### Video Transcript

In this video, we’ll learn about
impulse and momentum. These are physical quantities that
help us understand how objects move. In this lesson, we’ll learn the
relationship between the two and we’ll also see how impulse and force are
connected.

To get started, let’s remind
ourselves of what it means for an object to have momentum. Say we have an object right
here. We’ll say that it has a mass 𝑚 and
that initially it’s at rest. Remembering that the momentum 𝑝 of
an object is equal to that object’s mass times its velocity, we can say that because
this body isn’t moving, it has zero momentum. To change that, we would need to
apply a net force to it.

Say that we do that, applying a
constant force 𝐹 to the right. And say further that this is the
only force acting on our object in the horizontal direction. This force will make our mass start
to move. And after some amount of time, it
will have a velocity we can call 𝑣. Over the time interval during which
this force acts, we say that our object experiences an impulse. This impulse 𝐢 is equal to that
object’s mass multiplied by its change in velocity. Another way of saying this is to
say that impulse is equal to an object’s change in momentum, assuming that the mass
of that object stays the same.

Now, interestingly, we can also
understand impulse in terms of the force acting on a given object. In the case of our box, we had a
constant force 𝐹 acting on it for some amount of time. Graphically, that could look like
this, with the force acting on our object plotted against time. We see this force maintains a
constant value, and it lasts up until some time we’ll call 𝑡 sub f. At this exact moment, our velocity
has increased by 𝑣, which means that the mass has experienced an impulse of
𝑚Δ𝑣.

It turns out though that this
impulse is also equal to the area under this curve we see. In other words, it’s equal to the
net force acting on some body times the time interval during which that force
acts. And in fact this equation is true
even if the net force acting on our body is not constant in time.

For example, say that the force we
applied to our box looked like this. We could still find the impulse
experienced by the box by calculating the area under this curve. In this case though, our force is a
function of time. So to calculate the impulse, we
would need to compute an integral.

We’ll get some practice with these
ideas shortly. But for now, notice that we can
calculate the impulse of an object two different ways. It’s defined in terms of an
object’s change in momentum as well as the force the object experiences over some
amount of time. Notice also that we could deliver
the same impulse to a given object by acting on it with a small force over a long
amount of time or a large force over a short amount of time. Because impulse is equal to 𝐹
times Δ𝑡, these could both work out to the same value. Knowing all this, let’s look now at
a few practice exercises.

A smooth sphere of mass 1,412 grams
was moving horizontally in a straight line at 13.5 meters per second when it hit a
smooth vertical wall and rebounded at nine meters per second. Determine the magnitude of the
impulse exerted on the sphere.

Okay, so let’s say that this is our
vertical wall. And we’re told that our sphere is
initially moving toward the wall at 13.5 meters per second. It then runs into the wall and
bounces off, moving in the opposite direction at nine meters per second. Knowing all this and knowing that
the mass of the sphere is 1,412 grams, we want to calculate the magnitude of the
impulse exerted on the sphere.

To begin figuring this out, we can
recall that the impulse acting on an object is equal to that object’s mass
multiplied by its change in velocity. And it’s important that we specify
velocity rather than speed because velocity, recall, is a vector quantity, whereas
speed is a scalar. We can compute the change in
velocity our sphere undergoes by subtracting its initial velocity from its final
velocity.

We see though that these velocities
act in opposite directions. This means that one will be
positive and one will be negative. If we decide that motion to the
left is motion in the positive direction, then we can say our sphere’s initial
velocity is negative 13.5 meters per second and its final velocity is positive nine
meters per second. When we substitute these velocities
in for 𝑣 sub f and 𝑣 sub i, we see that Δ𝑣 equals nine meters per second minus
negative 13.5 meters per second. And this comes out to 22.5 meters
per second. And note that our answer would be
different if we were working with speeds rather than velocities.

Anyway, we now know Δ𝑣 and we just
need to multiply this by our object’s mass in order to calculate the impulse. Before we substitute the given mass
of our sphere in for 𝑚, let’s convert it into the SI base unit of mass. Currently, our mass is expressed in
units of grams, so let’s convert it to kilograms. We do this by recalling that 1,000
grams is equal to one kilogram, which tells us that 1,412 grams is equal to 1.412
kilograms. And this is the mass value we’ll
use to calculate 𝐼. Substituting in for 𝑚 and Δ𝑣, we
find their product is equal to exactly 31.77 kilograms meters per second. This is the magnitude of the
impulse exerted on the sphere.

Let’s now look at an example where
we calculate impulse using force.

