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)


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,




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)