Question

A ball of mass 0.2 kg rests on a vertical post of height 5 m. 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 distanceof 100 m from the foot of the post The initial velocity v of the bullet is

• A. 250 m/s
• B. $250 \sqrt m / s$
• C. 400 m/s.
• D. 500 m/s.

Answer 1

Answer: (D)

500 m/s.

Answer 2

R = $u \sqrt{ \frac{ 2 h}{ g}} \Rightarrow 20 = v_1 \sqrt{ \frac{ 2 \times 5 }{ 10 }} \, and \, 100 = v_2 \sqrt{ \frac{ 2 \times 5 }{ 10}}$
$\Rightarrow v_1 = 20 m/s, \, v_2 = 100 m/s.$
Applying momentum conservation just before and just after the collision
(0.01) (v) = (0.2) (20) + (0.01) (100)
v = 500 m/s