Relative velocity and momentum?

1. Jun 5, 2009

buckyball

1. The problem statement, all variables and given/known data
Some friends and I were working on some physics questions and when it came to this one, we all got different answers.

An 90 kg logger is standing on a 400 kg log floating at rest in a pond. The logger starts to
walk along the log at 1.5 m/s relative to the water. How fast is the log moving relative to
the water?

2. Relevant equations
One friend and I used conservation of momentum .

I used: (m1 + m2)(0) = m1(1.2) + m2(v')
She used: (m1+m2)(0) = m1(1.2) + (m1+m2)(v')

3. The attempt at a solution
I got 0.34 m/s and my friend got 0.28 m/s. Are either of us right?

Thank-you :)

2. Jun 5, 2009

LowlyPion

Welcome to PF.

Remember that the center of mass of the system will not move.

What condition then must be met such that this is accomplished? That is how fast must the center of mass of the log move counter to the movement of the logger?

3. Jun 5, 2009

buckyball

Are you saying that the velocity will be 1.5 m/s in the opposite direction of the logger?

4. Jun 7, 2009

buckyball

I've been thinking more about it. The logger and the log is at rest, so the center of mass of the system is zero. Once the logger moves in one direction along the length of the log, he is moving at 1.2 m/s.

I understand that if you use conservation of momentum, that the log must move in the opposite direction as the logger, but that is where my friends and I cannot agree on.

Do we use the mass of the logger and log moving in the opposite direction of the logger since the logger is still on the log? Or do you just use the mass of the log moving the other direction since the logger is moving in the other direction.

Any more clarification would be great! Thank-you.

DG.

5. Jun 7, 2009

LowlyPion

The center of mass of the system is made from the two centers of mass - the log, the runner.

If momentum is conserved, and the velocity of the runner is defined, then you know the momentum of the runner. If the center of mass remains constant then won't the center of mass of just the log need to move in the opposite direction at a speed that conserves momentum?

6. Jun 7, 2009

diazona

Here's something to think about: the total momentum of the system is the momentum of the log, plus the momentum of the logger. So when you use conservation of momentum, calculate the momentum of each separately, and add them up. All velocities should be taken relative to a common reference point (usually the ground).