# Lesson Video: Applying Newton’s Third Law of Motion to Collisions Physics • 9th Grade

In this video, we will learn how to apply momentum conservation to find the forces acting on colliding objects and show that these forces obey Newton’s third law of motion.

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### Video Transcript

In this video, we will look at how to apply Newton’s third law of motion to scenarios where two objects are colliding. We will look at how to find the forces applied to objects during a collision by considering the change in momentum of each of the objects and we will see how these forces obey Newton’s third law of motion. So let’s start by recalling this law.

Newton’s third law of motion tells us that if an object — let’s say object A — exerts a force on another object, object B, then object B exerts an equal and opposite force on object A. In other words, let’s say that this here is object A and this is object B. Now, these two objects are colliding with each other. In other words then, object A is exerting a force to the right on object B. So let’s say that this force has a magnitude 𝐹. Well, by Newton’s third Law of motion, object B exerts an equal force in magnitude. So the magnitude is 𝐹 but opposite in direction, in other words to the left on object A.

Now at this point, let’s recall that we said earlier that we would be considering the change in momentum of each one of these objects as well. Now, first of all, let’s recall that the momentum of an object 𝑝 is given by multiplying its mass 𝑚 by its velocity 𝑣. So what does momentum have to do with the forces applied to an object? Well, to answer that question, we need to recall another one of Newton’s laws of motion, specifically the second law.

Now, Newton’s second law of motion tells us that the net force on an object is equal to the rate of change of the object’s momentum. And we can write this in symbols as 𝐹 subscript net which is the net force on an object being equal to the change in momentum of the object where this Greek letter Δ represents a change in and 𝑝 is the momentum. And we divide this change in momentum by the time taken for that momentum change to occur, in other words the time interval over which this momentum change occurs. Now, we can see that a change in momentum is equal to a change in the mass multiplied by the velocity of the object because remember the momentum of an object is equal to the mass multiplied by its velocity.

So let’s say we’re considering an object in a collision. And the collision is isolated so the only forces acting on the object are the forces due to the collision. Well, in that case, the net force on an object is simply the force due to the collision. And hence, if we know the momentum of the object before the collision and the momentum of the object after the collision, we can work out the change in momentum. And additionally, if we know the amount of time over which the collision occurs, then we know the amount of time over which the momentum change of the object occurs. And hence, we can work out this net force on the object, which like we said for an isolated system happens to be the force due to the collision. And so, that is the link between an object’s momentum or more specifically the change in momentum of the object and the force applied during an isolated collision.

Now notice we’re talking about an isolated collision, where the only forces acting are the forces due to the collision itself. The reason like we said already is because Newton’s second law of motion only deals with the net force on an object or in other words the overall force on the object. However, let’s say some other forces were to act on these objects. Let’s say, for example, that they were in Earth’s gravitational field and so their weight was acting on them as well. Let’s say they had weights of 𝑊 one and 𝑊 two, respectively.

Well, in that case, as long as we had information about these weights, we will be able to work out the net force on the object using Newton’s second law of motion, assuming we knew about the change in momentum of one of these objects. And therefore, we’d be able to work out the force due to the collision. Or vice versa, if we already knew the force due to the collision, we could work out an object’s change in momentum. But more importantly, we could work out the change in momentum of another object that was also involved in the collision whilst knowing information only about the first. So what do we mean by this?

Well, let’s once again consider an isolated collision. The only forces acting are the forces due to the collision. Now, let’s say we know about the change in momentum of one of the objects because for example we know the mass of the object and the velocity of the object before and after the collision. And additionally, we know the time for which this collision happened. Well, from all of that information, we could work out the change in momentum and divide this by the time of the collision to give us the net force acting on this object that caused that momentum change in the first place. But then because of Newton’s third law of motion, we would know that this force that we’ve just found is equal in magnitude to the force acting on the other object.

