Question Video: Momentum and Force | Nagwa Question Video: Momentum and Force | Nagwa

Question Video: Momentum and Force Physics • First Year of Secondary School

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An object with an initial momentum of 20 kg⋅m/s is acted on by an average force of 2.5 N and comes to rest. For how long does the force act on the object?

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Video Transcript

An object with an initial momentum of 20 kilograms meters per second is acted on by an average force of 2.5 newtons and comes to rest. For how long does the force act on the object?

Okay, so in this question, we’re starting out with an object that has an initial momentum let’s say to the right arbitrarily. And we’ve been told that this initial momentum is 20 kilograms meters per second. So we can label that on our diagram. Then we’ve been told that the object is acted on by an average force of 2.5 newtons. And this results in the object coming to rest. In other words, the object, which was initially moving to the right, is going to eventually come to a point where it stops. It comes to rest. But then if the object is at rest, then the velocity of that object at that point is zero. And we can recall that the momentum of an object is given by multiplying the mass of that object by the velocity of the object.

Now even though we don’t know the mass of our object, we can work out the momentum of the object at the end point because we’ve been told that the object has come to rest. So its velocity is zero. And hence, its momentum is going to be zero. So we can say that, at the final position, the object’s momentum is zero kilograms meters per second.

Now this change in momentum is caused by a force of 2.5 newtons. And the only way that the momentum of the object can decrease, in other words the object decelerates because it loses velocity, is if the force was acting in the opposite direction to its initial motion. So we can say that the force is 2.5 newtons and is towards the left. Now, of course, the direction left is arbitrary. But the important thing is that it’s in the opposite direction to its initial momentum.

We’ve been asked to find for how long the force is acting on the object, in other words the time interval between here and here. Let’s call this time interval Δ𝑡. And let’s also arbitrarily decide that the object moving towards the right is moving in the positive direction. And therefore, anything towards the left is the negative direction.

Then we can recall that the force exerted on our object can be found by evaluating the change in momentum of the object, Δ𝑝, divided by the amount of time or the time interval over which that change in momentum occurs. Now in this case, we’ve already been given the force exerted on the object. It’s 2.5 newtons towards the left. Or in other words, it’s negative 2.5 newtons, because, remember, we said anything towards the left is negative. And as well as this, we’ve been given the initial and final momentum of the object. So we can work out the change in momentum caused by the force being exerted on the object.

We can recall that the change in momentum is given as the final momentum, which we’ll call 𝑝 subscript 𝑓, minus the initial momentum, which we’ll call 𝑝 subscript 𝑖. But then we see that the final momentum is zero kilograms meters per second, and the initial momentum is 20 kilograms meters per second. And so we subtract 20 kilograms meters per second from zero kilograms meters per second.

Evaluating the right-hand side of this equation, we find that this becomes negative 20 kilograms meters per second. And this is actually equal to Δ𝑝 or the change in momentum of the object. So we now know the value of Δ𝑝 and we know the value of 𝐹. All we need to do is to find the value of Δ𝑡. To do this, we can multiply both sides of the equation by Δ𝑡 over 𝐹. This way, on the left-hand side, the 𝐹 cancels. And on the right-hand side, the Δ𝑡s cancel. Hence, what we’re left with is that the time interval, Δ𝑡, is equal to Δ𝑝, the change in momentum, divided by 𝐹, the force.

We can therefore say that Δ𝑡 is equal to Δ𝑝, negative 20 kilograms meters per second, divided by 𝐹, negative 2.5 newtons. And this is why we take into account the directionality of the force and of the momenta. Because in order for the momentum of the object to go from 20 kilograms meters per second in this direction to be zero, that’s equivalent to a change in momentum in this direction. And that’s why we got a negative value for the change in momentum.

But, more importantly, this negative sign cancels with this one. And so what we’re going to be left with is a positive value for Δ𝑡. In other words, the time interval over which this change in momentum occurs is going to be positive, just as we need it to be.

As well as this, we can realize that we’ve got base units both in the numerator, kilograms meters per second being the base unit of momentum, and the denominator, newtons being the base unit of force. Therefore, our final answer for Δ𝑡 is going to be in its own base unit, which is seconds.

So all that’s left to do is to evaluate 20 divided by 2.5. And we find that that gives us a value of Δ𝑡 as eight seconds. Therefore, we found the final answer to our question. The 2.5-newton force acts on the object for a time interval of eight seconds.

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