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)


Correct option is


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,



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         v= – 10 m/s

So if both the particles move in same direction,

           vrel = 30 – 5 = 25 m/s

and if both move in opposite direction,

             vrel = 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




bullet of mass M is fired with a velocity 50 m/s at an angle θ with the horizontal. At the highest point of its trajectory, it collides head on with a bob of mass 3 M suspended by a massless string of length (10/3) m and gets embedded in the bob. After the collision the string moves through an angle 120°. Find: (a) The angle θ (b) the vertical and horizontal co-ordinates of the initial position of the bob with respect to the point of firing of the bullet. (g = 10 m/s2)