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 $m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$ or $\;\;0=m\times4+2m(u\cos\theta)$ or $-4=2u\cos\theta$or $-2=u\cos\theta$...(i) Again, applying law of conservation of linear momentum in Y-direction $0=m\times6+2m(u\sin\theta)$ ⇒ $\frac{-6}{2}=u\sin\theta$ or $-3 = u \sin \theta$ ...(ii) Squaring Eqs. (i) and (ii) and adding, $(4)+(9)=u^{2}\cos^{2}\theta+u^{2}\sin^{2}\theta$ $=u^{2}(\cos^{2}\theta+\sin^{2}\theta)$ or $\;\;13=u^{2}$ $\therefore\;\;u=\sqrt{13}\,\mathrm{m/s}$