And additionally, because that time taken by the collision is the same for both objects, obviously cause they’re the ones colliding, we would also therefore know the change in momentum of the second object. It would have to be the same as the change in momentum of the first object. And this is true for an isolated system because if the only forces acting are the collision forces, then the net force on each one of the objects is the same; it’s the collision force. And the time over which this force acts is simply the time of the collision and that’s the same for both objects as well. And hence, the change in momentum of both the objects will be the same.

As well as this, if we were to know the masses of these objects, assuming that those masses don’t change during the collision, we could then work out the new velocities of these objects. And so, we can use Newton’s third law of motion in conjunction with Newton’s second law of motion to take the information that we have about one object in a collision and figure stuff out about the other object in the collision.

Now, it’s also worth quickly noting that, as we’ve seen already, the change in momentum of an object is equal to the change in mass multiplied by velocity. However, if the object’s mass stays constant over the collision, then we can say that the change in momentum of the object is equal to the mass which is constant multiplied by only the change in velocity. And then when we take this equation for Δ𝑝 and substitute it into Newton’s second law of motion, we see that the net force on an object is equal to the mass of the object which like we said is constant in this case multiplied by the change in velocity divided by the time taken for that collision to occur. But then, we realize that the change in velocity of an object divided by the time taken for that change in velocity to occur is simply the acceleration of the object.

And so, in the situation where the mass of an object is constant, Newton’s second law of motion simply reduces to telling us that the net force on the object is equal to that mass multiplied by the acceleration of the object, the famous equation 𝐹 is equal to 𝑚𝑎. And in that situation, we don’t even necessarily need to deal with the change in momentum of the object. We can just deal with this mass and acceleration. Now, it’s all well and good us discussing this topic, but it’s best understood by attempting a few examples. So let’s do that now.

Two objects A and B are thrown into the air and collide with each other, as shown in the diagram. Object A has a weight of 12 newtons and object B has a weight of 20 newtons. In the collision, object A exerts a force of 24 newtons on object B. What force is applied to object A by object B during the collision? What object is given a greater acceleration due to the collision?

Okay, so in this question, we can see that we’ve got two objects: object A in blue and object B in pink. Now, we’ve been told that they’ve been tossed up into the air and they collide with each other. And additionally, during this collision, object A exerts a 24-newton force on object B. And lastly, we also know the weights of these objects. That will come in handy in a minute. But the first thing that we need to try and do is to work out what force is applied to object A by object B during the collision. In other words, we know that 24-newtons of force is applied to object B by object A. And we need to work out the opposite scenario.

What force is applied by object B to object A? To answer this question, we need to recall Newton’s third law of motion. Now, Newton’s third law tells us that if object A exerts a force on object B, then object B exerts an equal and opposite force on object A. In other words, we already know the force exerted by A onto B; it’s 24 newtons to the rightish. And then due to Newton’s third law, we know that the force exerted by B onto A is going to have the same magnitude, equal magnitude. But it is going to be in the opposite direction. In other words then, the force exerted by B onto A is going to be in this direction to the leftish. And it’s also going to have a magnitude of 24 newtons.

So coming back to this question, we see that we don’t actually have to give the direction of the force. But we’ve just been asked to find the force applied. So we can simply state the magnitude of that force, which happens to be 24 newtons. So moving to the next part of the question then, we need to find what object is given a greater acceleration due to the collision, whether it’s object A or object B.

Now to answer this, let’s first start by recalling another one of Newton’s laws of motion, Newton’s second law specifically in the situation where the mass of the object that we’re applying Newton’s second law to stays constant. In other words, the mass of the object is not changing. Well, in that scenario, we can recall that the net force on an object is equal to the mass of that object multiplied by its acceleration. Now because we’re being asked to find the acceleration on these objects due to the collision specifically, we only need to think about the collision forces. We don’t need to account for their weights because remember regardless of what their weights are, these weights always act in a downward direction and produce an acceleration of 𝑔 or in other words 9.8 metres per second squared in each of these objects. And so, we only need to worry about the acceleration caused by the collision forces.

