Video: Force as a Rate of Momentum Change

In this video, we will learn how to relate the rate of change of momentum of an object to the force acting on it, using the formula Δ𝑝 = 𝐹Δ𝑡.

15:38

Video Transcript

Newton’s second law of motion tells us that the force on an object is equal to the mass of that object multiplied by the acceleration experienced by that object. In other words, 𝐹 is equal to 𝑚𝑎. Now this is probably the most useful formulation of Newton’s second law of motion. But there is actually another way to state this famous law. And that is what we’ll be looking at in this video.

So what Newton’s second law actually says is that the force on an object is equal to the rate of change of the object’s momentum. In other words then, the force exerted on an object can be found by evaluating the change in that object’s momentum divided by the amount of time taken for that change in momentum to occur.

So say that we have an object here. This blob is our object. And it’s initially moving towards the right with a constant momentum 𝑝 one. Then the object experiences a force starting at this position over here. So a force 𝐹 is exerted on the object that causes the object to change momentum. And when it gets to this point, the object now has a momentum in the right direction of 𝑝 two.

If we want to calculate the force 𝐹, the force exerted on the object in order to change its momentum from 𝑝 one to 𝑝 two, then this force is equal to the change in the object’s momentum, which is given by finding 𝑝 two minus 𝑝 one, the final momentum minus the initial momentum. And we divide this by the amount of time taken to get from here to here.

Now we can say that the force 𝐹 was first exerted on the object at a time 𝑡 is equal to zero. And then the final time, the time at which the object has a momentum 𝑝 two, well let’s call that 𝑡 is equal to capital 𝑇. Well, in that case, the amount of time between when the force is first exerted and when it stops being exerted is capital 𝑇. So we divide by capital 𝑇. And in fact, more generally, we could just say that the force stops being exerted at a time lowercase 𝑡.

And so what we’ve done is to find the force exerted on the object that causes a change in momentum, Δ𝑝, where Δ𝑝 is 𝑝 two minus 𝑝 one. And we divide this by the amount of time taken for that change in momentum to occur.

Alternatively, we could also say that the force is exerted for the first time on the object at an arbitrary time, 𝑡 one. And it stops being exerted at another time, 𝑡 two. Well, in that case, the amount of time taken between 𝑡 one and 𝑡 two is given by 𝑡 two minus 𝑡 one. And so in the denominator, the time interval is 𝑡 two minus 𝑡 one. And another way to state this is Δ𝑡, just like 𝑝 two minus 𝑝 one is Δ𝑝.

So the point is, we might see this equation written as either Δ𝑝 divided by Δ𝑡 or even just Δ𝑝 divided by 𝑡 itself. In both cases, they’re referring to the same thing, the amount of time taken for that change in momentum to occur. And that’s what we need to remember most importantly.

Okay, but hold on a minute! Are we saying that this is Newton’s second law of motion? Haven’t we always been told that Newton’s second law of motion is 𝐹 is equal to 𝑚𝑎, the force is equal to the mass of the object multiplied by the acceleration?

Well, yes, so let’s take a look at where this 𝐹 is equal to 𝑚𝑎 actually comes from. To do this, we need to take a closer look at Δ𝑝. We just saw a few moments ago that the force exerted on our object is given by 𝑝 two minus 𝑝 one — that’s the same thing as Δ𝑝 — divided by 𝑡 two minus 𝑡 one — that’s Δ𝑡.

Now 𝑝 two and 𝑝 one refer to the final and initial momenta of the object, respectively. And at this point, we can recall that the momentum of an object is given by multiplying the mass of that object by the velocity with which it’s moving. So the final momentum of the object, 𝑝 two, can be written as 𝑚 two 𝑣 two. That’s the mass of the object at the final position multiplied by the velocity at the final position. And similarly, the initial momentum of the object can be written as 𝑚 one 𝑣 one, the initial mass of the object multiplied by the initial velocity.

