# Question Video: Newton’s Third Law of Motion in Collisions Physics • 9th Grade

A ball with a mass of 200 g collides with a wall. The impact of the ball on the wall exerts an average force of 7.5 N on the wall for a time of 0.025 s. The ball rebounds from the wall at a horizontal speed of 0.15 m/s. What was the speed of the ball when it hit the wall?

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### Video Transcript

A ball with a mass of 200 grams collides with a wall. The impact of the ball on the wall exerts an average force of 7.5 newtons on the wall for a time of 0.025 seconds. The ball rebounds from the wall at a horizontal speed of 0.15 meters per second. What was the speed of the ball when it hit the wall?

Okay, so in this question, we know that we’ve got a ball rebounding from a wall. So here’s a diagram showing our ball. And here is the wall. Now initially, the ball would have been travelling towards the wall with a certain speed which we’ll call 𝑢 because we don’t know what the speed is yet. And then at a certain later point of time, the ball actually collided with the wall. We’ve been told that this collision resulted in the ball exerting a force of 7.5 newtons onto the wall.

And as well as this, we were told that the force was exerted for time of 0.025 seconds. In other words, that’s how long the collision lasted. And then later still, the ball will have been moving away from the wall. This time, we’ve been told that it’s moving at a speed of 0.15 meters per second. And this is its horizontal speed. So based on all of this information, we need to find the speed of the ball when it hit the wall, in other words, the speed of 𝑢, which is the speed that the ball would’ve had throughout this entire journey before it collided with the wall.

So let’s start by labelling the first part of the diagram as before. This is representing before the collision. This is during the collision. And this part of the diagram is showing what happens after the collision. To answer this question then, we’re gonna start by focusing on the diagram that gives us the most information, the diagram representing what’s happening during the collision. Specifically, we’re told that this collision of the ball with the wall exerts a force of 7.5 newtons onto the wall. And if the ball is coming in from the left onto the wall, then the force exerted on the wall will be in the direction of the ball’s motion, in other words, towards the right as we’ve drawn it.

But then at this point, we can recall Newton’s third law of motion. 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 the second object, object B, exerts an equal and opposite force on object A. So in this case, we’ve got the ball as object A and the wall as object B. The ball is exerting a 7.5-newton force to the right as we’ve drawn in this diagram onto the wall. And so by Newton’s third law of motion, the wall will exert an equal in magnitude force on the ball. But it’ll be in the opposite direction. In other words, the wall exerts a 7.5-newton force to the left on the ball.

Now, this information is actually really useful because we’ve now got the force exerted on the ball. And what we’re trying to calculate in this question is the speed of the ball when it hit the wall. So in order to do this, we need to have as much information as is possible about what’s happening to the ball itself. It’s not so relevant what’s happening to the wall. But that information allows us to work out, using Newton’s third law, what’s actually happening to the ball during the collision.

Anyway, so at this point, we’ve figured out that, during the collision, there is a 7.5-newton force to the left on the ball and, of course, to the left is as we’ve drawn it. But the point is that the force on the ball is in the direction away from the wall. And then at this point, we can actually recall Newton’s second law of motion. This law tells us that the force on an object is equal to the rate of momentum change of the object. Mathematically, we can say that the force on an object 𝐹 is equal to the change in momentum of an object Δ𝑝 divided by the time interval over which this change in momentum occurs which we’ll call Δ𝑡. And this is why the force is the rate of change of momentum or rate of momentum change, because it’s the change in momentum divided by the time taken for that change to occur.

Now looking at this diagram once again, we can see that the force exerted on the ball is 7.5 newtons. And we also know the amount of time for which this force is exerted because we know that the ball exerts a 7.5-newton force on the wall for 0.025 seconds. And so the wall exerts an equal and opposite force for the same amount of time. And as well as this, at this point, we can make the arbitrary decision that anything moving in this direction is moving in a positive direction. And therefore, anything moving towards the wall is moving in a negative direction.

Hence, we can say that the force exerted on the ball is a positive force, whereas the force exerted on the wall is a negative force. So if we just consider the force exerted on the ball, then we can say that 7.5 newtons of force, that’s the force exerted on the ball, is equal to the change in momentum of the ball Δ𝑝, which we don’t know what it is yet, divided by the interval of time over which this change in momentum occurs. Well, this change in momentum is going to occur over the same period of time that the force is exerted on the ball. And this is because before and after the collision there are no forces acting on the ball. So the only momentum change that’s going to occur is during the collision.

And hence, we divide this Δ𝑝, this change in momentum of the ball, by 0.025 seconds, that’s the interval of collision. And at this point, we can rearrange the equation by multiplying both sides by 0.025 seconds which tells us that 7.5 newtons of force multiplied by 0.025 seconds is equal to the change in momentum of the ball. Now very quickly, we can see that we’ve got base units of newtons for the force and seconds for the time. And so when we evaluate the left-hand side of this equation, our right-hand side is going to be in its own base units. And the base units of momentum are kilograms meters per second.

So evaluating the left-hand side of this equation then, we see that the change in momentum of the ball due to this force is 0.1875 kilograms meters per second. And notice that this change in momentum, by the way, is positive. That’s because we’ve said that this force is positive. And the change in momentum is going to be in the same direction as the force exerted on the object. So let’s write down over here that the change in momentum of the ball during the collision is 0.1875 kilograms meters per second. And let’s try and work out what that actually means.

