Explanation

(b) Let particle with mass $m$, move with velocity u,and $v_{1}$and $v_{2}$be velocity after collision. Since , elastic collision is one in which the momentum is conserved , we have
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$\therefore mu = mv_{1}\cos{\theta_{1} + mv_{2}\cos\theta_{2}}$ …..(i)
In perpendicular direction
$0 = mv_{1}\sin{\theta_{2} - mv_{2}\sin\theta_{2}\ \ \ \ \ \ \ \ }$ …..(ii)
Also elastic collision occurs only if there is no conversion of kinetic energy into other from, Hence
$\frac{1}{2}mu^{2} = \frac{1}{2}mv_{1}^{2} + \frac{1}{2}mv_{2}^{2}$
${\ u}_{2} = v_{1}^{2} + v_{2}^{2}$ ….(iii)
Squaring Esq.(i)and (ii)and adding we get
$m^{2}u^{2} = m^{2}\left( v_{1}\cos{\theta_{1} + v_{2}\cos\theta_{2}} \right)^{2}$
$+ m^{2}\left( v_{1}\sin{\theta_{1} - v_{2}\sin\theta_{2}} \right)^{2}$
$u^{2} = v_{1}^{2} + v_{2}^{2} + 2v_{1}v_{2}\cos{\theta_{1}\cos{\theta_{2} - 2v_{1}v_{2}\sin{\theta_{1}\sin\theta_{2}}}}$
$u^{2} = v_{1}^{2} + v_{2}^{2} + 2v_{1}v_{2}\cos{(\theta_{1} + \theta_{2})}$
Using Eq.(iii),we get
$\ 2v_{1}v_{2}\cos{(\theta_{1} + \theta_{2})} = 0$
$since\ \ \ \ \ \ \ \ v_{1}v_{2} \neq 0$
Hence $\cos{\left( \theta_{1} + \theta_{2} \right) = 0}$
Or $\theta_{1} + \theta_{2} = 90{^\circ}$
When two identical particles collide elastically and obliquely,
One being at rest, then they fly off in mutually perpendicular directions.