Secondary 3 Physics Practice Paper 4

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

Subject: Physics
Level: Secondary 3 (Pure Physics)
Paper: Practice Paper — Mechanics (Version 4)
Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

Question 1 [1 mark]

Answer: B

Working:

• Initial velocity $u = 0$, final velocity $v = 20 \text{ m/s}$, time $t = 5.0 \text{ s}$
• Acceleration $a = \frac{v-u}{t} = \frac{20}{5} = 4 \text{ m/s}^2$
• Distance $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 4 \times 5^2 = 50 \text{ m}$

Alternatively: Average velocity = $\frac{0+20}{2} = 10 \text{ m/s}$, distance = $10 \times 5 = 50 \text{ m}$.

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Question 2 [1 mark]

Answer: B

Working:

• At maximum height, final velocity $v = 0$
• $v^2 = u^2 - 2gh$ (taking upward as positive)
• $0 = 15^2 - 2(10)h$
• $20h = 225$
• $h = 11.25 \text{ m}$

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Question 3 [1 mark]

Answer: B

Working:

• Net force $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 10 - 4 = 6 \text{ N}$
• $a = \frac{F_{\text{net}}}{m} = \frac{6}{2} = 3 \text{ m/s}^2$

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Question 4 [1 mark]

Answer: B

Working:

• For perpendicular forces: $F_{\text{resultant}} = \sqrt{F_1^2 + F_2^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N}$

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Question 5 [1 mark]

Answer: B

Working:

• Taking moments about the pivot (50 cm mark):
• Anticlockwise moment = $2.0 \times (50 - 20) = 2.0 \times 30 = 60 \text{ N cm}$
• Clockwise moment = $3.0 \times (x - 50)$
• For equilibrium: $3.0(x - 50) = 60$
• $x - 50 = 20$
• $x = 70 \text{ cm}$

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Question 6 [1 mark]

Answer: A

Working:

• Newton's law of gravitation: $F = \frac{GMm}{r^2}$
• $F \propto \frac{1}{r^2}$
• If $r$ doubles to $2r$, $F_{\text{new}} = \frac{F}{(2)^2} = \frac{F}{4}$

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Question 7 [1 mark]

Answer: A

Working:

• Moment = $F \times d_{\perp} = F \times (L \sin\theta)$
• $= 20 \times (0.25 \times \sin 30^\circ)$
• $= 20 \times (0.25 \times 0.5)$
• $= 20 \times 0.125 = 2.5 \text{ N m}$

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Question 8 [1 mark]

Answer: C

Working:

• Centripetal force $F_c = \frac{mv^2}{r} = \frac{1200 \times 25^2}{50} = \frac{1200 \times 625}{50} = 1200 \times 12.5 = 15000 \text{ N}$

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Question 9 [1 mark]

Answer: C

Working:

• Conservation of momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
• $(0.5 \times 10) + (1.0 \times 0) = (0.5 \times -2) + (1.0 \times v_2)$
• $5 = -1 + v_2$
• $v_2 = 6 \text{ m/s}$ (forward)

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Question 10 [1 mark]

Answer: B

Working:

• Work done = Gain in gravitational potential energy = $mgh = 500 \times 10 \times 20 = 100000 \text{ J}$
• Power = $\frac{\text{Work}}{\text{time}} = \frac{100000}{40} = 2500 \text{ W}$

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Section B: Structured Questions [30 marks]

Question 11 [4 marks]

(a) [2 marks]

• Description: The skydiver accelerates from rest with a constant acceleration (straight line on v-t graph) for the first 10 s.
• Explanation: Initially, the only significant force is weight ($mg$ downwards). Air resistance is small at low speeds. As speed increases, air resistance increases, reducing the net downward force and thus reducing acceleration. The gradient of the v-t graph decreases after 10 s, showing decreasing acceleration.

Marking: 1 mark for "constant acceleration" or "uniform acceleration"; 1 mark for explaining that air resistance increases with speed, reducing net force and acceleration.

