### Video Transcript

In this video, we will be
discussing a property of objects that have mass and are moving. This property is known as
momentum. So, let’s first start by
understanding what momentum is. If we start by thinking about an
object — let’s say this ball here — and we say that the ball has a mass 𝑚 and is
moving, in this case, toward the right at a velocity 𝑣. Then the momentum — which we will
call 𝑝 — of this ball is defined as the mass of the ball multiplied by its
velocity. This is a general definition of
momentum, the mass of an object multiplied by the velocity with which it’s
moving.

So, let’s now imagine that our ball
over here has a mass 𝑚 of two kilograms and a velocity 𝑣 of three meters per
second. In that case, we can say that the
momentum of this ball 𝑝 is equal to the mass, which is two kilograms, multiplied by
its velocity, which is three meters per second. This gives a numerical value of two
times three, which is six, and the unit is going to be kilograms times meters per
second. So, we find that the momentum of
the ball 𝑝 is equal to six kilogram meters per second. We use this dot here to signify
multiplication. In other words, we’re multiplying
the units of kilograms by the units of meters per second. So, this is how we calculate the
momentum of an object. We multiply its mass by its
velocity.

And we’ve also seen that the units
of momentum, at least in SI base units, are kilograms meters per second. However, it’s worth noting that we
may see momentum written in other units, units such as grams multiplied by meters
per second, or maybe kilograms multiplied by kilometers per hour, and so on and so
forth. As long as we’ve got a unit of mass
multiplied by a unit of velocity, the overall unit is going to be one of
momentum. And in base units, that happens to
be kilograms multiplied by meters per second.

So, let’s now imagine the following
scenario. Let’s say we’ve got this blue ball
here with a mass 𝑚 and a velocity 𝑣. We’ve already seen that the
momentum of this ball is going to be equal to its mass multiplied by its
velocity. And let’s now imagine that we’ve
got another ball, this time a pink one, which has the same mass as the blue ball but
a larger velocity. We’ll call this capital 𝑉. In this case, using the momentum
equation, we can see that the momentum of the pink ball is equal to the mass of the
pink ball, lower case 𝑚, multiplied by its velocity, uppercase 𝑉.

And since we’ve said that the
velocity uppercase 𝑉 is larger than the velocity lowercase 𝑣, this means that the
momentum 𝑚 multiplied by uppercase 𝑉 is larger than the momentum 𝑚 multiplied by
lowercase 𝑣. And so, what we’re saying here is
that the pink ball, which has the same mass as the blue ball but is moving at a
larger velocity than that of the blue ball, also has a larger momentum than the blue
ball.

Additionally, if we now think of a
different ball, this time having a mass capital 𝑀, which is larger than the mass of
the blue ball but it has the same velocity as the blue ball, lowercase 𝑣. Then, we can say that the momentum
of the orange ball is equal to the mass, uppercase 𝑀, multiplied by its velocity,
lowercase 𝑣. And once again, we see that,
because the mass of the orange ball is larger than the mass of the blue ball, even
though they’re traveling at the same velocities, the momentum of the orange ball is
larger than the momentum of the blue ball. So, what we’ve learnt is the
following. For two objects with the same mass,
the one with the larger velocity will have the larger momentum. And also, for two objects with the
same velocities, the one with the larger mass will have the larger momentum.

But this logic doesn’t necessarily
just have to apply to different objects. That is, if we were to now just
think about the blue ball and ignore the pink and the orange balls, we would see
that if the velocity of the blue ball were to increase to, say, capital 𝑉, then the
momentum of the blue ball would also increase to 𝑚 multiplied by capital 𝑉. And so, an object’s momentum can
change over time. If its velocity increases, its
momentum increases as well and vice versa. And this is also true for mass. For example, our ball could be
rolling on some flat ground. And as it rolls, it picks up some
dust, so its mass increases. Well, if its mass increases to,
let’s say, capital 𝑀 and still moving at a velocity capital 𝑉, then the momentum
of our ball, 𝑝, will be capital 𝑀 multiplied by capital 𝑉. And so, what we’re seeing here is
that if an object’s mass or velocity changes over time, then its momentum also
changes over time.

Now, here’s something
interesting. Let’s notice that, in the equation
for momentum, we’re using the velocity of an object and not its speed. We can recall that the velocity of
an object is a vector quantity. This means that it has both
magnitude, or size, and direction as well. So, if we return to the blue ball
that we saw earlier, it’s important to say that the velocity of this blue ball is 𝑣
toward the right. In other words, we include the
magnitude, that’s 𝑣, and the direction toward the right. And so, we’ve seen that velocity is
a vector quantity. Additionally, the mass of an object
is a scalar quantity. It only has magnitude and not
direction because we don’t talk about the mass of an object in a particular
direction. It just has this mass.

