## Question

### Solution

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,

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         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

#### SIMILAR QUESTIONS

Q1

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)