An Object Of Mass 5 Kg Is Projected With A Velocity Of 20 M/s 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 1 Kg And 4 Kg. The Fragments Separate Horizontally After The Explosion. The Explosion Releases Internal Energy Such That The Kinetic Energy Of The System At The Highest Point Is Doubled. Calculate The Separation Between The Two Fragments When They Reach The Ground. (g = 10 M/s2)

At the highest point of trajectory applying conservation of linear momentum along horizontal,

And by conservation of energy at highest point, according to given condition,

i.e.,                 

Substituting v₁ From Eqn. (1) in (2),

Which on solving gives

                 v₂ = 5 m/s     or    15 m/s

So from Eqn. (1),

For                v₂ = 5 m/s           v₁ = 30 m/s

And for         v₂ = 15 m/s         v₁ = – 10 m/s

So if both the particles move in same direction,

           v_rel = 30 – 5 = 25 m/s

and if both move in opposite direction,

             v_rel = 15 – (–10) = 25 m/s

i.e., fragments after explosion separate from each other horizontally 25 m per sec.

Now as time taken by fragments to reach ground

So the separation between two fragments when they reach the ground