Question

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 1

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`