# velocity and average force of a bullet fired?

* How do I find the velocity of the bullet as it exits the barrel?

* What was the avg force exerted on the bullet as it accelerated along the length of the barrel?

* What average force would be felt by the individual holding the gun for the time the bullet accelerates along the length of the barrel?

Favorite Answer

Anyway, momentum is defined as mass times velocity, or

p = m * v

Since we know the momentum, p, and the mass of the bullet, m, we can determine the velocity, but we must make sure we have consistent units. Let’s convert the mass of the bullet to kg from g.

26 g * 1 kg/100 g = 0.26 kg (units cancel out of fractions, just like numbers)

Then, we can insert our ‘knowns’ into our momentum equation:

p = m * v

6.2 kg m/s = 0.26 kg * v m/s divide both sides by the bullet’s mass and units (so the kg’s cancel out) to get…

v = 23.84 m/s

The average force on the bullet is its mass times its acceleration. We know the mass, 0.26 kg, so we need to find its acceleration, a.

The relationships between distance, d (32 cm or 0.32 m), v, a and time, t, are:

v = a * t (assuming a constant acceleration)

d = a * t * t / 2

So, we” solve for a in the upper and substitute it in the lower….

a = v / t

d = v / t * t * t / 2, or

d = v * t / 2 insert our knowns for d and v, and solve for t

0.32 m = 23.84 m/s * t /2

t = 0.0268 s

Use v = a * t to find a

23.84 m/s = a * 0.0268 s

a = 888.04 m/s^s

So if F = m * a, then the force on the bullet is…

F = 0.26 kg * 888.04 m/s^2

F = 230.89 kg m/s^2 (this mess of units in combination is called a Newton, and it has nothing to do with figgy cookies)

Since force is also a vector that is conserved, the person holding the gun would also feel 230.89 Newtons, just in the opposite direction.