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

Here, `m =5 kg, upsilon =20ms^(-1), theta =60^(@)`
Horizontal compt. Of velocity, `upsilon_(x)=upsiloncos 60^(@)=20xx(1)/(2)=10m//s`
Vertical compt. Of velocity, `upsilon_(y)=upsiloncos 60^(@)=20xx(sqrt(3))/(2)=10sqrt(3) m//s`
Time taken to reach the highest the highest point `=` time taken to reach the ground from the highest point
`t=(upsilonsintheta)/(g)=(upsilon_(y))/(g)=(10sqrt(3))/(9.8)=1.77s`
At the highest point, `m` splits up into two parts of masses `m_(1)=1 kg ` and `m_(2)=4kg`. If their velocities are `upsilon_(1)` and `upsilon_(2)` respectively, then applying the principle of conservation of linear momentum, we get
`m_(1)upsilon_(2)+m_(2)upsilon_(2)=m upsilon cos theta`
`upsilon_(1)+4upsilon_(2)=5xx10=50` ...(i)
Initial `K.E. =(1)/(2)m(upsilon cos theta)^(2)=(1)/(2)xx5(10)^(2)=250J`, Final `K.E. =2` (initial K.E.)`=2xx250=500J`
`:. (1)/(2)m_(1)upsilon_(1)^(2)+(1)/(2)m_(2)upsilon_(2)^(2)=500`
`(1)/(2)xx1v_(1)^(2)+(1)/(2)4upsilon_(2)^(2)=500` or `upsilon_(1)^(2)+4upsilon_(2)^(2)=1000` ...(ii)
On solving (i) and (ii), we get `upsilon_(1)=30m//s, upsilon_(2)=5m//s`.
`:.` Separation between the two fragments `=(upsilon_(1)-upsilon_(2))xxt=(30-5)xx1.77metre=44.25m`