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 \[u=\sqrt{13}m{{s}^{-1}}\]You need to login to perform this action.
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