Secondary 3 Physics Practice Paper 1

TuitionGoWhere Practice Paper — Physics Secondary 3

Answer Key — Mechanics Practice Paper, Version 1 of 5

Section A — Multiple Choice

1. B [1]
Average speed = total distance ÷ total time = 120 km ÷ 2 h = 60 km/h.
Common mistake: Multiplying instead of dividing (120 × 2 = 240 → option D).

2. D [1]
Velocity is a vector (has magnitude and direction). Distance, speed, and time are scalars.

3. C [1]
Using v² = u² + 2as: v² = 0 + 2(10)(20) = 400 → v = 20 m/s.
Common mistake: Using v = gt without finding t first, or using s = ½gt² and forgetting to then find v.

4. B [1]
Distance = area under v–t graph. The shape is a trapezium (or triangle + rectangle + triangle).
Area = area of trapezium from t = 0 to t = 10 (triangle: ½ × 10 × 12 = 60) + rectangle from t = 10 to t = 12 (2 × 12 = 24) — but the graph shows a triangle peaking at v = 12.
Area = ½ × base × height = ½ × 12 × 12 = 72 m.
Note: The graph is a single triangle with base 12 s and height 12 m/s.

5. B [1]
Net force = Applied force − Friction = 20 − 5 = 15 N.
a = F/m = 15/5 = 3 m/s².
Common mistake: Forgetting to subtract friction (20/5 = 4 → option C).

6. A [1]
Newton's First Law (Law of Inertia): A body remains at rest or in uniform motion unless acted on by a net external force. Option B is Newton's Second Law; C is the Third Law; D is an alternative statement of the Second Law.

7. B [1]
At maximum height, v = 0. Using v² = u² − 2gh: 0 = 30² − 2(10)h → h = 900/20 = 45 m.
Common mistake: Using h = ut = 30 × 3 = 90 m (confusing distance with the formula → option D).

8. B [1]
Resultant = √(3² + 4²) = √(9 + 16) = √25 = 5 N (Pythagoras' theorem).
Common mistake: Adding directly (3 + 4 = 7 → option C).

9. C [1]
Distance at constant speed = 6 × 10 = 60 m.
Distance during deceleration = average speed × time = (6 + 0)/2 × 4 = 12 m.
Total = 60 + 12 = 72 m.
Common mistake: Using only the constant-speed distance (60 m → option A).

10. C [1]
For constant acceleration from rest, s = ½at², so displacement is proportional to t² — a parabola (curve with increasing gradient, concave up).
Common mistake: Choosing B (straight line), which represents constant velocity, not constant acceleration.

Section B — Structured Questions

11.

(a) Displacement is the shortest straight-line distance from the starting point to the ending point, measured in a specific direction. [1]
Marking note: Must include both magnitude and direction idea. "Distance in a specific direction" or "straight-line distance with direction" are acceptable.

(b) Acceleration is the rate of change of velocity [1].
Marking note: "Change in velocity per unit time" is also acceptable. Simply saying "how fast something speeds up" is too vague — award 0.

12.

(a) Using v = u + at:
v = 0 + (2)(8) = 16 m/s [2]

• 1 mark for correct substitution, 1 mark for correct answer with unit.
  Common mistake: Using s = ut + ½at² instead (this gives distance, not velocity).

(b) Using s = ut + ½at²:
s = 0 + ½(2)(8²) = ½ × 2 × 64 = 64 m [2]

• 1 mark for correct substitution, 1 mark for correct answer with unit.
  Common mistake: Using average speed × time = (0 + 16)/2 × 8 = 64 m — this is also correct and should be awarded full marks.

13.

(a) The train accelerates uniformly from rest at a constant rate during the first 5 seconds, then travels at a constant velocity of 20 m/s from t = 5 s to t = 25 s, then decelerates uniformly to rest from t = 25 s to t = 30 s. [1]
Marking note: Award 1 mark for describing any one phase correctly. The question asks about the first 10 seconds only, so the answer should mention acceleration in the first 5 s and constant velocity from 5–10 s.

(b) Acceleration = gradient of v–t graph = rise/run = 20/5 = 4 m/s² [2]

• 1 mark for using gradient method, 1 mark for correct answer with unit.

(c) Total distance = area under the graph.
The graph is a trapezium: parallel sides are 20 (from t = 5 to t = 25) and 0 (at t = 0 and t = 30), but more accurately:

• Triangle (0–5 s): ½ × 5 × 20 = 50 m
• Rectangle (5–25 s): 20 × 20 = 400 m
• Triangle (25–30 s): ½ × 5 × 20 = 50 m
  Total = 50 + 400 + 50 = 500 m [3]
• 1 mark for identifying area method, 1 mark for correct calculation of each section, 1 mark for correct total.
  Common mistake: Forgetting the triangular sections and only calculating the rectangle (400 m).

14.

