Deriving the impulse-momentum theorem relies on knowledge of Newton’s Second Law. We use the second law and the definition of acceleration to find this important result.

\(\color{black}{\Sigma \vec{F} = m \vec{a}}\)

\(\color{black}{\displaystyle \Sigma \vec{F} = m \frac{\Delta \vec{v}}{\Delta t}}\)

If we now multiply each side by \(\color{black}{ \Delta t,}\) we find

\(\color{black}{\Sigma \vec{F} \Delta t = m \Delta \vec{v}.}\)

It is possible that the mass can also change, so we often rearrange to get

\(\color{black}{\Sigma \vec{F} \Delta t = \Delta m \vec{v}}\) or

\(\color{black}{ \Sigma \vec{F} \Delta t = \Delta \vec{p},}\) since \(\color{black}{\vec{p} = m \vec{v}.}\)

## The Impulse-Momentum Theorem

The left hand side of the equation, \(\color{black}{ \Sigma \vec{F} \Delta t}\) is known as the **impulse**. The right hand side is the change in momentum. The entire equation is known as the **impulse-momentum theorem**.

This result is important, because we can usually measure the speeds and masses of two objects after an interaction relatively easily. Measuring the forces can be extremely difficult. Using the impulse-momentum theorem, we can investigate the forces involved indirectly.

**Note:** In some textbooks, the impulse is represented with the letter *I*. I prefer to avoid this, because we will soon use *I* to represent the moment of inertia for spinning objects. Other textbooks use the symbol *J* to represent the impulse. Since the impulse equation is easy to remember, I prefer to just stick with the expression above.

Think about this…

- Suppose an object goes through a fixed momentum change in two different ways. In one case, the time is short. In the other, the time is long. How do the forces compare in these two cases?
- If you apply a force a long time, how should the momentum change?
- You apply a force that points to the left to an object. Which direction should its momentum change point?