(b) [1 mark]

• Acceleration = gradient of v-t graph at $t = 5 \text{ s}$
• Gradient = $\frac{50 - 0}{10 - 0} = 5 \text{ m/s}^2$

(c) [1 mark]

• Terminal velocity before parachute opens = $55 \text{ m/s}$ (horizontal section from 30 s to 40 s)

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Question 12 [5 marks]

(a) [2 marks] Forces on diagram (already shown in placeholder description):

1. Weight $W = mg$ vertically downwards
2. Normal reaction $N$ perpendicular to plane (outwards)
3. Friction $f$ down the plane (opposing motion up the plane)
4. Applied force $F$ up the plane

Marking: 1 mark for all four forces correctly drawn with correct directions; 1 mark for correct labels (W, N, f, F).

(b) [1 mark]

• $N = mg \cos\theta = 15 \times 10 \times \cos 30^\circ = 150 \times 0.866 = 129.9 \text{ N} \approx 130 \text{ N}$

(c) [1 mark]

• $f = \mu N = 0.25 \times 129.9 = 32.5 \text{ N}$

(d) [1 mark]

• Constant speed $\Rightarrow$ net force parallel to plane = 0
• $F = mg \sin\theta + f = 150 \times \sin 30^\circ + 32.5 = 150 \times 0.5 + 32.5 = 75 + 32.5 = 107.5 \text{ N}$

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Question 13 [4 marks]

(a) [2 marks]

• Anticlockwise moment (2.0 N weight): $2.0 \times 0.35 = 0.70 \text{ N m}$
• Clockwise moment (3.0 N weight): $3.0 \times 0.35 = 1.05 \text{ N m}$
• Wait — these are not equal! The rule is NOT in equilibrium with these positions as stated. Let me re-read the question...

Correction: The question states "The rule is balanced horizontally when..." but the distances given (15 cm and 85 cm from ends, pivot at 50 cm) give distances of 35 cm each side. Moments: $2.0 \times 0.35 = 0.70 \text{ N m}$ anticlockwise, $3.0 \times 0.35 = 1.05 \text{ N m}$ clockwise. These don't balance.

This appears to be an error in the question setup. For the rule to balance with these weights at equal distances from pivot, the weights would need to be equal. Let me adjust the answer to reflect what the question likely intended, or note the discrepancy.

Revised interpretation: Perhaps the weights are at 15 cm and 85 cm marks (from the zero end), so distances from pivot (50 cm) are 35 cm and 35 cm. For equilibrium, we need $W_1 d_1 = W_2 d_2$. With $d_1 = d_2$, we need $W_1 = W_2$. Since $2.0 \neq 3.0$, the rule cannot balance as described.

For the answer key, I'll assume the question meant the 2.0 N is at 20 cm mark (30 cm from pivot) and 3.0 N at 80 cm mark (30 cm from pivot) — but that still doesn't balance. Or perhaps the pivot is not at 50 cm? The question says "pivoted at its centre" which is 50 cm.

Let me provide the calculation as requested and note the issue.

Answer:

• Anticlockwise moment = $2.0 \text{ N} \times 0.35 \text{ m} = 0.70 \text{ N m}$
• Clockwise moment = $3.0 \text{ N} \times 0.35 \text{ m} = 1.05 \text{ N m}$
• Since $0.70 \neq 1.05$, the rule is not in equilibrium with the given positions. There appears to be an inconsistency in the question data.

Marking: 1 mark for correct moment calculations; 1 mark for stating they are not equal / rule not in equilibrium.

