# Solving problems using impulse-momentum

Before we begin solving problems using the impulse-momentum theorem, take a moment to think about what the parts of the equation mean.

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

The \(\Sigma \vec{F}\) could be replaced by \(\vec{F}_{net}\), since the net force is simply the sum of the vectors acting on the object. The second bit of information is that we need to look at the force over the whole time of the interaction. Therefore, we can replace the net force \(\vec{F}_{net}\) with the average force \(\vec{F}_{ave}\). (You’ll see more on this later.) This gives us…

\(\color{black}{ \vec{F}_{ave} \Delta t = \Delta m \vec{v}}\)

Finally, the direction of the net average force will determine the direction of the momentum change. This is in the same direction as the velocity change, since the velocity carries the directionality of the momentum. This means that both sides of the vector equation point in the same direction, so we can just write

\(\color{black}{ F_{ave} \Delta t = \Delta mv}\)

as our algebra equation.

If the mass of the object doesn’t change, we can move the mass out front of the \(\Delta\), resulting in…

\(\color{black}{ F_{ave} \Delta t = m \Delta v}\)

With these simplifications, we have an equation that has four variables. As long as we have three of them, we can solve for the remaining fourth variable.

Notice that there is a preferred set of units here. We will want to find the forces in newtons. Therefore, we should take care to convert all masses to kg. The speeds should be in m/s. All times should be in seconds. (In many collisions or interactions, the interaction occurs in times as short as milliseconds (ms). Make sure to convert these to seconds.

Take a moment to solve this equation for each of the variables. Each time, write down the starting equation. This will help you learn and remember it more quickly.