- #1

anotherghost

- 12

- 0

## Homework Statement

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. Let's 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.

## Homework Equations

I guess just conservation of momentum.

## 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.