(b) [2 marks]

• New position of 2.0 N weight: 10 cm mark $\Rightarrow$ distance from pivot = $50 - 10 = 40 \text{ cm} = 0.40 \text{ m}$
• Let $x$ be the distance of 3.0 N weight from pivot (on the right side)
• For equilibrium: $2.0 \times 0.40 = 3.0 \times x$
• $0.80 = 3.0x$
• $x = 0.267 \text{ m} = 26.7 \text{ cm}$ from pivot
• Position on rule = $50 + 26.7 = 76.7 \text{ cm}$ mark $\approx 77 \text{ cm}$ mark

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Question 14 [5 marks]

(a) [1 mark]

• $v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 2.0 = 40$
• $v = \sqrt{40} = 6.32 \text{ m/s}$ (downwards)

(b) [1 mark]

• Rebound: $v^2 = u^2 - 2gh$ (upward positive), at max height $v=0$
• $0 = u^2 - 2 \times 10 \times 1.2$
• $u^2 = 24$
• $u = \sqrt{24} = 4.90 \text{ m/s}$ (upwards)

(c) [1 mark]

• Take upward as positive.
• Initial momentum (before) = $m \times (-6.32) = 0.2 \times (-6.32) = -1.264 \text{ kg m/s}$
• Final momentum (after) = $m \times 4.90 = 0.2 \times 4.90 = 0.98 \text{ kg m/s}$
• Change in momentum = $p_f - p_i = 0.98 - (-1.264) = 2.244 \text{ kg m/s} \approx 2.24 \text{ kg m/s}$ (upwards)

(d) [2 marks]

• Average force $F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{2.244}{0.02} = 112.2 \text{ N}$ (upwards)
• Note: This is the net force. The force from the floor = net force + weight = $112.2 + (0.2 \times 10) = 114.2 \text{ N}$.
• However, typically in such questions, "average force exerted by the floor" means the normal reaction force, which includes supporting the weight. But many exam questions just ask for $\Delta p / \Delta t$ as the average force. I'll provide both with explanation.

Marking: 1 mark for $\Delta p / \Delta t = 112 \text{ N}$; 1 mark for adding weight to get $114 \text{ N}$ (or stating assumption).

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Question 15 [5 marks]

(a) [2 marks] Forces on car at maximum speed without skidding (up the bank):

1. Weight $mg$ vertically down
2. Normal reaction $N$ perpendicular to surface (up and left)
3. Friction $f = \mu N$ down the incline (towards centre of circle, preventing sliding up)

Marking: 1 mark for three forces with correct directions; 1 mark for correct labels and friction direction down the incline.

(b) [2 marks] Resolving horizontally (towards centre): $N \sin\theta + f \cos\theta = \frac{mv^2}{r}$

Resolving vertically: $N \cos\theta - f \mu f \sin\theta = mg$

Substitute $f = \mu N$: $N(\sin\theta + \mu \cos\theta) = \frac{mv^2}{r}$ ...(1) $N(\cos\theta - \mu \sin\theta) = mg$ ...(2)

Divide (1) by (2): $\frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta} = \frac{v^2}{rg}$

$v_{\text{max}}^2 = rg \frac{\sin\theta + \mu \cos\theta}{\cos\theta - \mu \sin\theta} = rg \frac{\tan\theta + \mu}{1 - \mu \tan\theta}$

(c) [1 mark]

• $\tan 15^\circ = 0.268$
• $v_{\text{max}}^2 = 80 \times 10 \times \frac{0.268 + 0.3}{1 - 0.3 \times 0.268} = 800 \times \frac{0.568}{1 - 0.0804} = 800 \times \frac{0.568}{0.9196} = 800 \times 0.6177 = 494.1$
• $v_{\text{max}} = \sqrt{494.1} = 22.2 \text{ m/s}$

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Question 16 [4 marks]

(a) [1 mark]

• Thrust $F = v_{\text{exhaust}} \times \frac{dm}{dt} = 2000 \times 50 = 100000 \text{ N} = 1.0 \times 10^5 \text{ N}$

(b) [2 marks]

• Initial weight = $mg = 5000 \times 10 = 50000 \text{ N}$
• Net force = Thrust - Weight = $100000 - 50000 = 50000 \text{ N}$
• Initial acceleration $a = \frac{F_{\text{net}}}{m} = \frac{50000}{5000} = 10 \text{ m/s}^2$

(c) [1 mark]

• As fuel is ejected, the mass of the rocket decreases while thrust remains constant.
• Since $a = \frac{F_{\text{thrust}} - mg}{m} = \frac{F_{\text{thrust}}}{m} - g$, as $m$ decreases, $\frac{F_{\text{thrust}}}{m}$ increases, so acceleration increases.

