A ball of mass $200 \mathrm{~g}$ rests on a vertical post of height $20 \mathrm{~m}$. A bullet of mass $10 \mathrm{~g}$, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance $30 \mathrm{~m}$ and the bullet at a distance of $120 \mathrm{~m}$ from the foot of the post. The value of initial velocity of the bullet will be (if $g=10 \mathrm{~m} / \mathrm{s}^{2}$) :
Solution
<p>$\because$ Time of flight of each ball and bullet</p>
<p>$= \sqrt {{{2H} \over g}} = \sqrt {{{2 \times 20} \over {10}}} = 2$ s</p>
<p>$\Rightarrow$ By applying linear momentum conservation</p>
<p>$$100u + 200(0) = 200\left( {{{30} \over 2}} \right) + 10\left( {{{120} \over 2}} \right)$$</p>
<p>$u = 360$ m/s</p>
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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