Discussion Forum

1)

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 distance of 100 m from the foot of the post. The initial velocity v of the bullet is

Answer:

Option D

Explanation:

Time taken by the bullet and ball to strike the ground is

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1s$

Let $v_{1}$and $v_{2}$ are the velocities of ball and bullet after collision.

Then applying

 x=vt 

 we have   $20=v_{1} \times 1$

 or $ v_{1}= 20ms^{-1}$

 $100=v_{2} \times 1$ or $v_{2}=100m/s^{-1}$

 Now, from conservation of linear momentum before and after collision

we have. 

$0.001 v= (0.2 \times 20) +(0.01 \times 100)$

 On solving, we get 

 $v= 500ms^{-1}$

 $\therefore$  Correct answer is (d).