JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 6)
A machine gun of mass $10 \mathrm{~kg}$ fires $20 \mathrm{~g}$ bullets at the rate of 180 bullets per minute with a speed of $100 \mathrm{~m} \mathrm{~s}^{-1}$ each. The recoil velocity of the gun is
$ 0.02 \mathrm{~m} / \mathrm{s}$
$1.5 \mathrm{~m} / \mathrm{s}$
$2.5 \mathrm{~m} / \mathrm{s}$
$0.6 \mathrm{~m} / \mathrm{s}$
Explanation
Momentum of bullets per unit time
$$ = {{180 \times {{20} \over {1000}} \times 100} \over {60}}$$ kg m/s$$^2$$
= 6 N
$$\Rightarrow$$ Force on gun = 6 N
We cannot calculate recoil velocity with the given data.
If we consider recoil velocity at $$t = 1$$ s, then
$${V_{\mathrm{recoil}}} = u + at$$
$$ = 0 + {6 \over {10}} \times 1$$ = 0.6 m/s
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