## Question

### Solution

Correct option is

(0.86, 31.25 m)

(a) At the highest point of trajectory horizontal force is zero (as mg cos 90° = 0); so by conservation of linear momentum in horizontal direction, i.e.,             V = (50/4) cos θ                                       …(1) Now equation of circular motion of the bob at B will be But at B,     T = 0   and     v ≠ 0                [as angle > 90°]

So that the above equation reduces to Now by conservation of mechanical energy between A and B (after collision)  Substituting the values of V and v from Eqns. (1) and (2) in (3), we get   (b) As initially the bob is at the highest point of trajectory,  #### SIMILAR QUESTIONS

Q1

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)