# Video: Impulse and Collisions

In this video we learn the definition of impulse for both constant and variable applied forces, as well as the Impulse-Momentum Theorem equating the impulse an object experiences with its change in momentum.

10:33

### Video Transcript

In this video, we’re going to learn about impulse and collisions. We’ll learn what impulse is, how it helps us understand collisions, and how impulse is related to momentum. To start out, we can consider the example of car manufacturers crash testing their vehicles using crash test dummies. As one of the outcomes of such testing, the airbag was developed as a safety mechanism for gradually slowing down drivers and passengers involved in auto accidents.

When a moving vehicle stops suddenly, say by running into a wall, all of the people and objects inside tend to wanna continue their forward motion. Before the airbag was developed and installed, drivers and passengers would tend to move forward and perhaps strike their heads suddenly on steering wheels or dashboards. But the airbag, when it’s deployed, cushions the driver’s and passenger’s head to a slower and gentler stop. To understand why an airbag is so helpful for passenger safety, we’ll want to learn a bit about impulse and collisions.

Impulse is a measure of applied force over an amount of time. Say that you throw a rubber ball against a flat wall. When the ball comes in contact with the wall, the wall exerts a force on the ball. In response to this force, the ball bounces back. If we call the amount of time that the wall and the ball are in contact Δ𝑡, then we can say the wall has delivered an impulse, symbolized by capital 𝐽, to the ball. And that impulse is equal to the force the wall applied on the ball multiplied by the time over which that force acted. This mathematical way of writing impulse is helpful if our force 𝐹 is constant in time. But in many real-world scenarios, force is not constant in time; it varies.

Say we had a car driving along a road and moving at a steady speed to the right. Then imagine that the driver saw something and tried to stop as quickly as possible by jamming on the brakes, so the car would accelerate to the left. If we were to plot the force applied by the brakes 𝐹 sub 𝑏 over the time interval 𝑡 during which the car comes to a stop, that force versus time might look something like this, where initially the force is high but then as the car approaches a stop, it tails off. In this case, our force is clearly variable with time. When the force applied changes over time, we can write impulse 𝐽 not as a linear product but as an integral of the force which varies with time times 𝑑𝑡.

This second integral expression for impulse covers both variable and constant forces. But when the force is constant in time, it can be helpful to write this simplified form. Knowing that in general impulse 𝐽 is equal to the integral of applied force times 𝑑𝑡, we can represent impulse graphically as the area under the curve 𝐹 versus 𝑡. Impulse is useful because it helps us understand collisions well. It also helps because it connects with another quantity we’ve learned about: momentum. Let’s recall for a moment Newton’s second law of motion, that the net force acting on an object is equal to the object’s mass times its acceleration.

We can rewrite the acceleration 𝑎 as a derivative with respect to time of the object’s velocity 𝑣. If we then multiply both sides of this equation by 𝑑𝑡, that expression on the right-hand side cancels out. Then we see that 𝑑𝑡 times 𝐹 is equal to 𝑚 times 𝑑𝑣 or Δ𝑡, the change in time, times 𝐹 is equal to 𝑚 times Δ𝑣, the change in velocity. But we can recall that for an object whose mass doesn’t change with time, 𝑚 Δ𝑣 is equal to the change in that object’s momentum, 𝑝. So Δ𝑡 times 𝐹 is equal to change in momentum. And since we’ve just seen that Δ𝑡 times 𝐹 is equal to impulse 𝐽, that means that impulse is equal to change in momentum Δ𝑝. This result is so useful to us that it has a name. It’s called the impulse–momentum theorem, which is a fancy way of saying impulse is equal to an object’s change in momentum. Let’s get some practice with this idea of impulse and how it connects with momentum through a couple of examples.

A rugby player of mass 105 kilograms is moving at 6.45 meters per second and collides with a padded goalpost. The collision exerts a force on the rugby player in the opposite direction to his motion of 15.0 kilonewtons for 72.8 milliseconds. At what speed does the rugby player move away from the goalpost?

