question_answer

A particle of mass 4 m which is at rest and explodes into three fragments. Two of the fragments each of mass \[m\] are found to move with a speed \[\upsilon \] in mutually perpendicular directions. The total energy released in the explosion is

A) \[\text{2m}{{\upsilon }^{\text{2}}}\]                     

B)         \[\frac{1}{2}\text{ m}{{\upsilon }^{\text{2}}}\]

C) \[\text{m}{{\upsilon }^{\text{2}}}\]                        

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

Correct Answer: D

Solution :

 Referring figure,                 \[mv\sqrt{2}=2m\times {{v}_{3}}\] \[\Rightarrow \]                     \[{{v}_{3}}=\frac{v}{\sqrt{2}}\] Total energy released in the explosion                 \[=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}m{{v}^{2}}+\frac{1}{2}\times 2m{{\left( \frac{v}{\sqrt{2}} \right)}^{2}}\]                 \[=\frac{\mathbf{3}}{\mathbf{2}}\mathbf{m}{{\mathbf{v}}^{\mathbf{2}}}\]