 = 400 - 20s$
• $s = 400 / 20 = 20 \text{ m}$

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5

Answer: B (173 J)

Working:

• Work done = Force × distance × cos θ
• $W = 50 \times 4.0 \times \cos 30°$
• $\cos 30° = \sqrt{3}/2 \approx 0.866$
• $W = 200 \times 0.866 = 173.2 \text{ J} \approx 173 \text{ J}$

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6

Answer: B (6000 N)

Working:

• Initial momentum = $mv = 1200 \times 25 = 30000 \text{ kg m/s}$
• Final momentum = 0
• Change in momentum = 30000 kg m/s
• Time = 5.0 s
• Average force = change in momentum / time = 30000 / 5 = 6000 N

Alternatively: $a = (v-u)/t = (0-25)/5 = -5 \text{ m/s}^2$, $F = ma = 1200 \times 5 = 6000 \text{ N}$

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7

Answer: B (0.8 N)

Working:

• Principle of moments: Clockwise moments = Anticlockwise moments (about pivot)
• Taking moments about the 30 cm mark (pivot):
• Anticlockwise moment: $2.0 \text{ N} \times (30 - 10) \text{ cm} = 2.0 \times 0.20 = 0.40 \text{ Nm}$
• Clockwise moment: $W \times (80 - 30) \text{ cm} = W \times 0.50 \text{ m}$
• $W \times 0.50 = 0.40$
• $W = 0.40 / 0.50 = 0.8 \text{ N}$

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8

Answer: D (4000 N)

Working:

• Pascal's principle: Pressure transmitted equally throughout fluid
• $F_1/A_1 = F_2/A_2$
• $F_2 = F_1 \times (A_2/A_1) = 100 \times (200/5) = 100 \times 40 = 4000 \text{ N}$

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9

Answer: C (10 N)

Working:

• Centripetal force $F_c = mv^2/r$
• $F_c = 0.5 \times 4.0^2 / 0.8 = 0.5 \times 16 / 0.8 = 8 / 0.8 = 10 \text{ N}$

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10

Answer: A (The satellite's kinetic energy is constant but its momentum changes.)

Explanation:

• In a circular orbit, speed is constant → kinetic energy ($\frac{1}{2}mv^2$) is constant.
• Velocity direction changes continuously → momentum ($mv$) changes direction → momentum vector changes.
• The gravitational force provides centripetal force, doing no work (force ⟂ displacement), so KE constant.

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Section B: Structured Questions [30 marks]

11

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

Answer: 10 N (for a 2.0 kg trolley) or $mg \sin \theta$ in general

Working:

• Weight $W = mg$ (mass not given in question, but typically trolley mass ~2 kg in such questions)
• Wait, mass of trolley not given in question stem. Let me re-read.
• Question says "A student investigates the motion of a trolley..." but doesn't give mass. This is an oversight. In the answer key, I'll assume a mass of 2.0 kg (common for such trolleys) or give the formula.

Better approach: The question likely expects the formula or a calculation with a given mass. Since mass isn't specified, I'll provide the method.

Method:

• Parallel component = $mg \sin \theta$
• $= m \times 10 \times \sin 30°$
• $= m \times 10 \times 0.5$
• $= 5m \text{ N}$

If mass = 2.0 kg (typical): $= 10 \text{ N}$

Marking: 1 mark for $mg \sin \theta$, 1 mark for correct substitution/answer.

(b) Acceleration down the plane [1 mark]

Answer: 5.0 m/s²

Working:

• $a = g \sin \theta = 10 \times \sin 30° = 10 \times 0.5 = 5.0 \text{ m/s}^2$
• (Independent of mass)

(c) Velocity at bottom [2 marks]

Answer: 4.47 m/s (or $\sqrt{20} \approx 4.5 \text{ m/s}$)

Working:

• $v^2 = u^2 + 2as$
• $u = 0$, $a = 5.0 \text{ m/s}^2$, $s = 2.0 \text{ m}$
• $v^2 = 0 + 2 \times 5.0 \times 2.0 = 20$
• $v = \sqrt{20} = 4.47 \text{ m/s}$

(d) Effect of doubling mass [2 marks]

Answer: Acceleration remains unchanged at 5.0 m/s².

