Physics Bootcamp

Subsection 7.2.1 Impulse as an Area

When you hit a baseball, you change direction and magnitue of the momentum of the ball in a very short duration of time. The change in momentum is caused by the force from the bat acting on the ball during the time they were in contact with each other.

If you look at the magnitude of the force on the ball, it will be initially zero till contact occurs, will increase as the bat crushes into the ball, and then, will decrease to zero as the ball leaves the bat. The stronger the force and/or the longer the force acts on the ball, the greater the change in the momentum of the ball.

That is, both the magnitude of the force and the time are important for causing a change in momentum. For this purpose, we define impulse, \(\vec J\text{,}\) during an infintesimal interval \(\Delta t\) by product of the force \(\vec F\) and the interval.

For constant force \(\vec F_0\text{,}\) such as weight, impulse over a finite interval \(t_1\le t \le t_2\) will be

\begin{equation} \vec J = \vec F_0 \;(t_2 - t_1).\tag{7.2.2} \end{equation}

If force is changing with time, as for example the force shown in Figure 7.2.2, we will need to integrate \(\vec F(t)\) with respect to time \(t\text{,}\) or, work with area under the curve of the components of force versus time. Thus, impulse is formally defined by the following “conceptual integral”

The Cartesian components of impulse will be integrals that you can perform from given funtions \(F_x(t)\text{,}\) \(F_y(t)\text{,}\) \(F_z(t)\text{.}\)

Previously, we have seen that another way to look at integrals is “area under the curve”. Thus, \(J_x\) during interval \(t_1\le t \le t_2\) will be area between \(F_x(t)\) and \(F_x=0\) when you plot \(F_x\) versus \(t\text{.}\)

From the components \((J_x, J_y, J_z)\text{,}\) we can obtain the mangitude and direction of impulse as we do for any other vector quantity.

Subsection 7.2.2 Net Impulse

Impulses from different forces add vectorially. Suppose force \(\vec F_1\) acts for a time \(\Delta t_1\) and \(\vec F_2\) for a time \(\Delta t_2\text{,}\) then the net impulse \(\vec J_{\text{net}}\) will be the vector sum of the impulses \(\vec J_1\) and \(\vec J_2\) of the two forces.

which must be evaluated as vector additions that you have learned in previous chapters. You can, of course, extend this to any number of forces in an obvious way.

Subsection 7.2.3 (Calculus) Impulse as an Integral

Consider an interval \(t=t_i\) to \(t=t_f\) during which we wish to compute the impulse of a force \(\vec F \text{,}\) whose direction is fixed in time, for simplicity. Here, we will show how the formula for the magnitude of impulse is derived.

Let us divide the interval into \(N \) subintervals: \((t_0=t_i, t_1)\text{,}\) \((t_1, t_2)\text{,}\) \((t_2, t_3)\text{,}\) \(\cdots\text{,}\) \((t_{N-2}, t_{N-1})\text{,}\) and \((t_{N-1}, t_N=t_f)\text{.}\) For simplicity, let the intervals be of equal length in time, \(\Delta t\text{,}\)

Now, let us compute the impulse magnitude \(J_k\) in the \(k^{\text{th}}\) subinterval. Let the magnitude of the force at the ends of this subinterval be \(F_{k-1} \) and \(F_{k} \text{.}\) If \(N \) is large enough, then \(F_{k-1} \) and \(F_{k} \) will be almost the same, and therefore, we can get a good approximation of impulse magnitude \(J_k\) by using the average force in the subinterval as

We now add up impulses from every subinterval to get the impulse over the original interval.

By increasing \(N \) as large as we like, we can make this aproximation as accurate as we are happy with. The limit \(N \rightarrow \infty \) is said to be the exact value, which is represented by the integral sign.

If we are dealing with a force whose magnitude as well as direction changes, then we will work with one component at a time. The same procedure as above will give us the components of the impulse vector.

From \((J_x, J_y, J_z)\) you can compute the magnitude and direction of the impulse in a usual way.