velocity and average force of a bullet fired?

First, start with conservation of momentum. If the gun and spring have a total momentum of 6.2 kg m/s, then the bullet must also have 6.2 kg m/s, just in the opposite direction. Normally, when dealing with such vectors, we would denote one as being positive and the other negative, thus making the conservation work. But we’ll stick to positives, knowing that they are just in opposite directions.

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.