Question

A bomb is projected with $20m/s$ at angle ${60}^{\circ }$ with horizontal. At the highest point, it explodes into three particles of equal masses. One goes vertically upward with velocity $100m/sec$, second particle goes vertically downward with the same velocity as the first. Then what is the velocity of the third one-

• A. $120m/sec$ with ${60}^{\circ }$ angle
• B. $200m/sec$ with ${30}^{\circ }$ angle
• C. $50m/sec$, in horizontal direction
• D. $30m/sec$, in horizontal direction

Answer 1

The correct option is D $30m/sec$, in horizontal direction

The horizontal velocity of the bomb is $v_h=20cos\left(60\right)=10m/s$
Horizontal momentum is $M=m\times 10$
The horizontal momentum must be conserved (at the highest point vertical velocity is 0, and that momentum is conserved by the two other parts)
$m/3\times v^{\prime }=m\times 10⟹v^{\prime }=30m/s$