Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I have a pretty simple homework question, but I can't rap my head around it.

In the question a swimmer of $55 \mbox{ } \mathrm{kg}$, jumps off a stationary raft of $210\mbox{ }\mathrm{kg} $. The swimmer jumps off the raft with a speed of $4.6 \mbox{ } \mathrm{ms}^{-1} $. I need to work out the recoil velocity of the raft.

So because Momentum before = Momentum after, I went:

$p_i = 0$

Therefore $0 = p_f$ and $p_f = mv$, $m = 210\mbox{ }\mathrm{ } $, so the $v$ would have to equal $0$. Making the recoil velocity equal $0$. However that doesn't seem right. Could use some clarification or help, thanks.

share|cite|improve this question
Seriously, Pi = 0? : ) P.S. Thank you for showing your work and what you've done so far . in a hw problem . –  Dimensio1n0 Aug 3 '13 at 11:26

2 Answers 2

up vote 2 down vote accepted

The total momentum of the whole Swimmer+Raft system is conserved, not of the raft only.

So your equation should be $$P_{i,system}=P_{f,system}$$ $$0=m_{raft}\vec v_{raft}+m_{swimmer}\vec v_{swimmer}$$

You cannot conserve momentum for the raft alone because there is a force on the raft(The swimmer pushing back with her legs, trying to jump forward).

Also note that $v_{raft}$ and $v_{swimmer}$ are the velocities in the ground frame. The $4.6 m/s$ of the swimmer might be with respect to the raft. So you'll have to convert that velocity to the ground frame velocity.

share|cite|improve this answer
Awesome. thanks, I knew I was missing something. –  user2396852 Aug 3 '13 at 11:31

Therefore $0=p_f$

NO! Don't you realise, that., if this were true, nothing would change momentum? You're applying conservation of momentum wrongly . What you should be doing is :

$$0=p_{1i}+p_{2i}=m_{1} v_{1i} + m_{2}v_{2i}=p_{1i}+p_{2i }=p_{1f}+p_{2f} =m_{1}v_{1f}+m_2v_{2f} $$

Substitute in your values $m_1=55$, $v_{1f}= 4.6$, $m_2=210$ and you have your answer immediattely.

share|cite|improve this answer
Thanks for your answer, but I already accepted udiboy's one. –  user2396852 Aug 3 '13 at 11:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.