Explanation:

• Acceleration down a frictionless incline: $a = g \sin \theta$
• This is independent of mass because both the driving force ($mg \sin \theta$) and inertia ($m$) are proportional to mass, so $m$ cancels out.
• Doubling mass doubles the parallel component of weight but also doubles the inertia, resulting in the same acceleration.

Marking: 1 mark for "unchanged", 1 mark for correct explanation referencing $a = g \sin \theta$ or force/inertia both proportional to mass.

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12

(a) Acceleration [1 mark]

Answer: 3.0 m/s²

Working:

• $a = (v - u)/t = (30 - 0)/10 = 3.0 \text{ m/s}^2$

(b) Resultant force [2 marks]

Answer: 4500 N

Working:

• $F = ma = 1500 \times 3.0 = 4500 \text{ N}$

(c) Resistive force during acceleration [2 marks]

Answer: 1500 N

Working:

• Resultant force = Driving force − Resistive force
• $4500 = 6000 - F_{\text{resistive}}$
• $F_{\text{resistive}} = 6000 - 4500 = 1500 \text{ N}$

(d) Resistive force at constant speed [2 marks]

Answer: 1500 N

Working:

• At constant speed, acceleration = 0, so resultant force = 0
• Driving force = Resistive force
• Power $P = Fv$ → $F = P/v = 45000 / 30 = 1500 \text{ N}$
• (Driving force = 1500 N, so resistive force = 1500 N)

Marking note: Many students incorrectly use the driving force from part (c). At constant speed, the engine power determines the driving force, which equals the resistive force.

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13

(a) Loss in GPE [1 mark]

Answer: 45 J

Working:

• $\Delta \text{GPE} = mgh = 3.0 \times 10 \times 1.5 = 45 \text{ J}$

(b) Gain in KE [1 mark]

Answer: 24 J

Working:

• $\Delta \text{KE} = \frac{1}{2}mv^2 - 0 = 0.5 \times 3.0 \times 4.0^2 = 1.5 \times 16 = 24 \text{ J}$

(c) Work done against friction [2 marks]

Answer: 21 J

Working:

• By work-energy principle: Loss in GPE = Gain in KE + Work against friction
• $45 = 24 + W_{\text{friction}}$
• $W_{\text{friction}} = 45 - 24 = 21 \text{ J}$

(d) Average frictional force [2 marks]

Answer: 5.25 N

Working:

• Work done against friction = Frictional force × distance along plane
• $21 = F_{\text{friction}} \times 4.0$
• $F_{\text{friction}} = 21 / 4.0 = 5.25 \text{ N}$

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14

(a) Principle of moments [1 mark]

Answer: For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point.

(b) Tension at B [3 marks]

Answer: 350 N

Working:

• Taking moments about A (clockwise positive):
• Clockwise moments: $(200 \times 2.0) + (600 \times 1.0) = 400 + 600 = 1000 \text{ Nm}$
• Anticlockwise moment: $T_B \times 4.0$
• $T_B \times 4.0 = 1000$
• $T_B = 1000 / 4.0 = 250 \text{ N}$

Wait, let me recalculate:

• Beam weight 200 N acts at centre (2.0 m from A) → moment = 200 × 2.0 = 400 Nm clockwise
• Painter weight 600 N at 1.0 m from A → moment = 600 × 1.0 = 600 Nm clockwise
• Total clockwise = 1000 Nm
• $T_B$ at 4.0 m from A → anticlockwise moment = $T_B \times 4.0$
• $T_B = 1000 / 4 = 250 \text{ N}$

Correct Answer: 250 N

(c) Tension at A [2 marks]

Answer: 550 N

Working:

• For vertical equilibrium: Upward forces = Downward forces
• $T_A + T_B = 200 + 600 = 800 \text{ N}$
• $T_A = 800 - T_B = 800 - 250 = 550 \text{ N}$

Alternative (moments about B):

• Anticlockwise: $T_A \times 4.0$
• Clockwise: $200 \times 2.0 + 600 \times 3.0 = 400 + 1800 = 2200$
• $T_A = 2200 / 4 = 550 \text{ N}$ ✓

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15

(a) Centripetal force at lowest point [2 marks]

Answer: 14.4 N

Working:

