Question

At a certain height a body at rest explodes into two equal fragments with one fragment receving a horizontal velocity of 10 ms–1 . The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is (g = 10ms–11)

Answer 1

The time interval after which both the velocity vector will be perpendicular to each other will be 1 sec

Explanation:

From the law of conservation of momentum

m x 0 = mv/2 + mu/2

=> v = -u

where u and v are velocity of both the fragments.

Given u = 10 m/s

=> |v| = |u| = 10 m/s

If Ф is the angle between both the velocity component, then

t = (√uv)/g x cotФ/2

given,

u = v = 10 m/s

g = 10 m/s²

Ф = 90 => cotФ/2 = cot 45 = 1

hence

t = (√10 x 10)/10 x 1

=> t = 10/10 = 1 sec

Hence the time interval after which both the velocity vector will be perpendicular to each other will be 1 sec