Question Video: Mass Ratios of Objects Colliding Inelastically

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 𝑀 and the mass of the smaller player lowercase 𝑚, we want to solve for the ratio capital 𝑀 over lowercase 𝑚. We’ll 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 we’ve called 𝑣 sub 𝑖 of 15 meters per second.

After a head-on collision, both players move in the same direction at the same speed. We’ve called it 𝑣 sub 𝑓 of 5.0 meters per second. The direction of 𝑣 sub 𝑓 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 can’t say, however, that kinetic energy is conserved. That’s only true for elastic collisions.

Knowing that momentum is a vector with both magnitude and direction, let’s write out the initial and final momentum of our two-hockey-player system. Recalling that an object’s 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, 𝑣 sub 𝑖, minus the mass of the smaller player times his speed, 𝑣 sub 𝑖. 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 𝑀 plus lowercase 𝑚, since the players move together, multiplied by the final speed, 𝑣 sub 𝑓, which is positive because it’s to the right. Applying the principle of conservation of momentum, we can write that the quantity capital 𝑀 minus lowercase 𝑚 times 𝑣 sub 𝑖 is equal to the quantity capital 𝑀 plus lowercase 𝑚 times 𝑣 sub 𝑓.

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 𝑣 sub 𝑖 and 𝑣 sub 𝑓, 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 it’s 2.0. That’s how many times more massive the larger player is compared to the smaller.