Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 1

TuitionGoWhere Practice Paper - Physics Secondary 3 SA2 Version 1 - Answer Key

Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

1. [1 mark] Answer: A

Working:
Micrometer reading = Main scale reading + (Thimble scale reading × Least count)
= 4.5 mm + (28 × 0.01 mm)
= 4.5 mm + 0.28 mm
= 4.78 mm

Wait - correction: The question states main scale 4.5 mm and thimble 28 divisions.
Main scale = 4.5 mm, Thimble = 28 × 0.01 = 0.28 mm
Total = 4.78 mm → Option B

Correction: The correct answer is B. 4.78 mm.
Common mistake: Forgetting to add the thimble reading to the main scale, or misreading the main scale as 4.5 cm.

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2. [1 mark] Answer: B

Explanation:

• Speed is a scalar (magnitude only), velocity is a vector (magnitude and direction).
• Mass and weight: mass is scalar, weight is vector (force) → one scalar, one vector ✓ but weight is a force, not typically paired this way
• Distance (scalar) and displacement (vector) → also correct pair
• Time (scalar) and acceleration (vector) → also correct pair

Best answer: B (Speed, Velocity) - this is the classic scalar/vector pair taught in kinematics. Both are measures of "how fast" but velocity includes direction.

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3. [1 mark] Answer: B

Working:
For uniform acceleration from rest:
$s = \frac{1}{2} a t^2$ and $v = at$
$a = \frac{v}{t} = \frac{20}{5} = 4 \text{ m/s}^2$
$s = \frac{1}{2} \times 4 \times 5^2 = 2 \times 25 = 50 \text{ m}$

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

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4. [1 mark] Answer: B

Working:
Resultant force = Applied force - Friction = 15 N - 5 N = 10 N
$a = \frac{F}{m} = \frac{10}{2} = 5.0 \text{ m/s}^2$

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5. [1 mark] Answer: C

Working:
At maximum height, $v = 0$. Using $v^2 = u^2 + 2as$:
$0 = 30^2 + 2(-10)h$
$0 = 900 - 20h$
$20h = 900$
$h = 45 \text{ m}$

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6. [1 mark] Answer: B

Working:
For perpendicular forces: $R = \sqrt{F_1^2 + F_2^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N}$
(3-4-5 triangle scaled by 2)

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7. [1 mark] Answer: B

Working:
$\Delta GPE = mgh = 0.5 \text{ kg} \times 10 \text{ N/kg} \times 2 \text{ m} = 10 \text{ J}$
Common mistake: Using mass in grams (500) instead of kg (0.5) → 10,000 J (not an option) or forgetting to convert.

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8. [1 mark] Answer: B

Working:
Moment = Force × Perpendicular distance = 20 N × 0.25 m = 5 N·m

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9. [1 mark] Answer: B

Working:
For equilibrium: Sum of clockwise moments = Sum of anticlockwise moments
Pivot at 30 cm. 2 N at 10 cm → distance from pivot = 20 cm (anticlockwise)
Moment = 2 N × 0.20 m = 0.4 N·m anticlockwise
Let 4 N be at distance $d$ from pivot (clockwise):
$4 \times d = 0.4$
$d = 0.1 \text{ m} = 10 \text{ cm}$ from pivot
Position = 30 cm + 10 cm = 40 cm mark

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10. [1 mark] Answer: C

Working:
Distance = Area under velocity-time graph

• 0-4 s: Triangle area = $\frac{1}{2} \times 4 \times 8 = 16 \text{ m}$
• 4-8 s: Rectangle area = $4 \times 8 = 32 \text{ m}$
• 8-12 s: Triangle area = $\frac{1}{2} \times 4 \times 8 = 16 \text{ m}$
  Total = 16 + 32 + 16 = 64 m

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Section B: Structured Questions [30 marks]

11. [5 marks]

(a) [2 marks]
Answer:
Between t = 0 s and t = 10 s, the skydiver accelerates downwards but with decreasing acceleration. The velocity increases from 0 to 40 m/s but the gradient of the graph decreases.
Explanation: Initially, the only force is weight (downwards), so acceleration = g. As velocity increases, air resistance increases (opposing motion). The resultant force (weight - air resistance) decreases, so acceleration decreases until it becomes zero at terminal velocity.

Marking points:

• Description: accelerates with decreasing acceleration / velocity increases but gradient decreases [1]
• Explanation: air resistance increases with speed, reducing resultant force [1]

(b) [1 mark]
Answer: 40 m/s (terminal velocity reached at t = 10 s)

(c) [2 marks]
Answer:
When the parachute opens, the surface area increases dramatically, causing a large increase in air resistance. The upward air resistance force becomes much larger than the downward weight, creating a large upward resultant force. This causes rapid deceleration (upward acceleration) until a new, lower terminal velocity is reached where air resistance again equals weight.

