### Video Transcript

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.