Q: A ball of mass 0.2 kg rests on a vertical post of height 5 cm. A bullet of mass 0.01 kg, travelling with a velocity v m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity v of the bullet is

(a)250 m/s

(b)250√2 m/s

(c)400 m/s

(d)500 m/s

Ans: (d)

Sol: Time taken by ball & bullet to hit the ground is

$\large t = \sqrt{\frac{2h}{g}}$

$\large t = \sqrt{\frac{2\times 5}{10}}$

t = 1 sec

v_{1} × t = 20 m/s

v_{2} × t = 100

v_{2} = 100 m/s

Applying momentum conservation before & after collision

0.01 × v = 0.2 × 20 + 0.01 × 100

v = 500 m/s