Secondary 1 Science Semestral Assessment 2 (End of Year) Paper 1

TuitionGoWhere Practice Paper - Science Secondary 1

SA2 Practice Paper Version 1 - Answer Key

Total Marks: 60

Section A: Multiple Choice Questions [10 marks]

Question 1 [1]

Answer: C
Working:
Work done = Force × Distance moved in direction of force
= 5 N × 1.2 m = 6.0 J

Explanation: When lifting at constant velocity, the upward force equals the weight (5 N). Work done = force × displacement = 5 × 1.2 = 6.0 J.

Question 2 [1]

Answer: A
Explanation: A compressed spring stores elastic potential energy. When released, this converts to kinetic energy of the toy car. No change in height means no gravitational potential energy change.

Question 3 [1]

Answer: B
Working:
Net force = Applied force − Frictional force = 15 N − 5 N = 10 N
Net work done = Net force × Distance = 10 N × 4 m = 40 J

Alternative: Work by applied force = 15 × 4 = 60 J; Work against friction = 5 × 4 = 20 J; Net work = 60 − 20 = 40 J.

Question 4 [1]

Answer: C
Working:
GPE lost = mgh = 2 × 10 × 10 = 200 J
By conservation of energy (no air resistance), KE at bottom = GPE lost = 200 J

Explanation: All gravitational potential energy converts to kinetic energy.

Question 5 [1]

Answer: C
Working:
Work output (useful) = Load × Distance moved by load = 500 N × 2 m = 1000 J
Work input = Effort × Distance moved by effort = 200 N × 6 m = 1200 J
Efficiency = (Work output / Work input) × 100% = (1000 / 1200) × 100% = 83.3%

Wait - recalculating: 1000/1200 = 5/6 = 83.3%. But option C is 66.7%. Let me check: 1000/1200 = 0.833 = 83.3%. Option D is 83.3%.
Correct Answer: D

Correction: Efficiency = (500 × 2) / (200 × 6) × 100% = 1000/1200 × 100% = 83.3%

Question 6 [1]

Answer: D
Explanation: Work done = Force × Distance × cos(θ). When force is perpendicular to motion (θ = 90°), cos(90°) = 0, so no work is done.

• A is incorrect: force must cause displacement in its direction.
• B is incorrect: work can be done when force has a component in direction of motion.
• C is incorrect: work is a scalar quantity.

Question 7 [1]

Answer: B
Explanation: Ignoring air resistance, mechanical energy is conserved. At X and Z, the bob is momentarily at rest (KE = 0), so total energy = GPE. Since height at Z equals height at X, GPE at Z = GPE at X.

• A: KE at Y = GPE at X (not greater).
• C: Total energy is conserved.
• D: KE at Z is zero (momentarily at rest).

Question 8 [1]

Answer: A
Working:
KE gained = ½mv² = ½ × 1000 × 20² = 200,000 J
Average power = Work done / Time = 200,000 J / 10 s = 20,000 W = 20 kW

Question 9 [1]

Answer: B
Explanation: On a rough inclined plane, gravitational potential energy converts to kinetic energy AND heat energy (due to work done against friction). Some GPE → KE + Heat.

Question 10 [1]

Answer: B
Working:
Work done = Force × Distance
50 J = Force × 5 m
Force = 50 / 5 = 10 N

Section B: Structured Questions [30 marks]

Question 11 [4]

(a) [1]
Weight = mg = 800 kg × 10 N/kg = 8000 N

(b) [2]
Tension = 8000 N, upwards
Explanation: Since the block moves at constant velocity, net force = 0 (Newton's First Law). The upward tension balances the downward weight. Tension = Weight = 8000 N, directed upwards.

(c) [1]
Work done by tension = Tension × Distance = 8000 N × 15 m = 120,000 J (or 120 kJ)

Question 12 [5]

(a) [1]
GPE at A = mgh = 500 kg × 10 N/kg × 30 m = 150,000 J (or 150 kJ)

(b) [2]
KE at B = 150,000 J (or 150 kJ)
Explanation: Track is frictionless, so mechanical energy is conserved. At A: Total energy = GPE = 150,000 J, KE = 0. At B: Height = 0, so GPE = 0. By conservation of energy, KE at B = Total energy = 150,000 J.

(c) [2]
At C: GPE = mgh = 500 × 10 × 20 = 100,000 J
Total energy = 150,000 J (conserved)
KE at C = Total energy − GPE at C = 150,000 − 100,000 = 50,000 J
KE = ½mv² → 50,000 = ½ × 500 × v²
v² = 100,000 / 500 = 200
v = √200 = 14.1 m/s (or 10√2 m/s)

Question 13 [4]

(a) [2]
Frictional force = 8 N, opposite to direction of motion
Reasoning: Constant velocity means zero acceleration, so net force = 0 (Newton's First Law). The pulling force (8 N forward) is balanced by the frictional force (8 N backward).

(b) [1]
Work done against friction = Frictional force × Distance = 8 N × 2 m = 16 J

(c) [1]
The block will accelerate (speed up) in the direction of the pull.
Explanation: Net force = 12 N − 8 N = 4 N forward. By Newton's Second Law (F = ma), a = F/m = 4/3 = 1.33 m/s² forward.

