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

`25sqrt(3)`

Explanation

The height `H = (u^2 sin^2 theta)/(2g) = (20 xx 20)/(2 xx 10) xx 3/4 = 15m`
At the highest point the projectile explodes and breaks into two particles which separate horizontally. As there is no external force acting in horizontal direction, hence linear momentum in horizontal direction should be conserved just before and just after explosion.
Applying conservation of linear momentum
`5 xx 20 cos 60^@ = 4 v_1 - 1 v_2`
`implies 4v_1 - 1v_2 = 50" ".....(i)`
The kinetic energy the highest point initially
`K_("initial") = 1/2 xx 5 xx 10 xx 10 = 250 J`
`:.` The kinetic energy after collision
`K_("final") = 2 xx 250 = 500 J`
`implies 1/2 xx 4 xx v_1^2 + 1/2 xx 1 xx v_2^2 = 500`
`implies 4 v_1^2 + v_2^2 = 1000 " "....(ii)`
After solving (i) and (ii) we get

`v_1 = 15 m//s "or" 5 m//s and v_2 = + 10m//s or –30 m//s`
Taking the pair `v_1 = 15 m//s and v_2 = +10 m//s`.
The relative velocity of separation of fragments in horizontal direction, `(v_("ret))_x = 25 m//s`. Let the time taken by each fragment to ground be t
Using `H = 1/2 gt^2 " " , " " 15= 1/2 xx 10 xx t^2 implies t = sqrt(3) "sec"`
Here the separation between the fragments when they reach the ground `x = (v_("ret"))_xt`
`implies x = 25sqrt(3)m`.