A bomb of mass `5 m` initially at rest explodes and breaks into three pieces of masses in the ratio `1 : 1 : 3`. The two pieces of equal mass fly off perpendicular to each other with a speed of `v_(0)`. What is the velocity of the heavier piece? Also , calculate the energy released in explosion.

Explanation

Total mass `5 m` is split in a ratio `1 : 1: 3` , hence masses of three pieces are `m_(1) = m , m_(2) = m` and `m_(3) = 3m`.
Let the velocity of heavier piece is `v` at an angle `theta`. Here , we have to conserve momentum in the `x`- direction and `y`- direction separately.
By momentum conservation
`x`-direction : `5 m xx 0 = mv_(0) + 3mv cos theta`
`v cos theta = -(v_(0))/(3)`
`y`-direction : `5 m xx 0 = mv_(0) + 3 mv sin theta`
`v sin theta = -(v_(0))/(3)`
The `-ve` sign indicates that the velocity is in opposite direction
`v = (v_(0))/(3) sqrt(2) , theta = 45^(@)`
The direction of the third piece is at an angle `135^(@)` from either of smaller pieces.
`Initial K.E. = 0`
Final `K.E. = (1)/(2) mv_(0)^(2) + (1)/(2) mv_(0)^(2) + (1)/(2) xx 3m ((sqrt(2) v_(0))/(3))^(2)`
` = mv_(0)^(2) + (1)/(3) mv_(0)^(2)`
`= (4)/(3) mv_(0)^(2)`
`K.E.` released in explosion `= (4)/(3) mv_(0)^(2)`
OR
This problem can be solved without conserving momentum conservation in `x`- and `y` directin separately.
Since the bomb is at rest , hence initial momentum is zero and hence the final momentum of system of pieces will be zero , due to the momentum conservation.
The resultant momentum of smaller pieces will be equal and opposite to the momentum of heavier piece.
Let `m v_(0) = p`
Resultant momentum `P = sqrt(2) p`
`theta = 45^(@)`
Hence the momentum of heavier piece is equal and opposite to `P`.
`P = sqrt(2) p = sqrt(2) mv_(0)`
`P = 3 mv`
`3 mv = sqrt(2) mv_(0)`
`v = (sqrt(2) v_(0))/(3)`
v : Velocity of heavier piece.