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Question 17 [4 marks]

(a) [1 mark]

• $h = L - L\cos\theta = L(1 - \cos\theta) = 1.0 \times (1 - \cos 30^\circ) = 1.0 \times (1 - 0.866) = 0.134 \text{ m}$

(b) [1 mark]

• Loss in GPE = Gain in KE
• $mgh = \frac{1}{2}mv^2$
• $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.134} = \sqrt{2.68} = 1.64 \text{ m/s}$

(c) [2 marks]

• At lowest point, forces on bob: Tension $T$ up, weight $mg$ down.
• Net upward force = centripetal force: $T - mg = \frac{mv^2}{L}$
• $T = mg + \frac{mv^2}{L} = 0.5 \times 10 + \frac{0.5 \times (1.64)^2}{1.0} = 5 + 0.5 \times 2.68 = 5 + 1.34 = 6.34 \text{ N}$

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Section C: Longer Structured Questions [20 marks]

Question 18 [7 marks]

(a) [2 marks] Forces on Block A (on table):

• Weight $W_A = 40 \text{ N}$ down
• Normal reaction $N = 40 \text{ N}$ up
• Tension $T$ to the right
• Friction $f = \mu N = 0.2 \times 40 = 8 \text{ N}$ to the left

Forces on Block B (hanging):

• Weight $W_B = 20 \text{ N}$ down
• Tension $T$ up

*Marking: 1 mark for all forces on A correct; 1

<stage5_exam_answers_md>

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

Subject: Physics
Level: Secondary 3 (Pure Physics)
Paper: Practice Paper — Mechanics (Version 4)
Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

Question 1 [1 mark]

Answer: B

Working:

• Initial velocity $u = 0$, final velocity $v = 20 \text{ m/s}$, time $t = 5.0 \text{ s}$
• Acceleration $a = \frac{v-u}{t} = \frac{20}{5} = 4 \text{ m/s}^2$
• Distance $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 4 \times 5^2 = 50 \text{ m}$

Alternatively: Average velocity = $\frac{0+20}{2} = 10 \text{ m/s}$, distance = $10 \times 5 = 50 \text{ m}$.

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Question 2 [1 mark]

Answer: B

Working:

• At maximum height, final velocity $v = 0$
• $v^2 = u^2 - 2gh$ (taking upward as positive)
• $0 = 15^2 - 2(10)h$
• $20h = 225$
• $h = 11.25 \text{ m}$

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Question 3 [1 mark]

Answer: B

Working:

• Net force $F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 10 - 4 = 6 \text{ N}$
• Acceleration $a = \frac{F_{\text{net}}}{m} = \frac{6}{2} = 3.0 \text{ m/s}^2$

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Question 4 [1 mark]

Answer: B

Working:

• Forces at $90^\circ$: resultant $R = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N}$

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Question 5 [1 mark]

Answer: B

Working:

• Anticlockwise moment = $2.0 \times (50 - 20) = 2.0 \times 30 = 60 \text{ N cm}$
• Clockwise moment = $3.0 \times (x - 50)$
• For equilibrium: $3.0(x - 50) = 60$
• $x - 50 = 20$
• $x = 70 \text{ cm}$

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Question 6 [1 mark]

Answer: A

Working:

• Gravitational force $F \propto \frac{1}{r^2}$
• At distance $2r$, force $F' = F \times \left(\frac{r}{2r}\right)^2 = \frac{F}{4}$

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Question 7 [1 mark]

Answer: A

Working:

• Moment = $F \times d_{\perp} = F \times (L \sin\theta)$
• $= 20 \times (0.25 \times \sin 30^\circ) = 20 \times (0.25 \times 0.5) = 20 \times 0.125 = 2.5 \text{ N m}$

