### Video Transcript

In this video, we’re going to learn
about linear momentum. We’ll learn what this term means,
how to calculate it, and how it relates to the net force on an object.

To get started, let’s contrast two
different scenarios involving object motion. In the first situation, say that
you and a friend are playing table tennis. You hit the table tennis ball back
and forth and it quickly and easily changes direction. In the second situation, imagine
that you’re in your car waiting at a railroad crossing, while a train passes by. The train is made of dozens, if not
hundreds, of railroad cars linked together and moves along at a steady speed.

If we compare the motion of the
table tennis ball in our first example to the motion of the train, we understand
that there is an essential difference between these two. While it is easy to change the
direction and speed of the table tennis ball, doing something similar with the train
will require quite a lot of force. To understand better the basic
difference between these two objects in motion, it will be helpful to learn about
linear momentum.

Linear momentum, or momentum for
short, describes an object’s capacity to resist changes to its motion. An object’s momentum is equal to
its mass multiplied by its velocity. This symbol we usually use to
represent momentum is a lowercase 𝑝. And notice that momentum is a
vector quantity. Momentum is equal to an object’s
mass multiplied by its velocity. And the direction of an object’s
momentum is the same as the direction of its velocity.

This mathematical relationship
helps us understand the difference between the motion of the table tennis ball and
the motion of the train. At any one moment, the magnitude of
the velocity of the table tennis ball and the train’s velocity may have been roughly
equal. But we know that the mass of the
table tennis ball is much, much, much smaller than the mass of the train. So, the train had much more
momentum than the table tennis ball and that’s why it’s motion was much harder to
change. The more linear momentum an object
has, the more force and/or the more time it takes to change that object’s
motion.

Now that we have a mathematical
relationship for momentum, let’s consider what will happen if we take the time
derivative of momentum. 𝑑𝑝 𝑑𝑡 is equal to the time
derivative of 𝑚 times 𝑣. And if we apply this operator to
the expression 𝑚 times 𝑣 using the chain rule, we find it’s equal to mass times
the time rate of change of velocity plus velocity times the time rate of change of
mass.

Considering the time rate of change
of mass, in a relativistic scenario, the mass of our object may change in time. But outside of that, the mass
typically is constant in time so that its time rate of change is zero. When that term is zero, it means
that 𝑑𝑝 𝑑𝑡 is equal to the object’s mass times its change in velocity with
respect to time. We can recall that 𝑑𝑣 𝑑𝑡 is
equal to an object’s acceleration.

Writing the expression this way, it
may remind us of Newton’s second law of motion, which says that an object’s mass
times its acceleration is equal to the net force acting on the object. So, an object’s time rate of change
in momentum is equal to the net force acting on it. And if we multiply both sides of
this equation by 𝑑𝑡, we find that 𝑑𝑝, the change in momentum, is equal to net
force times the change in time, or in a notation that may be more familiar, Δ𝑝 is
equal to 𝐹 net times Δ𝑡.

Now, we not only understand
momentum as a product of mass and velocity, but we see that an object’s change in
momentum is equal to the net force acting on it times the time over which that force
acts. Just as a side note, this
relationship as it’s written assumes that our net force is constant in time. Now that we know a bit about
momentum and how it can change, let’s get some practice with these ideas through a
couple of examples.

A hockey puck of mass 150 grams is
sliding due east on a frictionless table with a speed of 10 meters per second. Suddenly, a constant force of
magnitude 5.0 newtons and direction due north is applied to the puck for 1.5
seconds. Find the northward component of the
hockey puck’s momentum at the end of the 1.5-second interval. Find the eastward component of the
hockey puck’s momentum at the end of the 1.5-second interval.

We can refer to the northward
component of the puck’s momentum and the eastward component of the puck’s momentum
at the end of this interval as 𝑝 sub 𝑛 and 𝑝 sub 𝑒, respectively. So, we want to solve for 𝑝 sub 𝑛
and 𝑝 sub 𝑒, the northward and eastward components of the puck’s momentum , after
a time, we’ve called Δ𝑡, of 1.5 seconds has elapsed.

Before this time interval begins,
we’re told that the puck is moving to the east with a speed of 10 meters per
second. At time 𝑡 equals zero, a force 𝐹
is applied to the puck and lasts 1.5 seconds. Regarding momentum, there are two
relationships we can recall. The first is that an object’s
momentum is equal to its mass times its velocity. And the second is that the change
in an object’s momentum is equal to the net force acting on it times the time over
which that force acts.

Considering the northward component
of our puck’s momentum, we know that at the outset, it has no momentum in that
direction because its velocity to the north is zero. But after 1.5 seconds of the force
𝐹 being applied pointing to the north, referring to our second relationship for
momentum, we can write that 𝑝 sub 𝑛 is equal to the magnitude of 𝐹 multiplied by
Δ𝑡.

This change in momentum to the
north is equal to the final northward momentum component because the original
component was zero. Plugging in for these two values,
we find that 𝑝 sub 𝑛 is equal to 7.5 kilograms meters per second. That’s the northward component of
the puck’s momentum after a time Δ𝑡 has passed.

