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4-1.Newton's Laws of Motion
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A soldier with a machine gun, falling from an alrplane gets detached from his parachute. He is able to resist the downward acceleration, if he shoots $40$ bullets a second at the speed of $500 \,m / s$. If the weight of a bullet is $49 \,g$, the weight of the man .......... $ \,kg$ with the gun? Ignore resistance due to air and assume the acceleration due to gravity, $g=9.8 \,ms ^{-2}$.
A$50$
B$75$
C$100$
D$125$
(KVPY-2010)
Solution
(c)
In given case, rate of momentum change of bullets is equal to weight of soldier.
If $M=$ mass of soldier and his gun and $m=$ mass of bullet. Then,
$M g=\frac{N}{\Delta t} \times m(v-0)$
$\Rightarrow M =\frac{(N / \Delta t) \times m v}{g}$
where, $\frac{N}{\Delta t}$ is the number of bullets fired per second.
$\Rightarrow M =\frac{40 \times 49 \times 10^{-3} \times 500}{9.8}$
$=100 \,kg$
In given case, rate of momentum change of bullets is equal to weight of soldier.
If $M=$ mass of soldier and his gun and $m=$ mass of bullet. Then,
$M g=\frac{N}{\Delta t} \times m(v-0)$
$\Rightarrow M =\frac{(N / \Delta t) \times m v}{g}$
where, $\frac{N}{\Delta t}$ is the number of bullets fired per second.
$\Rightarrow M =\frac{40 \times 49 \times 10^{-3} \times 500}{9.8}$
$=100 \,kg$
Standard 11
Physics
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