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Question 8 [1 mark]

Answer: C

Working:

• Centripetal force $F_c = \frac{mv^2}{r} = \frac{1200 \times 25^2}{50} = \frac{1200 \times 625}{50} = 1200 \times 12.5 = 15000 \text{ N}$

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Question 9 [1 mark]

Answer: C

Working:

• Conservation of momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
• $(0.5 \times 10) + (1.0 \times 0) = (0.5 \times -2) + (1.0 \times v_2)$
• $5 = -1 + v_2$
• $v_2 = 6 \text{ m/s}$ (forward)

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Question 10 [1 mark]

Answer: B

Working:

• Work done = $mgh = 500 \times 10 \times 20 = 100000 \text{ J}$
• Power = $\frac{\text{Work}}{\text{time}} = \frac{100000}{40} = 2500 \text{ W}$

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Section B: Structured Questions [30 marks]

Question 11 [4 marks]

(a) Between $t = 0 \text{ s}$ and $t = 10 \text{ s}$, the skydiver accelerates from rest. The velocity increases uniformly from $0$ to $50 \text{ m/s}$, indicating constant acceleration. The acceleration is constant because the only significant force initially is weight (air resistance is negligible at low speeds), so the net force is constant ($mg$). [2 marks]

(b) Acceleration at $t = 5 \text{ s}$ = gradient of graph from $t=0$ to $t=10$
$a = \frac{\Delta v}{\Delta t} = \frac{50 - 0}{10 - 0} = 5.0 \text{ m/s}^2$ [1 mark]

(c) Terminal velocity before parachute opens = $55 \text{ m/s}$ (horizontal section from $t=30$ to $t=40 \text{ s}$) [1 mark]

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Question 12 [5 marks]

(a) Forces on the box (already shown in diagram):

• Weight $W = mg$ vertically downwards
• Normal reaction $N$ perpendicular to plane
• Friction $f$ down the plane (opposing motion)
• Applied force $F$ up the plane [2 marks]

(b) Normal reaction $N = mg\cos\theta = 15 \times 10 \times \cos 30^\circ = 150 \times 0.866 = 129.9 \text{ N} \approx 130 \text{ N}$ [1 mark]

(c) Frictional force $f = \mu N = 0.25 \times 129.9 = 32.5 \text{ N}$ [1 mark]

(d) For constant speed, net force parallel to plane = 0
$F = mg\sin\theta + f = 15 \times 10 \times \sin 30^\circ + 32.5 = 150 \times 0.5 + 32.5 = 75 + 32.5 = 107.5 \text{ N}$ [1 mark]

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Question 13 [4 marks]

(a) Anticlockwise moment = $2.0 \times 0.35 = 0.70 \text{ N m}$
Clockwise moment = $3.0 \times 0.35 = 1.05 \text{ N m}$
Wait — the rule is stated to be balanced, but moments are not equal!
Correction: The $2.0 \text{ N}$ weight is at $15 \text{ cm}$ mark (distance from pivot = $35 \text{ cm} = 0.35 \text{ m}$). The $3.0 \text{ N}$ weight is at $85 \text{ cm}$ mark (distance from pivot = $35 \text{ cm} = 0.35 \text{ m}$).
Anticlockwise moment = $2.0 \times 0.35 = 0.70 \text{ N m}$
Clockwise moment = $3.0 \times 0.35 = 1.05 \text{ N m}$
These are not equal, so the rule as described would not be in equilibrium.
However, if the question states it is balanced, there may be an error in the problem statement. Assuming the distances are correct as given, the moments are unequal.
For the purpose of this answer key, we note the discrepancy. [2 marks]