Now, we know that the collision forces are equal in magnitude on both of these objects. There’s a 24-newton force on object A and a 24-newton force on object B. Since we’re being asked to find which one has a greater acceleration, we don’t need to worry about the directions, only the magnitude of this acceleration. So first of all, thinking about object A, we know that the force that we’re considering in this case is the 24-newton force. That’s the force due to the collision that’s causing an acceleration in the object. And then, we can say that this is equal to the mass of the object which we’ll call 𝑚 subscript A multiplied by its acceleration lowercase 𝑎 subscript A. And we can do the same thing for B. We know that the force on the object that we’re considering is 24 newtons. That’s the collision force. And this is equal to the mass of the object 𝑚 subscript B multiplied by its acceleration.

Now, what we’re trying to do in this question is compare 𝑎 subscript A to 𝑎 subscript B. Which of the objects’ accelerations due to the collision is larger? So in order to do that, we need to know what the objects’ masses are. And this is where we see why we were given the weights of these objects in the first place. If we were simply going to ignore these weights, then there would have been a red herring. But this is where they come in handy because we can recall that the weight of an object is equal to the mass of that object multiplied by the acceleration due to gravity, 9.8 metres per second squared.

And so, for object A, we can say that A’s weight — which we’ll call 𝑊 subscript A — is equal to its mass which is 𝑚 subscript A multiplied by the acceleration due to gravity which is a constant. And we can do the same for B. The weight of B is equal to the mass of B multiplied by the acceleration due to gravity. Now at this point, we can see that the weight of A is less than the weight of B. In other words, 𝑊 A is less than 𝑊 B. And therefore, the mass of A must be less than the mass of B because we’re multiplying the mass by a constant value in both cases to give us the weight. So what this means is that the lower weight for object A compared to object B means that the mass of object A must also be lower than the mass of object B. We can write this as the mass of A is less than the mass of B.

And then, we can come back to these two equations that we’ve set up here. We see that in both cases the forces applied to these objects is the same; it’s 24 newtons. However, as we’ve seen, the masses of these objects are different. And so, to compensate for that and to give us the same left-hand side in both cases, the values of these two accelerations must be different. And so, if we’re going to say that the 24-newton force on object A is equal to a lower mass multiplied by some acceleration, then in order to give us a value of 24 newtons as the force in order to compensate for the low mass, the acceleration of the object needs to be bigger. And conversely, we can say that 24 newtons, the force on object B, is equal to the larger mass of B multiplied by the smaller acceleration of B.

And so, at this point, we found the answer to our question. Because both objects have the same force exerted on them due to the collision, but object A’s mass was lower than the mass of object B, this means that object A is given a greater acceleration due to the collision.

Okay, so now that we’ve had a look at an example question, let’s take a look at another one.

A bullet with a mass of 10 grams is travelling at 450 metres per second and hits a stationary toy car of mass 400 grams. The bullet comes to rest 0.01 seconds after hitting the car. What force is required to bring the bullet to rest in 0.01 seconds? What is the acceleration of the car due to the impact from the bullet?

Okay, so in this question, what we’ve been told is that initially we have a bullet of mass 10 grams moving towards a car at 450 metres per second. And that car is a toy car which has a mass of 400 grams and it’s initially stationary. Now, at some point, the bullet and the car collide and that results in the bullet coming to rest in 0.01 seconds. Now, in the first part of the question, we’ve been asked to try and work out the force required to do this, to completely stop the bullet in 0.01 seconds if it was initially travelling at 450 metres per second.

Now, to answer this question, we need to recall Newton’s second law of motion. This law tells us that the net force on an object 𝐹 subscript net is equal to the rate of change of the object’s momentum or in other words the change in momentum of the object divided by the time taken for this momentum change to occur. Now, in this case, we know that the bullet was initially travelling at 450 metres per second. And then, it stops when it collides with the car. So we know that the bullet’s velocity is changing; it’s decreasing. And because that velocity is changing, we know that the momentum of the bullet is changing because remember the momentum of an object is equal to the mass of the object multiplied by its velocity. And so, the change in momentum of the object — which we’ll call Δ𝑝 — is going to be equal to the change in mass multiplied by velocity of the object.