• $F_c = mv^2/r = 0.2 \times 6.0^2 / 0.5 = 0.2 \times 36 / 0.5 = 7.2 / 0.5 = 14.4 \text{ N}$

(b) Tension at lowest point [2 marks]

Answer: 16.4 N

Working:

• At lowest point: Tension acts upward, weight acts downward
• Net upward force = Centripetal force
• $T - mg = F_c$
• $T = F_c + mg = 14.4 + (0.2 \times 10) = 14.4 + 2 = 16.4 \text{ N}$

(c) Maximum speed without breaking [2 marks]

Answer: 6.71 m/s (or $\sqrt{45} \approx 6.7 \text{ m/s}$)

Working:

• Maximum tension $T_{\text{max}} = 20 \text{ N}$
• $T_{\text{max}} - mg = m v_{\text{max}}^2 / r$
• $20 - 2 = 0.2 \times v_{\text{max}}^2 / 0.5$
• $18 = 0.4 \times v_{\text{max}}^2$
• $v_{\text{max}}^2 = 18 / 0.4 = 45$
• $v_{\text{max}} = \sqrt{45} = 6.71 \text{ m/s}$

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16

(a) Weight of cube [2 marks]

Answer: 8.0 N

Working:

• Volume $V = (0.1)^3 = 0.001 \text{ m}^3$
• Mass $m = \rho V = 8000 \times 0.001 = 8.0 \text{ kg}$
• Weight $W = mg = 8.0 \times 10 = 80 \text{ N}$

Wait: 8000 kg/m³ × 0.001 m³ = 8 kg. Weight = 8 × 10 = 80 N. Not 8 N.

Correct Answer: 80 N

(b) Upthrust on cube [2 marks]

Answer: 12 N

Working:

• Upthrust = Weight of displaced fluid = $\rho_{\text{liquid}} V g$
• $= 1200 \times 0.001 \times 10 = 12 \text{ N}$

(c) Spring balance reading [2 marks]

Answer: 68 N

Working:

• Forces on cube: Weight down (80 N), Upthrust up (12 N), Tension up (T)
• Equilibrium: $T + \text{Upthrust} = \text{Weight}$
• $T = 80 - 12 = 68 \text{ N}$
• Spring balance reads tension = 68 N

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Section C: Longer Structured Questions [20 marks]

17

(a) Initial weight [1 mark]

Answer: 5000 N

Working:

• $W = mg = 500 \times 10 = 5000 \text{ N}$

(b) Thrust force [2 marks]

Answer: 1600 N

Working:

• Thrust = rate of mass ejection × exhaust velocity relative to rocket
• $F_{\text{thrust}} = (dm/dt) \times v_{\text{exhaust}} = 2.0 \times 800 = 1600 \text{ N}$

(c) Initial acceleration [2 marks]

Answer: -6.8 m/s² (or 6.8 m/s² downwards)

Working:

• Net force = Thrust − Weight = 1600 − 5000 = -3400 N
• $a = F_{\text{net}} / m = -3400 / 500 = -6.8 \text{ m/s}^2$
• Negative sign means acceleration is downwards. The rocket cannot lift off initially!

Marking note: This is a "trick" question showing that thrust (1600 N) is less than initial weight (5000 N), so the rocket doesn't move initially. Students should recognise this.

(d) Velocity after 50 s [3 marks]

Answer: Cannot be calculated as rocket doesn't lift off / 0 m/s (if assumed held down then released)

Explanation: Since initial acceleration is negative (thrust < weight), the rocket remains on the launch pad. The question assumes it launches, which is physically inconsistent with the given numbers.

Alternative interpretation if we assume the rocket is held down and released when thrust > weight:

• Mass after 50 s = 400 kg (given)
• Fuel mass = 100 kg, ejected at 2 kg/s → 50 s to exhaust fuel ✓
• But thrust (1600 N) < weight even at 400 kg (4000 N). Rocket never lifts off.

Marking: This question has inconsistent data. In a real exam, the numbers would be chosen so thrust > final weight at least. For this answer key, I'll note the inconsistency.

Expected exam answer (if numbers were consistent):

• Use rocket equation: $v = v_{\text{exhaust}} \ln(m_0/m) - gt$
• $v = 800 \ln(500/400) - 10 \times 50 = 800 \ln(1.25) - 500 = 800 \times 0.223 - 500 = 178 - 500 = -322 \text{ m/s}$ (still negative)

Conclusion: The given parameters don't allow lift-off. This would be flagged in a real paper.

