Question

the beretta model 92S (the standard issue U.S. army pistol) has a barrel 127 mm long. The bullets leave this barrel with a muzzle velocity of 336 m/s.

1. what is the acceleration of the bullet while it is in the barrel, assuming it to be constant? Express your answer in g's?

2. for how long is the bullet in the barrel (in seconds)? (two significant figures)

Answer 1

Initial velocity of the bullet, u = 0
Final velocity of the bullet, v = muzzle velocity = 336 m/s
Distance travelled, s = barrel length = 0.127 m

1)
Using the formula, v² - u² = 2as,
Where a is the acceleration.
a = [v² - u²] / 2s
= [(336)² - 0] / (2 x 0.127)
= 444472.44 m/s²
= 444472.44/9.81 g
= 45308.1 g

2)
Using the formula, v = u + at, where t is the time the bullet is accelerated.
t = (v - u) / a
= (336 - 0) / 444472.44
= 7.6 x 10^-4 s