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Answer:
The average recoil force exerted on the weapon during that burst is 36.65 N
Explanation:
Bullet Mass - m = 7.48 * [tex]10^{-3}[/tex] kg.
Bullet Speed - v = 294 m/s.
This way knowing that for each firing, the time elapsed is
Δt = 60 s / 1000 rounds = 0.06 s / round.
The average recoil force exerted on the weapon during that burst is given by F = -mv/Δt
F = -mv/Δt = -(7.48 * [tex]10^{-3}[/tex] kg)*(294 m/s) / 0.06s = -36.65 N.
The negative sign means it is directed in an opposite direction of the bullet.
The average recoil force exerted is 36.67N
Recoil Force:
It is given that the weapon fires 1000 rounds per minute. So the time taken to fire one round is given by:
[tex]\Delta t=\frac{60}{1000}=\frac{3}{50}s[/tex]
Now each bullet has mass m = 7.48g and velocity v = 294 m/s.
So the momentum of one bullet is:
[tex]\Delta p=mv=7.48\times 10^{-3}\times294=2.2\;kgm/s[/tex]
From Newton's Third law of motion, the recoil force will be equal and opposite to the force exerted on the bullet. The force exerted on the bullet is given by:
[tex]F=\frac{\Delta P}{\Delta t} =\frac{2.2\times50}{3}=36.67N[/tex]
So the recoil force of the weapon is 36.67N
Learn more about Laws of motion:
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