Marking points:

• Parachute increases surface area → large increase in air resistance [1]
• Upward air resistance > weight → upward resultant force → deceleration [1]

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12. [6 marks]

(a) [2 marks]
Working:
$KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 1200 \times 25^2 = 600 \times 625 = 375,000 \text{ J}$
Answer: 375,000 J (or 375 kJ)

(b) [2 marks]
Working:
Work done by braking force = Change in kinetic energy = 375,000 J
Work = Force × distance
$F = \frac{Work}{distance} = \frac{375,000}{50} = 7,500 \text{ N}$

Alternative using kinematics:
$v^2 = u^2 + 2as$
$0 = 25^2 + 2a(50)$
$100a = -625$
$a = -6.25 \text{ m/s}^2$
$F = ma = 1200 \times 6.25 = 7,500 \text{ N}$

Answer: 7,500 N

(c) [2 marks]
Answer:
The kinetic energy is converted primarily into thermal energy (heat) due to friction between the brake pads and brake discs/drums. Some energy is also transferred to the surroundings as sound energy. The total energy is conserved.

Marking points:

• Kinetic energy → thermal energy (heat) in brakes [1]
• Also sound energy / surroundings [1]

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13. [5 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) [2 marks]
Working:
Moment = Force × Perpendicular distance from pivot
Distance of 3 N weight from pivot (50 cm mark) = 50 - 20 = 30 cm = 0.30 m
Moment = 3 N × 0.30 m = 0.9 N·m (anticlockwise)

Answer: 0.9 N·m (anticlockwise)

(c) [2 marks]
Working:
New position of 2 N weight: 90 cm mark
Distance from pivot = 90 - 50 = 40 cm = 0.40 m
Clockwise moment = 2 N × 0.40 m = 0.8 N·m
Anticlockwise moment (unchanged) = 0.9 N·m

Since anticlockwise moment (0.9 N·m) > clockwise moment (0.8 N·m), the rule will not remain in equilibrium. The left side (side with 3 N weight) will tilt downwards.

Marking points:

• Calculation of new clockwise moment = 0.8 N·m [1]
• Comparison and conclusion: left side tilts down [1]

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14. [4 marks]

(a) [2 marks]
Forces to draw and label on diagram:

1. Weight (W) = 40 N, vertically downwards from centre of block
2. Normal reaction (R) perpendicular to plane, upwards
3. Tension (T) = 30 N, up the plane parallel to surface
4. Friction (f) down the plane parallel to surface

Marking points:

• All four forces correctly drawn and labelled [2]
• (1 mark for 3 correct forces, 0 for fewer)

(b) [2 marks]
Working:
Since constant velocity, resultant force parallel to plane = 0
Forces up the plane = Forces down the plane
Tension = Component of weight parallel to plane + Friction
$30 = (40 \sin 30°) + f$
$30 = (40 \times 0.5) + f$
$30 = 20 + f$
$f = 10 \text{ N}$ (down the plane)

Answer: 10 N down the plane

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15. [5 marks]

(a) [1 mark]
Working:
$W = mg = 500 \times 10 = 5,000 \text{ N}$

(b) [2 marks]
Working:
Resultant force = Thrust - Weight
$F_{res} = 8,000 - 5,000 = 3,000 \text{ N}$ (upwards)

(c) [2 marks]
Working:
$F = ma$
$a = \frac{F}{m} = \frac{3,000}{500} = 6 \text{ m/s}^2$ (upwards)

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16. [5 marks]

(a) [2 marks]
Working:
Loss in GPE = Gain in KE (conservation of energy)
$mgh = \frac{1}{2} mv^2$
$v^2 = 2gh = 2 \times 10 \times 0.5 = 10$
$v = \sqrt{10} = 3.16 \text{ m/s}$

Answer: 3.16 m/s (or $\sqrt{10}$ m/s)

(b) [2 marks]
Working:
Conservation of momentum (no external horizontal forces):
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
$0.2 \times 3.16 + 0.3 \times 0 = (0.2 + 0.3) v$
$0.632 = 0.5 v$
$v = 1.264 \text{ m/s}$

Answer: 1.26 m/s (in the same direction as the bob's initial motion)

(c) [1 mark]
Answer: Inelastic collision.
Explanation: The two objects stick together and move with a common velocity after collision. Kinetic energy is not conserved (some is converted to heat/sound/deformation).

