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A person holding a rifle (mass of person and rifle together is $100 \mathrm{~kg}$ ) stands on a smooth surface and fires 10 shots horizontally, in $5 \mathrm{~s}$. Each bullet has a mass of $10 \mathrm{~g}$ with a muzzle velocity of $800 \mathrm{~ms}^{-1}$. The final velocity acquired by the person and the average force exerted on the person are:
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Verified Answer
The correct answer is:
$-0.08 \mathrm{~ms}^{-1}, 16 \mathrm{~N}$
Using law of conservation of moments
$\begin{aligned}
& \text {Initial momentum }=m u_1+m u_2 \\
& \text {final momentum }=n m v_1+ \\
& (M-n m) v_2 \\
& \therefore m u_1+m u_2=n m v_1+(M-n m) v_2 \\
& 0=\frac{10 \times 10}{1000} \times 800 +\left(100-\frac{10 \times 10}{1000}\right) r^2
\end{aligned}$
$\begin{aligned}
-80 & =\frac{999}{10} v_2 \\
\Rightarrow v_2 & =-0.8 \mathrm{~ms}^{-1}
\end{aligned}$
And average force exerted on the person
$\begin{array}{l}
=\frac{n \times m \times u}{\Delta t} \\
=\frac{n \times m \times u}{\Delta t} \\
=\frac{10 \times \frac{1 \mathrm{G}}{1000} \times 800}{5} \\
=\frac{80}{5}=16 \mathrm{~N}
\end{array}$
$\begin{aligned}
& \text {Initial momentum }=m u_1+m u_2 \\
& \text {final momentum }=n m v_1+ \\
& (M-n m) v_2 \\
& \therefore m u_1+m u_2=n m v_1+(M-n m) v_2 \\
& 0=\frac{10 \times 10}{1000} \times 800 +\left(100-\frac{10 \times 10}{1000}\right) r^2
\end{aligned}$
$\begin{aligned}
-80 & =\frac{999}{10} v_2 \\
\Rightarrow v_2 & =-0.8 \mathrm{~ms}^{-1}
\end{aligned}$
And average force exerted on the person
$\begin{array}{l}
=\frac{n \times m \times u}{\Delta t} \\
=\frac{n \times m \times u}{\Delta t} \\
=\frac{10 \times \frac{1 \mathrm{G}}{1000} \times 800}{5} \\
=\frac{80}{5}=16 \mathrm{~N}
\end{array}$
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