Question

A 7.9 g bullet leaves the muzzle of a rifle with ayvaci (na25565) – Chapter 4 Basics – balantic – (APPhysPd4Balant) 2 a speed of 443.4 m/s. What constant force is exerted on the bullet while it is traveling down the 0.7 m length of the barrel of the rifle? Answer in units of N.

Answer 1

`1109.41` N

Step-by-step explanation:

`v_(o)` = initial velocity of the bullet = 0 m/s

`v_(f)` = final velocity of the bullet as it leaves = 443.4 m/s

`a` = acceleration of the bullet

`d` = length of the barrel of the rifle = 0.7 m

the kinematics equation we can use must include the variables in the above list, hence

`{v_(f)}^(2) = {v_(o)}^(2) + 2 a d`

`{443.4}^(2) = {0}^(2) + 2 (0.7)a`

`a = 140431.11` ms⁻²

`m` = mass of the bullet = 7.9 g = 0.0079 kg

Force exerted on the bullet is given as

`F = ma`

`F = (0.0079)(140431.11)`

`F = 1109.41` N