Centre of Mass and Collision Part 1

We can represent the given situation in figure asIt means linear momentum is conserved along X-axis.
According to the law of conservation of linear momentum,
pi=pf
m_{A} u_{A}+m_{B} u_{B}=m_{A} v_{A}+(m_{B} v_{B} \cos θ)
=mAvA+mB1vB1‌cos‌90∘+mB2vB2‌cos‌θ
⇒10×10√3+20×0=10×0+10×10×0+10×20‌cos‌θ
⇒‌‌10×10√3=200‌cos‌θ
where, cos‌θ being the horizontal component i.e., along X-axis
⇒‌‌cos‌θ=√3/2
⇒‌‌θ=30∘‌‌[∵cos‌30∘=√3/2]