We can label the speed 𝑣 and start off by sketching the scenario. In this scenario, our rugby player of mass 𝑚, 105 kilograms, approaches a padded goalpost with a speed we’ve called 𝑣 sub 𝑖 of 6.45 meters per second. When the player runs into the goalpost, the post exerts a force we’ve called 𝐹 on the player in opposite direction of their motion for a time period we’ve called Δ𝑡. The result of this force over this time is that the player’s motion is reversed in direction. It’s this speed 𝑣 that we want to solve for. Since we’re given information about the player’s mass and speed, that may make us think of momentum. And since we’re also told about the force and the time over which the force acts of the goalpost on the player, that makes us think of impulse.

Bringing these two terms together, we can recall the impulse–momentum theorem which says that the impulse acting on an object is equal to its change in momentum. So the impulse acting on the rugby player is equal to the player’s change in momentum. When we recall further that impulse for a constant force 𝐹 is equal to the product of that force 𝐹 times the time over which it acts and that momentum 𝑝 is equal to an object’s mass times its velocity, we can write that the magnitude of the force 𝐹 times Δ𝑡 is equal to the player’s mass times 𝑣 sub 𝑖 plus 𝑣 where 𝑣 sub 𝑖 and 𝑣 are the speeds of the incoming and outgoing player relative to the post.

Rearranging this expression to solve for 𝑣, we find it’s equal to 𝐹Δ𝑡 over 𝑚 minus 𝑣 sub 𝑖. We’ve been given 𝑚, 𝑣 sub 𝑖, 𝐹, and Δ𝑡 in our problem statement, so we’re ready to plug in and solve for 𝑣. When we plug in for these values, we convert our force into units of newtons and our change in time Δ𝑡 units of seconds. We do this so the units of these two terms are consistent with the units in the rest of our expression. We find that, to three significant figures, 𝑣 is 3.95 meters per second. That’s the speed at which the rugby player rebounds from the goalpost.

Now, let’s look at an example where the applied force changes in time.

The 𝑥-component of the force applied on a golf ball of mass 38 grams by a golf club is shown in the diagram. What is the 𝑥-component of the impulse applied on the golf ball between zero milliseconds and 50 milliseconds? What is the 𝑥-component of the impulse applied on a golf ball between 50 milliseconds and 100 milliseconds?

We can call these 𝑥-components of the impulse 𝐽 sub 𝑥 one and 𝐽 sub 𝑥 two, respectively. When we look at the diagram, we see it’s a graphical representation of the force in the 𝑥-direction applied to the golf ball versus the time in milliseconds over which that force is applied. Over a time span of zero to 100 milliseconds, the force varies from zero to 300 newtons. Since we want to solve for components of impulse, we’ll consider that impulse in general is equal to the integral of the applied force times 𝑑𝑡. We know that this integral represented graphically would equal area under the curve of force versus time. So on our diagram, we can graphically represent what 𝐽 sub 𝑥 one and 𝐽 sub 𝑥 two are.

𝐽 sub 𝑥 one, which is the impulse applied in the 𝑥-direction to the golf ball from zero to 50 milliseconds, is equal to the area under the curve for that time interval. Similarly, 𝐽 sub 𝑥 two, it’s the area under the curve from 50 to 100 milliseconds. We can write then that 𝐽 sub 𝑥 one is equal to the area of the triangle shown, one-half the base of 50 times 10 to the negative third seconds or 50 milliseconds times the height of 300 newtons. When we multiply these three values together, we find a result of 7.5 kilograms-meters per second. That’s the 𝑥-component impulse delivered to the golf ball from zero to 50 milliseconds.

Next, we want to solve for 𝐽 sub 𝑥 two, the area under the curve from 50 to 100 milliseconds. That area is equal to the area of the rectangle with base 100 minus 50, or simply 50 times 10 to the negative third seconds, multiplied by the height of 300 newtons. This results in a value of 15 kilograms-meters per second. That’s the 𝑥-component of the impulse delivered to the golf ball from 50 to 100 milliseconds.

Let’s summarize what we’ve learned so far about impulse and collisions. We’ve seen that impulse measures the force applied to an object over the time that that force is applied. When the applied force is constant with time, impulse 𝐽 is equal to the product of that force times Δ𝑡. But when the applied force 𝐹 is variable with time, as it is in many real-life applications, we can write impulse 𝐽 as the integral of that force times 𝑑𝑡 from the initial time 𝑡 one to the final time 𝑡 two. And finally, we’ve seen that the impulse delivered to an object is equal to its change in momentum. This result is called the impulse–momentum theorem.
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