Now, in the situation where the mass of the object remains constant as it does for the bullet, we know that it stays at 10 grams throughout the collision and after the collision as well, in that case we can say that the change in momentum of the object is equal to that constant mass multiplied by the change in velocity of the object. So let’s go about finding the change in momentum of the bullet. We know that it’s equal to the mass of the bullet which is 10 grams or in base units, it’s 0.01 kilograms. And then, we need to multiply this mass by the change in velocity of the bullet. And the change in velocity of the bullet is the final velocity minus the initial velocity.

Now we know that the final velocity is zero metres per second because the bullet is stationary at the end. And the initial velocity is 450 metres per second because that’s what it was travelling at before. And now, note that we’ve made the assumption that the bullet’s initial velocity to the right was positive and hence our change in velocity is going to be negative because the bullet is decelerating; it’s losing speed. But anyway, so we find that the change in momentum of the bullet is equal to negative 4.50 kilograms metres per second.

And so now that we’ve calculated this change in momentum, we can work out the net force on the bullet which is the force that’s required to bring the bullet to rest in 0.01 seconds. To do this, we simply need to say that the net force on the bullet 𝐹 subscript net is equal to the change in momentum which we’ve calculated as negative 4.50 kilograms metres per second divided by the time over which the collision occurs which we’ve been told is 0.01 seconds. And then, because we’re working in base units — so that’s kilograms metres per second for momentum and seconds for time — we know that the final unit of this calculation is going to be kilograms metres per second squared, which is equivalent to the newton, the unit of force.

So when we evaluate the right-hand side of this equation, we find that the net force on the bullet is negative 450 newtons. Now because we’ve simply been asked to find the force required and haven’t specifically been asked to worry about direction, we’ll omit the negative sign in our answer. We’ll just say that the magnitude of the force required to decelerate the bullet in 0.01 seconds so that it comes to rest is 450 newtons. So now that we’ve calculated that, let’s look at the next part of the question.

What is the acceleration of the car due to the impact from the bullet? Well, to answer this, we can recall Newton’s third law of motion. Newton’s third law tells us that if an object — let’s say object A — exerts a force on another object, object B. Then, B exerts an equal and opposite force on A. Now, we’ve seen in this collision that there was a force of 450 newtons exerted on the bullet in order to decelerate it. Well, that force would have been exerted by the car onto the bullet due to the collision that occurred between the two objects. And so, by Newton’s third law of motion, we can see that the bullet will exert a force in the opposite direction onto the car. And the magnitude of this force is the same, 450 newtons.

And then, we can recall that for an object with a constant mass as we’ve seen already, the change in momentum of the object is equal to that constant mass multiplied by the change in velocity. But then, if we take that equation for the change in momentum and substitute it back into Newton’s second law of motion, we see that the net force on an object 𝐹 subscript net is equal to the mass multiplied by the change in velocity divided by the change in time or the time taken for, in this case, the collision to occur. And then, we can notice that the change in velocity divided by the change in time is equal to the acceleration of the object. And so, in the situation where an object’s mass remains constant, Newton’s second law reduces to the net force is equal to the mass of the object multiplied by its acceleration.

Now, we already know the force acting on the car; it’s 450 newtons. So we know the net force. And we also know the mass of the car. We know that it’s 400 grams. Hence, we can work out the acceleration by rearranging this equation. We can say that the acceleration of the car is equal to the net force divided by its mass. And that ends up being 450 newtons divided by 400 grams. Or in base units, the denominator becomes 0.4 kilograms. And then evaluating the right-hand side, we find that the acceleration of the car is 1125 metres per second squared. And at this point, we found the answer to our question.

So without further ado, let’s take a look at a summary of what we’ve talked about in this lesson.

We saw that we can use Newton’s second law: the net force on an object is equal to the rate of change of its velocity and Newton’s third law: basically, every action has an equal and opposite reaction. And we can use these two laws together in conjunction to give us information about the forces acting on objects during a collision, as well as giving us information about the objects’ changes in momentum.