### Video Transcript

An object increases its momentum by 10 kilogram meters per second in a time of five seconds. What average force was applied to the object in that time?

Okay, so in this question, we’ve been told that we’ve got an object which increases its momentum by 10 kilograms meters per second. So in other words, what we can say is that the change in momentum of the object, which we’ll call Δ𝑝, is equal to 10 kilograms meters per second, because if the objective is to increase its momentum, then the momentum has to change. And we know that it’s an increase in momentum because we’ve got a positive 10 kilograms metres per second, not a negative.

Now we also know that the object is increasing its momentum in a time interval of five seconds. So what we’ll say is that Δ𝑡, which is what we call the time interval, is equal to five seconds.

Now why are we saying a time interval of five seconds rather than a time of five seconds? Well, it’s because let’s say that this is our object and it has a certain initial momentum, which we’ll call 𝑝 one. It’s at this point that the momentum of the object starts to increase. Then a few moments later, it has a new momentum, which we’ll call 𝑝 two.

Now it was increasing its momentum in a time interval which we’ll call Δ𝑡. And generally, these time intervals in questions like this tend to be very short. That’s why we use Δ because Δ often represents a small change in a quantity and Δ𝑡 represents a small change in time, or specifically the small time interval over which the object changes its momentum.

Now in this question, we’ve been asked to find the force that was applied to the object in that time. To do this, we need to recall that an impulse is defined as the force applied on an object multiplied by the time interval for which that force acts, which we’ll call Δ𝑡. Surprise, surprise!

Now looking back at the diagram, how is this object changing its momentum from 𝑝 one to 𝑝 two? Well, the way it’s doing this is because there’s a force exerted on that object, which we’ll call 𝐹. But that force is only being exerted for the time interval Δ𝑡. And it’s within that time interval due to the force 𝐹 being exerted that the object’s momentum changes from 𝑝 one to 𝑝 two.

So we can also therefore recall that 𝐹Δ𝑡, or the impulse, is equal to the change in momentum of the object, Δ𝑝, at which point we have an expression for 𝐹 and Δ𝑡 in terms of Δ𝑝. And we can actually rearrange this equation by dividing both sides of the equation by Δ𝑡 to give us an expression for 𝐹, because the Δ𝑡s on the left-hand side cancel out. This leaves us with 𝐹 is equal to Δ𝑝 divided by Δ𝑡. And at this point, we can substitute in the values that we’ve been given here.

So we say that 𝐹 is equal to Δ𝑝, which is 10 kilogram meters per second, divided by Δ𝑡, which is five seconds. And when we evaluate the fraction on the right-hand side of the equation, we find that the average force exerted on the object during that time period of five seconds was two newtons. And this is our final answer.