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A Level H1 Physics Mechanics Quiz
Free A Level H1 Physics Mechanics quiz, Exam version, with questions, answers, and A Level-style practice for Singapore students.
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A-Level Physics H1 Quiz - Mechanics (Answer Key)
Section A: Short Answer Questions [8 marks]
1. State the principle of conservation of linear momentum. [2]
Answer: In a closed/isolated system, the total momentum remains constant [B1] provided no external forces act on the system [B1]
Accept: "momentum before = momentum after" for first mark Accept: "in the absence of external forces" for second mark
2. Write down expressions for: [2]
(a) momentum p in terms of mass m and velocity v [1] Answer: p = mv [B1]
(b) kinetic energy K in terms of mass m and velocity v [1] Answer: K = ½mv² [B1]
Common error: Writing K = mv² (missing factor of ½)
3. Two parallel current-carrying wires [2]
Answer: Arrows should point toward each other (attractive force) [B1 for each arrow]
Explanation: Parallel currents in same direction attract each other Common error: Drawing arrows pointing away (repulsive)
4. Speed-time graph with air resistance [2]
Answer:
- Curve starting steep, gradually flattening to horizontal [B1]
- Eventually reaching constant speed (terminal velocity) [B1]
The graph should be concave down, not a straight line
Section B: Structured Questions [15 marks]
5. Hockey puck momentum and energy [5]
(a) Calculate the mass of the puck. [3]
Working: From p = mv: v = p/m [M1] From K = ½mv²: substitute v = p/m K = ½m(p/m)² = p²/2m [M1] Rearranging: m = p²/2K = (2.4)²/(2 × 7.2) = 5.76/14.4 = 0.40 kg [A1]
Answer: Mass = 0.40 kg
(b) Calculate the velocity of the puck. [2]
Working: v = p/m = 2.4/0.40 = 6.0 m s⁻¹ [M1][A1]
Answer: Velocity = 6.0 m s⁻¹
Check: K = ½ × 0.40 × (6.0)² = 7.2 J ✓
6. Plank equilibrium [5]
(a) Draw and label all forces [2] Answer:
- Weight of plank (200 N downward at center, 2.0 m from A) [B1]
- Weight of person (600 N downward at distance x from A) [B1]
- Reaction forces R₁ and R₂ upward at P and Q respectively
(b) Moment equation about P when x = 1.5 m [3]
Working: Taking moments about P (clockwise positive): Weight of person: 600 × (1.5 - 1.0) = 600 × 0.5 [M1] Weight of plank: 200 × (2.0 - 1.0) = 200 × 1.0 [M1] Reaction at Q: R₂ × (3.0 - 1.0) = R₂ × 2.0 (anticlockwise)
Equation: 600 × 0.5 + 200 × 1.0 = R₂ × 2.0 [A1]
Or: 300 + 200 = 2R₂, giving R₂ = 250 N
Section C: Data Analysis [12 marks]
7. Trolley motion analysis [12]
(a)(i) Initial velocity = 8.0 m s⁻¹ [B1] Read from y-intercept
(a)(ii) Time to stop = 10 s [B1] Read from x-intercept
(a)(iii) Acceleration [2] Working: a = (v - u)/t = (0 - 8.0)/10 = -0.80 m s⁻² [M1][A1]
The negative sign indicates deceleration
(b) Friction force [2] Working: F = ma = 2.5 × 0.80 = 2.0 N [M1][A1]
Magnitude only required, so positive value
(c) Distance travelled [3] Method 1: Using area under graph Distance = ½ × base × height = ½ × 10 × 8.0 = 40 m [M1][M1][A1]
Method 2: Using kinematic equation s = ut + ½at² = 8.0 × 10 + ½ × (-0.80) × 10² = 80 - 40 = 40 m
Answer: Distance = 40 m
(d) Explanation for straight line [3] Answer: The friction force is constant [B1] Therefore the acceleration is constant (F = ma) [B1] Constant acceleration produces a linear velocity-time graph [B1]
Accept: "No other forces acting" or "uniform deceleration"
Total: 35 marks