A Bomb is projected with 200 m/s at an angle 60 degree with horizontal. At hte highest point, it explodes into three particles of equal masses. One goes vertically upward with velocity 100 m/sec, second particle goes vertically downward with the same velocity as the first. Then what is the velocity of third one
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Answer:
Option (4) 300 m/s in the horizontal direction
Explanation:
At the highest point, the velocity will be only the horizontal component, its vertical component will be zero
The horizontal component of the velocity
m/s
Let after exploding, the masses of the three parts be m
Then total mass of the bomb = 3m
The velocity of the part that goes upwards with 100 m/s
The velocity of the part that goes downwards with 100 m/s
Let the velocity of the third part be
Then from the conservation of linear momentum
Thus the velocity of the third particle is 300 m/s in the horizontal direction.
Hope this helps.
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