Secondary 3 Physics Practice Paper 3

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

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

---

Section A: Multiple Choice Questions [10 marks]

Question 1 [1 mark]

Answer: B

Working:
$a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4 \text{ m/s}^2$

Key concept: Acceleration is the rate of change of velocity. For uniform acceleration from rest, $a = v/t$.

---

Question 2 [1 mark]

Answer: C

Working:
Distance = Area under velocity-time graph
= Area of triangle (0–3 s) + Area of rectangle (3–7 s) + Area of triangle (7–10 s)
= $\frac{1}{2} \times 3 \times 6 + 4 \times 6 + \frac{1}{2} \times 3 \times 6$
= $9 + 24 + 9 = 42 \text{ m}$

Wait, let me recalculate:
Triangle 1: $\frac{1}{2} \times 3 \times 6 = 9$
Rectangle: $4 \times 6 = 24$
Triangle 2: $\frac{1}{2} \times 3 \times 6 = 9$
Total = $9 + 24 + 9 = 42 \text{ m}$

Answer: D (42 m)

Correction: The correct answer is D. The area calculation gives 42 m.

Key concept: Distance travelled = area under velocity-time graph. For uniform acceleration/deceleration phases, use triangle area; for constant velocity, use rectangle area.

---

Question 3 [1 mark]

Answer: B

Working:
Net force $F_{\text{net}} = 15 - 5 = 10 \text{ N}$
$a = \frac{F_{\text{net}}}{m} = \frac{10}{2} = 5 \text{ m/s}^2$

Key concept: Newton's Second Law: $F_{\text{net}} = ma$. The net force is the applied force minus friction.

---

Question 4 [1 mark]

Answer: B

Working:
At maximum height, $v = 0$.
Using $v^2 = u^2 + 2as$:
$0 = 25^2 + 2(-10)h$
$0 = 625 - 20h$
$20h = 625$
$h = 31.25 \text{ m}$

Key concept: For vertical motion under gravity, use $v^2 = u^2 + 2as$ with $a = -g$. At maximum height, final velocity is zero.

---

Question 5 [1 mark]

Answer: B

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

Key concept: Resultant of two perpendicular forces is found using Pythagoras' theorem: $R = \sqrt{F_1^2 + F_2^2}$.

---

Question 6 [1 mark]

Answer: B

Working:
$m = 500 \text{ g} = 0.5 \text{ kg}$
$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times 4^2 = 0.25 \times 16 = 4 \text{ J}$

Key concept: Kinetic energy $KE = \frac{1}{2}mv^2$. Always convert mass to kg.

---

Question 7 [1 mark]

Answer: B

Working:
Moment = Force $\times$ perpendicular distance = $20 \times 0.25 = 5 \text{ N m}$

Key concept: Moment of a force = force $\times$ perpendicular distance from pivot. Unit is N m (not J).

---

Question 8 [1 mark]

Answer: B

Working:
Taking moments about the pivot (50 cm mark):
Clockwise moment = Anticlockwise moment
$2 \times (50 - 20) = 4 \times (x - 50)$
$2 \times 30 = 4(x - 50)$
$60 = 4x - 200$
$4x = 260$
$x = 65 \text{ cm}$

Key concept: For equilibrium, sum of clockwise moments = sum of anticlockwise moments about any point. Distances measured from pivot.

---

Question 9 [1 mark]

Answer: B

Working:
Centripetal force $F_c = \frac{mv^2}{r} = \frac{1000 \times 20^2}{50} = \frac{1000 \times 400}{50} = 8000 \text{ N}$

Key concept: Centripetal force $F_c = \frac{mv^2}{r} = m\omega^2r$. Directed towards centre of circle.

---

Question 10 [1 mark]

Answer: A

Explanation:
In uniform circular motion, speed is constant, so kinetic energy ($\frac{1}{2}mv^2$) is constant. However, velocity direction changes continuously, so momentum ($mv$) changes direction (though magnitude is constant). Momentum is a vector quantity.

Key concept: Kinetic energy is a scalar (depends on speed only); momentum is a vector (depends on velocity). In uniform circular motion, speed is constant but velocity direction changes.

