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,

We need v2f

Substituting the known values in the equation,

`40*2+20*2=40*4+20* v_(2f)`

Simplyfing,

`120=160+20* v_(2f)`

On rearranging the above equation we get,

`v_(2f)=(120-160)/(20)=(-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.