Now if we assume that the object is moving such that the mass of the object is not changing over this entire period of time, then essentially what we get left with is that 𝑚 one is equal to 𝑚 two. And for simplicity, we can just call this 𝑚, the mass of the object. In that case, we can get rid of the subscripts beside both the 𝑚s. And then we can see that we’ve got a common factor of 𝑚 in the numerator.

So we can factorize this. When we do this, what we get left with is the mass of the object multiplied by 𝑣 two minus 𝑣 one. But then the thing inside parentheses, 𝑣 two minus 𝑣 one, can be written as Δ𝑣, the change in velocity of the object. And this velocity change occurs over the period of time that the force is exerted, assuming that the mass of the object does not change. That’s this assumption we made earlier.

So we can say that the numerator is equal to the mass of the object multiplied by the change in velocity of the object, Δ𝑣. And as we’ve seen earlier already, 𝑡 two minus 𝑡 one can be written as Δ𝑡. So Newton’s second law of motion, under the assumption that the mass of the object is constant, becomes 𝑚 multiplied by Δ𝑣 divided by Δ𝑡.

But then if we just look at this section of the right-hand side of the equation, Δ𝑣 divided by Δ𝑡, we can spot a rather familiar expression. We can recall that the acceleration of an object is defined as the change in velocity of that object divided by the amount of time taken for that change in velocity to occur. In other words, this whole thing, Δ𝑣 divided by Δ𝑡, is equal to the acceleration of the object. And so when we substitute that in, we get that the force on the object is equal to the mass of the object multiplied by the acceleration of the object. This is where we get 𝐹 is equal to 𝑚𝑎. In other words, 𝐹 is equal to 𝑚𝑎 is a formulation of Newton’s second law of motion under the assumption that the mass of the object is not changing.

However, this is most commonly used and most commonly seen because it’s the most useful version of Newton’s second law of motion. Very rarely do we deal with objects whose mass is changing. And moreover, it’s actually easier to deal with masses and accelerations than it is to deal with changes in momentum. So 𝐹 is equal to 𝑚𝑎 is not wrong by any means. It’s just a special case of Newton’s second law of motion. But it’s probably the most commonly used and most commonly required special case.

However, in many cases, we’re better off not using 𝐹 is equal to 𝑚𝑎 and instead using 𝐹 is equal to the change in momentum divided by the time interval for that change to occur. The reason for this is because the question might give us information about the change in momentum, for example. And in fact, let’s take a look at an example question that does exactly that.

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. Okay, let’s take a look at another example now.

A frog of mass 30 grams jumping into the air from a resting position accelerates to a speed of 12 centimeters per second in 0.025 seconds. How many newtons of force did the frog’s legs exert in the jump?

Okay, so here’s our frog in its initial resting position. And we’ve been told that it has a mass of 30 grams. And as well as this, it’s just about to jump. In fact, here he is a few moments later, having just jumped. We’ve been told that, in this position, the frog is moving at a speed of 12 centimeters per second. So since the frog jumped from here to here, let’s say that the frog is moving in this direction at 12 centimeters per second. The direction at the moment is arbitrary. But it is important that we label it just for the sake of drawing a complete diagram.

Now, as well as this, we’ve been told that the time taken between the frog being at rest and the frog traveling at 12 centimeters per second is 0.025 seconds. So that’s how long it took for the frog to accelerate from zero meters per second here to traveling at 12 meters per second here.

We’ve been asked to find how many newtons of force the frog’s legs exerted in this jump. Now this is a really useful sentence because not only is it telling us that we need to find the force exerted by the frog’s legs, but we also need to find it in newtons. And, as we know, newtons is the base unit of force. So regardless of whatever equations we have to use to find the force, if we want to find it in its base unit, we need to convert everything else to base units as well.

So the mass of the frog given to us as 30 grams needs to be converted to kilograms. And similarly, the speed of 12 centimeters per second needs to be converted to meters per second. However, the 0.025 seconds are already in their own base unit, seconds. So we don’t need to worry about that one. So let’s focus on converting the mass to kilograms and the speed to meters per second. we can start by recalling that one gram is equivalent to one thousandth of a kilogram. And therefore, 30 grams is equivalent to thirty thousandths of a kilogram, or in other words 0.03 kilograms. So let’s replace our label here on the diagram with 0.03 kilograms.

