Question

A boy on a 2.2 kg skateboard initially at rest tosses an 8.7 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 1.0 m/s, find the boy's mass. Round to the hundredths place.

Answer 1

Given,

The mass of the skateboard, m₁=2.2 kg

The mass of the jug, m₂=8.7 kg

The speed of the jug after he throws it, v=3.0 m/s

The speed of the boy and skateboard, u=-1.0 m/s

From the law of conservation of momentum, the total momentum of the boy, skateboard, and the jug before the boy throws the jug should be equal to the total momentum of the boy, skateboard, and the jug after he throws it.

As all the objects were at rest before the boy throws the jug, the total momentum initially was equal to zero.

Thus,

`(m_1+m_3)u+m_2v=0`

Where m₃ is the mass of the boy.

On rearranging the equation,

`\begin{gathered} m_1+m_3=(-m_2v)/(u) \\ m_3=(-m_2v)/(u)-m_1 \end{gathered}`

On substituting the known values in the above equation,

`\begin{gathered} m_3=(-8.7*3.0)/(-1.0)-2.2 \\ =26.10-2.2 \\ =23.90\text{ kg} \end{gathered}`

Thus the mass of the boy is 23.90 kg