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Section C: Longer Structured Questions [20 marks]

17. [7 marks]

(a) [3 marks]
Graph requirements:

• Axes labelled with quantities and units: Force / N (x-axis), Acceleration / m/s² (y-axis) [1]
• Suitable scales covering at least 50% of grid [1]
• All 5 points plotted correctly (± half a small square) [1]
• Best-fit straight line through origin [1]
  (Note: 3 marks total - typically 1 for axes/scales, 1 for plotting, 1 for line)

(b) [2 marks]
Working:
Gradient = $\frac{\Delta a}{\Delta F}$
Using points on best-fit line (e.g., (0,0) and (2.5, 5.0)):
Gradient = $\frac{5.0 - 0}{2.5 - 0} = 2.0 \text{ m/s}^2 \text{ per N}$ or 2.0 kg⁻¹

Unit: m/s² per N or kg⁻¹ (since N = kg·m/s²)

(c) [2 marks]
Working:
Theoretical gradient = $\frac{1}{m} = \frac{1}{0.5} = 2.0 \text{ kg}^{-1}$
Experimental gradient = 2.0 kg⁻¹ (from graph)
Percentage difference = $\frac{|Experimental - Theoretical|}{Theoretical} \times 100\%$
= $\frac{|2.0 - 2.0|}{2.0} \times 100\% = 0\%$

Note: If student's graph gives slightly different gradient (e.g., 1.95), calculate accordingly.
Answer: 0% (or calculated value based on student's gradient)

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18. [7 marks]

(a) [2 marks]
Working:
Constant velocity → resultant force = 0
Tension = Weight = $mg = 800 \times 10 = 8,000 \text{ N}$

Answer: 8,000 N

(b) [2 marks]
Working:
Work done = Force × Distance in direction of force
$W = T \times h = 8,000 \times 15 = 120,000 \text{ J}$ (or 120 kJ)

Answer: 120,000 J

(c) [2 marks]
Working:
Power = $\frac{Work}{Time}$
Time = $\frac{Distance}{Speed} = \frac{15}{2} = 7.5 \text{ s}$
$P = \frac{120,000}{7.5} = 16,000 \text{ W}$ (or 16 kW)

Alternative: Power = Force × Velocity = 8,000 × 2 = 16,000 W

Answer: 16,000 W

(d) [1 mark]
Working:
Efficiency = $\frac{Output Power}{Input Power} \times 100\%$
$0.80 = \frac{16,000}{P_{in}}$
$P_{in} = \frac{16,000}{0.80} = 20,000 \text{ W}$

Answer: 20,000 W (or 20 kW)

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19. [6 marks]

(a) [1 mark]
Answer:
The total momentum of a closed system remains constant if no external resultant force acts on the system.

(b) [3 marks]
Working:
Initial momentum = 0 (both at rest)
Final momentum = $m_A v_A + m_B v_B = 0$
$60 v_A + 40 \times 3 = 0$
$60 v_A = - = -120$
$v_A = -2 \text{ m/s}$

Answer: 2 m/s in the opposite direction to skater B (or away from skater B)

Marking points:

• Conservation of momentum equation set up correctly [1]
• Correct magnitude 2 m/s [1]
• Correct direction (opposite to B) [1]

(c) [2 marks]
Working:
$KE_{total} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2$
$= \frac{1}{2} \times 60 \times 2^2 + \frac{1}{2} \times 40 \times 3^2$
$= 30 \times 4 + 20 \times 9$
$= 120 + 180 = 300 \text{ J}$

Answer: 300 J

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20. [6 marks]

(a) [2 marks]
Working:
$u_x = u \cos \theta = 40 \cos 30° = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \approx 34.6 \text{ m/s}$
$u_y = u \sin \theta = 40 \sin 30° = 40 \times 0.5 = 20 \text{ m/s}$

Answer: Horizontal: 34.6 m/s, Vertical: 20 m/s

(b) [2 marks]
Working:
At maximum height, $v_y = 0$
$v_y^2 = u_y^2 + 2 a_y s_y$
$0 = 20^2 + 2(-10)H$
$0 = 400 - 20H$
$H = 20 \text{ m}$

Answer: 20 m

(c) [2 marks]
Working:
Time to reach max height: $v_y = u_y + a_y t_{up}$
$0 = 20 - 10 t_{up}$
$t_{up} = 2 \text{ s}$

Time of flight = $2 \times t_{up} = 4 \text{ s}$ (symmetrical trajectory, launch and landing at same level)