---

Section B: Structured Questions [30 marks]

Question 11 [4 marks]

(a) [1 mark]
Answer: $20 \text{ m/s}$
Working: $v = u + at = 0 + 0.5 \times 40 = 20 \text{ m/s}$

(b) [1 mark]
Answer: $400 \text{ m}$
Working: $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 0.5 \times 40^2 = 0.25 \times 1600 = 400 \text{ m}$

(c) [2 marks]
Answer: $3400 \text{ m}$
Working:
Distance during constant velocity: $s_2 = v \times t = 20 \times 120 = 2400 \text{ m}$
Distance during deceleration: $s_3 = \frac{v+u}{2} \times t = \frac{20+0}{2} \times 30 = 10 \times 30 = 300 \text{ m}$
Total distance = $400 + 2400 + 300 = 3100 \text{ m}$

Wait, let me recalculate:
$s_1 = 400 \text{ m}$
$s_2 = 20 \times 120 = 2400 \text{ m}$
$s_3 = \text{average velocity} \times \text{time} = \frac{20+0}{2} \times 30 = 300 \text{ m}$
Total = $400 + 2400 + 300 = 3100 \text{ m}$

Answer: $3100 \text{ m}$

Marking notes:

• 1 mark for correct constant velocity distance
• 1 mark for correct deceleration distance and total

---

Question 12 [5 marks]

(a) [1 mark]
Answer: $2.0 \text{ m/s}$
Working: Velocity = card length / time = $0.10 / 0.05 = 2.0 \text{ m/s}$

(b) [1 mark]
Answer: $4.0 \text{ m/s}$
Working: Velocity = card length / time = $0.10 / 0.025 = 4.0 \text{ m/s}$

(c) [2 marks]
Answer: $5.0 \text{ m/s}^2$
Working:
Using $v^2 = u^2 + 2as$:
$4.0^2 = 2.0^2 + 2 \times a \times 0.80$
$16 = 4 + 1.6a$
$1.6a = 12$
$a = 7.5 \text{ m/s}^2$

Wait, let me check: $v^2 = u^2 + 2as$
$4^2 = 2^2 + 2 \times a \times 0.8$
$16 = 4 + 1.6a$
$12 = 1.6a$
$a = 7.5 \text{ m/s}^2$

Answer: $7.5 \text{ m/s}^2$

Marking notes:

• 1 mark for correct formula/substitution
• 1 mark for correct answer with unit

(d) [1 mark]
Answer: The acceleration is uniform (constant) between the two light gates. / The card passes through each light gate at constant velocity (negligible acceleration during gate passage).

---

Question 13 [4 marks]

(a) [1 mark]
Answer: $75 \text{ N}$
Working: Component down plane = $mg \sin\theta = 15 \times 10 \times \sin 30^\circ = 150 \times 0.5 = 75 \text{ N}$

(b) [2 marks]
Answer: $95 \text{ N}$
Working:
At constant speed, net force parallel to plane = 0
$F = mg \sin\theta + \text{friction} = 75 + 20 = 95 \text{ N}$

Marking notes:

• 1 mark for recognising equilibrium (net force = 0)
• 1 mark for correct calculation

(c) [1 mark]
Answer: $475 \text{ J}$
Working: Work done = $F \times s = 95 \times 5 = 475 \text{ J}$

---

Question 14 [5 marks]

(a) [1 mark]
Answer: $6.32 \text{ m/s}$ (or $2\sqrt{10} \text{ m/s}$)
Working:
$v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 2.0 = 40$
$v = \sqrt{40} = 2\sqrt{10} \approx 6.32 \text{ m/s}$

(b) [1 mark]
Answer: $4.90 \text{ m/s}$ (or $\sqrt{24} \text{ m/s}$)
Working:
$v^2 = u^2 + 2gh = 0 + 2 \times 10 \times 1.2 = 24$
$v = \sqrt{24} = 2\sqrt{6} \approx 4.90 \text{ m/s}$

(c) [2 marks]
Answer: $2.24 \text{ kg m/s}$ (or $0.2(\sqrt{40} + \sqrt{24})$)
Working:
Change in momentum = $m(v_{\text{after}} - v_{\text{before}})$
Taking upward as positive:
$v_{\text{before}} = -6.32 \text{ m/s}$ (downward)
$v_{\text{after}} = +4.90 \text{ m/s}$ (upward)
$\Delta p = 0.2 \times (4.90 - (-6.32)) = 0.2 \times 11.22 = 2.244 \text{ kg m/s}$

Marking notes:

