Let third mass particle (2m) moves making angle θ with X-axis. The horizontal component of velocity of 2m mass particle =u‌cos‌θ and vertical component =usin‌θ 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, we get (4)+(9)‌=u2cos2θ+u2sin‌2θ ‌=u2(cos2θ+sin‌2θ) or 13‌=u2 or u‌=√13ms−1