(e) Why acceleration increases [2 marks]

Answer:

• Thrust remains constant

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TuitionGoWhere Practice Paper - Physics Secondary 3 SA2 Version 3 - Answer Key

Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

1

Answer: A (4.76 mm)

Working:

• Main scale reading = 4.5 mm
• Thimble scale reading = 28 × 0.01 mm = 0.28 mm
• Observed reading = 4.5 + 0.28 = 4.78 mm
• Zero error = +0.02 mm (positive zero error means reading is larger than actual)
• Actual diameter = Observed reading − Zero error = 4.78 − 0.02 = 4.76 mm

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2

Answer: B (1.0 m/s²)

Working:

• Acceleration = gradient of velocity-time graph
• Between t = 2 s and t = 6 s: velocity changes from 2 m/s to 6 m/s
• Time interval = 6 − 2 = 4 s
• Acceleration = (6 − 2) / 4 = 4 / 4 = 1.0 m/s²

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3

Answer: B (6.0 m/s²)

Working:

• Newton's second law: $F = ma$
• $a = F/m = 12 / 2.0 = 6.0 \text{ m/s}^2$

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4

Answer: B (20 m)

Working:

• At maximum height, final velocity $v = 0$
• Using $v^2 = u^2 + 2as$: $0 = 20^2 + 2(-10)h$
• $400 = 20h \Rightarrow h = 20 \text{ m}$

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5

Answer: B (173 J)

Working:

• Work done = $F \cos\theta \times d$
• $W = 50 \times \cos 30° \times 4.0 = 50 \times 0.866 \times 4.0 = 173.2 \text{ J} \approx 173 \text{ J}$

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6

Answer: B (6000 N)

Working:

• Acceleration $a = \frac{v-u}{t} = \frac{0-25}{5} = -5 \text{ m/s}^2$
• Force $F = ma = 1200 \times 5 = 6000 \text{ N}$ (magnitude)

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7

Answer: B (0.8 N)

Working:

• Taking moments about pivot (30 cm mark):
• Clockwise moment = $2.0 \times (30-10) = 2.0 \times 20 = 40 \text{ N cm}$
• Anticlockwise moment = $W \times (80-30) = W \times 50$
• For equilibrium: $50W = 40 \Rightarrow W = 0.8 \text{ N}$

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8

Answer: D (4000 N)

Working:

• Pascal's principle: $\frac{F_1}{A_1} = \frac{F_2}{A_2}$
• $F_2 = F_1 \times \frac{A_2}{A_1} = 100 \times \frac{200}{5} = 100 \times 40 = 4000 \text{ N}$

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9

Answer: C (10 N)

Working:

• Centripetal force $F_c = \frac{mv^2}{r} = \frac{0.5 \times 4.0^2}{0.8} = \frac{0.5 \times 16}{0.8} = \frac{8}{0.8} = 10 \text{ N}$

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10

Answer: A (The satellite's kinetic energy is constant but its momentum changes.)

Working:

• In circular orbit, speed is constant → kinetic energy constant
• Direction of velocity changes continuously → momentum (vector) changes

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Section B: Structured Questions [30 marks]

11

(a) Component of weight parallel to plane = $mg \sin\theta = m \times 10 \times \sin 30° = 5m \text{ N}$ [1] Since mass not given, answer in terms of $m$: $5m \text{ N}$ or if mass assumed from context, but question asks for component of weight → $mg \sin 30° = 0.5mg$ [2]

Better working:

• Weight = $mg$
• Parallel component = $mg \sin 30° = mg \times 0.5 = 0.5mg$ [2]

(b) Acceleration $a = g \sin\theta = 10 \times \sin 30° = 5.0 \text{ m/s}^2$ [1]

(c) Using $v^2 = u^2 + 2as$: $v^2 = 0 + 2 \times 5.0 \times 2.0 = 20$ $v = \sqrt{20} = 4.47 \text{ m/s}$ [2]

(d) Acceleration remains the same (5.0 m/s²). [1] Explanation: On a frictionless incline, acceleration $a = g \sin\theta$ is independent of mass. [1]

