A) \[1/4m{{V}^{2}}\]
B) \[3/2m{{V}^{2}}\]
C) \[1/2m{{V}^{2}}\]
D) \[3/4m{{V}^{2}}\]
Correct Answer: B
Solution :
From the law of conservation of momentum From question, initial momentum = 0, so that final momentum should be zero. Let V be the velocity of mass 2m then \[\therefore \] \[\sqrt{{{(mV)}^{2}}+{{(mV)}^{2}}}=2m{{V}_{1}}\] or \[{{V}_{1}}=\frac{V}{\sqrt{2}}\] .....(i) \[\therefore \] K.E. Total = K.E (First piece) + K.E. (Second piece) + K.E (Third piece) \[K\,E=\frac{1}{2}m{{V}^{2}}+\frac{1}{2}(2m){{V}_{1}}^{2}\] \[=2\left( \frac{1}{2}m{{V}^{2}} \right)+m\left( \frac{{{V}^{2}}}{2} \right)\] \[\Rightarrow \] \[K.{{E}_{after\,\,\exp losion}}=\frac{3}{2}m{{V}^{2}}\] Also, \[K.{{E}_{Before\,\,\exp losion}}=0\] Hence, Piece in K.E \[=\frac{3}{2}m{{V}^{2}}\]You need to login to perform this action.
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