Given m1=m2 We will apply the principle of conservation of momentum in the mutually perpendicular dirn Along x-axis m1u1=m1v1cosθ+m2v2cosϕ or u1=v1cosθ+v2cosϕ....(1) Along y-axis 0=m1v1sinθ−m2v2sinϕ or 0=v1sinθ−v2cosϕ......(2) Again for elastic collision, kinetic energy is conserved ⇒
1
2
mu12=
1
2
mv12+
1
2
mv22 or u12=v12+v22 ....(iii) Squaring and adding (i) & (ii), we get
u12=v12(cos2θ+sin2θ)+v22(cos2ϕ+sin2ϕ)+2v1v2cosθcosϕ−2v1v2sinθsinϕ or u12=v12+v22+2v1v2cos(θ+ϕ)......(iv)
Using (iii) & (iv), we get
cos(θ+ϕ)=0=cos
π
2
⇒θ+ϕ=
π
2
Note : This is a standard case of oblique collision.