# Conservation of momentum and relative velocity

1. Nov 23, 2009

### anotherghost

1. The problem statement, all variables and given/known data

So you've got two people standing on a boat, everything at rest, and the masses of both people as well as the boat are known. Lets call them Richard and Sandra, so the variables are Mr (mass of Richard), Ms (mass of Sandra), and Mb (mass of boat). When one of these people jumps off the side of the boat, they have a velocity of 3.5m/s relative to the boat.

The problem asks to find the final speed of the boat Mbf after:
a) Both people jump off in the same direction at the same time.
b) First Richard jumps off, then Sandra a few seconds later, both in the same direction.
c) First Sandra jumps off, then Richard a few seconds later, both in the same direction.
d) First Richard jumps off to the right, then a few seconds later Sandra jumps off to the left.

2. Relevant equations

I guess just conservation of momentum.

3. The attempt at a solution

a) 0 = Mb * Vbf + (Mr + Ms)(3.5 + Vbf)

b) 0 = (Mb + Ms) * Vbs + Mr * (3.5 + Vbs)
(Mb + Ms) * Vbs = Mb * Vbf + Ms * (3.5 + Vbf)

c) 0 = (Mb + Mr) * Vbr + Ms * (3.5 + Vbr)
(Mb + Mr) * Vbr = Mb * Vbf + Mr * (3.5 + Vbf)

d) 0 = (Mb + Ms) * Vbs + Mr * (3.5 + Vbs)
(Mb + Ms) * Vbs = Mb * Vbf + Ms * (-3.5 + Vbs)

After these equations, of course, you solve for Vbf. I can do that part easily but I want to make sure the physics of these equations are right. My rationale is that I can first equate 0 (because the system starts at rest) with the first jump, and then equate the system without the person who jumped off with the second jump. I solve the first equation for the velocity needed in the left side of the second equation, then I solve the second equation for Vbf.

I am not sure if these are even remotely right or not - they make sense to me, but I'm not sure! Relative velocity is confusing. Please somebody give me a hand, even if it's only on one part.

Thanks.

2. Nov 23, 2009

### Delphi51

Looks good to me. Very tricky business.