Video: Momentum and Force

An average force of 12.5 N is applied by moving air to a bicycle and its rider, of total mass 80 kg, while it is windy that day. The force is applied throughout a time interval of 0.8 s as the gust of wind is very brief. The wind blows from directly behind the bicycle. What is the total momentum change?

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Video Transcript

An average force of 12.5 newtons is applied by moving air to a bicycle and its rider, of total mass 80 kilograms, while it is windy that day. The force is applied throughout a time interval of 0.8 seconds as the gust of wind is very brief. The wind blows from directly behind the bicycle. What is the total momentum change?

This is a typical long question with lots of information to pick out. Let’s start by underlining some of it. So we know that we’ve got a force, an average force of 12.5 newtons, that is applied by moving air to a bicycle and its rider. The total mass of the bicycle and its rider is 80 kilograms. And we know that it’s windy that day. Now I won’t underline that because even though it’s important to know the weather, it’s probably not that important to helping us solve the problem. Anyway, so the force is applied throughout a time interval of 0.8 seconds because we know the gust of wind is very brief. Again, not the most important thing to know. The wind blows from directly behind the bicycle. Now that is an important piece of information. So yes, the directionality of the weather is an important piece of information, whereas the weather itself, who cares. That’s because the direction of this wind is going to directly affect the momentum of the bicycle and the rider. If the wind is coming from ahead of the bicycle, then it’s going to slow down the rider. Whereas, in this case, if it’s coming from directly behind, that is going to help the rider speed up.

So we’ve been asked to calculate the total momentum change. We can start by labelling quantities. The average force applied by the wind is 12.5 newtons. And we’ll label this 𝐹. The total mass of the rider and the bicycle together is 80 kilograms. We’ll label this lowercase 𝑚. The very short time interval for which the forces applied on the bicycle list or rider, should I say, is 0.8 seconds. We’ll label this Δ𝑡 because Δ means very small or change in. In this case, we’ve got a very small period of time. But also, we can think about it as the change in time between when the force starts supplying and finishes applying. This might seem a little bit complicated and convoluted. But we’ll see why we use Δ𝑡 in a second.

Now what we’ve been asked to calculate is the total change in momentum Δ𝑝. 𝑝 stands for momentum and Δ stands for change in. Therefore, Δ𝑝 is change in momentum. So here’s our bicycle rider. And I would like you to pretend that it looks something like a bicycle rider at least. And let’s say he’s moving in this direction, to the right. He’s moving with some speed. So he’s bound to have some momentum. But what that momentum is doesn’t matter because what we’re looking for in this question is the change in momentum. So it doesn’t matter what he started out with. All that matters is how much it changes by.

Now, here comes the gust of wind. And it helps him out by pushing him to the right as well. We know it pushes him in the same direction because we’ve been told that the wind blows from directly behind the bicycle. If it’s blowing from directly behind, then it’ll be blowing in the direction that he is going. Therefore, this wind will help him speed up. It’s providing a force in the same direction as he’s travelling. So it’s going to help him out. So if he’s speeding up, then we know that his momentum is going to increase. Without doing any calculations, we know that the change in momentum of the rider and the bike is going to be positive.

Now let’s actually try and work out what this change in momentum is. To do this, we need to recall impulses. An impulse is when a force is applied to an object for a short period of time. This is exactly what we have here. The impulse is defined as the force multiplied by the short period of time for which it applies. That’s 𝐹Δ𝑡. But we can also recall that an impulse is the same as the change in momentum. That’s Δ𝑝. So we have 𝐹Δ𝑡 is equal to Δ𝑝. And these quantities are very much the same as what we’ve been given in the question. So all we need to do is substitute. Δ𝑝 is equal to 𝐹Δ𝑡 which looks like 12.5 times 0.8. And putting that into our calculator, we get 10. But that’s not our final answer. 10 what? What is the change in momentum? We need the units. If it’s a change in momentum, then the units must be those of momentum. Momentum is defined as mass multiplied by velocity. So its units must be those of mass, kilograms, multiplied by that of velocity, meters per second. And so we can stick that in front of our 10. Finally, our answer is that the total change in momentum is 10 kilograms meters per second.

By the way, we didn’t use the mass. We’ve gone through this entire calculation without even mentioning the fact that the rider and the bike have a total mass of 80 kilograms. But in this case, we didn’t need the mass. And sometimes, you will come across questions where you’ll be given quantities that you don’t need. In that case, it’s very important to think about what we’ve been given. And whether or not it’s actually necessary for our calculation. Of course, the mass is very useful if we wanted to calculate the change in velocity of the rider because we’ve got the change in momentum and we’ve got the mass. So we’d have to divide the momentum by the mass to find the velocity. However, since we’ve not been asked to do anything like that, we don’t need to know the mass. So this question goes to show that sometimes we don’t need everything given to us in a question.

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