A) \[3v\]
B) \[\frac{1}{3}v\]
C) \[2v\]
D) \[\frac{1}{2}v\]
Correct Answer: B
Solution :
Key Idea: In the given case inelastic collision occurs. When particles stick after collision, then momentum of the system remains conserved. Given, \[{{m}_{1}}=m,\,\,{{v}_{1}}=v,\,{{v}_{2}}=0\] (stationary), \[{{m}_{2}}=2\,m\] Momentum before collision = momentum after collision \[{{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}=({{m}_{1}}+{{m}_{2}})v\] \[m\times v+2m\times 0=(m+2\,m)\,v\] \[\Rightarrow \] \[v=\frac{v}{3}\] Note: Velocity of combined mass is reduced.
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