JEE MAIN - Physics (2023 - 30th January Morning Shift - No. 17)
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 $$g=10 \mathrm{~m} / \mathrm{s}^{2}$$) :
120 m/s
360 m/s
400 m/s
60 m/s
Explanation
$$\because$$ Time of flight of each ball and bullet
$$ = \sqrt {{{2H} \over g}} = \sqrt {{{2 \times 20} \over {10}}} = 2$$ s
$$\Rightarrow$$ By applying linear momentum conservation
$$100u + 200(0) = 200\left( {{{30} \over 2}} \right) + 10\left( {{{120} \over 2}} \right)$$
$$u = 360$$ m/s
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