Well, what we know is that initially the ball was travelling in this direction. And then during the collision, its momentum increases in this direction, in the positive direction as we’ve called it, by 0.1875 kilograms meters per second. And this results in the ball afterwards having a velocity of 0.15 meters per second to the left. So what we need to do now is to link the initial velocity of the ball with the change in momentum of the ball and then the final velocity of the ball.

To do this, we can start by recalling that the momentum of an object 𝑝 is equal to the mass of the object multiplied by the velocity of the object. And it’s also worth-remembering that momentum and velocity are both vector quantities. In other words, the direction in which they point is very important. But anyway, so if the momentum of an object is equal to the mass of that object multiplied by its velocity, then the change in momentum of an object, Δ𝑝, is equal to the change in mass multiplied by velocity.

However, in this particular situation, we’ve been told that the mass of the ball is 200 grams. That mass is constant. The mass of the ball is not changing. And so in this particular situation, because the mass of the ball is constant, we can therefore say that the change in momentum of the ball is equal to the mass of the ball multiplied by the change in velocity because, in this case, the change in mass multiplied by velocity is the same thing as the mass of the ball, which is not changing, multiplied by whatever the velocity change of the ball is.

So let’s write down over here that the change in momentum of the ball is equal to the mass of the ball multiplied by its change in velocity. And let’s now recall that the change in velocity of an object is equal to the final velocity of the object, which we’ll call 𝑣 subscript fin, minus the initial velocity of the object, which we’ll call 𝑣 subscript init. Now at this point, we once again need to remember that velocity is a vector quantity. We said earlier that the initial speed of the ball was 𝑢 towards the right. And the final speed of the ball was 0.15 meters per second towards the left. And as well as this, we’ve arbitrarily decided earlier that left was the positive direction. And right was the negative direction.

So based on this information, we can say that the change in velocity of an object is equal to the final velocity which is 0.15 meters per second. And it’s positive because it’s towards the left. And so we can say that the final velocity is positive 0.15 meters per second. And from this, we need to subtract the initial velocity which is actually negative 𝑢 because it’s in this direction. And it’s at this point that we can see that if we can calculate what the value of Δ𝑣 is, then we can solve for 𝑢. But then we’ve written down this equation earlier.

We already know the value of Δ𝑝, that’s 0.1875 kilograms meters per second. And we’ve been told the mass of the ball in the question itself. So let’s start by rearranging this equation by dividing both sides of the equation by 𝑚, the mass of the ball. This way it cancels on the right-hand side. And what we’re left with is that the change in momentum of the ball divided by the mass of the ball is equal to the change in velocity of the ball. And so we can say that the change in velocity is equal to the change in momentum which is 0.1875 kilograms meters per second. And we need to divide this by the mass of the ball.

However, we’ve been told the mass of the ball in grams. But we want it in the base unit of kilograms. To do this, we can recall that one gram is equivalent to one thousandth of a kilogram. And, therefore, 200 grams is equivalent to 200 thousandths of a kilogram or simply 0.2 kilograms. And so we divide this change in momentum of the ball by 0.2 kilograms which is the mass of the ball. As we evaluate this fraction on the right, we can see that the units of kilograms in the numerator will cancel with the unit of kilograms in the denominator. And we’re going to be left with the quantity in meters per second which is good because we’re looking for a change in velocity on the left-hand side.

So evaluating the fraction on the right, we find that the change in velocity of the ball due to the collision is 0.9375 meters per second. And once again, we know that this is a positive value. Therefore, what we’re saying is that the velocity of the ball has increased in this direction towards the left, whereas initially, it was travelling in this direction towards the right. But anyway, so at this point, we can take this value for Δ𝑣 and substitute it into this equation.

And so we can say that 0.9375 meters per second, the change in velocity of the ball, is equal to 0.15 meters per second, the final velocity of the ball, minus the initial velocity of the ball which is negative 𝑢. And then at this point, we can multiply this negative sign by this negative sign. And so what we’re left with is plus 𝑢. And then we can move forward by subtracting 0.15 meters per second from both sides of the equation. This way, 0.15 meters per second cancels on the right-hand side. And so we’re just left with 𝑢 on the right. And on the left, we’re left with 0.7875 meters per second.

Therefore, we now know the value of 𝑢 which is 0.7875 meters per second. But at this point, we might be wondering why this value of 𝑢 is not a negative value because remember 𝑢 is representing the initial speed of the ball towards the right. But the point is that the way we’ve defined 𝑢, 𝑢 is actually only the magnitude of the velocity of the ball. The direction was encoded for later because we said that the initial velocity of the ball was negative 𝑢. And hence, 𝑢 is just the magnitude. And the negative sign gives the direction.

And in fact, this actually ends up being a good thing for us because what we’ve been asked to find in this question is the speed of the ball when it hit the wall not its velocity. In other words, the speed of the ball is just a magnitude of the velocity of the ball when it hit the wall. And so all we need to do is to find the value of 𝑢. We don’t need to worry about its sign. And since we’ve already calculated the value of 𝑢, we can now say that the speed of the ball when it hit the wall was 0.7875 meters per second.