• 1 mark for correct velocities with signs
• 1 mark for correct change in momentum calculation

(d) [1 mark]
Answer: $44.9 \text{ N}$ (or $44.88 \text{ N}$)
Working:
Average force = $\frac{\Delta p}{\Delta t} = \frac{2.244}{0.05} = 44.88 \text{ N}$

---

Question 15 [4 marks]

(a) [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) [3 marks]
Answer: $462 \text{ N}$ (or $\frac{800}{\sin 30^\circ} = 1600 \text{ N}$? Let me recalculate)

Working:
Taking moments about hinge A:
Clockwise moments:

• Beam weight: $200 \times 2.0 = 400 \text{ N m}$
• Load: $300 \times 1.0 = 300 \text{ N m}$
  Total clockwise = $700 \text{ N m}$

Anticlockwise moment from tension:
Vertical component of tension = $T \sin 30^\circ = 0.5T$
Moment = $0.5T \times 4.0 = 2T$

For equilibrium: $2T = 700$
$T = 350 \text{ N}$

Wait, let me check the geometry. Cable at 30° to beam (horizontal). Vertical component = $T \sin 30^\circ$. Perpendicular distance from A = 4.0 m.
Moment = $T \sin 30^\circ \times 4.0 = T \times 0.5 \times 4 = 2T$

Yes, $2T = 700$, so $T = 350 \text{ N}$.

Answer: $350 \text{ N}$

Marking notes:

• 1 mark for correct moment equation (clockwise = anticlockwise)
• 1 mark for correct vertical component of tension
• 1 mark for correct final answer with unit

---

Question 16 [4 marks]

(a) [1 mark]
Answer: $4000 \text{ N}$ opposite to the direction of motion (or backwards/horizontally backwards).

Explanation: At constant velocity, net force = 0. Driving force = resistive force = $4000 \text{ N}$.

(b) [1 mark]
Answer: $100,000 \text{ W}$ (or $100 \text{ kW}$ kW}$) **Working:** Power = Force$\times$velocity =$4000 \times 25 = 100,000 \text{ W}$

(c) [2 marks]
Answer: $14,732 \text{ N}$ (or approximately $14,700 \text{ N}$)
Working:
Component of weight down slope = $mg \sin 5^\circ = 1000 \times 10 \times \sin 5^\circ = 10000 \times 0.08716 = 871.6 \text{ N}$
New driving force = Resistive force + Component of weight down slope
= $4000 + 871.6 = 4871.6 \text{ N}$

Wait, the question says "resistive force remains unchanged" at 4000 N. But the driving force was 4000 N to overcome resistive force on horizontal. Now on slope, driving force must overcome resistive force PLUS component of weight down slope.

$F_{\text{new}} = 4000 + 10000 \sin 5^\circ = 4000 + 871.6 = 4871.6 \text{ N}$

Answer: $4870 \text{ N}$ (or $4872 \text{ N}$)

Marking notes:

• 1 mark for calculating weight component down slope
• 1 mark for adding to resistive force

---

Question 17 [4 marks]

(a) [1 mark]
Answer: $0.134 \text{ m}$ (or $1 - \frac{\sqrt{3}}{2} \text{ m}$)
Working:
$h = L - L\cos\theta = L(1 - \cos\theta) = 1.0 \times (1 - \cos 30^\circ) = 1 - \frac{\sqrt{3}}{2} \approx 1 - 0.866 = 0.134 \text{ m}$

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

Marking notes:

• 1 mark for energy conservation equation
• 1 mark for correct calculation

(c) [1 mark]
Answer: Gravitational potential energy → Kinetic energy → Thermal energy (heat) and sound energy (dissipated to surroundings).

---

Section C: Longer Structured Questions [20 marks]

Question 18 [7 marks]

(a) [1 mark]
Answer: $4000 \text{ kg}$
Working: Mass lost = $50 \times 20 = 1000 \text{ kg}$
Mass remaining = $5000 - 1000 = 4000 \text{ kg}$

(b) [2 marks]
Answer: $100,000 \text{ N}$
Working: Thrust = rate of mass ejection $\times$ exhaust speed = $50 \times 2000 = 100,000 \text{ N}$

Marking notes:

• 1 mark for correct formula (thrust = $\frac{dm}{dt} \times v_{\text{exhaust}}$)
• 1 mark for correct calculation with unit

