Conservation of Momentum: Change in mass

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Homework Help Overview

The problem involves two identical buggies, each with a man, moving along parallel rails. The scenario describes a situation where the men exchange positions by jumping perpendicular to the direction of motion, resulting in a change in the velocities of the buggies. The objective is to find the initial velocities of the buggies based on the conservation of momentum.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the implications of momentum conservation when mass changes due to the men jumping between buggies. Questions arise regarding the reference frame in which the jumps are considered and how this affects the perceived velocities of the buggies and men.

Discussion Status

Some participants have offered insights into the importance of the reference frame and the implications of treating the buggy and man as a system. There is recognition of the complexity introduced by the change in mass and the need for clarity in the problem's wording. Multiple interpretations of the system dynamics are being explored.

Contextual Notes

There is a noted ambiguity in the problem statement regarding the frame of reference for the jumps, which may affect the understanding of momentum conservation in this scenario. Additionally, the initial conditions of the buggies' velocities are not explicitly defined, leading to further questions about the assumptions being made.

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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