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12

(a) Acceleration $a = \frac{v-u}{t} = \frac{30-0}{10} = 3.0 \text{ m/s}^2$ [1]

(b) Resultant force $F = ma = 1500 \times 3.0 = 4500 \text{ N}$ [2]

(c) Resultant force = Driving force − Resistive force $4500 = 6000 - F_{\text{resistive}}$ $F_{\text{resistive}} = 6000 - 4500 = 1500 \text{ N}$ [2]

(d) At constant speed, driving force = resistive force Power $P = Fv \Rightarrow F = \frac{P}{v} = \frac{45000}{30} = 1500 \text{ N}$ [2]

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13

(a) Loss in GPE = $mgh = 3.0 \times 10 \times 1.5 = 45 \text{ J}$ [1]

(b) Gain in KE = $\frac{1}{2}mv^2 = 0.5 \times 3.0 \times 4.0^2 = 0.5 \times 3.0 \times 16 = 24 \text{ J}$ [1]

(c) Work-energy principle: Loss in GPE = Gain in KE + Work against friction $45 = 24 + W_{\text{friction}}$ $W_{\text{friction}} = 21 \text{ J}$ [2]

(d) Work against friction = Frictional force × distance $21 = F_{\text{friction}} \times 4.0$ $F_{\text{friction}} = 5.25 \text{ N}$ [2]

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14

(a) Principle of moments: For a body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments about the same point. [1]

(b) Taking moments about A: Clockwise moments = $(200 \times 2.0) + (600 \times 1.0) = 400 + 600 = 1000 \text{ N m}$ Anticlockwise moment = $T_B \times 4.0$ $T_B \times 4.0 = 1000 \Rightarrow T_B = 250 \text{ N}$ [3]

(c) For vertical equilibrium: $T_A + T_B = 200 + 600 = 800 \text{ N}$ $T_A = 800 - 250 = 550 \text{ N}$ [2]

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15

(a) Centripetal force $F_c = \frac{mv^2}{r} = \frac{0.2 \times 6.0^2}{0.5} = \frac{0.2 \times 36}{0.5} = \frac{7.2}{0.5} = 14.4 \text{ N}$ [2]

(b) At lowest point: $T - mg = F_c$ $T = F_c + mg = 14.4 + (0.2 \times 10) = 14.4 + 2 = 16.4 \text{ N}$ [2]

(c) Maximum tension = 20 N $T_{\text{max}} = \frac{mv_{\text{max}}^2}{r} + mg$ $20 = \frac{0.2 \times v_{\text{max}}^2}{0.5} + 2$ $18 = 0.4 v_{\text{max}}^2$ $v_{\text{max}}^2 = 45$ $v_{\text{max}} = \sqrt{45} = 6.71 \text{ m/s}$ [2]

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16

(a) Volume of cube = $(0.1)^3 = 0.001 \text{ m}^3$ Mass = density × volume = $8000 \times 0.001 = 8 \text{ kg}$ Weight = $mg = 8 \times 10 = 80 \text{ N}$ [2]

(b) Upthrust = weight of displaced liquid = $\rho_{\text{liquid}} V g = 1200 \times 0.001 \times 10 = 12 \text{ N}$ [2]

(c) Spring balance reading = Tension = Weight − Upthrust = $80 - 12 = 68 \text{ N}$ [2]

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Section C: Longer Structured Questions [20 marks]

17

(a) Initial weight = $mg = 500 \times 10 = 5000 \text{ N}$ [1]

(b) Thrust = rate of mass ejection × exhaust velocity = $2.0 \times 800 = 1600 \text{ N}$ [2]

(c) Resultant force = Thrust − Weight = $1600 - 5000 = -3400 \text{ N}$ Acceleration = $\frac{F}{m} = \frac{-3400}{500} = -6.8 \text{ m/s}^2$ [2] (Note: Negative means rocket doesn't lift off initially! This is a trick question showing thrust < weight.)

(d) This scenario is physically impossible as rocket cannot lift off with thrust < weight. However, if we assume it could: Using rocket equation: $v = u \ln\left(\frac{m_0}{m}\right) - gt$ $v = 800 \ln\left(\frac{500}{400}\right) - 10 \times 50$ $v = 800 \ln(1.25) - 500 = 800 \times 0.223 - 500 = 178.4 - 500 = -321.6 \text{ m/s}$ [3] (Again negative, confirming rocket doesn't lift off.)

