IGCSE Physics 1.6 — Momentum Numericals (Solutions)

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IGCSE Physics 0625 · Section 1.6 · Momentum — Numerical Solutions · Extended only

Momentum — Numerical Practice: Solutions

A. Basic Momentum

A1

5 kg at 4 m/s. Find momentum.

1
p = mv
5 × 4
✓20 kg m/s
A2

p = 150 kg m/s, m = 30 kg. Find v.

1
v = p/m
150 ÷ 30
✓5 m/s
A3

p = 84 kg m/s, v = 12 m/s. Find m.

1
m = p/v
84 ÷ 12
✓7 kg
A4

2000 kg at 3 m/s vs 60 kg at 8 m/s.

1
Car: p = mv
2000 × 3 = 6000 kg m/s
2
Cyclist: p = mv
60 × 8 = 480 kg m/s
✓The car has more momentum (6000 vs 480 kg m/s)
A5

0.01 kg bullet at 400 m/s. Find momentum.

1
p = mv
0.01 × 400
✓4 kg m/s

B. Impulse and Force

B1

50 N for 0.2 s. Find impulse.

1
impulse = FΔt
50 × 0.2
✓10 kg m/s (10 N s)
B2

Impulse 30 kg m/s over 0.05 s. Find force.

1
F = impulse/Δt
30 ÷ 0.05
✓600 N
B3

0.4 kg ball, 2→18 m/s in 0.01 s. Find average force.

1
Δp = m(v−u)
0.4 × (18−2) = 6.4 kg m/s
2
F = Δp/Δt
6.4 ÷ 0.01
✓640 N
B4

25 N for 4 s. Find change in momentum.

1
impulse = FΔt = Δp
25 × 4
✓100 kg m/s
B5

Why does bending knees reduce landing force?

✓Bending your knees increases the time (Δt) over which your momentum reduces to zero. Since F = Δp/Δt and the momentum change (Δp) is fixed by your mass and landing speed, a larger Δt produces a smaller force F on your legs.

C. Conservation of Momentum

C1

2 kg at 5 m/s hits stationary 3 kg, they stick. Find combined velocity.

1
Momentum before = momentum after
(2×5)+(3×0) = 10 kg m/s
2
v = total momentum / combined mass
10 ÷ (2+3)
✓2 m/s
C2

4 kg at 6 m/s hits stationary 2 kg; 4 kg trolley ends at 3 m/s. Find 2 kg trolley’s velocity.

1
Momentum before
(4×6)+(2×0) = 24 kg m/s
2
Momentum after: (4×3) + (2×v₂) = 24
12 + 2v₂ = 24 → v₂ = 12 ÷ 2
✓6 m/s
C3

1 kg at 8 m/s right, 1 kg at 4 m/s left, stick together. Find combined velocity.

1
Taking right as positive: momentum before
(1×8)+(1×−4) = 4 kg m/s
2
v = total momentum / combined mass
4 ÷ (1+1)
✓2 m/s, in the direction of the ball that was originally moving at 8 m/s (i.e. to the right)
C4

40 kg cannon fires 1 kg ball at 100 m/s. Find recoil velocity.

1
Momentum before = 0, so momentum after = 0
(1×100) + (40×v) = 0
2
Solve for v
100 + 40v = 0 → v = −100 ÷ 40
✓2.5 m/s in the opposite direction to the cannonball
C5

3.5 kg rifle fires 0.025 kg bullet at 350 m/s. Find recoil.

1
Bullet’s momentum = rifle’s recoil momentum
0.025 × 350 = 8.75 kg m/s
2
v = momentum / mass
8.75 ÷ 3.5
✓2.5 m/s
C6

50 kg and 70 kg skaters push apart from rest; 50 kg moves at 2.8 m/s. Find 70 kg’s speed.

1
Total momentum before = 0, so after must also = 0
(50×2.8) + (70×v) = 0
2
Solve for v
140 + 70v = 0 → v = −140 ÷ 70
✓2 m/s, in the opposite direction to the 50 kg skater
C7

1500 kg at 20 m/s hits stationary 1000 kg, lock together. Find combined velocity.

1
Momentum before
1500 × 20 = 30000 kg m/s
2
v = momentum / combined mass
30000 ÷ 2500
✓12 m/s
C8

3000 kg at 10 m/s hits stationary 1000 kg; truck ends at 7 m/s. Find car’s velocity.

1
Momentum before
3000 × 10 = 30000 kg m/s
2
Momentum after: (3000×7) + (1000×v) = 30000
21000 + 1000v = 30000 → v = 9000 ÷ 1000
✓9 m/s

Mastery Test — Answer Key

  1. 24 kg m/s (8 × 3)
  2. 8 m/s (200 ÷ 25)
  3. 20 kg m/s (40 × 0.5)
  4. 500 N (15 ÷ 0.03)
  5. 3 m/s — momentum before = 3×4 = 12; v = 12 ÷ (3+1)
  6. 3 m/s — momentum before = 5×6 = 30; v = 30 ÷ (5+5)
  7. 4.5 m/s, in the opposite direction to the canoeist — momentum before = 0, so 60×1.5 + 20×v = 0 → v = −4.5 m/s
  8. Momentum is a vector because it depends on velocity, which is itself a vector (has direction as well as magnitude) — so momentum always has a direction too.
  9. 0 m/s — taking right as positive: momentum before = (2×6)+(2×−6) = 0, so the combined object is at rest after the collision
  10. In any collision or interaction where no external force acts, the total momentum before the interaction equals the total momentum after.

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