EAMCET Engineering 2009 Solved Paper

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