# Conservation of the impulse, center of the masses

## Homework Statement

Two men of equal masses m stand on a platform of mass M. Both men jump with the same horizontal velocity with respect to the platform. Find the relation between velocities of the platform if they jump simultaneously and if they jump one after the other.

## The Attempt at a Solution

Impulse is constant. At the initial point, impulse is 0. Thus: mv + mv = MV, if they jump simultaneously. If they don’t jump at the same time, mv = (M+m)V’. I think that my reasoning for the second case is not correct. Is this problem easier if it is considered from the point of the center of the masses?

At the initial point, impulse is 0. Thus: mv + mv = MV, if they jump simultaneously. If they don’t jump at the same time, mv = (M+m)V’

You are correct till there. You have got the velocity of the platform after one man has jumped. Lets say after some time the second man jumps, with the velocity 'v' w.r.t the platform, so again apply the conservation of linear momentum, (since in both the cases there is no external force acting on the system in the horizontal direction) to determine the final velocity of the cart after the second man has jumped. This should be the required answer.

Is this problem easier if it is considered from the point of the center of the masses?

You can consider the motion of the bodies in terms of the center of mass of the system as well. The basic essence of using it is that there is no external force acting on the system, so its center of mass will have no velocity. So if you differentiate the equation defining the COM of the system and equate the velocity of the COM to zero, you will end up having the same equations!!