(c) [2 marks]
Answer: $10 \text{ m/s}^2$
Working:
Initial weight = $5000 \times 10 = 50,000 \text{ N}$
Net force = Thrust - Weight = $100,000 - 50,000 = 50,000 \text{ N}$
Initial acceleration = $\frac{F_{\text{net}}}{m} = \frac{50,000}{5000} = 10 \text{ m/s}^2$

Marking notes:

• 1 mark for net force calculation
• 1 mark for acceleration calculation

(d) [2 marks]
Answer: As fuel is ejected, the mass of the rocket decreases

<stage5_exam_answers_md>

TuitionGoWhere Practice Paper - Physics Secondary 3 (Answer Key)

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

---

Section A: Multiple Choice Questions [10 marks]

Question 1 [1 mark]

Answer: B

Working:
$a = \frac{v - u}{t} = \frac{20 - 0}{5} = 4 \text{ m/s}^2$

Key concept: Acceleration is the rate of change of velocity. For uniform acceleration from rest, $a = v/t$.

---

Question 2 [1 mark]

Answer: D

Working:
Distance = Area under velocity-time graph
= Area of triangle (0–3 s) + Area of rectangle (3–7 s) + Area of triangle (7–10 s)
= $\frac{1}{2} \times 3 \times 6 + 4 \times 6 + \frac{1}{2} \times 3 \times 6$
= $9 + 24 + 9 = 42 \text{ m}$

Key concept: Distance travelled = area under velocity-time graph. For uniform acceleration/deceleration, area is triangular; for constant velocity, area is rectangular.

---

Question 3 [1 mark]

Answer: B

Working:
Net force = Applied force – Friction = $15 - 5 = 10 \text{ N}$
$a = \frac{F_{\text{net}}}{m} = \frac{10}{2} = 5 \text{ m/s}^2$

Key concept: Newton's Second Law: $F_{\text{net}} = ma$. Friction opposes motion.

---

Question 4 [1 mark]

Answer: B

Working:
At maximum height, $v = 0$.
$v^2 = u^2 + 2as$
$0 = 25^2 + 2(-10)s$
$625 = 20s$
$s = 31.25 \text{ m}$

Key concept: For vertical motion under gravity, use $v^2 = u^2 + 2as$ with $a = -g$.

---

Question 5 [1 mark]

Answer: B

Working:
Resultant force = $\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N}$

Key concept: For perpendicular forces, resultant magnitude = $\sqrt{F_1^2 + F_2^2}$ (Pythagoras' theorem).

---

Question 6 [1 mark]

Answer: B

Working:
$m = 500 \text{ g} = 0.5 \text{ kg}$
$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5 \times 4^2 = 0.25 \times 16 = 4 \text{ J}$

Key concept: Kinetic energy $KE = \frac{1}{2}mv^2$. Mass must be in kg.

---

Question 7 [1 mark]

Answer: B

Working:
Moment = Force × Perpendicular distance = $20 \times 0.25 = 5 \text{ N m}$

Key concept: Moment of a force = Force × Perpendicular distance from pivot. Unit: N m.

---

Question 8 [1 mark]

Answer: C

Working:
Taking moments about pivot (50 cm mark):
Clockwise moment = Anticlockwise moment
$2 \times (50 - 20) = 4 \times (x - 50)$
$2 \times 30 = 4(x - 50)$
$60 = 4x - 200$
$4x = 260$
$x = 65 \text{ cm}$

Key concept: Principle of moments: For equilibrium, sum of clockwise moments = sum of anticlockwise moments about any pivot.

---

Question 9 [1 mark]

Answer: B

Working:
$F_c = \frac{mv^2}{r} = \frac{1000 \times 20^2}{50} = \frac{1000 \times 400}{50} = 8000 \text{ N}$

Key concept: Centripetal force $F_c = \frac{mv^2}{r} = m\omega^2 r$. Directed towards centre of circle.

---

Question 10 [1 mark]

Answer: A

Explanation:
In 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.

Key concept: Kinetic energy is a scalar (constant if speed constant). Momentum is a vector (changes if direction changes).

