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.

Explanation

At the highest point of trajectory applying conservation of linear momentum along horizontal,
`5xx20cos 60^(@)=1xxv_(1)+4v_(2)`
i.e., `v_(1)+4v_(2)=50` ......(i)
And by conservation of energy at highest point, according to given condition,
`((1)/(2)xx5xx10^(2))xx2=(1)/(2)xx1v_(1)^(2)+(1)/(2)4v_(2)^(2)`,
i.e., `v_(1)^(2)+4v_(2)^(2)=1000` ........(ii)
Substituting `v_(1)` from EQn. (i) in (ii)
`(50-4v_(2))^(2)+4v_(2)^(2)=1000, i.e., v_(2)^(2)-20v_(2)+75=0`
whch on solving gives
`v_(2)=5m//s` or 15m/s
So, from Eqn. (i),
for `vF_(2)=5m//s v_(1)=30m//s`
and for `v_(2)=15m//s v_(1)=10m//s`
So, if both the particles move in same direction,
`V_("rel")=30-5=25m//s`
and if both move in opposite direction,
`v_("rel")=15-(-10)=25m//s`
i.e., fragments after explosion separate from each other horizontally 25m per sec.
Now as time taken by fragments to reach ground
`t=(T)/(2)=(u sin theta)/(g)=(20xxsqrt(3))/(10xx2)=sqrt(3)s`
So, the separation between two fragments when they reach the ground.
`d=v_("rel")xxt=25xxsqrt(3)=43.3m`