A ball moving with speed v hits another identical ball at rest. The two balls stick together after collision. If specific heat of the material of the balls is S, the temperature rise resulting from the collision is 

Initial momentum = mv
Final momentum = 2mV
By the conservation of momentum, $\mathrm{mv}=2\mathrm{mV}$
$\Rightarrow \mathrm{V}=\frac{\mathrm{v}}{2}$

K.E. of the system after the collision =  $\frac{1}{2}\left(2\mathrm{m}\right)\mathrm{\hspace{0.33em}}{\left(\frac{\mathrm{v}}{2}\right)}^2$
$\therefore$ loss in K.E. $=\frac{1}{2}{\mathrm{mv}}^2-\frac{1}{4}{\mathrm{mv}}^2=\frac{1}{4}{\mathrm{mv}}^2$
This loss in K.E. will increase the temperature 
$\therefore 2\mathrm{m}\times \mathrm{s}\times \mathrm{\Delta t}=\frac{1}{4}{\mathrm{mv}}^2\Rightarrow \mathrm{\Delta t}=\frac{{\mathrm{v}}^2}{8\mathrm{s}}$