Q44 / 60 · JEE Advanced 2011 (Paper 2) · Physics

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 Figure

  • A
    $250 \mathrm{~m} / \mathrm{s}$
  • B
    $250 \sqrt{2} \mathrm{~m} / \mathrm{s}$
  • C
    $400 \mathrm{~m} / \mathrm{s}$
  • Correct answer
    D
    $500 \mathrm{~m} / \mathrm{s}$

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.

Hence, the correct answer is option: 'D'.

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