NEETPhysicsCenter of Mass, Momentum and Collision
A projectile of mass M is fired from the ground with an initial speed u at an angle of 60^ to the horizontal. At the highest point of its trajectory, it explodes into two fragments of mass M 3 and 2M 3 . If the lighter fragment comes to rest immediately after the explosion, the extra energy released during the explosion is
Options
- AM u^2 4
- B3 M u^2 16
- CM u^2 16
- DM u^2 8
Correct answer
C. M u^2 16
Step-by-step solution
At the highest point of the trajectory, the vertical component of velocity is zero. The horizontal component of velocity is: v_x = u 60^ = u 2 The kinetic energy of the projectile just before the explosion is: K_i = 1 2 M ( u 2 )^2 = M u^2 8 The momentum of the projectile just before the explosion is entirely horizontal: p_i = M ( u 2 ) = M u 2 Let v_f be the velocity of the heavier fragment (mass 2M 3 ) after the explosion. Since the lighter fragment (mass M 3 ) comes to rest, its momentum is zero. By conservation