---

Section B: Structured Questions [30 marks]

Question 11 [4 marks]

(a) Maximum velocity:
$v = u + at = 0 + 0.5 \times 40 = 20 \text{ m/s}$
Answer: 20 m/s [1 mark]

(b) Distance during acceleration:
$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 0.5 \times 40^2 = 0.25 \times 1600 = 400 \text{ m}$
Answer: 400 m [1 mark]

(c) Total distance:
Distance at constant velocity = $v \times t = 20 \times 120 = 2400 \text{ m}$
Distance during deceleration:
Deceleration $a = \frac{0 - 20}{30} = -\frac{2}{3} \text{ m/s}^2$
$s = ut + \frac{1}{2}at^2 = 20 \times 30 + \frac{1}{2} \times (-\frac{2}{3}) \times 30^2 = 600 - 300 = 300 \text{ m}$
Total = $400 + 2400 + 300 = 3100 \text{ m}$
Answer: 3100 m [2 marks]

---

Question 12 [5 marks]

(a) Velocity at first light gate:
$v_1 = \frac{\text{card length}}{\text{time}} = \frac{0.10}{0.05} = 2.0 \text{ m/s}$
Answer: 2.0 m/s [1 mark]

(b) Velocity at second light gate:
$v_2 = \frac{0.10}{0.025} = 4.0 \text{ m/s}$
Answer: 4.0 m/s [1 mark]

(c) Acceleration:
Using $v^2 = u^2 + 2as$:
$4.0^2 = 2.0^2 + 2 \times a \times 0.80$
$16 = 4 + 1.6a$
$1.6a = 12$
$a = 7.5 \text{ m/s}^2$
Answer: 7.5 m/s² [2 marks]

(d) Assumption: The acceleration is uniform (constant) between the two light gates.
Answer: The acceleration is uniform/constant between the light gates. [1 mark]

---

Question 13 [4 marks]

(a) Component of weight down the plane:
$W_{\parallel} = mg \sin\theta = 15 \times 10 \times \sin 30^\circ = 150 \times 0.5 = 75 \text{ N}$
Answer: 75 N [1 mark]

(b) Force $F$ (constant speed → net force = 0):
$F = W_{\parallel} + \text{Friction} = 75 + 20 = 95 \text{ N}$
Answer: 95 N [2 marks]

(c) Work done by $F$:
$W = F \times s = 95 \times 5 = 475 \text{ J}$
Answer: 475 J [1 mark]

---

Question 14 [5 marks]

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

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

(c) Change in momentum:
Take upward as positive.
$p_{\text{before}} = m(-v_{\text{before}}) = 0.2 \times (-6.32) = -1.264 \text{ kg m/s}$
$p_{\text{after}} = m(+v_{\text{after}}) = 0.2 \times 4.90 = 0.98 \text{ kg m/s}$
$\Delta p = p_{\text{after}} - p_{\text{before}} = 0.98 - (-1.264) = 2.244 \text{ kg m/s}$
Answer: 2.24 kg m/s (upwards) [2 marks]

(d) Average force:
$F_{\text{avg}} = \frac{\Delta p}{\Delta t} = \frac{2.244}{0.05} = 44.88 \text{ N}$
Answer: 44.9 N (upwards) [1 mark]

---

Question 15 [4 marks]

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

(b) Taking moments about hinge A:
Clockwise moments:
Beam weight: $200 \times 2.0 = 400 \text{ N m}$
Load: $300 \times 1.0 = 300 \text{ N m}$
Total clockwise = $700 \text{ N m}$

Anticlockwise moment from cable tension $T$:
Vertical component of $T = T \sin 30^\circ = 0.5T$
Moment arm = $4.0 \text{ m}$
Anticlockwise moment = $0.5T \times 4.0 = 2.0T$

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

---

Question 16 [4 marks]

(a) Constant speed → net force = 0.
Resistive force = Driving force = $4000 \text{ N}$, opposite to direction of motion.
Answer: 4000 N, opposite to direction of motion (backwards). [1 mark]

(b) Power = Force × Velocity = $4000 \times 25 = 100,000 \text{ W} = 100 \text{ kW}$
Answer: 100,000 W (or 100 kW) [1 mark]

(c) On incline:
Component of weight down slope = $mg \sin 5^\circ = 1000 \times 10 \times \sin 5^\circ = 10000 \times 0.08716 = 871.6 \text{ N}$
New driving force = Resistive force + Weight component = $4000 + 871.6 = 4871.6 \text{ N}$
Answer: 4870 N (or 4872 N) [2 marks]

---

Question 17 [4 marks]