(a) Using F = ma:
a = F/m = 12/2 = 6 m/s² [2]

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

(b) Net force = Applied force − Friction = 12 − 4 = 8 N
a = F_net/m = 8/2 = 4 m/s² [2]

• 1 mark for finding net force, 1 mark for correct acceleration with unit.
  Common mistake: Using the original 12 N without subtracting friction.

15. Newton's Second Law of Motion states that the net force acting on a body is directly proportional to the rate of change of its momentum, and the force acts in the direction of the change in momentum. [2]
Alternatively: The acceleration of a body is directly proportional to the net force acting on it and inversely proportional to its mass (F = ma). [2]

• 1 mark for stating the relationship between force and acceleration/momentum, 1 mark for mentioning the direction or the inverse mass relationship.
  Marking note: "Force equals mass times acceleration" alone is acceptable for 1 mark; the second mark requires the proportional reasoning or direction.

Section C — Free Response

16.

(a) At maximum height, v = 0.
Using v = u − gt: 0 = 25 − 10t → t = 2.5 s [2]

• 1 mark for correct substitution, 1 mark for correct answer with unit.

(b) Using v² = u² − 2gh: 0 = 25² − 2(10)h → h = 625/20 = 31.25 m [2]

• 1 mark for correct substitution, 1 mark for correct answer with unit.
  Alternative: h = ut − ½gt² = 25(2.5) − ½(10)(2.5²) = 62.5 − 31.25 = 31.25 m.

(c) By symmetry (no air resistance), time to go up = time to come down.
Total time = 2 × 2.5 = 5.0 s [2]

• 1 mark for using symmetry or solving s = 0 = 25t − 5t², 1 mark for correct answer.
  Alternative method: s = ut − ½gt² → 0 = 25t − 5t² → t(25 − 5t) = 0 → t = 0 or t = 5 s.

(d) Velocity-time graph:

• Starts at v = +25 m/s at t = 0
• Straight line with gradient −10 m/s² (negative slope)
• Crosses v = 0 at t = 2.5 s
• Continues to v = −25 m/s at t = 5.0 s

v (m/s)
 25 |*
    | \
    |  \
    |   \
  0 |----*-------- t (s)
    |     \      2.5  5.0
    |      \
-25 |       *
    +------------------------ t (s)
      0    2.5   5.0

[2]

• 1 mark for correct shape (straight line with negative gradient), 1 mark for correct key values (25, 0, −25 m/s and 2.5, 5.0 s).
  Marking note: The gradient must be constant (straight line, not curved). Values must be labelled.

17.

(a) Horizontal component = F cos θ = 50 × cos 30° = 50 × 0.866 = 43.3 N [2]

• 1 mark for using cosine, 1 mark for correct answer.
  Common mistake: Using sine instead of cosine (50 × sin 30° = 25 N).

(b) Net horizontal force = Horizontal component − Friction = 43.3 − 15 = 28.3 N [1]

(c) a = F_net/m = 28.3/10 = 2.83 m/s² (or 2.8 m/s² to 2 s.f.) [1]
Marking note: Accept answers consistent with the student's answer to (b).

18. When a person walks, their foot pushes backward on the ground (action force). By Newton's Third Law, the ground exerts an equal and opposite forward force on the foot (reaction force). This forward reaction force propels the person forward. [2]

• 1 mark for identifying the action (foot pushes ground backward), 1 mark for identifying the reaction (ground pushes foot forward) and linking it to motion.
  Marking note: Students must mention both forces and that they are equal, opposite, and on different bodies.

19. Both objects hit the ground at the same time [1]. In the absence of air resistance, all objects fall with the same acceleration due to gravity (g = 10 m/s²), regardless of their mass [1]. Since they are dropped from the same height with zero initial velocity, the time of fall depends only on the height and g, not on mass.
Marking note: Award 1 mark for the correct conclusion and 1 mark for the correct physics reasoning. Simply saying "gravity is the same" without explanation earns only 1 mark.

20.

(a) Using v = u + at: 0 = 20 + a(5) → a = −20/5 = −4 m/s² (deceleration = 4 m/s²) [2]

• 1 mark for correct substitution, 1 mark for correct answer with unit.
  Marking note: Accept "4 m/s²" if the question asks for deceleration (magnitude). Award full marks either way if working is clear.

(b) Using F = ma: F = 1000 × 4 = 4000 N [2]

• 1 mark for using F = ma, 1 mark for correct answer with unit.
  Common mistake: Using the velocity (20 m/s) directly in F = ma without finding acceleration first.

(c) Using v² = u² + 2as: 0 = 20² + 2(−4)s → s = 400/8 = 50 m [2]

• 1 mark for correct substitution, 1 mark for correct answer with unit.
  Alternative: s = average speed × time = (20 + 0)/2 × 5 = 50 m. Also acceptable.

End of Answer Key