Conservation of Momentum: Change in mass

AI Thread Summary
The discussion focuses on a physics problem involving two identical buggies and the conservation of momentum during a mass exchange when the men jump between them. The key point is that when one man jumps from buggy one to buggy two, the system's momentum must remain constant, leading to changes in velocity due to the change in mass. The confusion arises from the choice of reference frames, as the initial momentum calculations depend on whether the buggies or the ground is considered as the system. It is clarified that if the buggies are the system, the initial momentum is zero, but this is only valid if their initial velocities are also zero. Understanding the correct frame of reference is crucial for solving the problem accurately.
Titan97
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Homework Statement


Two identical buggies 1 and 2 of mass ##M## with one man of mass ##m## in each, move without friction due to inertia towards along two parallel rails. When the buggies are opposite to each other, the men exchange positions by jumping in a direction perpendicular to motion of buggy. As a consequence, buggy 1 stops, while buggy 2's velocity becomes v. Find initial velocities ##v_1## nad ##v_2## of buggy 1 and 2.

Homework Equations


Conservation of momentum

The Attempt at a Solution


When the man jumps from buggy one, no force acts on the man+buggy system along horizontal. Since mass changes, velocity changes accordingly to make momentum constant. So,
##(m+M)v_1=Mu_1##
##u_1=(\frac{m}{M}+1)v_1##
This means velocity of buggy increased.
When the second man jumps to buggy one, again no force acts along horizontal. Hence,
##Mu_1=(m+M)u'_1##
But here is the problem. ##u'_1=0##. This means ##u_1=0## which means ##v_1=0##. This is wrong.
What is my mistake?
 
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Titan97 said:
Since mass changes, velocity changes accordingly to make momentum constant.
You have to take care in what system you are considering. If you consider the buggy, there is no force acting on it in horizontal so the velocity remains constant (until the other man comes in). It does not increase.

Part of your problem may also be based in the wording. It is not specified in which frame the jump is perpendicular to the direction of motion. I suspect the intended meaning is in the rest frame of the buggies (otherwise it will be impossible for one buggy to stop).
 
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I made a mistake. The man acquires a horizontal velocity as well with respect to ground.
So ##(m+M)v_1=Mu_1+mu_1## along horizontal. Hence ##v_1=u_1##.
Yes. Since I took man+buggy as system, there is no change in mass.
But if I take the buggy as the system, initial momentum is zero right? Because with respect to buggy, the velocity of man is zero.
 
Titan97 said:
But if I take the buggy as the system, initial momentum is zero right?
This depends on which system you are considering. With respect to the ground system, this is only true if the initial velocity is zero.
 
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