Let third mass particle (2m) moves making angle θ with X-axis. The horizontal component of velocity of 2m mass particle = u cos θ. And vertical component = u sin θ
From conservation of linear momentum in X-direction m1u1+m2u2=m1v1+m2v2 or ‌‌0=m×4+2m(u‌cos‌θ) or −4=2u‌cos‌θ or −2=u‌cos‌θ...(i) Again, applying law of conservation of linear momentum in Y-direction 0=m×6+2m(u‌sin‌θ) ⇒ −
6
2
=u‌sin‌θ or −3=u‌sin‌θ...(ii) Squaring Eqs. (i) and (ii) and adding, (4)+(9)=u2cos2θ+u2sin2θ =u2(cos2θ+sin2θ) or ‌‌13=u2 ∴‌‌u=√13ms−1