Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,
[tex]F\times t=mv\\\\F=\dfrac{mv}{t}[/tex]
For 8 shot burst, average recoil force on the gun is :
[tex]F=\dfrac{8mv}{t}\\\\F=\dfrac{7.5}{1000}\cdot\dfrac{300}{0.4}\cdot8\\\\F=45\ N[/tex]
So, the average recoil force on the gun during that 0.40 s burst is 45 N.