Question

A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a 650-N man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle.

Answer 1

Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$.

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$.

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle,  $m_{rifle}$ is the mass of the rifle,$u_{man},u{bullet}$  are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

$m_{bullet}=4.5 g$

$m_{bullet}=.0045 kg$

Put $m_{bullet}=.0045 kg$,$m_{man,rifle}=68.88 kg$ , $u_{man,rifle}=0$, $v_{bullet}= 240 ms^{-1}$ and $u_{bullet}=0$.

$m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+1.08$

$(68.88)v_{man,rifle}=\frac{-1.08}{68.88}$

$v_{man,rifle}=-0.016 ms^{-1}$  

Therefore, the recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$.