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- Thread starter chrizzle
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1. We take into consideration, the conservation of momentum, besides kinetic energy (KE) conservation.

2. We assume the collision to be 1D collision; ie, it is a head-on collision and NOT an oblique collision. This assumption avoids the complications that are associated with elastic scatter process, which is not same as elastic collision process.

3. In the center of mass(CM) reference frame, the velocity of CM is zero. each mass changes the sign of its velocity (but not magnitude), after the collision thereby conserving its KE.

4. In the laboratory reference frame (where one of the masses is at rest before collision), CM moves with a velocity equal in magnitude and opposite in sign to that of the velocity with which the mass at rest would move when the collision is viewed in CM frame. (This gives the reason why the mass at rest appears to be at rest in the laboratory frame).

Now the answer to the question:

When the body (say, m2), at rest before collision decreases its mass slightly, the velocity of CM increases slightly (compared to the value it had when the masses were equal), in order that m2 appears to be at rest.

After the collision the velocity of CM remains the same. Velocity of the mass m2 increases (compared to the value it had after the collision when the masses were equal); consequently, KE of m2 increases when its mass decreases.

The other body, (m1) (which came to rest(zero velocity) after the collision when the masses were equal) moves with a non-zero velocity. Consequently, its KE also increases compared to the value it had (zero) when the masses were equal).

The solution is very simple in the geometrical form (and is due to Huygens). The algebraic solution can be obtained by solving the equations of conservation of mass, momentum and KE.

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