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

TuitionGoWhere Practice Paper — Physics Secondary 3

Answer Key — Version 1 of 5

Assessment: SA2 (End-of-Year Examination) Paper: Paper 2 — Structured & Free Response Total Marks: 50

Section A — Multiple Choice [10 marks]

1. (C) Velocity [1]

• Marking note: Velocity has both magnitude and direction, making it a vector. Speed, distance, and time are scalars.

2. (B) 60 km/h [1]

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

3. (B) The resultant force on the object is zero. [1]

• Marking note: By Newton's first law, an object moving at constant velocity has zero resultant force. It does not mean no forces act on it — forces may be balanced.

4. (C) 20 m/s [1]

• Working: Using v² = u² + 2as: v² = 0 + 2(10)(20) = 400, so v = √400 = 20 m/s

5. (B) F = m × a [1]

• Marking note: Newton's second law states that the resultant force equals mass multiplied by acceleration.

6. (B) 3 m/s² [1]

• Working: Resultant force = 20 N − 5 N = 15 N; a = F/m = 15/5 = 3 m/s²

7. (B) Weight is mass multiplied by gravitational field strength. [1]

• Marking note: W = mg. Mass is measured in kg and does not depend on gravity. Weight is measured in newtons.

8. (B) 4 m/s² [1]

• Working: a = (v − u)/t = (30 − 10)/5 = 20/5 = 4 m/s²

9. (C) distance travelled [1]

• Marking note: The gradient of a velocity–time graph gives acceleration; the area under it gives displacement (or distance if no change in direction).

10. (C) 20 N·s [1]

• Working: Impulse = F × t = 10 × 2 = 20 N·s

Section B — Structured Questions [25 marks]

11.

(a) Speed is the distance travelled per unit time (or rate of change of distance). [1]

(b) Velocity is the speed in a given direction (or rate of change of displacement). [1]

• Marking note: Must mention direction to distinguish from speed.

(c) Acceleration is the rate of change of velocity. [1]

• Accept: change in velocity per unit time.

12.

(a) Average speed = total distance ÷ total time [1] for formula/method = (300 + 200) ÷ (60 + 40) = 500 ÷ 100 = 5 m/s [1] for correct answer

(b) Displacement = 300 m north − 200 m south = 100 m north [1] Average velocity = displacement ÷ total time = 100 ÷ 100 = 1 m/s north [1]

• Marking note: Direction must be stated for full marks.

13.

(a) Resultant force = 12 N − 4 N = 8 N [1]

(b) Using F = ma: a = F/m = 8/2 = 4 m/s² [1] for method, [1] for answer

(c) Using v = u + at: v = 0 + (4)(3) = 12 m/s [1] for method, [1] for answer

14.

(a) At maximum height, v = 0. Using v² = u² − 2gs (taking upward as positive, a = −g): 0 = 15² − 2(10)s [1] for correct substitution 20s = 225 s = 11.25 m [1] for correct answer, [1] for unit

(b) Time to reach max height: v = u − gt → 0 = 15 − 10t → t = 1.5 s [1] Total time = 2 × 1.5 = 3.0 s [1]

• Accept: Using s = ut − ½gt² with s = 0 to get t = 0 or t = 3 s directly.

15.

(a) Acceleration = gradient = (20 − 0)/(4 − 0) = 5 m/s² [1] for method, [1] for answer

(b) Total distance = area under graph:

• Triangle (0–4 s): ½ × 4 × 20 = 40 m
• Rectangle (4–7 s): 3 × 20 = 60 m
• Triangle (7–10 s): ½ × 3 × 20 = 30 m [1] for any two areas correct Total = 40 + 60 + 30 = 130 m [1] for correct total

(c) The car has zero acceleration during the interval 4 s to 7 s [1] because the velocity is constant (gradient of v–t graph is zero), meaning there is no change in velocity. [1]

Section C — Free Response [15 marks]

16. [4 marks]

• Newton's first law: An object remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force. [1]

• Explanation: When the bus is moving, the passengers are also moving forward at the same speed. When the bus suddenly brakes, the bus decelerates, but the passengers' bodies tend to continue moving forward due to inertia (their tendency to maintain their state of motion). [1] This causes them to lurch forward relative to the bus. [1]

• Role of seat belts: Seat belts provide a backward force on the passengers, decelerating them along with the bus, preventing them from being thrown forward. [1]

17.

(a) Weight = mg = 60 × 10 = 600 N. The scale reads 600 N (or 60 kg if the scale is calibrated in mass units). [1]

(b) When the lift accelerates upward, the resultant force on the student is upward. Using Newton's second law: R − mg = ma [1] for correct equation] R = m(g + a) = 60(10 + 2) = 60 × 12 = 720 N [1] for correct answer] The scale reads 720 N (or 72 kg). [1] Explanation: The normal reaction (scale reading) must exceed the student's weight to provide the upward resultant force needed for upward acceleration. [1]

(c) The student would feel heavier than normal [1] because the floor of the lift exerts a greater normal force on the student's feet than when stationary. This increased upward force from the floor is what the student perceives as increased weight. [1]

18.

(a) Total mass = 3 + 5 = 8 kg Using F = ma: a = 24/8 = 3 m/s² [1] for method, [1] for answer]

(b) Consider box B alone. The only horizontal force on B is the contact force from A (call it P). For box B: P = mB × a = 5 × 3 = 15 N [1] for isolating box B, [1] for correct substitution, [1] for answer]

• Alternative method: Force on A = 24 − P = mA × a = 3 × 3 = 9 N, so P = 24 − 9 = 15 N. Award full marks for correct alternative.

(c) The acceleration would decrease [1] because the frictional force would oppose the applied force, reducing the resultant force on the system. Since a = Fresultant/m, a smaller resultant force means smaller acceleration. [1]

19.

(a) KE = ½mv² = ½ × 1000 × 25² = ½ × 1000 × 625 = 312 500 J (or 312.5 kJ) [1] for substitution, [1] for answer]

(b) Work done by braking force = 312 500 J [1] Explanation: By the work-energy principle (or conservation of energy), the kinetic energy of the car is converted into work done against the braking force (which becomes thermal energy in the brakes). The work done by the braking force equals the initial kinetic energy of the car. [1]

(c) Work done = Force × distance [1] 312 500 = F × 50 F = 312 500 / 50 = 6250 N [1]

• Accept: Using v² = u² + 2as to find deceleration (6.25 m/s²), then F = ma = 1000 × 6.25 = 6250 N.

20.

(a) Vertical motion only determines time of flight: s = ½gt² (initial vertical velocity = 0) [1] 45 = ½ × 10 × t² t² = 9 t = 3 s [1] for correct answer, [1] for unit]

(b) Horizontal distance = horizontal velocity × time = 8 × 3 = 24 m [1] for method, [1] for answer]

(c) The time of flight would remain the same [1] because the vertical motion (which determines the time of flight) is independent of the horizontal speed. The initial vertical velocity is zero in both cases, and the vertical displacement and acceleration are unchanged. [1]

— End of Answer Key —

Mark Summary