Three forces — 𝐅 one equals
negative five 𝐢 minus two 𝐣 plus two 𝐤 newtons, 𝐅 two equals 𝐣 minus three 𝐤
newtons, and 𝐅 three equals negative 𝐢 minus five 𝐣 minus two 𝐤 newtons, where
𝐢, 𝐣, and 𝐤 are three mutually perpendicular unit vectors — acted on a body for
three seconds. Find the magnitude of their
combined impulse on the body.

In this exercise, we have some
body, and we’re told it’s being acted on at the same time by three forces: 𝐅 one,
𝐅 two, and 𝐅 three. These forces are all constant, and
we’re told that they act over a time interval we’ll call Δ𝑡 of three seconds. We want to know the magnitude of
the impulse on the body from these three forces acting together.

We could calculate the impulse due
to each of the three forces individually and then add those impulses together. But a shorter way is to add our
three forces together as vectors to solve for the net force acting on our body. So we write out 𝐅 one, 𝐅 two, and
𝐅 three, lining them up by their 𝐢-, 𝐣-, and 𝐤-components. We then add them by their
components, starting with the 𝐢-component. Negative five 𝐢 minus 𝐢 gives a
total of negative six 𝐢. Then negative two 𝐣 plus 𝐣 minus
five 𝐣 adds up to negative six 𝐣. And lastly, two 𝐤 minus three 𝐤
minus two 𝐤 equals negative three 𝐤. The units of these components are
all newtons. And this vector overall is equal to
the net force acting on our body.

Knowing this, let’s recall a
mathematical relationship for the impulse experienced by an object due to a force
acting on it. Impulse 𝐢 is equal to 𝐅 times
Δ𝑡. In other words, the impulse an
object experiences equals the net force acting on that object times the time
interval over which that force acts. In our case, we know 𝐅 sub net and
we also know Δ𝑡. That was given as three
seconds.

But let’s recall that we want to
solve for impulse magnitude. That means we’ll need to calculate
the magnitude of our net force and use that in our equation. Clearing a bit of space, we can
recall that given a three-dimensional vector — we’ll call this example vector 𝐯 —
the magnitude of 𝐯 is equal to the square root of its 𝐢-component squared plus its
𝐣-component squared plus its 𝐤-component squared.

Applying this relationship to our
net force, we see it’s equal to the square root of negative six squared plus
negative six squared plus negative three squared newtons. This is equal to the square root of
36 plus 36 plus nine newtons or the square root of 81 newtons, which is nine
newtons. That’s the magnitude of the net
force acting on our body. To calculate 𝐼 then, we substitute
this value in for the magnitude of 𝐅 sub net and Δ𝑡 is equal to three seconds. So when we calculate 𝐼, we find a
result of 27 newton seconds. That’s the magnitude of the total
impulse acting on our body.

Now let’s look at an example where
we calculate impulse based on a graph.

The given figure shows a force–time
graph for a force acting in a constant direction on a body moving along a smooth
horizontal plane. Using the information provided,
calculate the magnitude of the force’s impulse.

In this scenario then, we have a
smooth horizontal plane and a body moving across it. The force that this body
experiences is plotted against time in our graph. At the outset, there’s zero
force. But then, over the next 20 seconds,
the force applied to the body increases at a constant rate until it reaches 90
newtons. This force is held constant for the
next 50 seconds up until the 70-second mark. At that point, the applied force
begins to decrease at a constant rate until at a time of 80 seconds it returns to
zero.

With this in mind, we want to
calculate the magnitude of the impulse experienced by this body. We can clear a bit of space to work
and then recall that the impulse experienced by an object is equal to the net force
acting on the object times the time over which that force acts. In our graph, we’re shown the net
force acting on our body over time. So to calculate the impulse our
body experiences, we’ll multiply each individual force value by the corresponding
time interval for that force. This means that the total impulse
is equal to the area under this force-versus-time curve.

If we calculate that area, then
we’ll have solved for 𝐼. To make this easier, we can divide
up this overall area into common shapes. We see that from zero to 20
seconds, this area is a right triangle. Then, from 20 to 70 seconds, it’s a
rectangle, and then from 70 to 80 seconds another right triangle. If we call these areas 𝐴 one, 𝐴
two, and 𝐴 three, then the impulse magnitude we want to solve for is actually equal
to their sum.

We can start by calculating 𝐴
one. This is the area of a right
triangle, which we can recall is equal to one-half the triangle’s base times its
height. The base of 𝐴 one we can see from
our graph is 20 seconds, and its height we can also see is 90 newtons. This is equal in total to 900
newton seconds, which we then substitute in for 𝐴 one.

