A particle A suffers an oblique elastic collision with a particle B that is at rest initially. If their masses are the same, then, after the collision

c)      

a)By law of conservation of momentum

i) along x axis

$m{u}_1+0=m{v}_1+\cos \varphi +m{v}_2\cos \theta$ ……………(i)

ii)along y axis

$0=m{v}_1\sin \varphi -m{v}_2\sin \varphi$ ……………………..(ii)

b)By law of conservation of energy

$\frac{1}{2}m{u}_1^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2\Rightarrow u_1^2=v_1^2+v_2^2\mathrm{..............}\left(iii\right)$

From eq (i) & (ii) we get

$v_1^2=v_1^2+v_2^2+2v_1{v}_2\cos \left(\varphi +\theta \right)$

Now use eq (iii) in eq (iv) we get

$\cos \left(\varphi +\theta \right)=0\to \theta +\varphi ={90}^0$

i.e, after collision they will move in mutually perpendicular directions.