Consider a body at rest in the L-Frame, which explodes into fragments of masses `m_(1)` and `m_(2)`. Calculate energies of the fragments of the body.

Explanation

Since the body is initially at rest, its total momentum is zero. As a result of the explosion, the two fragments will separate in opposite directions with momentum `P_(1)` and `P_(2)` such that `P'_(1) +P'_(2) =0` , magnitude of momentum will be same, i.e., `P'_(1)=P'_(2)`.
On substituing,
`E'_(K) =(P'_(1)^(2))/(2 m_(1)) + (P'_(2)^(2))/(2) m_(2)=(1)/(m_(1)) + (1)/(m_(2))P'_1`
`E_(K)=0`
In Eq. (iv) of kinematics of reaction, we get
`(1)/(2)((1)/(m_(1)) +(1)/(m_(2)))P'_(1)^(2) =Q`
On rearranging this equation, we get
`P'_(1)=((2m_(1)m_(2))/(m_(1)+m_(2))Q)^(1//2)=(2mu_(12)Q)^(1//2)`
Where `mu_(12)` is the reduced mass of the system. The kinetic energies of the fragmnets are `(p'_(1)^(2))/(2m_(1))` and `(p'_(1)^(2))/(2m_(2))`with `p'_(1)=p'_(2)`. Hence,
`E'_(k,1)=(p'_(1)^(2))/(2m_(1))=(m_(1)Q)/(m_(1)+m_(2))` and `E'_(k,2)=(p'_(2)^(2))/(2m_(2)) =(m_(2)Q)/(m_(1)+m_(2))`.