# Trying to prove that Impulse = change in momentum

1. Oct 29, 2008

### GRB 080319B

1. The problem statement, all variables and given/known data

I was trying to prove that Impulse = change in momentum. I end up proving it, but I'm not sure if the two assumptions I made are true. I also am not sure if my initial equations are completely true (I got them from a conceptual physics book which isn't heavy on equations themselves: I'm not sure if the equations they are using are instantaneous or average). I apologize for the notation; I don't know how to use latex. Thank you.

2. Relevant equations

I=Ft
F=ma
a=(delta v/ delta t)
p=mv

3. The attempt at a solution

I=Ft
I=(ma)t
I=m(delta v/ delta t)t
I=m(delta v) Here I made an assumption that the time in the initial impulse function was equivalent to the delta time for the velocity in order to cancel the two to get velocity by itself.
I=m(v1 - v0) The 1s and 0s signify final and initial values respectively.
I=m((p1/m1) - (p0/m0))
I=m(1/m)(p1 - p0) Here I made an assumption that the initial and final masses were equivalent to each other and to the mass from the 2nd law substitution I performed earlier in order to cancel the two to get the momentum by itself.
I=(p1 - p0)
I= delta p

2. Oct 29, 2008

### kmikias

Re: Impulse

you should include mass in to your derivative because somethimes mass makes the different so J = d(mv)/dt ........so mv=p .....then J = d(p)/dt...

3. Oct 31, 2008

### GRB 080319B

Re: Impulse

What notation do you mean by J, do you mean joules? Also, do these equations involve derivatives or just difference quotients? My first assumption was based on treating the acceleration as delta v / delta t, not dv/dt. However, you seem to be using the derivative, which leaves me in a quandary: I want to end up with delta p instead of a derivative, but can't seem to do so. Is acceleration treated as instantaneous, and are the other parts right?

Last edited: Oct 31, 2008
4. Oct 31, 2008

### heth

Re: Impulse

J is sometimes used to represent impulse.

5. Oct 31, 2008

### GRB 080319B

Re: Impulse

On wikipedia, gives the equation as I = F(delta t) = m(delta v) = delta p , assuming that the mass is constant. That's where become confused, because I'm not sure if I can assume that the mass doesn't change according to my work above. Should the mass from the 2nd law and the mass of the momentum be equivalent?

6. Nov 1, 2008

### heth

Re: Impulse

The 'universal' definition that you can use for billiard balls going at everyday speeds, and also for something like a rocket (mass changes as fuel burns) and also for photons (which have zero mass but do have momentum) and also for relativity, is:

Impulse = Fdt = dp

(since F = dp/dt)

Strictly speaking, this is the 'correct' form of Newton's 2nd Law, F = ma is what you get if the mass is constant.

As you say, if you're dealing with everyday objects, you can assume that the mass is constant.

It's not clear from the problem statement which equations you're supposed to start with to "prove" this.