VITEEE 2009 Paper

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