Then we can focus on converting the speed to meters per second. To do this, we can first recall that one centimeter is equivalent to one hundredth of a meter. And therefore, if we divide both sides of the equation by the unit seconds, we can see that one centimeter per second is equivalent to one hundredth meters per second. Then we multiply both sides of the equation by 12 to see that 12 centimeters per second is equivalent to twelve one hundredth meters per second, or in other words 0.12 meters per second. Therefore, on our diagram, we can replace the final speed of the frog with 0.12 meters per second.

Now at this point, we’ve converted all of the quantities we’ve been given into their base units. So let’s now actually look at finding the number of newtons of force exerted by the frog’s legs to get it from here to here and moving at 0.12 meters per second. We can recall that the force exerted on an object is given by finding the change in momentum of that object, Δ𝑝, divided by the amount of time taken for that change in momentum to occur. We’ll call that time interval Δ𝑡.

Now the change in momentum of an object is given by finding the final momentum of the object, which we’ll call 𝑝 subscript 𝑓, minus the initial momentum, which we’ll call 𝑝 subscript 𝑖. In this case, the object is the frog. So all we need to do to find the change in momentum of the frog is to find the final momentum of the frog and subtract from this the initial momentum of the frog.

To do this, we can further recall that the momentum of an object, 𝑝, is given by multiplying the mass of that object by the velocity with which it’s traveling at that point in time. So the momentum of the frog here in its initial position is going to be given by multiplying the mass of the frog, 0.03 kilograms, by its velocity at that point in time.

But then, at that point in time, the frog is in a resting position. That’s what we’ve been told in the question. Therefore, its velocity is zero. And hence, its momentum — that’s its initial momentum 𝑝 subscript 𝑖 — is also equal to zero.

Then we can move on to working out the momentum of the frog when it’s already jumped and it’s moving at 0.12 meters per second. We can say that the final momentum of the frog, 𝑝 subscript 𝑓, which is the momentum at this point, is given by multiplying the mass of the frog, which is 0.03 kilograms still, by its velocity, which is 0.12 meters per second. And of course, the velocity is in this direction. But that’s not really relevant to us right now, cause all we’re trying to find is the magnitude or size of this momentum.

So evaluating the right-hand side of this equation, we find that it’s 0.0036 kilograms meters per second, at which point we can go back to finding the change in momentum of the frog, Δ𝑝. We can see that this is equal to the final momentum, 0.0036 kilograms meters per second, minus the initial momentum, which is zero kilograms meters per second, which then simplifies to 0.0036 kilograms meters per second. That’s how much the momentum of the frog has increased because of it jumping.

And we’ve also been told that this momentum increase occurs over a time period of 0.025 seconds. Therefore, the question already gives us the value of Δ𝑡. So now we can sub in the values of Δ𝑝 — that’s 0.0036 kilograms meters per second — and Δ𝑡 — that’s 0.025 seconds — to give us a value for the force exerted by the frog’s legs to get it from here to here. And that value for force is going to be in newtons by the way. This is because we earlier converted everything into base units, which means our answer for the force is going to be in its own base unit, which is the newton.

So when we evaluate the right-hand side, we find that it’s equal to 0.144 newtons. And hence, we have our final answer to the question. The frog’s legs exert 0.144 newtons of force in the jump.

Okay, so now that we’ve had a look at a couple of examples, let’s summarize what we’ve talked about in this video. We firstly saw that the force on an object is given by finding its rate of change of momentum. Mathematically, this can be written as 𝐹, the force, is equal to the change in momentum, Δ𝑝, divided by Δ𝑡, the time interval over which this change in momentum occurs. And secondly, we saw that, assuming the object’s mass does not change, the equation 𝐹 is equal to Δ𝑝 divided by Δ𝑡 becomes 𝐹 is equal to 𝑚𝑎, Newton’s famous second law of motion.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.