And so, in order to get momentum,
we’re multiplying a scalar quantity by a vector quantity. And this means that momentum itself
must also be a vector quantity. It must have magnitude and
direction as well. After all, the velocity of an
object is a quantity that contains information both about the speed of the object
and the direction in which it’s moving. So, when we multiply this velocity
by a mass, we get momentum. And therefore, momentum must also
contain information about the direction in which that object is moving. We therefore say that an object
which has a mass 𝑚 and is moving toward the right with a velocity 𝑣 has a momentum
𝑝, which has a magnitude or size of 𝑚 multiplied by 𝑣. And crucially, this momentum is to
the right, in other words, in the same direction as the velocity.

So, the whole point of this is for
us to remember that when we calculate an object’s momentum, we can use 𝑝 is equal
to 𝑚𝑣 to calculate the momentum’s magnitude or size. And we also separately have to
account for the direction of the momentum. This becomes most useful when we
start thinking about the momentum of not just one object but the total momentum of
more than one object. Let’s imagine that we’re now
thinking about two balls, a blue one and a pink one. Let’s say that both have a mass of
one kilogram each, and that both are moving toward the right with a velocity of one
meter per second each. In this case, what’s the total
momentum of both of these balls combined? Well, we can say that this total
momentum — we’ll call this 𝑝 subscript tot — is equal to the momentum of the blue
ball — we’ll call this 𝑝 subscript b — plus the momentum of the pink ball — we’ll
call this 𝑝 subscript p.

And then, we can recall that the
momentum of any particular object is equal to its mass multiplied by its
velocity. And so we can say that the momentum
of the blue ball is equal to its mass, one kilogram, multiplied by its velocity, one
meter per second. And to this we added the momentum
of the pink ball, which is also one kilogram — which is its mass — multiplied by its
velocity of one meter per second. For the momentum of the blue ball,
the numerical value becomes one times one. And the unit, of course, is
kilograms times meters per second, or kilogram meters per second. Which means that the momentum of
the blue ball is one kilogram meter per second. And here, we’re doing exactly the
same calculation for the pink ball. And so, the momentum of the pink
ball is one kilogram meter per second as well.

This means that the total momentum
of the two balls combined is equal to one kilogram meter per second plus one
kilogram meter per second. Now, one plus one is two. And since we’ve got the same units
for both quantities, this means we can add them together. This gives us a total momentum for
both the blue and the pink balls combined of two kilograms meters per second. Now as an aside, by the way, the
reason we find the total momentum of the blue and the pink balls combined is because
sometimes we need to consider the effect of the motion of multiple objects. For example, instead of two balls,
we may’ve had an orange object that was made up of two components, the blue one and
the pink one. And the overall motion of our
orange object was dependent on the motion of the blue and the pink objects. So, that’s one reason we may have
to find the total momentum of more than one object combined.

Anyway, so in this case, we found
that the total momentum of two objects combined, the blue ball and the pink ball, as
two kilograms meters per second. However, let’s now imagine the
following scenario. Let’s imagine that the blue ball
stays exactly as it is, a mass of one kilogram and a velocity of one meter per
second to the right. Let’s also imagine that the pink
ball has the same mass as before of one kilogram, but this time is moving toward the
left at one meter per second. This is where we need to be careful
about the vector nature of both velocity and momentum. Because now that we’ve got
velocities of two objects in opposite directions, one is moving to the right and the
other is moving to the left, we need to account for this by saying that one of these
velocities is positive and the other is negative.

So, for argument’s sake, let’s just
say that we choose that any velocity toward the right is positive and any velocity
toward the left is negative. This means that our blue object is
moving to the right with a speed of one meter per second and has a velocity of one
meter per second to the right and hence has a velocity of positive one meter per
second. Whereas our pink ball has a speed
of one meter per second as well, but a velocity of one meter per second to the left
or, in other words, a velocity of negative one meter per second. Therefore, when we go to find the
total momentum of both the blue and the pink balls combined. In this particular case, we find
that the momentum of the blue ball is equal to its mass, one kilogram, multiplied by
its velocity of one meter per second, since it’s positive, plus the momentum of the
pink ball. Which is its mass, one kilogram,
multiplied by its velocity, which is now negative one meter per second.

