Question

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. $\frac{5}{2}m{\upsilon }^2$

Answer 1

Answer: (C)

Total mass = 4m Two pieces, each has mass $=m$ 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. $=\frac{1}{2}m{\upsilon }^2+\frac{1}{2}m{\upsilon }^2+\frac{1}{2}2m{\upsilon }_1^2$ $m{\upsilon }^2+m{\left(\frac{\upsilon }{\sqrt{2}}\right)}^2=m{\upsilon }^2+\frac{m{\upsilon }^2}{2}=\frac{3}{2}m{\upsilon }^2$