Secondary 3 Physics Semestral Assessment 2 (End of Year) Paper 4

TuitionGoWhere Practice Paper - Physics Secondary 3

SA2 (Version 4 of 5) — Answer Key

Section A: Multiple Choice Questions

1. C
Reasoning: Velocity has both magnitude and direction, making it a vector. Speed, distance, and time are scalars.

2. B
Working: Average speed = total distance ÷ total time = 120 km ÷ 2 h = 60 km/h

3. B
Reasoning: By Newton's First Law, an object moving with constant velocity has zero net force acting on it.

4. C
Reasoning: Throughout the motion, the only force acting is gravity, so the acceleration is always 9.8 m/s² downward, even at the highest point (where velocity is momentarily zero).

5. D
Working: Normal force = weight = mg = 5 × 10 = 50 N

6. C
Reasoning: Newton's Third Law states that forces occur in equal and opposite pairs between two interacting objects.

7. B
Working: F = ma → a = F/m = 20 ÷ 4 = 5 m/s²

8. C
Working: v = u + at = 0 + (10 × 3) = 30 m/s

9. B
Working: Conservation of momentum: (2 × 3) + (4 × 0) = (2 + 4) × v → 6 = 6v → v = 1.0 m/s

10. B
Working: Net force = applied force − friction = 30 − 10 = 20 N

Section B: Structured Questions

11.
(a) Displacement is the shortest distance from one point to another in a specified direction. [1]
(b) Acceleration is the rate of change of velocity. [1]

12.
(a) v = u + at = 0 + (2.5 × 8) = 20 m/s [2]
(1 mark for correct substitution, 1 mark for correct answer)

(b) s = ut + ½at² = 0 + ½(2.5)(8²) = ½ × 2.5 × 64 = 80 m [2]
(1 mark for correct substitution, 1 mark for correct answer)

13.
(a) The object accelerates uniformly from rest to 15 m/s in the first 4 seconds. [1]

(b) Acceleration = gradient of v-t graph = (15 − 0) ÷ (4 − 0) = 3.75 m/s² [2]
(1 mark for method, 1 mark for answer)

(c) Distance = area under v-t graph:
Area = area of triangle (0–4 s) + area of rectangle (4–10 s) + area of triangle (10–12 s)
= ½ × 4 × 15 + 6 × 15 + ½ × 2 × 15
= 30 + 90 + 15 = 135 m [2]
(1 mark for correct area method, 1 mark for correct answer)

14.
(a) Reading = mg = 60 × 10 = 600 N [1]

(b) When accelerating upward, the net force = ma:
R − mg = ma
R = m(g + a) = 60 × (10 + 1.5) = 60 × 11.5 = 690 N [3]
(1 mark for stating Newton's Second Law, 1 mark for correct substitution, 1 mark for correct answer)
The scale reads the normal contact force (reaction force), which must exceed the weight to provide the upward net force for acceleration.

15.
(a) Vertical motion: h = ½gt² → 45 = ½ × 10 × t² → t² = 9 → t = 3 s [2]
(1 mark for substitution, 1 mark for answer)

(b) Horizontal distance: s = vt = 10 × 3 = 30 m [2]
(1 mark for method, 1 mark for answer)

16.
(a) Resultant magnitude: R = √(12² + 5²) = √(144 + 25) = √169 = 13 N
Direction: θ = tan⁻¹(5/12) = 22.6° north of east (or bearing 067.4°) [3]
(1 mark for magnitude, 1 mark for direction calculation, 1 mark for correct angle)

(b) Equilibrium force = 13 N (equal in magnitude but opposite in direction to the resultant) [1]

17.
(a) Horizontal component = 40 cos 30° = 40 × 0.866 = 34.6 N (or 20√3 N) [1]

(b) Frictional force = 34.6 N (acting opposite to the direction of motion) [2]
Explanation: Since the block moves at constant velocity, the net horizontal force is zero. Therefore, friction must balance the horizontal component of tension. (1 mark for value, 1 mark for explanation)

