Question

A particle of mass 4 m which is at rest explodes into three fragments. Two of the fragments each of mass m are found to move with a speed v each in mutually perpendicular directions. The total energy released in the process of explosion is ............
A. `(3)/(2) mv^(2)`
B. `2 mv^(2)`
C. `4 mv^(2)`
D. `mv^(2)`

Answer 1

Correct Answer - A
Conserving momentum , we get
`0 = sqrt((mv)^(2) + (mv)^(2)) - 2mV = mv sqrt(2) - 2mV`
`V = (v)/(sqrt(2))`
`K.E.` generated due to explosion
`K = (1)/(2) mv^(2) + (1)/(2) mv^(2) + (1)/(2) 2m V^(2)`
`= mv^(2) + m((v)/(sqrt(2)))^(2) = (3)/(2) mv^(2)`