Question

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

Answer 1

what is your exact question

Answer 2

Kinetic energy of a body before collision = $K_{i}=\frac{1}{2}\times m_{1}\times (u_{1})^{2}$

Kinetic energy of a body before collision = $K_{f}=\frac{1}{2}\times m_{1}\times (v_{1})^{2}$

Kinetic energy transferred from one body to another

$\therefore\Delta K=\frac{1}{2}\times m_{1}\times (u_{1})^{2}-\frac{1}{2}\times m_{1}\times (v_{1})^{2}=\frac{1}{2}\times m_{1}\times [(u_{1})^{2}-(v_{1})^{2}]$

Therefore,

Fractional decrease in Kinetic energy

$\frac{\Delta K}{K}=\frac{\frac{1}{2}\times m_{1}\times [(u_{1})^{2}-(v_{1})^{2}]}{\frac{1}{2}\times m_{1}\times (u_{1})^{2}}=1-\frac{(v_{1})^{2}}{(u_{1})^{2}}$

These equations may be solved directly to find $v_{1} \text{ and } v_{2}$} when $u_{1}\text{ and }u_{2}}$ are known:

$\therefore v_{1}={\frac {m_{1}-m_{2}}{m_{1}+m_{2}}}u_{1}+{\frac {2m_{2}}{m_{1}+m_{2}}}u_{2}$

$v_{2}={\frac {m_{2}-m_{1}}{m_{1}+m_{2}}}u_{2}+{\frac {2m_{1}}{m_{1}+m_{2}}}u_{1}$

If the target is as rest then $u_{2}=0$

$\therefore v_{1}={\frac {m_{1}-m_{2}}{m_{1}+m_{2}}}u_{1}\\\\\Rightarrow\frac{ v_{1}}{u_{1}}={\frac {m_{1}-m_{2}}{m_{1}+m_{2}}}$

$\therefore\frac{\Delta K}{K}=1-\frac{(v_{1})^{2}}{(u_{1})^{2}}\\\\\Rightarrow\frac{\Delta K}{K}=1-({\frac {m_{1}-m_{2}}{m_{1}+m_{2}}})^{2}$

The transfer of the kinetic energy is maximum when $\frac{\Delta K}{K}=1$

therefore,

$({\frac {m_{1}-m_{2}}{m_{1}+m_{2}}})^{2}=0\\\\ m_{1}-m_{2}=0\\\\\therefore m_{1}=m_{2}$

So we said that for one dimensional elastic collision between two bodies the energy transfer is maximum when their masses are equal.