Respuesta :
Answer:
The recoil speed of the man and rifle is [tex]v_{man}=0.016 ms^{-1}[/tex].
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.
[tex]m=\frac{F}{g}[/tex]
Calculate the combined mass of the man and rifle.
[tex]m_{man,rifle}=\frac{650+25}{g}[/tex]
Put [tex]g=9.8 ms^{-2}[/tex].
[tex]m_{man,rifle}=\frac{650+25}{9.8}[/tex]
[tex]m_{man,rifle}=68.88 kg[/tex]
The expression for the conservation of momentum is as follows as;
[tex]m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}[/tex]
Here, [tex]m_{man,rifle}[/tex] is the mass of the man and rifle, [tex]m_{rifle}[/tex] is the mass of the rifle,[tex]u_{man},u{bullet}[/tex] are the initial velocities of the man and bullet and [tex]v_{man},v{man,rifle}[/tex] 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.
[tex]m_{bullet}=4.5 g[/tex]
[tex]m_{bullet}=.0045 kg[/tex]
Put [tex]m_{bullet}=.0045 kg[/tex],[tex]m_{man,rifle}=68.88 kg[/tex] , [tex]u_{man,rifle}=0[/tex], [tex]v_{bullet}= 240 ms^{-1}[/tex] and [tex]u_{bullet}=0[/tex].
[tex]m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)[/tex]
[tex]0=(68.88)v_{man,rifle}+(.0045)(240)[/tex]
[tex]0=(68.88)v_{man,rifle}+1.08[/tex]
[tex](68.88)v_{man,rifle}=\frac{-1.08}{68.88}[/tex]
[tex]v_{man,rifle}=-0.016 ms^{-1}[/tex]
Therefore, the recoil speed of the man and rifle is [tex]v_{man}=0.016 ms^{-1}[/tex].
