### Video Transcript

A snowball rolls down a snowy hill side, and as more snow sticks to the snowball, it increases in mass. The graph shows how the snowball’s mass and velocity change. How much does the momentum of the snowball increase from when its velocity is five meters per second to when its velocity is 10 meters per second? By how much would the momentum of the snowball have increased when its velocity changed from five meters per second to 10 meters per second if no snow stuck to it as it rolled?

Okay, so this question is about a snowball that’s rolling down a hill. We’re told that this snowball picks up more and more snow as it rolls so that it increases in mass. We’re given a graph of the snowball’s velocity and its mass as it rolls down the hill. We can see from this graph that as the velocity of the snowball increases, its mass also increases. So what’s happening here is that the snowball is getting faster and faster as it rolls further down the hill. And we know that its mass is also increasing because more and more snow sticks to it on the way.

With that in mind, let’s turn our attention to the first part of the question. How much does the momentum of the snowball increase from when its velocity is five meters per second to when its velocity is 10 meters per second? To answer this, we need to recall that the momentum of an object, which we’ve labeled as 𝑃, is equal to the object’s mass 𝑚 multiplied by its velocity 𝑣. We’re being asked to find the increase in the momentum of the snowball between two points. The first of these two points is when its velocity is five meters per second, and we’ll label this with a subscript 𝑖 because this is the initial velocity we’re considering.

The second point is when the velocity is 10 meters per second. And we’ve labeled this with a subscript 𝑓 because this is the final velocity of the snowball during the interval we’re asked to consider. The change in the snowball’s momentum between these two points, which we’ve labeled as Δ𝑃, is equal to the final momentum 𝑃 subscript 𝑓, which is the momentum of the snowball when its velocity is 𝑣 subscript 𝑓, minus the initial momentum 𝑃 subscript 𝑖, that’s the momentum when its velocity is 𝑣 subscript 𝑖. So what we need to do is to find the values of 𝑃 subscript 𝑓 and 𝑃 subscript 𝑖 so that we can use these values to calculate the change in momentum Δ𝑃.

Since momentum is equal to mass multiplied by velocity, then we need to find the mass of the snowball at each of these two values of velocity. We can find these values using the graph that we’re given. At a velocity of 𝑣 subscript 𝑖 equal to five meters per second, then by tracing upward from five meters per second on the velocity axis, we see that the mass of the snowball at this velocity is two kilograms. We’ll label this mass as 𝑚 subscript 𝑖. Now let’s do the same thing for the velocity 𝑣 subscript 𝑓, which is 10 meters per second. Tracing upward from 10 meters per second on the velocity axis and then tracing across until we get to the vertical mass axis, we find that at a velocity of 10 meters per second, the snowball has a mass of 3.5 kilograms. Let’s label this as 𝑚 subscript 𝑓.

We are now ready to calculate the initial momentum and final momentum of the snowball. To calculate the initial momentum 𝑃 subscript 𝑖, we need to substitute our values for the initial velocity 𝑣 subscript 𝑖 and the initial mass 𝑚 subscript 𝑖 into this equation. When we do this, we find that 𝑃 subscript 𝑖 is equal to two kilograms multiplied by five meters per second, which works out as 10 kilogram-meters per second. To get the final momentum 𝑃 subscript 𝑓, we instead use our values for the final velocity 𝑣 subscript 𝑓 and the final mass 𝑚 subscript 𝑓. We get that 𝑃 subscript 𝑓 is equal to 3.5 kilograms multiplied by 10 meters per second. This works out as a value of 35 kilogram-meters per second.

We can now take our values for the final momentum 𝑃 subscript 𝑓 and the initial momentum 𝑃 subscript 𝑖 and substitute them into this equation in order to calculate the change in the snowball’s momentum Δ𝑃. Let’s clear ourselves some space to do this. When we substitute our values for 𝑃 subscript 𝑖 and 𝑃 subscript 𝑓 into this equation, we get that the change in momentum Δ𝑃 is equal to the final momentum 35 kilogram-meters per second minus the initial momentum of 10 kilogram-meters per second. This works out as 25 kilogram-meters per second. Since the value for the change in the snowball’s momentum is positive, this means that its momentum has increased.

If we think about the definition of momentum as mass multiplied by velocity, then this makes sense because we can see that as the snowball rolls further down the hill, both its velocity and its mass increase. So our answer to this first part of the question is that the momentum of the snowball increases by 25 kilogram-meters per second.

Let’s now clear ourselves some space so that we can look at the second part of the question.

By how much would the momentum of the snowball have increased when its velocity changed from five meters per second to 10 meters per second if no snow stuck to it as it rolled?

Now the main text of the question told us that it’s because more snow sticks to the snowball as it rolls, that the snowball increases in mass. In the second bit of the question, we’re asked to consider the situation where no snow sticks to the snowball as it rolls. That means that the mass of the snowball isn’t going to change.

We know from our graph that at a velocity of five meters per second, the snowball has a mass of two kilograms. And if the mass doesn’t increase as it rolls, then rather than the blue line shown on the graph, we would actually have this orange line, which shows that as the velocity increases, the mass remains constant at two kilograms. As with the first part of the question, the two equations we’re going to need are that momentum is equal to mass multiplied by velocity and that the change in momentum Δ𝑃 is equal to the final momentum 𝑃 subscript 𝑓 minus the initial momentum 𝑃 subscript 𝑖. Like before, we’re considering an initial velocity 𝑣 subscript 𝑖 of five meters per second and a final velocity 𝑣 subscript 𝑓 of 10 meters per second.

However, this time we’ve just got the one value of mass, which we’ve labeled as 𝑚, because we know that the mass of the snowball remains constant at two kilograms. The initial momentum of the snowball 𝑃 subscript 𝑖 is equal to the mass 𝑚 multiplied by the initial velocity 𝑣 subscript 𝑖. Substituting in our values for 𝑚 and 𝑣 subscript 𝑖, we have that 𝑃 subscript 𝑖 is equal to two kilograms multiplied by five meters per second, which is 10 kilogram-meters per second. In the same way, the final momentum 𝑃 subscript 𝑓 is equal to the mass 𝑚 multiplied by the final velocity 𝑣 subscript 𝑓. Using our values for 𝑚 and 𝑣 subscript 𝑓, we have that 𝑃 subscript 𝑓 is equal to two kilograms multiplied by 10 meters per second, which is 20 kilogram-meters per second.

Let’s clear ourselves some space so that we can take our values for the initial and final momentum and substitute them into this equation. Substituting in these values, we find that the change in the snowball’s momentum Δ𝑃 is equal to its final momentum of 20 kilogram-meters per second minus its initial momentum of 10 kilogram-meters per second. This works out as a change of 10 kilogram-meters per second. And again, since this value is positive, it represents an increase in momentum. So our answer to the second part of the question is that if snow didn’t stick to the snowball as it rolled, then its momentum would increase by 10 kilogram-meters per second.