Question

A boy on a 2.0 kg skateboard initially at rest tosses an 8.0 kg jug of water in the forward direction. If the jug has a speed of 3.0 m/s relative to the ground and the boy and skateboard move in the opposite direction at 0.60 m/s, find the boy’s mass.

Answer 1

According to the momentum conservation law, the magnitude of boy with skateboard momentum must be equal to the magnitude of jug's momentum:

$$(m_b+m_s)v_1 = m_jv_2$$

where $v_1 = 0.6m/s, v_2 = 3m/s,m_s = 2kg,m_j=8kg$ and $m_b$ is boy's mass. Thus. obtain:

$$m_b = \dfrac{m_jv_2-m_sv_1}{v_1} = \dfrac{8\cdot 3 - 2\cdot 0.6}{0.6} = 38kg$$

Answer. 38kg.