# How to work out momentum when there are velocity and mass changes

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.

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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

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.

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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.

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