IB PHYSICS SL (Standard level)- 2024 – Practice Questions- All Topics
Topic 2.1 – Motion
Topic 2 Weightage : 13 %
All Questions for Topic 2.1 – Distance and displacement , Speed and velocity , Acceleration , Graphs describing motion , Equations of motion for uniform acceleration , Projectile motion , Fluid resistance and terminal speed
▶️ Answer/Explanation
Markscheme
a.
- Solution: t = √(2×0.24/10) = 0.219 s ≈ 0.2 s ✓
- Explanation: Using vertical motion equation s = ½gt² with s = 0.24 m and g = 10 m/s².
b.
- Graph features: ✓✓
- Straight line with positive gradient (10 m/s²)
- Starting at (0,0) and ending at (0.22 s, 2.2 m/s)
- Note: Must use given g = +10 m/s² (positive convention)
c.
- Time to net midpoint: t = 1.37 m / 12 m/s = 0.114 s ✓
- Vertical fall: y = ½×10×(0.114)² = 0.065 m ✓
- Height above net: 0.24 – 0.065 = 0.175 m > 0.15 m ✓
- Alternative method: Calculate horizontal distance during 0.09 m fall (1.6 m) which exceeds 1.37 m
d.i.
- Method 1: KE = ½mv² + mgh ✓
- = ½×0.0027×10.5² + 0.0027×9.81×0.18 ≈ 0.15 J ✓
- Method 2: Find resultant velocity (√(10.5² + 1.88²)) then KE = ½mv²
d.ii.
- Velocity change: Δv = 2×10.5 = 21 m/s (elastic collision) ✓
- Force: F = mΔv/Δt = (0.0027×21)/0.010 ✓
- = 5.67 N → 5.7 N (2 SF) ✓
- Key point: Elastic collision means velocity reverses, giving double the momentum change
▶️ Answer/Explanation
Markscheme
a.
- Answer: 0 N (zero) ✓
- Explanation: When hovering, the aircraft is in equilibrium. The upward lift force exactly balances the downward weight, so the resultant force is zero.
b.
- Blades push air downward (action force) ✓
- Air exerts equal upward force on blades (reaction force) ✓
- Key point: Must mention both action-reaction pair and Newton’s Third Law explicitly for full marks.
c.
- Total mass = 0.95 + 0.45 = 1.4 kg
- Weight = 1.4 × 9.81 = 13.734 N
- Momentum change per second: 1.7v = 13.734
- v = 13.734/1.7 ≈ 8.0788 m/s
- Final answer: 8.1 m/s (to 2 significant figures) ✓✓✓
- Note: 2 SF because given masses have 2 SF. Accept 8.2 if using g=10 m/s².
d.
- Lift force remains 13.734 N
- New weight = 0.95 × 9.81 = 9.3195 N
- Net force = 13.734 – 9.3195 = 4.4145 N
- Acceleration = 4.4145/0.95 ≈ 4.6468 m/s²
- Final answer: 4.6 m/s² upward ✓✓
- Why upward? The unchanged lift force now exceeds the reduced weight.
▶️ Answer/Explanation
Markscheme
a.
- Answer: \(11 \, \text{m/s}^2\)
b.
- Change in velocity (\(\Delta v\)) = \(236 \, \text{m/s}\)
- Acceleration (\(a\)) = \(\frac{236}{8} = 29.5 \, \text{m/s}^2\)
- Force (\(F = ma\)) = \(1.1 \times 10^4 \times 29.5 = 3.2 \times 10^5 \, \text{N}\)
- Note:
- Award [2 max] if initial speed is omitted (incorrect answer: 390 kN).
- Award [3] for a correct, unsupported answer.
c.
- Phase 1 distance: 88 m
- Phase 2 distance: 1296 m
- Total distance: 1400 m (or 1384 m if not rounded)
- Note:
- Watch for significant figure penalties.
- Award [1 max] if only \(\frac{1}{2}at^2\) is correctly substituted for phase 1 but distances are not calculated.
- Award [1 max] for correct addition of incorrect phase distances.