Physics - W.E.P. Question with Solution | TestHub

Let${\mathrm{u}}_1$and${\mathrm{v}}_1$be the initial and final velocities for ball$1$and${\mathrm{u}}_2$and${\mathrm{v}}_2$be the initial and final velocities for ball$2$. Here,${\mathrm{u}}_2=0$and${\mathrm{v}}_1=0$.

${\text{K}}_{\text{i}}=\frac{1}{2}{\text{mu}}_1^2+\frac{1}{2}{\text{mu}}_2^2=\frac{1}{2}{\text{mu}}_1^2$

${\text{K}}_{\text{f}}=\frac{1}{2}{\text{mv}}_1^2+\frac{1}{2}{\text{mv}}_2^2=\frac{1}{2}{\text{mv}}_2^2$

Loss of$\text{K.E}=\frac{1}{2}{\text{mu}}_1^2-\frac{1}{2}{\text{mv}}_2^2$

According to question,

$\frac{1}{2}\left(\frac{1}{2}{\text{mu}}_1^2\right)=\frac{1}{2}{\text{mu}}_1^2-\frac{1}{2}{\text{mv}}_2^2$

(as half of$\mathrm{K}.\mathrm{E}.$is lost by the impact)

${\therefore \text{u}}_1^2=2{\text{v}}_2^2\text{or}{\text{v}}_2=\frac{{\text{u}}_1}{\sqrt{2}}$

$\therefore \text{e}=\frac{{\text{v}}_2-{\text{v}}_1}{{\text{u}}_1-{\text{u}}_2}=\frac{{\text{v}}_2}{{\text{u}}_1}=\frac{1}{\sqrt{2}}$