Question

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

Answer 1

Answer:

Velocity of the third particle is 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.

Answer 2

Answer:

Therefore the answer for the given question is 300 m/s