Two equal masses  moving along the same straight line with velocities $\text{+\hspace{0.17em}3\hspace{0.17em}m}/\text{s}$ and  $-5\text{\hspace{0.17em}m}/\text{s}$ respectively collide elastically. their velocities after the collision will be respectively

Given, ${\text{u}}_{\text{1}}=3\text{m}/\text{s},{\text{\hspace{0.17em}\hspace{0.17em}u}}_{\text{2}}=-5\text{m}/\text{s}$

${\text{m}}_1={\text{m}}_2=\text{m}$

According to principle of conservation of linear momentum

${\text{m}}_{\text{1}}{\text{u}}_{\text{1}}+{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}={\text{m}}_{\text{1}}{\text{υ}}_{\text{1}}+{\text{m}}_{\text{2}}{\text{υ}}_{\text{2}}$

$\text{m}\times 3-\text{m}\times 5={\text{mυ}}_{\text{1}}+{\text{mυ}}_{\text{2}}$

or     ${\text{υ}}_1+{\text{υ}}_2=-2$ ….. (i)

In an elastic collision,

$\text{e}=\frac{{\text{υ}}_2-{\text{υ}}_1}{{\text{u}}_1-{\text{u}}_2}$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_2-{\text{υ}}_1=\text{e}\left({\text{u}}_1-{\text{u}}_2\right)$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_2-{\text{υ}}_1=\left(1\right)\left(3+5\right)\left(\because \text{e}=1\right)$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_1-{\text{υ}}_2=-8$…. (ii)

Adding Equations (i) and (ii), we obtain

$2{\text{\hspace{0.17em}υ}}_1=-10$

$\Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}υ}}_1=-5\text{m}/\text{s}$

From Equation (i),

${\text{υ}}_2=-2-{\text{υ}}_1$

$=$$-2+5=3\text{\hspace{0.17em}m}/\text{s}$

Thus,   ${\text{υ}}_1=-5\text{m}/\text{s},{\upsilon }_2=+3\text{m}/\text{s}$