### Video Transcript

A child sleds down a hill and collides at 5.6 meters per second into a stationary sled that is identical to his. The child is launched forward at the same speed, leaving behind the two sleds that lock together and slide forward more slowly. What is the speed of the two sleds after this collision?

Weβre told in this problem statement that the child sleds down a hill and at 5.6 meters per second runs into a second identical stationery sled. Weβll call that initial speed π£ sub π. We want to solve for the speed of the two locked together sleds after the collision occurs. Weβll call that π£ sub π.

Letβs draw a sketch of this scenario to orient ourselves to the problem. In this scenario, we have a child on a sled sliding down a hill, which weβve shown here at different snapshots in time. The child on the sled approaches another stationery sled and when colliding with that sled then continues to move on ahead at the same initial speed π£ sub π. The two sleds lock together and move ahead in the same direction, but at a slower speed; thatβs π£ sub π that we want to solve for.

In this exercise, weβll rely on the fact that momentum is conserved throughout this interaction. That conservation means that the initial mass of the system, π sub π, times the initial speed of that mass is equal to the final mass of the system, π sub π, times the final speed of the system, π£ sub π.

For the mass of the child, letβs use a shorthand notation π sub π, and weβll call π sub π the mass of the sled. So before the two sleds collide, the initial momentum of our system can be written as the mass of the child plus the mass of the sled, since they ride together, times π£ sub π, the initial speed of both the child and the sled.

After the collision, the systemβs momentum can be written the mass of the child times the initial speed, because remember the child moves ahead at that same initial speed, plus two times the mass of the sled, because now there are two sleds, times the final speed of that two-sled system, π£ sub π. And itβs π£ sub π we want to solve for.

Letβs do this; letβs multiply through on the left-hand side so that, instead of having a π£ sub π term factored, we have two separate terms. Now having done that, we see something interesting: both sides of the equation have the term π sub π π£ sub π, the mass of the child times the initial speed.

Therefore, we can cancel that term out. And having done that, we see thereβs even more cancelation that can happen. This simplified version of that equation has π sub π , the mass of the sled on both sides. Therefore, we cancel that term out.

Now our equation for momentum conservation simply says that the initial speed, π£ sub π, is equal to two times the final speed, π£ sub π. In other words, π£ sub π is equal to π£ sub π divided by two.

When we plug in the given value for π£ sub π, 5.6 meters per second, and calculate this value, we find that π£ sub π is equal to 2.8 meters per second. Thatβs the speed of the two sleds as they move together after the collision.