Question

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

• A. they will move in opposite directions
• B. $A$ continues to move in the original direction while $B$ remains at rest
• C. they will move in mutually perpendicular directions
• D. $A$ comes to rest and $B$ starts moving in the direction of the original motion of $A$

Answer 1

Answer: (C)

as collision is elastic
$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}{v}_2^2$
$u^2=v_1^2+v_2^2$ ......... $(1)$
Momentum is conserved
$m\overrightarrow{u}=m{\overrightarrow{v}}_1+m{\overrightarrow{v}}_2$
$\overrightarrow{u}={\overrightarrow{v}}_1+{\overrightarrow{v}}_2$
$u^2=v_1^2+v_2^2+2v_1.v_{2}cos\theta$ ....... $(2)$
Solving $(1)$ and $(2)$
$2v_1.v_{2}cos\theta =0$
$cos\theta =0$
$\theta =90^o$
i.e. $v_1$ &$v_2$ are mutually perpendicular.