Next, we wanna solve for the
eastward component of the momentum after that same time interval. Unlike to the north, to the east
our puck does have an initial speed. And because the force applied to
the puck is perpendicular to that direction, it doesn’t affect the puck’s speed to
the east. This means the puck’s momentum to
the east is a constant value. It doesn’t change in time and is
always equal to the puck’s mass times 𝑣 sub 𝑖.

We’re told that the mass of the
puck is 150 grams. But when we enter it into our
expression, we write it in units of kilograms to be consistent with the units in the
rest of the expression. Calculating this product, we find
that 𝑝 sub 𝑒 is 1.5 kilograms meters per second. That’s the puck’s eastward momentum
component after Δ𝑡 has passed or at any time.

Next, let’s look at an example
involving momentum and projectile motion.

A tennis ball of mass 150 grams is
thrown at an angle of 40 degrees above the horizontal and a speed of 18 meters per
second. What is the momentum of the ball
after it has been in flight for 0.36 seconds?

We can call the ball’s momentum
after it’s been in flight for this amount of time 𝑝 and begin on our solution by
drawing a diagram of this situation. Our sketch represents that we’ve
thrown this ball at an angle, we’ve called 𝜃, of 40 degrees above the horizontal at
an initial speed, we’ve called 𝑣 sub 𝑖, of 18 meters per second. After a time in flight, we’ve
called Δ𝑡, of 0.36 seconds, we want to know the momentum of the ball.

If we let upward motion be motion
in the positive 𝑦-direction and motion to the right be in the positive 𝑥 and if we
recall that momentum 𝑝 is equal to an object’s mass times its velocity, then we can
write that momentum 𝑝 is equal to our ball’s mass 𝑚 multiplied by the square root
of 𝑣 sub 𝑥 squared plus 𝑣 sub 𝑦 squared. Here, 𝑣 sub 𝑥 is the ball’s speed
in the 𝑥-direction at a time value of 0.36 seconds. And likewise, 𝑣 sub 𝑦 is the
ball’s vertical speed at that same time.

Since we’ve been told the ball’s
mass 𝑚, that means if we can solve for 𝑣 sub 𝑥 and 𝑣 sub 𝑦, we’ll be able to
solve for momentum 𝑝. Starting with 𝑣 sub 𝑥, when we
look at our diagram, we see that we can represent the speed of the ball in the
𝑥-direction in terms of 𝑣 sub 𝑖 and 𝜃. 𝑣 sub 𝑥 which, recall, is the
ball’s 𝑥-velocity component at time 𝑡 equals 0.36 seconds is equal to the ball’s
initial launch speed 𝑣 sub 𝑖 times the cosine of the angle 𝜃.

Plugging in for these values, when
we calculate 𝑣 sub 𝑥, we find it’s approximately 13.79 meters per second. This is the ball’s speed in the
𝑥-direction and is constant throughout the ball’s flight. Now that we know 𝑣 sub 𝑥, we move
on to solving for 𝑣 sub 𝑦, the ball’s vertical speed component at time 𝑡 equal
0.36 seconds.

Unlike the ball’s 𝑥-component
speed, its 𝑦-component speed is not constant in time, but constantly changes under
the influence of gravity. As a projectile, the ball’s motion
can be described by the kinematic equations. In particular, we recall the
relationship final speed is equal to initial speed plus acceleration times time. When we write this equation in
terms of our variables, we can write that 𝑣 sub 𝑦 is equal to the initial speed of
the ball in the 𝑦-direction minus the acceleration due to gravity 𝑔 times Δ𝑡.

We’ll treat gravitational
acceleration 𝑔 as exactly 9.8 meters per second squared. Looking again at our diagram, we
see that we can write the expression for 𝑣 sub 𝑦𝑖, the initial speed of the ball
in the vertical direction, as the product of 𝑣 sub 𝑖, the ball’s initial speed,
times the sin of the angle 𝜃. Since we’re given 𝑣 sub 𝑖, 𝜃,
and Δ𝑡, and 𝑔 is a constant, we’re ready to plug in and solve for 𝑣 sub 𝑦.

When we do and enter this
expression on our calculator, we find that the 𝑣 sub 𝑦 is approximately 8.042
meters per second. Now that we know 𝑣 sub 𝑥 and 𝑣
sub 𝑦, we’re ready to plug in and solve for momentum 𝑝. When we do plug in, we’re careful
to convert our mass into units of kilograms. And we find that 𝑝 is equal to 2.4
kilograms meters per second. That’s the overall momentum of this
ball 0.36 seconds after its been launched.

Let’s summarize what we’ve learned
about linear momentum. We’ve seen that an object’s
momentum describes its capacity to resist changes to its motion. The motion of our table tennis ball
was easy to change and it had a small momentum, while our train’s momentum was very
large indicating that its motion was difficult to change. We’ve also seen that an object’s
momentum is equal to its mass multiplied by its velocity.

Mathematically, we write that
momentum, symbolized by a lowercase 𝑝, is equal to 𝑚 times 𝑣. And we’ve noted that momentum is a
vector quantity with magnitude as well as direction. And finally, we consider that the
change in an object’s momentum Δ𝑝 is equal to the net force acting on that object
times the time Δ𝑡 over which that force acts. Linear momentum is a helpful tool
for understanding differences in object motion.