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An object of mass 5 kg is projecte with ...

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.

Text Solution

Verified by Experts

The correct Answer is:
`44.44.25 m`
Let `v_(1)` and `v_(2)` be the velocities after explosion in the direction shwon in figure. From conservation of linear momentum, we have
`5(20 cos 60^(@)) = 4v_(1) - 1 xx v_(2)`
or `4v_(1)-v_(2) = 50` …(1)
Just before explosion
Just after explosion
Further it is given that, kinetic energy after explosion becomes two times. Therefore,
`(1)/(2)xx 4 xx v_(1)^(2)+(1)/(2)xx 1 xx v_(2)^(2) = 2[(1)/(2)xx 5 xx (20 cos 60^(@))^(2)]` or
`4v_(1)^(2) + v_(2)^(2) = 1000` ....(2)
Solving Eqs. (1) and (2) we have
`v_(1) = 15 m//s , v_(2) = -30 m//s`
In both the cases relative velocity of separation in horizontal direction is `25 m//s`.

`\x = 25 t =` distance between them when they strike the ground
Here, `t = (T)/(2) (T = "time of flight of projectile") =`
`(u sin theta)/(g) = (20 sin 60^(@))/(9.8) = 1.77 s`
`:. x = 25 xx 1.77 m = 44.25 m`
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