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A rifle of $20 \mathrm{~kg}$ mass can fire 4 bullets per second. The mass of each bullet is $35 \times 10^{-3} \mathrm{~kg}$ and its final velocity is $400 \mathrm{~ms}^{-1}$. Then what force must be applied on the rifle so that it does not move backwards while firing the bullets ?
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Verified Answer
The correct answer is:
$-56 \mathrm{~N}$
From conservation of momentum
$\begin{aligned} & M V+(4 m v)=0 \\ \Rightarrow \quad V & =-\frac{4 m v}{M}=-\frac{4 \times 35 \times 10^{-3} \times 400}{20} \\ = & -2.8 \mathrm{~ms}^{-1}\end{aligned}$
Force applied on the rifle,
$\begin{aligned} F & =\frac{M V}{t}=-\frac{20 \times 2.8}{1} \\ & =-56 \mathrm{~N}\end{aligned}$
$\begin{aligned} & M V+(4 m v)=0 \\ \Rightarrow \quad V & =-\frac{4 m v}{M}=-\frac{4 \times 35 \times 10^{-3} \times 400}{20} \\ = & -2.8 \mathrm{~ms}^{-1}\end{aligned}$
Force applied on the rifle,
$\begin{aligned} F & =\frac{M V}{t}=-\frac{20 \times 2.8}{1} \\ & =-56 \mathrm{~N}\end{aligned}$
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