question_answer

A mass 'm' moves with a velocity 'v' and collides inelastically with another identical mass. After collision the Ist mass moves with velocity \[\frac{v}{\sqrt{3}}\] in a direction perpendicular to the initial direction of motion. Find the speed of the 2nd mass after collision                                      [AIEEE 2005]

A)          \[\frac{2}{\sqrt{3}}v\]

B)                        \[\frac{v}{\sqrt{3}}\]    

C)          v   

D)          \[\sqrt{3}\,v\]

Correct Answer: A

Solution :

                                        Let mass A moves with velocity v and collides in elastically with mass B, which is at rest. According to problem mass A moves in a perpendicular direction and let the mass B moves at angle q with the horizontal with velocity v. Initial horizontal momentum of system (before collision)   = mv                                  ....(i) Final horizontal momentum of system (after collision)    \[~~=mVcos\theta \]                      ....(ii) From the conservation of horizontal linear momentum      mv = mV cosq Þ v = V cosq                   ...(iii) Initial vertical momentum of system (before collision) is zero. Final vertical momentum of system \[\frac{mv}{\sqrt{3}}-mV\sin \theta \] From the conservation of vertical linear momentum \[\frac{mv}{\sqrt{3}}-mV\sin \theta =0\]Þ\[\frac{v}{\sqrt{3}}=V\sin \theta \]                 ...(iv) By solving (iii) and (iv) \[{{v}^{2}}+\frac{{{v}^{2}}}{3}={{V}^{2}}({{\sin }^{2}}\theta +{{\cos }^{2}}\theta )\]   Þ \[\frac{4{{v}^{2}}}{3}={{V}^{2}}\]  Þ \[V=\frac{2}{\sqrt{3}}v\].