Question Video: Finding the Impulse and the Speed of a Body Formed by the Collision of Two Bodies Mathematics

Video Transcript

Two smooth spheres of masses 83 grams and 37 grams were moving in a straight line. At time 𝑡 seconds, where 𝑡 is greater than or equal to zero, the spheres’ displacements relative to a fixed point are given by 𝐬 one equals 165𝑡 times 𝐜 centimeters and 𝐬 two equals negative 195𝑡 times 𝐜 centimeters, respectively, where 𝐜 is a fixed unit vector. Given that the two spheres collided and coalesced into one body, determine the speed 𝑣 of this composite body and the magnitude of the impulse 𝐼 between the spheres.

Okay, so we can see that this is a question about a collision between two spheres. One of these spheres, which we’ll call sphere one, we’re told has a mass of 83 grams. And the other sphere, which we’ll call sphere two, has a mass of 37 grams. Now we’re told in the question that these two spheres collide and coalesce into a single composite body. And it’s our job to determine the speed of this composite body as well as the impulse that’s exerted between the spheres when they collide. In order to do this, we need to know exactly how the spheres are moving when they collide, and understanding this is the real challenge of this question.

The information that we’re given about the two spheres’ motion is really limited to these two expressions. These describe the displacement of each sphere. Let’s start by looking at the expression that describes the displacement 𝐬 one of sphere one. We can see that the displacement of sphere one is given by a constant 165 multiplied by time 𝑡 multiplied by some vector 𝐜. And this gives us the displacement measured in centimeters. Now the first thing that we’re gonna note here is that the expression is a linear function of 𝑡. In other words, it contains a factor of 𝑡. This means that the displacement of the sphere is proportional to the amount of time that elapses.

So, for example, when time 𝑡 is equal to one, the displacement of sphere one is equal to 165 times one times 𝐜, which is equal to 165𝐜. When time 𝑡 is equal to two, its displacement is equal to 165 times two times 𝐜, which is equal to 330𝐜. And when 𝑡 is equal to three, the displacement of 𝐬 one is equal to 165 times three times 𝐜 or 495𝐜. So we can see that for every second that elapses, the displacement of sphere one increases by a fixed amount, in this case, by 165 times the vector 𝐜. In other words, the factor of 𝑡 in this expression just tells us that sphere one is moving at a constant velocity.

Now, the next thing we can notice about this expression is that this vector 𝐜 is a fixed unit vector. Fixed means that it doesn’t change over time, and the fact that it’s a unit vector means it has a magnitude of one. We can also signify that 𝐜 is a unit vector by using hat notation instead of the half arrow notation that we use for other vectors. Now, because 𝐜 is a unit vector with a magnitude of one, the factor of 𝐜 in this expression doesn’t actually change the magnitude of the expression. Really 𝐜 exists just to tell us the direction that the displacement vector 𝐬 one points. So if we say that the unit vector 𝐜 points in this direction, for example, then we know that for every second that elapses, sphere one moves 165 centimeters in that direction.

So now we’re narrowing in on what this expression actually means. We’ve shown that for every second that passes, the displacement of sphere one increases by 165𝐜. Since 𝐜 has a magnitude of one and these displacements are expressed in units of centimeters, all of this just means that sphere one is moving at 165 centimeters per second in the direction of 𝐜. Looking at the question, we’re not told the actual direction that 𝐜 acts in, which means that we’re free to draw this unit vector pointing in whatever direction we like in our diagrams. So to make our lives easier, let’s say that 𝐜 points horizontally to the right.

Now, let’s look at the expression for the displacement of sphere two. We can see that the expression for 𝐬 two follows a very similar format to 𝐬 one. It consists of a constant multiplied by time 𝑡 multiplied by the unit vector 𝐜, and again it’s measured in centimeters. So we can interpret 𝐬 two in exactly the same way as we interpreted 𝐬 one but with one crucial difference. In 𝐬 two, the constant is negative. We can recall that multiplying a vector by a negative number results in a vector that points in the opposite direction. So if the unit vector 𝐜 looks like this, then negative 𝐜 looks like this. So we can see that sphere two is traveling at a constant speed just like sphere one. But it’s moving in the opposite direction, this time with the speed of 195 centimeters per second.

So now our question is starting to look a bit more like a normal collision problem. We have two spheres with different masses, and they’re moving directly towards each other with different speeds. We’re also told that when they collide, they coalesce into one body. The mass of this composite body, which we can call 𝑚 𝑐, is equal to the sum of the masses of sphere one and sphere two. This is equal to 83 grams plus 37 grams, which makes 120 grams in total. We’re told that this composite body is moving with some speed 𝑣. So for now, we’ll draw a velocity vector acting in this direction.

Okay, now these two diagrams contain all of the information that we need to solve the problem. The way that we do this is by using the principle of the conservation of momentum. This tells us that, within any isolated system, the total momentum is constant. In other words, the total initial momentum of the system, which we can write 𝑝 𝑖, is always equal to the total final momentum of the system 𝑝 𝑓. We can use this equation along with the fact that momentum is equal to mass times velocity to find the answer to this problem.