Question 14 [5]

(a) [1]
Work done by effort = Effort × Distance moved by effort
270 J = E × 1.5 m
E = 270 / 1.5 = 180 N

(b) [1]
Work done on load = Load × Distance moved by load = 120 N × 1.5 m = 180 J

(c) [1]
Efficiency = (Work output / Work input) × 100% = (180 / 270) × 100% = 66.7%

(d) [1]
Work done against friction in the pulley bearings / Work done against the weight of the moving parts of the pulley / Energy lost as heat and sound.
(Any one valid reason)

(e) [1]
The effort required would decrease (become one-third of the original effort for an ideal system, or proportionally less for the same efficiency).
Explanation: Velocity ratio = 3 means effort moves 3 times the distance of the load. For the same work output and efficiency, Work input = Work output / efficiency. Since Work input = Effort × Distance_effort, and Distance_effort increases by factor of 3, Effort decreases by factor of 3.

Question 15 [4]

(a) [1]
Initial KE = ½mv² = ½ × 0.2 kg × (15 m/s)² = 0.1 × 225 = 22.5 J

(b) [2]
At max height, KE = 0, all initial KE → GPE
mgh = 22.5 J
0.2 × 10 × h = 22.5
2h = 22.5
h = 11.25 m

(c) [1]
Kinetic energy at maximum height = 0 J
Explanation: At maximum height, the ball is momentarily at rest before falling down, so velocity = 0, KE = 0.

Question 16 [4]

(a) [1]
Electrical energy = Power × Time = 60 W × 4 s = 240 J

(b) [1]
GPE gained = mgh = 2 kg × 10 N/kg × 5 m = 100 J

(c) [1]
Efficiency = (Useful energy output / Energy input) × 100% = (100 / 240) × 100% = 41.7%

(d) [1]
The remaining energy is converted to heat and sound (dissipated to the surroundings).
Explanation: Energy losses in the motor (resistance heating in coils, friction in bearings, sound) account for the difference.

Question 17 [4]

(a) [1]
Force is directly proportional to extension (Hooke's Law: F = kx). The graph is a straight line passing through the origin.

(b) [2]
From graph: At extension = 10 cm = 0.10 m, Force = 20 N
Spring constant k = F / x = 20 N / 0.10 m = 200 N/m

(c) [1]
Elastic potential energy = ½kx² = ½ × 200 N/m × (0.06 m)² = 100 × 0.0036 = 0.36 J
Alternative: Area under graph = ½ × base × height = ½ × 0.06 m × (200 × 0.06) = ½ × 0.06 × 12 = 0.36 J

Section C: Longer Structured Questions [20 marks]

Question 18 [7]

(a) [1]
GPE lost = mgh = 80 kg × 10 N/kg × 25 m = 20,000 J (or 20 kJ)

(b) [1]
KE at bottom = ½mv² = ½ × 80 × 18² = 40 × 324 = 12,960 J

(c) [1]
Work done against friction and air resistance = GPE lost − KE gained = 20,000 − 12,960 = 7,040 J

(d) [2]
Work done against resistive force = Resistive force × Distance
7,040 J = F_resistive × 100 m
F_resistive = 7,040 / 100 = 70.4 N

(e) [2]
Going up second hill:
GPE gained = mgh = 80 × 10 × 15 = 12,000 J
KE at top = ½ × 80 × 5² = 40 × 25 = 1,000 J
Total energy at top = 12,000 + 1,000 = 13,000 J
Energy at bottom of second hill = KE at bottom of first hill = 12,960 J (assuming no loss on flat)
Work done by cyclist = Energy at top − Energy at bottom = 13,000 − 12,960 = 40 J

Wait - this seems too small. Let me reconsider.
Actually, the cyclist starts at bottom of first hill with 12,960 J KE. To reach top of second hill (15 m high) with 5 m/s:
Energy needed at top = GPE + KE = 12,000 + 1,000 = 13,000 J
Energy available at bottom = 12,960 J
Work done by cyclist = 13,000 − 12,960 = 40 J

This is correct - the cyclist only needs to add a small amount because the KE at bottom is almost enough. The small difference is due to rounding (18² = 324, ½×80×324 = 12,960 exactly).

Question 19 [7]

(a) [1]
Component of weight down plane = mg sin θ = 5 × 10 × sin 30° = 50 × 0.5 = 25 N

(b) [2]
Normal reaction R = mg cos θ = 5 × 10 × cos 30° = 50 × 0.866 = 43.3 N
Frictional force = μR = 0.2 × 43.3 = 8.66 N (or 5√3 ≈ 8.66 N)

(c) [1]
Net force parallel to plane = Applied force − Weight component down − Friction
= 40 − 25 − 8.66 = 6.34 N (up the plane)

(d) [1]
Acceleration a = F_net / m = 6.34 / 5 = 1.27 m/s² (up the plane)

(e) [1]
Work done by applied force F = F × s = 40 N × 4 m = 160 J

(f) [1]
Vertical height gained = s sin θ = 4 × sin 30° = 4 × 0.5 = 2 m
GPE gained = mgh = 5 × 10 × 2 = 100 J

Question 20 [6]

(a) [1]
GPE lost per second = mass rate × g × h = 200 kg/s × 10 N/kg × 50 m = 100,000 J/s (or 100 kW)

(b) [1]
Power input = Rate of GPE loss = 100 kW (or 100,000 W)

(c) [1]
Efficiency = (Power output / Power input) × 100% = (80 kW / 100 kW) × 100% = 80%

(d) [2]
Any two of:

• Heat energy (due to friction in turbines, generators, and water turbulence)
• Sound energy (from moving water and machinery)
• Kinetic energy of water leaving the system (water still has kinetic energy after passing turbines)
• Electrical resistance losses in generator windings and transmission lines

(e) [1]
Increase the efficiency of the turbines/generators (e.g., better blade design, reduced friction, improved generator efficiency).
Alternative: Reduce energy losses in the system (any valid method that increases efficiency without changing flow rate or height).

End of Answer Key