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A marksman at rest fires a 4.00-kg gun that expels a bullet of mass 0.014 kg with a velocity of 181 m/s. The marksman's mass is 81 kg. What is the marksman's velocity after firing the gun?

Answer :

The marksman's velocity after firing the gun is about -0.63 m/s (in the opposite direction of the bullet's velocity), as calculated using the conservation of momentum principle.

The conservation of momentum principle can be used to resolve this issue. The initial momentum of the system (marksman plus gun) is zero, and the final momentum is also zero because the bullet and the marksman have equal and opposite momenta. Therefore, we can write:

initial momentum = final momentum

[tex]0 = (M + m) * V + m * v[/tex]

where M is the mass of the gun, m is the mass of the bullet, V is the velocity of the gun after firing, and v is the velocity of the bullet after firing (which is 181 m/s in this case).

Solving for V, we get:

[tex]V = - m * v / (M + m)[/tex]

= [tex]\frac{- 0.014 kg * 181 m/s}{(4.00 kg + 0.014 kg)}[/tex] ≈ -0.63 m/s

Therefore, the marksman's velocity after firing the gun is about -0.63 m/s (in the opposite direction of the bullet's velocity).

Learn more about velocity here:

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