Question

prove that in one dimensional elastic collison between two bodies the energy transfer is maximum when their masses are equal.​

Answer 1

The coefficient of restitution ,

e=(v2-v1)/(u1-u2).

v2 is velocity of target body after collision.

v1 is velocity of projected body after collision.

u1 is velocity of projected body before collision.

u2 is velocity of target body before collision .

For elastic collision e=1. Therefore u1-u2=v2-v1.

If we take target body at rest before collision then u2=0.

When velocity is exchanged between projected and target bodies, u1=v2 and we have v1=0.

Thus, velocity is exchanged only when there is one dimensional elastic collision between two bodies of equal mass because, according to conservation of momentum in the present case mu1=mv2.

Answer 2

$v_{2}=(1+e)u_{CM}$Let there be a body 1 of mass $m_{1}$ and body 2 of mass $m_{2}$ are going with the initial velocities $u_{1}$and $u_{2}$ respectively in the same direction, $u_{1}\neq u_{2}$

Let their final velocities after the collision is $v_{1}$ and respectively.

Then, by conservation of momentum $m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$ ....(1)

$e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}\\\Rightarrow v_{2}=e(u_{1}-u_{2})+v_{1}$ .....(2)

Solving Eqns (1) and (2)

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}\{e(u_{1}-u_{2})+v_{1}\}\\\Rightarrow v_{1}(m_{1}+m_{2})=m_{1}u_{1}(1+e)+m_{2}u_{2}(1+e)-eu_{1}(m_{1}+m_{2})\\\Rightarrow \therefore v_{1}=(1+e)\{ \frac{m_{1}u_{1}+m_{2}u_{2}}{m_{1}+m_{2}}\} -eu_{1}\\\Rightarrow \therefore v_{1}=(1+e)u_{CM}-eu_{1}$

Similarly, $v_{2}=(1+e)u_{CM}-eu_{2}$

Now, the energy transfer will be maximum for the largest $v_{2}$

For largest $v_{2}$, $(1+e)u_{CM$ should be largest which in turns means, $u_{CM}$ should be largest.

It will be largest when $m_{1}=m_{2}$

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