question_answer

A body of mass 4m at rest explodes into three pieces. Two of the pieces each of mass m move with a speed v each in mutually perpendicular directions. The total kinetic energy released is:

A) \[\frac{1}{2}m{{v}^{2}}\]             

B)                         \[m{{\upsilon }^{2}}\]                  

C)  \[\frac{3}{2}m{{\upsilon }^{2}}\]                             

D)        \[\frac{5}{2}m{{\upsilon }^{2}}\]

Correct Answer: C

Solution :

Total mass = 4m Two pieces, each has mass \[=\text{ }m\] Third piece has a mass \[=4m-m-n\] By law of conservation of momentum \[2m{{\upsilon }_{1}}=m\upsilon \cos {{45}^{o}}+m\upsilon \cos {{45}^{o}}\] \[=\frac{m\upsilon }{\sqrt{2}}+\frac{m\upsilon }{\sqrt{2}}=\frac{2m\upsilon }{\sqrt{2}}\]                 \[\therefore \]  \[{{\upsilon }_{1}}=\frac{\upsilon }{\sqrt{2}}\] Therefore, K.E. \[=\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}m{{\upsilon }^{2}}+\frac{1}{2}2m\upsilon _{1}^{2}\] \[m{{\upsilon }^{2}}+m{{\left( \frac{\upsilon }{\sqrt{2}} \right)}^{2}}=m{{\upsilon }^{2}}+\frac{m{{\upsilon }^{2}}}{2}=\frac{3}{2}m{{\upsilon }^{2}}\]