First, we can write down an expression for the total initial momentum of the system. This is equal to all the momentum in the system before the collision takes place. Specifically, it’s equal to the momentum of sphere one plus the momentum of sphere two. So we could write that it’s equal to 𝑝 one plus 𝑝 two. Now, 𝑝 one, the momentum of sphere one, is equal to the mass of sphere one, which we can call 𝑚 one, multiplied by the velocity of sphere one, which we can call 𝑣 one. And 𝑝 two is equal to the mass of sphere two multiplied by the velocity of sphere two. Now, here it’s important to remember that both momentum and velocity are vector quantities.

Now, in this first diagram that we drew, we’ve actually already drawn our positive vectors as acting to the right and negative vectors as acting to the left. So we need to bear this convention in mind as we work through the problem. Looking at this expression, we know that 𝑚 one is equal to 83 grams and 𝑣 one is positive 165 since it points in the positive direction. We also need to remember that the velocity of sphere one is multiplied by the unit vector 𝐜. We then add 𝑚 two, which is 37 grams, multiplied by 𝑣 two, which is negative 195𝐜. Simplifying this expression, 83 times 165𝐜 is equal to 13695𝐜 and 37 times negative 195𝐜 is negative 7215𝐜. Subtracting the second term from the first term gives us 6480𝐜, which is the initial momentum of the system measured in gram centimeters per second.

The next step is to come up with an expression for the final momentum of the system 𝑝 𝑓. After the collision, we just have a single composite object with a mass of 120 grams and an unknown velocity with magnitude 𝑣, which we assumed was acting in the positive direction. This means that the total momentum of the system after the collision can be expressed as 120 times 𝑣. And once again, since this is acting in the direction of the unit vector 𝐜, we’ll include this vector of 𝐜. Now that we have expressions for the initial and final momentum of the system, we’re ready to apply the principle of conservation of momentum. This enables us to directly equate the initial momentum and the final momentum, which gives us the equation 6480𝐜 equals 120𝑣𝐜.

We can cancel the factors of 𝐜 on both sides and then solve for 𝑣 by dividing both sides of the equation by 120. This gives us 6480 divided by 120 so that the speed of this composite body is 54 centimeters per second. We should note that when we drew this velocity vector on our diagram, we weren’t actually sure about which way the composite body was moving. Instead, we just assumed that it was moving in the positive direction using a velocity of positive 𝑣𝐜 in the equation. Because this resulted in a positive answer, this means that our assumption turned out to be correct, whereas if we’d obtained a negative value here, we’d know that in fact 𝑣 acted in the negative direction. Either way, the question actually only asks us for the speed of the composite body. So for our final answer, we just take the positive value of this quantity no matter which way it acts.

Now that we found this part of our answer, the next thing we need to do is find the magnitude of the impulse 𝐼 between the spheres. Let’s recall that impulse is equivalent to a change in momentum. We can express this with the equation 𝐼 equals Δ𝑝. So when an object experiences an impulse of a certain size, its momentum will change by that exact amount. It might seem slightly strange to think about the impulse between the spheres. Usually, we would think about an impulse exerted on or experienced by an object. So when our spheres collide, they each experience an impulse. Sphere one exerts an impulse on sphere two that acts to the right, and sphere two exerts an impulse on sphere one that acts to the left.

So what is the magnitude of the impulse between the spheres? Well, much like with forces, any time an object experiences an impulse, it also exerts an impulse of the same size in the opposite direction. This idea of an equal and opposite reaction tells us that the impulse exerted on sphere two by sphere one is exactly the same size as the impulse exerted on sphere one by sphere two. So both of these impulses on our diagram have the same magnitude 𝐼. This means that to work out the answer to this question, we can either choose to look at the impulse experienced by sphere one or the impulse experienced by sphere two as they have the same magnitude.

So let’s choose to calculate the impulse experienced by sphere one. Because an impulse is equal to a change in momentum, the impulse we’re looking for here is simply the change in momentum of sphere one. And this is equal to the final momentum of sphere one, which we’ll call 𝑝 𝑓 one, minus the initial momentum of sphere one, which we’ll call 𝑝 𝑖 one. Now, talking about the final momentum of sphere one seems a bit strange as after the collision sphere one forms part of a composite body. However, all of its mass still remains, and it’s the momentum of this mass that we’re interested in. So the final momentum of sphere one is equal to the mass of sphere one, that’s 83 grams, multiplied by the velocity of the composite body, which we calculated to be 54 centimeters per second, multiplied by the unit vector 𝐜.

The initial momentum of sphere one is again equal to the mass of sphere one, 83 grams, but this time multiplied by its initial velocity, 165 centimeters per second, times the unit vector 𝐜. 83 times 54𝐜 simplifies to 4482𝐜 and 83 times 165𝐜 is 13695𝐜. And subtracting these values gives us an impulse of negative 9213𝐜 gram centimeters per second.

One final thing to note is that the question just asks us for the magnitude of this quantity. This means that we can ignore the minus sign and the unit vector 𝐜, since we’re not concerned with the direction. So now we fully answered the question. If two smooth spheres collide as described in the question and coalesce into one body, the speed of this composite body 𝑣 is equal to 54 centimeters per second, and the magnitude of the impulse between the spheres is 9213 gram centimeters per second.