Alternative: $s_y = u_y t + \frac{1}{2} a_y t^2$
$0 = 20t - 5t^2$
$5t(4 - t) = 0$
$t = 0$ or $t = 4 \text{ s}$

Answer: 4

<stage3_exam_answers_md>

TuitionGoWhere Practice Paper - Physics Secondary 3

TuitionGoWhere Secondary School (AI)

Subject: Physics
Level: Secondary 3 (Pure Physics)
Paper: SA2 Version 1 - Answer Key
Total Marks: 60

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Section A: Multiple Choice Questions [10 marks]

Question	Answer	Explanation
1	A	Main scale: 4.5 mm, Thimble: 28 × 0.01 mm = 0.28 mm. Total = 4.5 + 0.28 = 4.78 mm
2	B	Speed is scalar, velocity is vector. Mass (scalar), weight (vector). Distance (scalar), displacement (vector). Time (scalar), acceleration (vector).
3	B	$u = 0$, $v = 20 \text{ m/s}$, $t = 5 \text{ s}$. $a = \frac{v-u}{t} = 4 \text{ m/s}^2$. $s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(4)(25) = \textbf{50 m}$
4	B	Resultant force = 15 N - 5 N = 10 N. $a = \frac{F}{m} = \frac{10}{2} = \textbf{5.0 m/s}^2$
5	C	$v^2 = u^2 + 2as$. At max height $v = 0$. $0 = 30^2 + 2(-10)s$. $s = \frac{900}{20} = \textbf{45 m}$
6	B	$R = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = \textbf{10 N}$
7	B	$m = 0.5 \text{ kg}$, $h = 2 \text{ m}$, $g = 10 \text{ N/kg}$. $GPE = mgh = 0.5 \times 10 \times 2 = \textbf{10 J}$
8	B	Moment = Force × perpendicular distance = $20 \times 0.25 = \textbf{5 N·m}$
9	B	Clockwise moment = $2 \times (30-10) = 40 \text{ N·cm}$. Anticlockwise moment = $4 \times (x-30)$. For equilibrium: $4(x-30) = 40 \Rightarrow x-30 = 10 \Rightarrow x = \textbf{40 cm}$
10	C	Area under graph: Triangle (0-4s) = $\frac{1}{2} \times 4 \times 8 = 16$. Rectangle (4-8s) = $4 \times 8 = 32$. Triangle (8-12s) = $\frac{1}{2} \times 4 \times 8 = 16$. Total = $16 + 32 + 16 = \textbf{64 m}$

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Section B: Structured Questions [30 marks]

11. [5 marks]

(a) Between t = 0 s and t = 10 s, the skydiver accelerates downwards but with decreasing acceleration. Initially, only weight acts, so acceleration = g. As velocity increases, air resistance increases, reducing the resultant force ($W - R$), hence acceleration decreases until it reaches zero at terminal velocity. [2]

(b) 40 m/s [1]

(c) When the parachute opens, the surface area increases dramatically, causing a large increase in air resistance. The upward air resistance becomes much greater than the weight, resulting in a large upward resultant force and rapid deceleration. [2]

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12. [6 marks]

(a) $KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 1200 \times 25^2 = 600 \times 625 = \textbf{375,000 J}$ (or 375 kJ) [2]

(b) Work done by braking force = Loss in KE = 375,000 J.
$F \times d = 375,000 \Rightarrow F \times 50 = 375,000 \Rightarrow F = \textbf{7,500 N}$ [2]

(c) The kinetic energy is converted to thermal energy (heat) due to friction between brake pads and discs/drums, and between tyres and road. Some energy is also dissipated as sound energy. [2]

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13. [5 marks]

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

(b) Moment = Force × perpendicular distance from pivot = $3 \text{ N} \times 0.30 \text{ m} = \textbf{0.9 N·m}$ (clockwise) [2]

(c) New anticlockwise moment = $2 \text{ N} \times 0.40 \text{ m} = 0.8 \text{ N·m}$.
Clockwise moment = 0.9 N·m (unchanged).
Since clockwise moment > anticlockwise moment, the rule will not remain in equilibrium. The left side (3 N side) will tilt downwards. [2]

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14. [4 marks]

(a) Forces on diagram:

• Weight ($W = mg = 40 \text{ N}$) vertically downwards
• Normal reaction ($R$) perpendicular to plane
• Tension ($T = 30 \text{ N}$) up the plane, parallel to surface
• Friction ($f$) down the plane, parallel to surface [2]

