# Two people jumping out of a boat in different directions - momentum problem

1. Mar 23, 2009

### wizzle

1. The problem statement, all variables and given/known data
A woman and her husband simultaneously dive from a 120 kg raft that is initially
at rest. The woman (60 kg) jumps from the boat with a horizontal speed of 1.9
m/s due south, while her husband (77 kg) jumps with a horizontal speed of 1.5
m/s due west. Calculate the magnitude and direction of the boat’s velocity
immediately after their dives.

2. Relevant equations

P=mv
C^2=a^2+b^2
Pfinal-Pinitial=0

3. The attempt at a solution
I calculated the momentum of the woman to be (60)(1.9)=114 kg*m/s south
and the man's momentum to be (77)(1.5)=115.5 kg*m/s west. Since it was at rest, the initial momentum of the boat and the two people would have been zero. I'm confused about how to link the mass of the boat and its velocity to these equations. Will the final momentum of the boat be equal to their two momenta? MbVb=MmanVman+MwomanVwoman? I'm thinking I'll have to use the pythagorean theorem to calculate the direction of the displacement. Any help would be greatly appreciated!

2. Mar 23, 2009

### alphysicist

Hi wizzle,

This is close, but there are a couple of issues. First, momentum is conserved for the entire system, so we would say that the total momentum before they jump equals the total momentum after they jump (and you said the initial momentum was zero). Also, momentum is a vector quantity, so you need:

$$0 = M_b \vec v_b + M_m \vec v_m + M_w \vec v_w$$

since the action in this problem is not taking place in a straight line. So you'll get an equation for the x direction, and an equation for the y direction, and that will allow you to find the x and y velocities of the boat.

The Pythagorean theorem is the easiest way to get the magnitude; then you can use trig to find the direction.

3. Mar 23, 2009

### wizzle

Wow that was helpful! Thank you so much, I've been able to work it out :)

4. Mar 23, 2009