### 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.