Knowing this, we’ll move on to
calculate 𝐴 two. This is the area of a rectangle,
which is the base of that rectangle multiplied by its height. The rectangle’s base is 70 seconds
minus 20 seconds, or 50 seconds, and its height is 90 newtons. This means that 𝐴 two is 4,500
newton seconds, which we can then substitute in to our equation for 𝐼.

Then, finally, we’ll calculate 𝐴
three. This is the area of a right
triangle whose base is 80 seconds minus 70 seconds, or 10 seconds, and its height is
once again 90 newtons. 𝐴 three then comes out to 450
newton seconds.

We can now calculate 𝐼. And just as a side note, notice
that we could have treated the area under our curve as one single trapezoid and
solved for it that way. In any case, when we compute the
impulse, we find a result of 5,850 newton seconds. This is the magnitude of the
impulse experienced by our body.

Let’s look now at one last example
where we calculate impulse for a time-varying force.

The given figure shows a force–time
graph. At time 𝑡 seconds, where 𝑡 is
greater than or equal to zero, the force is given by 𝐹 is equal to the quantity 𝑡
minus two squared newtons. Find the impulse over the first
four seconds.

Taking a look at this graph, we see
that it shows us force versus time and that this force starts out at four newtons,
decreases to zero newtons at a time of two seconds, and then increases at an
increasing rate until it reaches nine newtons at a time of five seconds. Focusing on the first four seconds,
we want to calculate the impulse of this force.

Graphically, this is equal to the
area under our curve from 𝑡 equals zero to 𝑡 equals four seconds. But note that we’re given the
algebraic equation for this force in terms of the time 𝑡. Knowing that in our case 𝐹 of 𝑡
is equal to the quantity 𝑡 minus two squared newtons, we can recall that when we
have a time-varying force, the impulse due to that force is equal to the integral of
the force over time. In this expression, 𝑡 one is the
initial time being considered and 𝑡 two is the final time. That is, we’re integrating 𝐹 of 𝑡
over the time interval of interest.

Applying this relationship to our
scenario, we’re working from a time interval of 𝑡 equals zero seconds up to 𝑡
equals four seconds. We’ll integrate 𝐹 of 𝑡 with
respect to time. And we’ll keep track of the units
involved.

Our first step is to multiply out
this expression. Quantity 𝑡 minus two squared is
equal to 𝑡 squared minus four 𝑡 plus four. And now we want to integrate each
of these three separate terms with respect to 𝑡. The integral with respect to 𝑡 of
𝑡 squared is 𝑡 cubed over three. The integral of negative four 𝑡 is
negative two 𝑡 squared. And that of four is four times
𝑡. And we’ll evaluate this expression
from 𝑡 equals four seconds to 𝑡 equals zero seconds.

Note that when we substitute in
zero seconds for 𝑡, all three of these terms will equal zero. So we only really need to
substitute in four seconds for the time 𝑡. Before we substitute in this time
though, let’s divide it up into its numerical value, that’s four, and its unit,
that’s seconds. We know that, in the end, our
impulse will have units of newton seconds. That’s what it means to multiply a
force by a time.

Therefore, when we substitute in
our time of four seconds, we won’t substitute in the unit. But we’ll instead pull that out and
group it with our other unit, the newton. So now, to finish evaluating this
integral, we’ll substitute in four for 𝑡 here, here, and here. Doing that gives us this
expression.

And now we just want to
simplify. Four cubed over three is sixty-four
thirds. Negative two times four squared is
negative 32. And four times four is 16. To combine these values, we’ll
first want to get a common denominator for all three. We’ll choose that denominator to be
three so that negative 32 becomes negative 96 over three and positive 16 becomes
positive 48 over three. Combining all these fractions, we
come up with a result of sixteen thirds newton seconds. This then is the impulse due to our
force over the first four seconds.

Let’s take a moment now to review
some key points from this lesson. We’ve seen that the impulse
experienced by an object equals its change in momentum. Written as an equation, we can say
that 𝐼 is equal to 𝑚 times Δ𝑣. And we saw that this relationship
assumes that the mass of our object is constant in time.

Related to this, we saw that an
object’s impulse is equal to the net force acting on it times the time over which
that force acts. We saw that this equation is true
whether the force 𝐹 is constant or varies in time. When the force does vary, we can
write an integral expression for the impulse.

Lastly, we saw that when we’re
presented with a force-versus-time graph, the impulse delivered by that force over
some time interval is equal to the area under the curve.