An object of mass 5 kg is projecte with a velocity of `20ms^(-1)` at an angle of `60^(@)` to the horizontal. At the highest point of its path , the projectile explodes and breaks up into two fragments of masses 1kg and 4 kg. The fragments separate horizontally after the explosion, which releases internal energy such that `K.E.` of the system at the highest point is doubled. Calculate the separation betweent the two fragments when they reach the ground.

Answer

`44.25 m`

Explanation

`COLM : 1 xx v_(1) + 4 xx v_(2) = 5 xx 20 cos 60^(@) = 50` …(i)

`COME rArr (1)/(2) xx 1 xx v_(1)^(2) + (1)/(2) xx 4 xx v_(2)^(2)`
`= [(1)/(2) 5 xx (10)^(2)] xx 2` …..(ii)
`rArr v_(1) = -10` or `m//s` & `v_(2) = 15` or `5 m//s`
`rArr Deltav = 25 m//s`
Time to fall down to ground `= sqrt((2h)/(g))= (sqrt3)/(0.98)sec`
`:.` Separation between particles `= Deltav.t = 44.2 m`