Thank you for reporting, we will resolve it shortly
Q.
A particle $A$ suffers an oblique elastic collision with a particle $B$ which is initially at rest. If their masses are same, then after the collision
NTA AbhyasNTA Abhyas 2020
Solution:
as collision is elastic
$\frac{1}{2}mu^{2}=\frac{1}{2}mv_{1}^{2}+\frac{1}{2}v_{2}^{2}$
$u^{2}=v_{1}^{2}+v_{2}^{2}$ ......... $\left(\right.1\left.\right)$
Momentum is conserved
$m\overset{ \rightarrow }{u}=m\overset{ \rightarrow }{v}_{1}+m\overset{ \rightarrow }{v}_{2}$
$\overset{ \rightarrow }{u}=\overset{ \rightarrow }{v}_{1}+\overset{ \rightarrow }{v}_{2}$
$u^{2}=v_{1}^{2}+v_{2}^{2}+2v_{1}.v_{2}cos \theta $ ....... $\left(\right.2\left.\right)$
Solving $\left(\right.1\left.\right)$ and $\left(\right.2\left.\right)$
$2v_{1}.v_{2}cos \theta =0$
$cos \theta =0$
$\theta =90^{o}$
i.e. $v_{1}$ &$v_{2}$ are mutually perpendicular.