(b) New position of $2.0 \text{ N}$ weight: $10 \text{ cm}$ mark → distance from pivot = $40 \text{ cm} = 0.40 \text{ m}$
Let $x$ be the distance of $3.0 \text{ N}$ weight from pivot (on the right side).
For equilibrium: $2.0 \times 0.40 = 3.0 \times x$
$0.80 = 3.0x$
$x = 0.267 \text{ m} = 26.7 \text{ cm}$ from pivot
Position on rule = $50 + 26.7 = 76.7 \text{ cm}$ mark [2 marks]

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Question 14 [5 marks]

(a) Speed just before impact: $v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 2.0 = 40$
$v = \sqrt{40} = 6.32 \text{ m/s}$ (downwards) [1 mark]

(b) Speed just after rebound: $v^2 = u^2 + 2gh$ (upwards, $v=0$ at max height)
$0 = u^2 - 2 \times 10 \times 1.2$
$u^2 = 24$
$u = \sqrt{24} = 4.90 \text{ m/s}$ (upwards) [1 mark]

(c) Change in momentum = $m(v_{\text{final}} - v_{\text{initial}})$
Take upward as positive:
$v_{\text{initial}} = -6.32 \text{ m/s}$, $v_{\text{final}} = +4.90 \text{ m/s}$
$\Delta p = 0.2 \times (4.90 - (-6.32)) = 0.2 \times 11.22 = 2.24 \text{ kg m/s}$ [1 mark]

(d) Average force $F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{2.24}{0.02} = 112 \text{ N}$ (upwards) [2 marks]

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Question 15 [5 marks]

(a) Forces on car at maximum speed (no skidding up the bank):

• Weight $mg$ vertically down
• Normal reaction $N$ perpendicular to surface
• Friction $f$ down the incline (prevents sliding up) [2 marks]

(b) Resolving vertically: $N\cos\theta - f\sin\theta = mg$
Resolving horizontally (towards centre): $N\sin\theta + f\cos\theta = \frac{mv^2}{R}$
With $f = \mu N$ (maximum friction):
$N(\cos\theta - \mu\sin\theta) = mg$ → $N = \frac{mg}{\cos\theta - \mu\sin\theta}$
Substitute into horizontal:
$\frac{mg}{\cos\theta - \mu\sin\theta} (\sin\theta + \mu\cos\theta) = \frac{mv_{\text{max}}^2}{R}$
$v_{\text{max}}^2 = gR \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}$
$v_{\text{max}} = \sqrt{gR \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}}$ [2 marks]

(c) $\theta = 15^\circ$, $\mu = 0.3$, $R = 80 \text{ m}$, $g = 10 \text{ m/s}^2$
$\sin 15^\circ \approx 0.259$, $\cos 15^\circ \approx 0.966$
Numerator: $0.259 + 0.3 \times 0.966 = 0.259 + 0.290 = 0.549$
Denominator: $0.966 - 0.3 \times 0.259 = 0.966 - 0.078 = 0.888$
$v_{\text{max}}^2 = 10 \times 80 \times \frac{0.549}{0.888} = 800 \times 0.618 = 494.6$
$v_{\text{max}} = \sqrt{494.6} \approx 22.2 \text{ m/s}$ [1 mark]

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Question 16 [4 marks]

(a) Thrust $F_{\text{thrust}} = \text{rate of mass ejection} \times \text{exhaust speed relative to rocket}$
$= 50 \text{ kg/s} \times 2000 \text{ m/s} = 100000 \text{ N}$ [1 mark]

(b) Initial weight = $mg = 5000 \times 10 = 50000 \text{ N}$
Net force = Thrust - Weight = $100000 - 50000 = 50000 \text{ N}$
Initial acceleration $a = \frac{F_{\text{net}}}{m} = \frac{50000}{5000} = 10 \text{ m/s}^2$ [2 marks]

(c) As fuel is consumed, the mass of the rocket decreases while thrust remains constant. Since $a = \frac{F_{\text{net}}}{m}$ and $F_{\text{net}} = \text{Thrust} - mg$, both the decreasing mass and decreasing weight cause the net force to increase relative to mass, so acceleration increases. [1 mark]

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Question 17 [4 marks]

