Question

A 40 kg surfer jumps off the front of a 20 kg surfboard moving forward. Find the surfboard’s velocity immediately after the girl jumps, assuming that the surfboard’s initial velocity is 2 m/s and the girl’s velocity when jumping off the front is 4 m/s.

Answer 1

According to the law of conservation of momentum,

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}_{}_{}$

Where,

$m_1=40\operatorname{kg},\text{ }m_2=20\operatorname{kg},v_{1i}=2m/s,v_{2i}=2m/s,\text{ }v_{1f}=4ms^{-1}_{}$

We need v2f

Substituting the known values in the equation,

$40\times2+20\times2=40\times4+20\times v_{2f}$

Simplyfing,

$120=160+20\times v_{2f}$

On rearranging the above equation we get,

$v_{2f}=\frac{120-160}{20}=\frac{-40}{20}$

i.e.

$v_{2f}=-2\text{ m/s}$

The negative sign indicates that the surfboard will move backward with a velocity of 2m/s after the surfer jumps off of it.