(b) Constant velocity ⇒ resultant force parallel to plane = 0.
Component of weight parallel to plane = $mg \sin 30° = 40 \times 0.5 = 20 \text{ N}$ (down plane).
$T - mg \sin 30° - f = 0 \Rightarrow 30 - 20 - f = 0 \Rightarrow f = \textbf{10 N}$ (down the plane) [2]

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15. [5 marks]

(a) $W = mg = 500 \times 10 = \textbf{5000 N}$ [1]

(b) Resultant force = Thrust - Weight = $8000 - 5000 = \textbf{3000 N}$ (upwards) [2]

(c) $a = \frac{F_{\text{net}}}{m} = \frac{3000}{500} = \textbf{6 m/s}^2$ (upwards) [2]

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16. [5 marks]

(a) Loss in GPE = Gain in KE. $mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.5} = \sqrt{10} = \textbf{3.16 m/s}$ [2]

(b) Conservation of momentum: $m_1u_1 + m_2u_2 = (m_1 + m_2)v$
$(0.2 \times 3.16) + (0.3 \times 0) = (0.5)v \Rightarrow 0.632 = 0.5v \Rightarrow v = \textbf{1.26 m/s}$ [2]

(c) Inelastic collision. The two objects stick together and move with a common velocity after collision. Kinetic energy is not conserved (some is lost to heat/sound/deformation). [1]

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Section C: Longer Structured Questions [20 marks]

17. [7 marks]

(a) Graph requirements:

• Axes labelled with units: Force / N (x-axis), Acceleration / m/s² (y-axis)
• Suitable scales (e.g., 1 cm = 0.5 N on x-axis, 1 cm = 1 m/s² on y-axis)
• All 5 points plotted correctly: (0.5, 1.0), (1.0, 2.1), (1.5, 3.0), (2.0, 4.1), (2.5, 5.0)
• Best-fit straight line passing through origin (0,0) [3]

(b) Gradient = $\frac{\Delta a}{\Delta F} = \frac{5.0 - 0}{2.5 - 0} = \textbf{2.0 m/s}^2 \text{ per N}$ (or 2.0 kg⁻¹) [2]

(c) Theoretical gradient = $\frac{1}{m} = \frac{1}{0.5} = 2.0 \text{ kg}^{-1}$.
Experimental gradient = 2.0 kg⁻¹.
Percentage difference = $\frac{|2.0 - 2.0|}{2.0} \times 100\% = \textbf{0\%}$ [2]

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18. [7 marks]

(a) Constant velocity ⇒ resultant force = 0.
Tension = Weight = $mg = 800 \times 10 = \textbf{8000 N}$ [2]

(b) Work done = Force × distance in direction of force = $T \times h = 8000 \times 15 = \textbf{120,000 J}$ (or 120 kJ) [2]

(c) Power = $\frac{\text{Work done}}{\text{time}}$. Time = $\frac{\text{distance}}{\text{velocity}} = \frac{15}{2} = 7.5 \text{ s}$.
Power = $\frac{120,000}{7.5} = \textbf{16,000 W}$ (or 16 kW) [2]

(d) Efficiency = $\frac{\text{Output power}}{\text{Input power}} \times 100\%$.
$0.80 = \frac{16,000}{\text{Input power}} \Rightarrow \text{Input power} = \frac{16,000}{0.80} = \textbf{20,000 W}$ (or 20 kW) [1]

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19. [6 marks]

(a) The total momentum of a closed system remains constant if no external resultant force acts on the system. [1]

(b) Initial momentum = 0 (both at rest).
Final momentum = $m_A v_A + m_B v_B = 0$.
$60 v_A + 40 \times 3 = 0 \Rightarrow 60 v_A = -120 \Rightarrow v_A = \textbf{-2 m/s}$.
Skater A moves at 2 m/s in the opposite direction to skater B. [3]

(c) $KE_{\text{total}} = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 = \frac{1}{2}(60)(2^2) + \frac{1}{2}(40)(3^2) = 120 + 180 = \textbf{300 J}$ [2]

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20. [6 marks]

(a) $u_x = u \cos \theta = 40 \cos 30° = 40 \times \frac{\sqrt{3}}{2} = \textbf{34.6 m/s}$
$u_y = u \sin \theta = 40 \sin 30° = 40 \times 0.5 = \textbf{20 m/s}$ [2]

(b) Time to max height: $v_y = u_y - gt = 0 \Rightarrow t = \frac{u_y}{g} = \frac{20}{10} = 2 \text{ s}$.
Total time of flight = $2 \times 2 = \textbf{4 s}$ [2]

(c) Range $R = u_x \times \text{total time} = 34.6 \times 4 = \textbf{138.4 m}$ (or 138.6 m using exact values) [2]

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End of Answer Key