(a) Vertical height raised:
$h = L - L\cos\theta = L(1 - \cos\theta) = 1.0 \times (1 - \cos 30^\circ) = 1.0 \times (1 - 0.8660) = 0.134 \text{ m}$
Answer: 0.134 m [1 mark]

(b) Maximum speed at lowest point (conservation of energy):
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.134} = \sqrt{2.68} = 1.64 \text{ m/s}$
Answer: 1.64 m/s [2 marks]

(c) Energy conversion:
Gravitational potential energy → Kinetic energy → (work done against air resistance) → Thermal energy (heat/sound)
Answer: Gravitational potential energy is converted to kinetic energy, which is then dissipated as thermal energy (heat and sound) due to air resistance. [1 mark]

---

Section C: Longer Structured Questions [20 marks]

Question 18 [7 marks]

(a) Mass after 20 s:
Mass loss rate = $50 \text{ kg/s}$
Mass lost = $50 \times 20 = 1000 \text{ kg}$
Mass remaining = $5000 - 1000 = 4000 \text{ kg}$
Answer: 4000 kg [1 mark]

(b) Thrust force:
Thrust = Rate of mass ejection × Exhaust speed relative to rocket
$F_{\text{thrust}} = 50 \times 2000 = 100,000 \text{ N}$
Answer: 100,000 N [2 marks]

(c) Initial acceleration:
Initial weight = $5000 \times 10 = 50,000 \text{ N}$
Net force = Thrust – Weight = $100,000 - 50,000 = 50,000 \text{ N}$
$a = \frac{F_{\text{net}}}{m} = \frac{50,000}{5000} = 10 \text{ m/s}^2$
Answer: 10 m/s² [2 marks]

(d) Explanation:
Thrust remains constant at 100,000 N. However, the mass of the rocket decreases continuously as fuel is ejected (from 5000 kg to 4000 kg over 20 s). Weight also decreases. 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.
Answer: The thrust force remains constant but the mass of the rocket decreases as fuel is ejected. Since acceleration = (Thrust – Weight)/mass, and both mass and weight decrease while thrust stays constant, the acceleration increases. [2 marks]

---

Question 19 [7 marks]

(a) Forces on Block A (3 kg on table):

• Weight $W_A = 30 \text{ N}$ downwards
• Normal reaction $N_A = 30 \text{ N}$ upwards
• Tension $T$ to the right (horizontal)

Forces on Block B (2 kg hanging):

• Weight $W_B = 20 \text{0} \text{ N}$ downwards
• Tension $T$ upwards

**Answer:

Generated diagram for this question.

** [2 marks]

(b) Acceleration of system:
For Block A: $T = 3a$
For Block B: $20 - T = 2a$
Adding: $20 = 5a$ → $a = 4 \text{ m/s}^2$
Answer: 4 m/s² [2 marks]

(c) Tension:
$T = 3a = 3 \times 4 = 12 \text{ N}$
Answer: 12 N [1 mark]

(d) After Block B hits floor:
Block B stops. String goes slack. Block A continues moving horizontally at constant velocity (since table is smooth, no friction, no horizontal force) until it reaches the pulley or edge of table.
Answer: Block A continues moving at constant velocity (4 m/s × 1.5 s = 6 m/s) horizontally with no acceleration because the string goes slack and there is no friction on the table. [2 marks]

---

Question 20 [6 marks]

(a) Deceleration:
$v^2 = u^2 + 2as$
$0 = 30^2 + 2 \times a \times 75$
$0 = 900 + 150a$
$a = -6 \text{ m/s}^2$
Deceleration = $6 \text{ m/s}^2$
Answer: 6 m/s² [2 marks]

(b) Average braking force:
$F = ma = 1000 \times 6 = 6000 \text{ N}$
Answer: 6000 N [1 mark]

(c) Work done by braking force = Initial kinetic energy of car
$KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \times 30^2 = 500 \times 900 = 450,000 \text{ J}$

Energy per brake disc = $\frac{450,000}{4} = 112,500 \text{ J}$

$Q = mc\Delta\theta$
$112,500 = 2.5 \times 500 \times \Delta\theta$
$112,500 = 1250 \times \Delta\theta$
$\Delta\theta = \frac{112,500}{1250} = 90 \text{ K}$
Answer: 90 K [3 marks]

---

End of Answer Key