### Video Transcript

Two cars of equal masses approach a
frictionless perpendicular intersection. Car A travels north at 22 meters
per second and car B travels east. The cars collide and stick
together, traveling at 17 degrees north of east. What was the initial speed of car
B?

We can label car Bβs initial speed
π£ sub B and begin our solution by drawing a sketch of the situation. In this scenario, initially, we
have two cars approaching one another at an intersection. Car A is traveling north at a speed
given as 22 meters per second, and car B is traveling to the east at an unknown
speed. At the center of the intersection,
the cars collide and, finally, they move together as one mass in a direction given
as 17 degrees north of east. Based on this information, we want
to solve for the initial speed of car B.

Weβll use the principle that the
total initial momentum in a system is equal to the total final momentum in that
system. In our scenario, initially and
finally, we have momenta pointing in two different directions. In the northern direction, we have
π sub A times π£ sub A. And in the eastern direction,
orthogonal to the northβsouth axis, we have π sub B times π£ sub B.

If we then consider the final
momentum in our system, again both in the north and eastern directions separately,
to the north we have our combined mass π sub A plus π sub B, since the cars have
stuck together, times their final speed, whatever that is, weβll call it π£ sub π,
multiplied by the sine of the angle π³. And to the east, we have the same
expression except now we use the cosine of that angle to point in the direction east
of north.

Applying our conservation
principle, we can write in the northern direction that π sub Aπ£ sub A equals π
sub A plus π sub B times π£ sub π times the sine of π³. And separately, in the eastern
direction, π sub Bπ£ sub B is equal to the sum of the two masses times their final
velocity times the cosine of π³. We recall that we want to solve for
π£ sub B, the initial speed of car B. And very importantly, weβre told in
the problem statement that the masses of the two cars are equal. π sub A is equal to π sub B. So, if we replace every π sub A
and π sub B we see in our two equations simply with the mass π, we see then that
that mass π cancels out from both sides of both equations.

Looking at the equation in the
northern direction, we can use the fact that π£ sub A and π³ are given information
to solve for π£ sub π, the final speed of the combined car mass. π£ sub π is equal to π£ sub A over
two times the sine of π³, or 22 meters per second divided by two times the sine of
17 degrees. If we substitute this expression
for π£ sub π into π£ sub π in our equation for π£ sub B and enter our value of 17
degrees for π³. Then we see our expression
simplifies a bit and that factors of two cancel and the cosine of one angle divided
by the sine of the same angle is equal to the cotangent of that angle.

So, entering these values on our
calculator, we find a result for π£ sub B which, to two significant figures, is 72
meters per second. Thatβs the initial speed of car B
in this scenario.