How is impulse related to momentum?

An impulse represents a change in momentum, ##J = F \Delta t## is the definition of an impulse and we can say that ##F \Delta t = \Delta p##. In fact have the same units kg m/s or N/s. If we are talking in the language of calculus we see that:

##J = \int_0^t F(t) dt' = \int_(p_1)^(p_2) dp = p_2 - p_1 = m(v_1 - v_2)## The last step can only be certain if ##m ## is constant.

This follows directly from whereby a force is related to ##F = ma##.

Of course acceleration is a measure of the change in velocity so what we are really saying is that a force is related to the change in an objects momentum:

## F = m (dv)/(dt) = (d(mv))/(dt) = (dp)/(dt) ##

Whereby the integral equation above represents the solution to the differential equation we crafted above.

As an example, consider the impulse given to a bullet being shot by a hand gun. An average 9mm round weighs 7.5 grams and has a top speed of 400 m/s.

We find then that the bullet carries a momentum of ##p_1 = mv_1 = 400## m/s ##* 7.5 g *1/1000## kg/g## = 3## kg m/s

This doesn't seem like so much momentum. If I had this much momentum ( weighing 100kg) I would only be going 0.03 m/s or 0.108 km/h! This is slower than a dead crawl! Why can bullets do so much damage then?

Let's see how much impulse a bullet with that much momentum requires to slow down to a dead halt. From the top of my response:

##J = p_2 - p_1 = -3## kg m/s

as ## p_2 = 0## when the bullet is fully stopped.

If it takes 0.001 seconds for the bullet to completely stop we can find the force required to stop the bullet:

##F = J/(\Delta t) = -3/0.001 N = -3000N##

That is an incredible amount of force! Especially when we consider that tells us that the object stopping the bullet must have an equal and opposite force upon it.

Good summary of the subject with additional info here http://physics.info/momentum/summary.shtml