The correct answer is C because:

The boy jumps from stationary and hits the floor with a maximum velocity

To calculate the maximum velocity method 1: Energy

There is no rebound, so all the gravitational potential energy from the jump is transferred to kinetic energy
- $E_{p} = m g h = m \times 10 \times 5 = 50 m$
- $E_{p} = E_{k}$
- $50 m = \frac{1}{2} m v^{2}$
- $v^{2 =} \frac{2 \times 50 m}{m}$
- $v = \sqrt{100} = 10 m s^{- 1}$

To calculate the maximum velocity method 2: Kinematics

The acceleration is uniform g, therefore $v^{2} = u^{2} + 2 a s$ can be used
Where:
- Initial velocity, u = 0
- Acceleration, a = g ≈ 10 m s⁻¹
- Displacement, s = h = 5 m
Therefore,
- $v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10$

Calculate change in momentum:
- Change in momentum = momentum after – momentum before
- ∆ρ = mv – mu
- ∆ρ = 10m – 0
- ∆ρ = 10m

Use the answer options to find an estimate for an appropriate mass of the boy:
Option C: 4 × 10²kg m s^–1= 10m
- Rearrange to calculate mass m: $\frac{4 \times 10^{2}}{10}$ = m
- So, m = 40 kg
This is a suitable size for the mass of a boy, so the correct answer is C, 4 × 10²kg m s^–1

A is incorrect as when change in momentum = 4 × 10⁰ kg m s^–1= 4 kg m s^–1, then mass = $\frac{4}{10}$ = 0.4 kg which is too small to be the mass of the boy

B is incorrect as when change in momentum = 4 × 10¹ kg m s^–1= 40 kg m s^–1, then mass = $\frac{40}{10}$ = 4 kg which is still too small to be the mass of the boy

D is incorrect as when change in momentum = 4 × 10³ kg m s^–1= 4000 kg m s^–1, then mass = $\frac{4000}{10}$ = 400 kg which is too large to be the mass of the boy

Physical Quantities (CIE AS Physics)

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