And so, we find that, for the blue
ball, we’ve got one kilogram multiplied by one meter per second, which is simply one
kilogram meter per second. And for the pink ball, we’ve got
one kilogram multiplied by negative one meter per second, which is simply negative
one kilogram meter per second. And so, in order to find the total
momentum, we’re adding one kilogram meter per second to negative one kilogram meter
per second. These quantities add together to
give zero. And so, we find that the total
momentum of the orange object consisting of the blue ball and the pink ball is
actually zero kilogram meters per second, which actually makes some sense when we
think about it.

In this particular instance, we’ve
got a blue ball which has a momentum of one kilogram meter per second to the
right. And that momentum, in this case, is
perfectly canceled out by the momentum of the pink object, which is one kilogram
meter per second to the left or, equivalently, negative one kilogram meter per
second to the right. Since we said that anything toward
the right is positive and anything toward the left is negative. And because the momenta of the two
objects cancel out, the momentum of the system, as we call it, which consists of the
blue and the pink ball, is zero kilogram meters per second. So, we found a special case where
the total momentum of our system, the orange object, is zero because the momenta of
its components cancel each other out. And this is where the
directionality of both velocity and momentum are really important. In other words, the fact that they
are vector quantities is really important.

So, now that we’ve understood a bit
about what momentum is and how we can apply it to certain objects, let’s take a look
at an example question.

A cat has a mass of three
kilograms. The cat moves four meters in a
straight line in a time of two seconds. What is the momentum of the
cat?

Okay, so in this question, we’ve
been told that we’ve got a cat which has a mass, which we will call 𝑚, of three
kilograms. We’ve also been told that the cat
moves four meters in a straight line in a time of two seconds. So, let’s say that the cat starts
here and ends up here. And we can say that the distance
moved by the cat is four meters, and it does this in a time of two seconds. Based on this information, we’ve
been asked to find the momentum, which we will call 𝑝, of the cat. So, let’s first start by recalling
that the momentum of an object 𝑝 is defined as the mass of that object multiplied
by its velocity. So, if we want to find the momentum
𝑝 of our cat, we need to know its mass and its velocity. Well, we already know its mass. We know that it’s three
kilograms.

However, we haven’t been given its
velocity in this question. What we’ve been given, instead, is
enough information to calculate this cat’s velocity. Let’s recall that the velocity of
an object is defined as the displacement 𝑠 of the object divided by the time taken
for the object to travel that displacement. Now, the displacement 𝑠 of an
object is simply the distance between its start point and its finish point in a
straight line. And luckily, we’ve been told that
the cat moves in a straight line. Therefore, in this particular case,
the cat’s displacement is four meters. And the time taken to move that
displacement 𝑡 is two seconds. Hence, we can say, first of all,
that the velocity of our cat is equal to the displacement, which is four meters,
divided by the time taken to travel that displacement, which is two seconds. And this gives us a numerical value
of four divided by two and a unit of meters divided by seconds or meters per
second.

And so, we find that the velocity
of our cat is two meters per second. And because velocity is a vector
quantity — which, in other words, means that it has magnitude and direction — we can
say that the velocity is in the same direction as the cat’s displacement, which is
also a vector quantity. In this case, we’ve drawn it as
moving toward the right. However, we haven’t actually been
given this information in the question. We just sort of assumed that the
cat was moving toward the right when we drew our diagram. So here, we don’t need to worry too
much about the direction in which the cat is moving. And therefore, we don’t need to
worry about the direction of the cat’s velocity. All we care about is that the cat’s
velocity has a magnitude of two meters per second.

This means that we can now
calculate the cat’s momentum 𝑝, which happens to be the mass of the cat — which we
know is three kilograms from the question — multiplied by its velocity — which we’ve
just calculated to be two meters per second in the line above. And so, we find that the momentum
𝑝 of the cat has a numerical value of three times two, which is six, and a unit of
kilogram times meters per second, which is kilogram meters per second. Hence, our answer is that the cat
has a momentum of six kilogram meters per second. And once again, even though
momentum is a vector quantity — that is, it has magnitude and direction — because we
haven’t explicitly been told the direction in which the cat is actually moving, we
don’t need to give the direction in our answer. We can simply say that our cat has
a momentum of six kilogram meters per second.

So, now that we’ve had a look at an
example question, let’s summarize what we’ve talked about in this lesson. Starting off, we saw that the
momentum, 𝑝, of an object with a mass 𝑚 and a velocity 𝑣 is given by 𝑝 is equal
to 𝑚𝑣. Secondly, we saw that momentum is a
vector quantity. This means that it has magnitude,
or size, and direction. And finally, we also saw that
momentum has units of kilograms meters per second in SI base units, but can also be
written in grams meters per second or kilograms kilometers per hour and other units
of mass multiplied by velocity.