# Person-jumping-out-of-a-moving-boat problem

by redshift
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 P: 199 You're right that this is a conservation of momentum problem. It also has to do with relative velocity. Using P for person, B for boat, you should be able to write a conservation of momentum equation, where the total momentum before the jump equals the total momentum after the jump: $$1350 = 50v_P + 400v_B$$ Recognize $v_P$ and $v_B$ are the velocities after the jump relative to the still water. Also keep in mind that it should come out that $v_P$ is negative if $v_B$ is positive, since they are in opposite directions. You're given the velocity of the boat with respect to the person, $v_{B/P}$. This is how I always remember the relationship between relative velocities: $$v_B=v_P+v_{B/P}$$ With this you should have enough information with which to solve for both $v_B$ and $v_P$.
 P: 199 Person-jumping-out-of-a-moving-boat problem Remembering relative velocity in that way has saved me many headaches. For me, it's just easier to remember one catchall than to reason out a relationship for each problem. This relative velocity expression can be expressed even more generally in vector form: $$\vec{v}_B=\vec{v}_A+\vec{v}_{B/A}$$ Where you can make B and A represent anything.