### Video Transcript

Two hockey players of unequal
masses collide with each other head-on, each moving at a speed of 15 meters per
second. After the collision, the hockey
players move in the same direction as each other, each at the speed of 5.0 meters
per second. How many times greater is the mass
of the more massive hockey player than the mass of the other hockey player?

If we call the mass of the larger
player capital π and the mass of the smaller player lowercase π, we want to solve
for the ratio capital π over lowercase π. Weβll start on our solution by
drawing a diagram. When we sketch out both the initial
and final conditions, that is, before and after the collision, we see that,
initially, the larger and the smaller player approach one another at the same speed
weβve called π£ sub π of 15 meters per second.

After a head-on collision, both
players move in the same direction at the same speed. Weβve called it π£ sub π of 5.0
meters per second. The direction of π£ sub π is the
same as the original direction of the larger player. Based on all this, we want to solve
for how many times more massive the larger player is than the small player.

Since this example involves a
collision, the first thing we can ask ourselves is whether the collision is elastic
or inelastic. Since the two players effectively
stick together after the collision has occurred, we know that this is an inelastic
collision. In a collision of this type, we
know that the initial momentum involved is equal to the final momentum of the
system. We canβt say, however, that kinetic
energy is conserved. Thatβs only true for elastic
collisions.

Knowing that momentum is a vector
with both magnitude and direction, letβs write out the initial and final momentum of
our two-hockey-player system. Recalling that an objectβs momentum
is equal to its mass times its velocity, if we decide that motion to the right is
motion in the positive direction, we can write that the initial momentum of our
system is equal to the mass of the larger player times his speed, π£ sub π, minus
the mass of the smaller player times his speed, π£ sub π. The reason for the minus sign is
because our second player, the smaller one, is moving in the negative direction.

We can write the final momentum of
our system as capital π plus lowercase π, since the players move together,
multiplied by the final speed, π£ sub π, which is positive because itβs to the
right. Applying the principle of
conservation of momentum, we can write that the quantity capital π minus lowercase
π times π£ sub π is equal to the quantity capital π plus lowercase π times π£
sub π.

If we rewrite this equation so that
instead of grouping it by speed we group by the mass of the two different players,
we see that the mass of the larger player times the difference in the initial and
final speeds equals the mass of the smaller player times the sum of these two
speeds. If we cross-multiply to get the
mass ratio, the larger mass over the smaller mass, we see this is equal to the sum
of the initial and final speeds divided by their difference.

Since we know both π£ sub π and π£
sub π, we can plug in and solve for the ratio of the masses of the players. Plugging these values in, we see
that the units of meters per second cancel out and that our numerator is equal to 20
and our denominator equal to 10. When we calculate this ratio then,
we find itβs 2.0. Thatβs how many times more massive
the larger player is compared to the smaller.