Video: Identifying Collisions That Does Not Conserve Momentum

Two balls of equal masses collide with each other, as shown in the Before part of the diagram. Which of the four ways that the two balls could be moving as shown in the After part of the diagram does not conserve the balls’ total momentum?

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Video Transcript

Two balls with equal masses collide with each other, as shown in the Before part of the diagram. Which of the four ways that the two balls could be moving as shown in the After part of the diagram does not conserve the balls’ total momentum?

Okay, so what this question is asking us is to find the option A, B, C, or D that shows the motion of the two balls after their collision. That’s the total momentum of the balls after the collision is not conserved. Now, in order for the total momentum of the two balls to be conserved, the momentum of the two balls after the collision must be the same as the momentum of the two balls before the collision. That’s what it means for the momentum to be conserved. So essntially, what we’re trying to find is which one of these four options gives the total momentum of the two balls to be different to what it was before the collision.

So to answer this, let’s first work out the total momentum of the two balls before the collision. We can recall that the momentum 𝑝 of an object is defined as the mass of the object multiplied by the velocity with which it’s moving. And so we can say that the total momentum of the two balls combined before the collision, which we’ll call 𝑝 subscript before, is equal to the momentum of the blue ball before the collision plus the momentum of the orange ball before the collision. Now, we’ve been told at the beginning of the question that the two balls have equal masses. We don’t know what these masses actually are. So let’s just say that each one of them has a mass of 𝑚 kilograms.

We don’t know what the value of 𝑚 is. All we know is that the two balls have the same mass. And then we can go on to work out the total momentum of the two balls before the collision. We see that this is equal to the mass of the first ball, the blue ball, which is 𝑚 kilograms multiplied by its velocity which is two metres per second, in this case to the right, plus the momentum of the orange ball which is 𝑚 kilograms once again, because that’s also the orange ball’s mass, multiplied by its velocity which is one metre per second, in this case to the right as well.

Now, saying that these velocities, two metres per second and one metre per second, are positive means that we’ve implicitly assumed that the direction towards the right is positive. And, therefore, anything moving towards the left must be negative. This will become important later. But for now, we can see that the total momentum of the two balls before the collision is equal to 𝑚 kilograms multiplied by two metres per second plus 𝑚 kilograms multiplied by one metre per second. And this momentum ends up being three 𝑚 kilograms metres per second. So basically, this momentum is what we’re aiming for when we go to find the total momentum of the balls after the collision.

So let’s start by working out the total momentum of the two balls in option A. Let’s say that the total momentum here is 𝑝 subcript Aft, for the momentum after, superscript A cause it’s option A. And then we can see that 𝑝 Aft A is equal to the mass of the first ball, that’s the blue ball, 𝑚 kilograms once again, multiplied by its velocity, which is now one metre per second to the right. And we keep that velocity positive because we’ve said that to the right is a positive value. And then to this, we add the momentum of the second ball which is 𝑚 kilograms once again, that’s the mass of the second ball that’s not changing, multiplied by two metres per second.

And then when we evaluate the right-hand side of this equation, we get a value for the total momentum of the two balls after the collision in option A as three 𝑚 kilograms metres per second. Now, this momentum is same as the momentum before the collision. And hence, if option A were to show the motion of the balls after the collision, then in that situation, momentum would be conserved because the momentum before and momentum after are the same. However, in this question, we’re looking for the diagram that does not conserve the balls total momentum. And hence, option A is out of the question.

Let’s look at option B then. We can say that the total momentum of the balls after the collision in option B is equal to the momentum of the blue ball, that’s 𝑚 kilograms multiplied by zero metres per second because this ball is not moving, plus the momentum of the orange ball, that’s 𝑚 kilograms multiplied by three metres per second. And when we evaluate this. It once again ends up being three 𝑚 kilograms metres per second. Therefore, this option, option B, is another way of conserving momentum. And hence, this is out of the question as well.

Option C then, we can see that the total momentum after the collision for this diagram is equal to the momentum of the blue ball, that’s 𝑚 kilograms multiplied by negative one metre per second this time because, remember, anything travelling towards the left has to be given a negative velocity. And so we say that this is the momentum of the blue ball. And to this, we add the momentum of the orange ball. And that ends up being 𝑚 kilograms multiplied by two metres per second. And so when we evaluate the right-hand side we find that the total momentum here is 𝑚 kilograms metres per second not three 𝑚 kilograms metres per second. And so option C shows a possible diagram where the total momentum of the balls is not conserved. And hence, that may be our answer.

But let’s just confirm that option D is not the answer to our question as well. The total momentum in option D ends up being 𝑚 kilograms multiplied by 1.5 metres per second, that’s the momentum of the blue ball plus 𝑚 kilograms multiplied by 1.5 metres per second for the momentum of the orange ball. And all of this evaluates to three 𝑚 kilograms metres per second. Therefore, the momentum in option D is conserved. And so this is not the answer to our question either. Hence, option C is the answer that we’re looking for.

However, it is worth noting that option C is not the answer simply because the blue ball is moving towards the left. In fact, the blue ball is perfectly free to move at one metre per second to the left. And if the yellow ball had ended up moving at four metres per second to the right, then the total momentum of the system would still have been conserved. Try out for yourself. However, it just so happens that, in this situation, the momentum is not conserved. And hence, the answer to our question is that option C is the diagram that does not conserve the balls’ total momentum.

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