Video: Conservation of Momentum

During a game of pool, a cue ball that has a mass of 170 g moves along a pool table at a constant speed of 60 cm/s. The cue ball hits another ball, which has a mass of 160 g and is also moving along the table and in the same direction as the cue ball but at a constant speed of only 15 cm/s, as shown in the diagram. After the collision, the balls move together at the same constant speed. What is the speed of the balls after the collision? Round your answer to the nearest centimeter per second.

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Video Transcript

During a game of pool, a cue ball that has a mass of 170 grams moves along a pool table at a constant speed of 60 centimeters per second. The cue ball hits another ball, which has a mass of 160 grams and is also moving along the table and in the same direction as the cue ball but at a constant speed of only 15 centimeters per second, as shown in the diagram. After the collision, the balls move together at the same constant speed. What is the speed of the balls after the collision? Round your answer to the nearest centimeter per second.

Okay, so, we can see here in this diagram that we’ve got a cue ball, mass 170 grams, moving towards the orange ball, the other ball at 60 centimeters per second. As well as, this we’ve got the other ball, 160 grams, moving right at 15 centimeters per second. Now, eventually, the cue ball is going to collide with the orange ball, as we’ve been told in the question. And in fact, after the question, both the cue ball and the orange ball move together at the same speed. So, we can actually draw a diagram showing the motion of the two balls after the collision.

So, our diagram for after the collision shows the two balls moving towards the right. And they’re both moving at the same speed, which we’ve called 𝑉. We can also, of course, keep in mind that the mass of the cue ball and the mass of the orange ball are not changing. And so, at this point, we can use our two diagrams to work out the value of 𝑉. As well as this, we can label the first diagram to say that it’s before the collision. And the second diagram shows what happens after.

So, to find the value of the speed 𝑉 that the two balls move at after the collision, we need to recall the law of conservation of momentum. This law tells us that in an isolated system, momentum is conserved. In other words, in a system where there’re no external influences or external forces acting, the total momentum of the entire system before an event, which in this case is a collision, is equal to the total momentum of the entire system after that event. This is what it means for momentum to be conserved.

As well as this, we can recall that the momentum of an object, which we’ll call 𝑝, is given by multiplying the mass of that object by the velocity with which it’s travelling. So, we can use these two pieces of information to work out the value of the speed 𝑉. This is because the system that we have is indeed an isolated system because there’s no external forces acting on any of the objects within the system. Therefore, momentum conservation must apply. So, let’s start by finding the total momentum of the system before the collision.

Let’s call this total momentum 𝑝 subscript bef, for before the collision. And this total momentum is going to be equal to the momentum of the cue ball plus the momentum of the orange ball. Now the momentum of the cue ball is the mass of the cue ball, 170 grams, multiplied by its velocity, which is 60 centimeters per second to the right. And to this, we add the momentum of the orange ball, that’s 160 grams multiplied by 15 centimeters per second.

Now notice that by doing this, we’ve implicitly assumed that anything travelling towards the right is moving in the positive direction. This is why we’ve simply used the positive velocities. And another thing to note here is that we haven’t converted everything into base units. The reason is because if we’re consistent with the units that we use, then we don’t really need to convert to base units.

What we mean by that is that if for the entire calculation we stay consistent with the units we use, for mass for example, that is we always use masses in grams, then it doesn’t matter. And similarly, of all our speeds are in centimeters per second rather than meters per second, then this isn’t too much of a problem. We just need to keep track of the units that we’re using.

So, the momentum of the cue ball ends up being 170 times 60, that’s 10200 grams centimeters per second. And similarly, the momentum of the orange ball ends up being 160 times 15, that’s 2400 grams centimeters per second. Then, we can add these two values to get that the total momentum of the entire system before the collision is 12600 grams centimeters per second. So, at this point we can go on to work out an expression for the total momentum of the system after the collision.

We can say that the momentum after the collision, 𝑝 subscript aft, is equal to, once again, the momentum of the cue ball plus the momentum of the orange ball. But this time the momentum of the cue ball is given by the mass of the cue ball 170 grams multiplied by the velocity of the cue ball, which happens to be 𝑉. And it’s to the right, so it’s going to be positive. This is, once again, using the same sign convention as before, where anything travelling towards the right is moving in the positive direction.

And then, to all of this, we add the momentum of the orange ball, which happens to be 160 grams, that’s the mass of the orange ball, multiplied by 𝑉 again. Because the orange ball is also moving at 𝑉 towards the right.

Now at this point, we can see that on the right-hand side, we’ve got a common factor of 𝑉. We can, therefore, factorise this. What we get is that the total momentum of the system after the collision is equal to 𝑉 multiplied by 170 grams plus one 160 grams. But then, 170 grams plus 160 grams is equal to 330 grams. And at this point, we can’t simplify expression any further. So, we then come back to our law of conservation of momentum.

Now since we said earlier that our system is an isolated system, we, therefore, know that momentum must be conserved. In other words, the total momentum of the system before the collision must be equal to the total momentum of the system after the collision. Hence, 12600 grams centimeters per second is equal to 𝑉 times 330 grams. And this is what equating the two looks like.

At this point, we can divide both sides of the equation by 330 grams, which means that it cancels on the right-hand side. And on the left, we have 12600 divided by 330. But if we look at the units, there’s a unit of grams in the numerator that cancels with the grams in the denominator. And what we’re left with is centimeters per second, which is perfectly fine because we’re trying to find a velocity. But since we know that the velocity is towards the right because it’s a positive value, we’re essentially finding the speed of the two balls combined after the collision.

Now we can clean up the right-hand side because all that’s left on the right is 𝑉. And on the left, we’ve got 12600 divided by 330, centimeters per second. Now evaluating the left-hand side give us a value of 38.1818 recurring centimeters per second. However, this is not our final answer because, remember, we’ve been told to round our answer to the nearest centimeter per second.

In other words, we need to round this value. Either that value is going to stay the same, or it’s going to round up. And that depends on the first decimal place. Now that value, the value at the first decimal place, is a one. One is less than five. And therefore, this eight is going to stay the same. It’s not going to round up. And hence, we found our value of 𝑉 as 38 centimeters per second. So, now we have our final answer. The speed of the balls after the collision is 38 centimeters per second to the nearest centimeter per second.

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