Physics, asked by farhin9865, 11 months ago

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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Answered by sonuvuce
2

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

v_x=200\cos60^\circ\hat i

\implies v_x=200\times\frac{1}{2}\hat i

v_x=100\hat i 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 =100\hat j

The velocity of the part that goes downwards with 100 m/s =-100\hat j

Let the velocity of the third part be \vec v

Then from the conservation of linear momentum

3mv_x=m\times 100\hat j+m\times (-100\hat j)+m\vec v

3m\times 100\hat i=m\vec v

3m\times 100\hat i=m\vec v

\implies \vec v=3\times 100\hat i

\implies \vec v=300\hat i

Thus the velocity of the third particle is 300 m/s in the horizontal direction.

Hope this helps.

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