question_answer

A particle of mass \[4m\] explodes into three pieces of masses\[m\], \[m\] and \[2m.\] the equal masses move along X-axis and V-axis with velocities \[4\text{ }m{{s}^{-1}}\] and \[6m{{s}^{-1}}\] respectively. The magnitude of the velocity of the heavier mass is

A)  \[\sqrt{17}m{{s}^{-1}}\]

B)  \[2\sqrt{13}\,m{{s}^{-1}}\]

C)  \[\sqrt{13}\,m{{s}^{-1}}\]

D)  \[\frac{\sqrt{13}}{2}\,m{{s}^{-1}}\]  

Correct Answer: C

Solution :

Let third mass particle \[(2m)\] moves making angle O with X-axis. The horizontal component of velocity of \[2m\] mass particle \[u=\cos \,\theta \] and vertical component \[u=sin\,\theta \] 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             \[0=m\times 4+2m(\mu \,\,cos\,\theta )\] or            \[-4=2u\,\cos \theta \] or          \[-2=u\,\,\cos \theta \]           ...(i) Again, applying law of conservation of linear momentum in Y-direction. \[0=m\times 6+2m(u\,\sin \theta )\] \[\Rightarrow \] \[-\frac{6}{2}=u\,\sin \,\theta \] or \[-3=u\,\sin \theta \]             ?..(ii) Squaring Eqs. (i) and(ii) and adding, we get \[(4)+(9)={{u}^{2}}\,{{\cos }^{2}}\theta +{{u}^{2}}\,{{\sin }^{2}}\theta \] \[={{u}^{2}}({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )\] or \[13={{u}^{2}}\] or \[u=\sqrt{13}m{{s}^{-1}}\]