Question

A ball of mass 0.2 kg is situated on the top of a pillar of height 5 m. A bullet of mass 0.01 kg moving horizontally with velocity of v m s^-1 strikes the ball at its centre, the ball strikes the ground at a distance of 20 m and bullet at 100 m. The initial velocity v of the bullet is:

(a) 250 ms^-1

(b) \(250\sqrt{2}\ ms^{-1}\)

(c) 400 ms^-1

(d) 500 ms^-1

Answer 1

Correct option is : (d) 500 ms^-1

Time taken by the bullet and ball in reaching the ground

⇒ v₂ = 100 ms^-1

Now according to the principle of conservation of linear momentum

m x v = mv₁ + mv₂

or 0.01 x v = 0.2 x 0.2 + 0.01 x 100 = 4 + 1 = 5

\(\therefore\ v = \frac{5}{0.01 } = 500\ ms^{-1}\) 

∴ Option (d) is correct.