A 22g bullet is fired at a muzzel velocity of 509m/s from an 8.75 kg gun with a barrel 72cm long. the force on the bullet while it is in the barrel is about?
To find the force on the bullet while it is in the barrel, we can apply the concept of impulse and use Newton's second law of motion.
First, we need to determine the time the bullet spends in the barrel. Since the barrel length is given as 72 cm (or 0.72 m) and the bullet's velocity is given as 509 m/s, we can use the equation v = d/t, where v is the velocity, d is the distance, and t is the time.
t = d / v
t = 0.72 m / 509 m/s
t = 0.001413 s
Now, we can calculate the initial momentum of the bullet using the formula p = m * v, where p is the momentum, m is the mass, and v is the velocity.
p = (22 g) * (509 m/s)
= 0.022 kg * 509 m/s
= 11.198 kg·m/s
The change in momentum (impulse) experienced by the bullet is equal to the initial momentum of the bullet. Therefore, by Newton's second law of motion, we know that the force, F, is equal to the change in momentum divided by the time:
F = Δp / t
F = 11.198 kg·m/s / 0.001413 s
F ≈ 7919.39 N
Therefore, the force on the bullet while it is in the barrel is approximately 7919.39 Newtons.