A ball of mass $0.2 \mathrm{~kg}$ rests on a vertical post of height $5 \mathrm{~m}$. A bullet of mass $0.01…
A ball of mass $0.2 \mathrm{~kg}$ rests on a vertical post of height $5 \mathrm{~m}$. A bullet of mass $0.01 \mathrm{~kg}$, travelling with a velocity $v \mathrm{~m} / \mathrm{s}$ 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 $20 \mathrm{~m}$ and the bullet at a distance of $100 \mathrm{~m}$ from the foot of the post. The initial velocity $v$ of the bullet is
$250 \mathrm{~m} / \mathrm{s}$
$250 \sqrt{2} \mathrm{~m} / \mathrm{s}$
$400 \mathrm{~m} / \mathrm{s}$
$500 \mathrm{~m} / \mathrm{s}$
Solution
Time taken by the bullet and ball to strike the ground is
$
t=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \times 5}{10}}=1 \mathrm{~s}
$
Let $v_1$ and $v_2$ are the velocities of ball and bullet after collision.
Then applying
We have, $20=v_1 \times 1$
or $v_1=20 \mathrm{~ms}^{-1}$
$
100=v_2 \times 1 \text { or } v_2=100 \mathrm{~m} / \mathrm{s}^{-1}
$
Now, from conservation of linear momentum before and after collision we have,
$0.01 v=(0.2 \times 20)+(0.01 \times 100)$
On solving, we get
$
v=500 \mathrm{~ms}^{-1}
$
$\therefore$ Correct answer is (d).
Analysis of Question
Question is moderately lengthy from calculation point of view, otherwise it is simple