(c) Vertical component of tension = 40 sin 30° = 40 × 0.5 = 20 N (upward)
Normal contact force: N + 20 = mg → N = (4 × 10) − 20 = 40 − 20 = 20 N [2]
(1 mark for vertical component, 1 mark for normal force)

18.
Newton's First Law: An object at rest stays at rest, and an object in motion continues in uniform motion in a straight line, unless acted upon by a net external force. [1]

When the car brakes suddenly, the car decelerates, but the passenger's body tends to continue moving forward at the original speed (due to inertia). This causes the passenger to lurch forward relative to the car. [2]
(1 mark for identifying inertia/Newton's First Law applies, 1 mark for explaining the relative motion of the passenger's body)

Section C: Application Questions

19.
(a) Net force = 5000 − 2000 = 3000 N [1]

(b) a = F/m = 3000 ÷ 2000 = 1.5 m/s² [2]
(1 mark for substitution, 1 mark for answer)

(c) v = u + at = 0 + 1.5 × 10 = 15 m/s [2]
(1 mark for substitution, 1 mark for answer)

(d)(i) Using v² = u² + 2as: 0 = 15² + 2a(150) → 0 = 225 + 300a → a = −225/300 = −0.75 m/s²
Deceleration = 0.75 m/s² [2]
(1 mark for substitution, 1 mark for answer)

(d)(ii) Net force = ma = 2000 × (−0.75) = −1500 N
Net force = Braking force + Resistive force (both oppose motion)
1500 = Braking force + 2000 → This gives a contradiction; re-evaluate:
Net retarding force = 1500 N. Resistive force already provides 2000 N, which alone would decelerate the van at 1 m/s². Since the required deceleration is only 0.75 m/s², the resistive force alone is more than sufficient. The braking force is therefore 0 N — the resistive force alone causes a greater deceleration than needed, so the van would stop in less than 150 m.

Re-interpretation (more consistent): The problem intends that the van is travelling at 15 m/s and must stop in 150 m. Required deceleration = 0.75 m/s². Net retarding force needed = 2000 × 0.75 = 1500 N. Since resistive force = 2000 N alone exceeds this, the braking force = 0 N (the van would actually stop in 112.5 m with resistive force alone).

Alternative consistent interpretation: If the van is no longer under driving force and brakes are applied:
Braking force + resistive force = ma → Braking force + 2000 = 2000 × 0.75 → This yields a negative braking force, which is unphysical.

Marking note: Award full marks for correct method. Accept: Braking force = 500 N if the problem is interpreted as total retarding force = 1500 N and braking force = 1500 − 2000 = −500 N (not physically meaningful). The most sensible interpretation: the van must decelerate at 0.75 m/s², requiring net retarding force of 1500 N. Since friction alone is 2000 N, no additional braking force is needed; the van stops in 112.5 m. Award [2] for correct method and logical conclusion.

Simplified marking scheme:

• Correct deceleration from (d)(i): confirmed 0.75 m/s² [already awarded]
• Braking force calculation: Net retarding force = 1500 N. Braking force = 1500 − 2000 → 0 N (friction alone is sufficient) [2]
  (1 mark for calculating net retarding force, 1 mark for braking force)

20.
(a) Weight = mg = 500 × 10 = 5000 N [1]

(b)(i) Tension = 5000 N [2]
Explanation: At constant velocity, acceleration = 0, so net force = 0. The tension must equal the weight. (1 mark for value, 1 mark for explanation)

(b)(ii) Work done = Force × distance = 5000 × 12 = 60 000 J (or 60 kJ) [2]
(1 mark for formula/substitution, 1 mark for answer)

(c)(i) Using Newton's Second Law: T − mg = ma
T = m(g + a) = 500 × (10 + 2) = 500 × 12 = 6000 N [2]
(1 mark for correct equation, 1 mark for answer)

(c)(ii) Work done in (c) = T × d = 6000 × 12 = 72 000 J, which is greater than in (b)(ii). [1]
Explanation: The tension is larger during the accelerated lift because the cable must not only support the weight but also provide the net upward force for acceleration. Since work = force × distance and the distance is the same, the greater tension means more work is done.