Question

A 65 kg man stands on a surfboard holding a 3 kg ball at rest. If the man throws the ballforward to his friend at 9 m/s, how fast will the man move and in what direction?

Answer 1

Given data:

* The initial velocity of the man is u_1 = 0 m/s.

* The initial velocity of the ball is u_2 = 0 m/s.

* The final velocity of the ball is v_2 = 9 m/s.

* The mass of the man is m_1 = 65 kg.

* The mass of the ball is m_2 = 3 kg.

Solution:

The initial momentum of the system is,

$p_1=m_1u_1+m_2u_2$

Substituting the known values,

$\begin{gathered} p_1=0+0 \\ p_1=0\text{ kgm/s} \end{gathered}$

The final momentum of the system is,

$p_2=m_1v_1+m_2v_2$

where v_1 is the final velocity of the man,

Substituting the known values,

$\begin{gathered} p_2=65v_1+3\times9 \\ p_2=65v_1+27 \end{gathered}$

According to the law of conservation of momentum,

$\begin{gathered} p_1=p_2 \\ 0=65v_1+27 \\ 65v_1=-27 \\ v_1=\frac{-27}{65} \end{gathered}$

By simplifying,

$v_1=-0.42\text{ m/s}$

Here, the negative sign indicates the direction of motion of man is opposite to the direction of motion of ball.

Thus, the speed of the man is 0.42 meter