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 ball hits the ground at a distance of 100 m from the foot of the post. The initial velocity $v$ of the bullet is [image:5_image5.jpeg]

Question

Dr. Abubacker · Accepted Answer

(d) Time taken by the bullet and ball to strike the ground is
$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 5}{10}} = 1\ s$
Let $v_{1}$ and $v_{2}$ are the velocities of ball and bullet after collision. Then applying
$x = vt$
$We\ have,\ 20 = v_{1} 	imes 1$
$or\ \ v_{1} = 20\ m/s$
$100 = v_{2} 	imes 1\ or\ v_{2} = 100\ m/s$
Now, from conservation of linear momentum before and after collision we have,
$0.01v = (0.2 	imes 20) + (0.01 	imes 100)$
On solving, we get
$v = 500\ m/s$

Explanation

Reviewed by Dr. Abubacker

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