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

1. ### redshift

53
I'd appreciate a hint on how to start this one.

"A person weighing 50kg is riding in a 400kg boat heading due east at 3.0 m/s. The person jumps horizontally out of the boat in the other direction (i.e., westward). Just after he jumps, the velocity of the boat as seen from the person is 4.5 m/s. Determine the velocity of the boat relative to the water surface just after he jumps."

I assume the water is motionless. This seems like another conservation of momentum problem. The initial momentum of the boat with the person riding it is 1,350 kgms. Would this equal the difference between the man's momentum after jumping and the boat's momentum after he jumped?

2. ### TALewis

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 $v_P$.

3. ### redshift

53
Glad to know i started off right. The relative velocity equation was what i tripped up on. Many thanks.

4. ### TALewis

199
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.