(e) Acceleration increases because mass decreases while thrust remains constant ($a = \frac{F_{\text{thrust}} - mg}{m}$). As $m$ decreases, the thrust-to-weight ratio increases and the denominator decreases, both increasing acceleration. [2]

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18

(a) Height $h = L(1 - \cos\theta) = 1.2 \times (1 - \cos 30°) = 1.2 \times (1 - 0.866) = 1.2 \times 0.134 = 0.1608 \text{ m}$ [2]

(b) Loss in GPE = Gain in KE (assuming no air resistance for max speed) $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.1608} = \sqrt{3.216} = 1.79 \text{ m/s}$ [2]

(c) Initial energy = $mgh = 0.5 \times 10 \times 0.1608 = 0.804 \text{ J}$ Final energy at 0.08 m height = $mg \times 0.08 = 0.5 \times 10 \times 0.08 = 0.4 \text{ J}$ Energy lost = $0.804 - 0.4 = 0.404 \text{ J}$ [2]

(d) Graph description:

• Period $T = 2\pi\sqrt{\frac{L}{g}} = 2\pi\sqrt{\frac{1.2}{10}} = 2.18 \text{ s}$
• Max KE = 0.804 J at $t = T/4, 3T/4, \dots$
• Min KE = 0 at $t = 0, T/2, T, \dots$
• After each half-swing, max KE decreases by 0.404 J
• Sketch: Periodic peaks decaying linearly in amplitude [3]

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19

(a) Forces on diagram:

• Weight $mg$ vertically downward
• Normal reaction $N$ perpendicular to track surface
• Friction $f$ parallel to track surface (direction down the slope for maximum speed) [2]

(b) Ideal speed (no friction): $\tan\theta = \frac{v^2}{rg}$ $v = \sqrt{rg \tan\theta} = \sqrt{80 \times 10 \times \tan 15°} = \sqrt{800 \times 0.268} = \sqrt{214.4} = 14.64 \text{ m/s} \approx 14.5 \text{ m/s}$ [3]

(c) Maximum speed without skidding up: Resolving horizontally: $N \sin\theta + f \cos\theta = \frac{mv^2}{r}$ Resolving vertically: $N \cos\theta - f \sin\theta = mg$ With $f = \mu N = 0.3N$ Substituting and solving: $N(\sin\theta + 0.3\cos\theta) = \frac{mv^2}{r}$ $N(\cos\theta - 0.3\sin\theta) = mg$ Dividing: $\frac{\sin\theta + 0.3\cos\theta}{\cos\theta - 0.3\sin\theta} = \frac{v^2}{rg}$ $\frac{\tan\theta + 0.3}{1 - 0.3\tan\theta} = \frac{v^2}{rg}$ $\frac{0.268 + 0.3}{1 - 0.3 \times 0.268} = \frac{0.568}{0.9196} = 0.6176$ $v^2 = 0.6176 \times 80 \times 10 = 494.1$ $v = 22.2 \text{ m/s}$ [4]

(d) The car will skid up the bank (move outward/up the slope). [1]

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20

(a) Plot points: (0,0), (0.2,10), (0.4,20), (0.6,30), (0.8,40), (1.1,50), (1.5,60) Best-fit straight line through first 5 points (0 to 40 N), curve deviates after. [3]

(b) Spring constant $k = \frac{F}{x}$ (gradient of linear region) Using first 5 points: $k = \frac{40}{0.8 \times 10^{-3}} = 50000 \text{ N/m}$ [2]

(c) Young modulus $E = \frac{FL}{Ax}$ $A = \pi r^2 = \pi (0.25 \times 10^{-3})^2 = 1.96 \times 10^{-7} \text{ m}^2$ Using linear region: $E = \frac{kL}{A} = \frac{50000 \times 2.0}{1.96 \times 10^{-7}} = 5.1 \times 10^{11} \text{ Pa}$ [3]

(d) Deviation indicates the wire has exceeded its limit of proportionality / elastic limit. The wire is undergoing plastic deformation and will not return to its original length when unloaded. [2]

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End of Answer Key