(a) Vertical height fallen: $h = L(1 - \cos\theta) = 1.0 \times (1 - \cos 30^\circ) = 1 - 0.866 = 0.134 \text{ m}$ [1 mark]

(b) Loss in GPE = Gain in KE: $mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.134} = \sqrt{2.68} = 1.64 \text{ m/s}$ [1 mark]

(c) At lowest point, centripetal force required = $\frac{mv^2}{L}$
Tension $T - mg = \frac{mv^2}{L}$
$T = mg + \frac{mv^2}{L} = 0.5 \times 10 + \frac{0.5 \times (1.64)^2}{1.0} = 5 + 1.34 = 6.34 \text{ N}$ [2 marks]

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Section C: Longer Structured Questions [20 marks]

Question 18 [7 marks]

(a) Forces on Block A (on table):

• Weight $W_A = 40 \text{ N}$ down
• Normal reaction $N$ up
• Tension $T$ to the right
• Friction $f$ to the left

Forces on Block B (hanging):

• Weight $W_B = 20 \text{ N}$ down
• Tension $T$ up [2 marks]

(b) Friction on A: $f = \mu N = \mu m_A g = 0.2 \times 4.0 \times 10 = 8.0 \text{ N}$ [1 mark]

(c) Block A (horizontal): $T - f = m_A a$ → $T - 8 = 4a$
Block B (vertical): $W_B - T = m_B a$ → $20 - T = 2a$ [2 marks]

(d) Add equations: $(T - 8) + (20 - T) = 4a + 2a$
$12 = 6a$ → $a = 2.0 \text{ m/s}^2$
Substitute: $T - 8 = 4 \times 2 = 8$ → $T = 16 \text{ N}$ [2 marks]

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Question 19 [7 marks]

(a) Horizontal component: $u_x = u\cos\theta = 40 \cos 30^\circ = 40 \times 0.866 = 34.6 \text{ m/s}$
Vertical component: $u_y = u\sin\theta = 40 \sin 30^\circ = 40 \times 0.5 = 20 \text{ m/s}$ [2 marks]

(b) Time to max height: $v_y = u_y - gt = 0$
$t = \frac{u_y}{g} = \frac{20}{10} = 2.0 \text{ s}$ [1 mark]

(c) Max height: $h = \frac{u_y^2}{2g} = \frac{20^2}{2 \times 10} = \frac{400}{20} = 20 \text{ m}$ [1 mark]

(d) Total time of flight = $2 \times \text{time to max height} = 2 \times 2.0 = 4.0 \text{ s}$ [1 mark]

(e) Horizontal range = $u_x \times \text{total time} = 34.6 \times 4.0 = 138.4 \text{ m}$ [1 mark]

(f) Velocity-time graph for vertical component:

• Straight line with negative gradient ($-g$) from $(0, 20)$ to $(2, 0)$
• Continues with same gradient to $(4, -20)$
• Axes: $v_y$ (m/s) vs $t$ (s)
• Key points labelled: $(0, 20)$, $(2, 0)$, $(4, -20)$ [1 mark]

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Question 20 [6 marks]

(a) Taking moments about A (clockwise positive):

• Beam weight: $200 \times 2.0 = 400 \text{ N m}$ (clockwise)
• Load: $300 \times 1.0 = 300 \text{ N m}$ (clockwise)
• Cable tension: $T \sin 30^\circ \times 4.0 = T \times 0.5 \times 4.0 = 2.0T$ (anticlockwise)

Equilibrium: $2.0T = 400 + 300 = 700$
$T = 350 \text{ N}$ [3 marks]

(b) Horizontal forces: $H = T \cos 30^\circ = 350 \times 0.866 = 303 \text{ N}$ (towards wall)
Vertical forces: $V + T \sin 30^\circ = 200 + 300 = 500$
$V + 350 \times 0.5 = 500$
$V + 175 = 500$
$V = 325 \text{ N}$ (upwards) [3 marks]

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End of Answer Key