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JEE Advanced · Physics · 7. COM & Collisions

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

  1. A \(250 \mathrm{~m} / \mathrm{s}\)
  2. B \(250 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
  3. C \(400 \mathrm{~m} / \mathrm{s}\)
  4. D \(500 \mathrm{~m} / \mathrm{s}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(500 \mathrm{~m} / \mathrm{s}\)

Step-by-step Solution

Detailed explanation

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.
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