A) \[m{{v}^{2}}\]
B) \[\frac{3}{2}m{{v}^{2}}\]
C) \[2m{{v}^{2}}\]
D) \[4m{{v}^{2}}\]
Correct Answer: D
Solution :
According to question, the third part of mass 2m will move as shown in the figure, because the total momentum of the system after explosion must remain zero. Let the velocity of third part is v'. From the conservation of momentum \[\sqrt{2}(mv)=(2m)\times v'\Rightarrow v'=\frac{v}{\sqrt{2}}\] \[\Rightarrow \] So total kinetic energy generated by the explosion \[=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}+\frac{1}{2}(2m)v{{'}^{2}}\] \[=m{{v}^{2}}+m\times {{\left( \frac{v}{\sqrt{2}} \right)}^{2}}\] \[=m{{v}^{2}}+\frac{m{{v}^{2}}}{2}=\frac{3}{2}m{{v}^{2}}\]You need to login to perform this action.
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