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The Conservation of Momentum Recall from chapter four that the conservation of energy states that the initial energy of a system equals the final energy. This is what conservation means. So, it follows that the conservation of momentum means the initial momentum of a system equals the final momentum. This can actually be proven from Newton's Third Law. Newton's Third Law from chapter three says that for every force there is an equal and opposite force. Consider the car and truck below.  The concept equation shows that the car will have the greatest acceleration because it has the smallest mass. If we take the concept equation and plug in the change in velocity over time equation for acceleration we have,  Notice that both sides of the equation has time. The times are equal because both vehicles are in the same collision, so they share the same impact time. Therefore, the times can be canceled.  We can now distribute the masses to the velocities and get,  If we multiply everything by negative one and move all of the initial velocities to the left side, and keep the final velocities on the right, we have the conservation equation of momentum!  We can now instead of using Newton's Third Law, use momentum conservation to illustrate the examples used in chapter three.  Before the rifle is shot, the momentum of the rifle and bullet is zero. Therefore, the momentum of the rifle-bullet system must be zero after the shot. So, the momentum of the rifle going to the left equals the momentum of the bullet to the right. Since they are in the opposite direction, they cancel to make the total momentum zero. Expressing this with the conservation equation requires the entire initial side to be zero.  I made the final velocity of the rifle negative because it is moving to the left, which is why there is a negative sign in front of the final momentum term. So, when you move it to the other side it becomes positive and it shows that the two momenta are equal. Whenever two objects are at rest initially, the initial momentum side will always be zero, which therefore makes the two final momenta equal and opposite. Another example used was the rocket. You can now explain why a rocket moves up with momentum conservation.  The initial momentum before the rocket takes off is zero. When the boosters fire, the gas is given a downward momentum. In order to keep the final momentum zero, the rocket has to go up with the same momentum as the gas. Recall the two people sitting in chairs holding a rope from chapter three.  Before the pull, their momenta is zero. If the man on the right has a mass of 50.0-kg, and the man on the left has a mass of 75.0-kg, and they pull on the rope so that the 50.0-kg man has a speed of 3.0 m/s, we can find the speed of the other man by setting up the conservation equation up as follows. Solving for Vf yields a negative 2.0 m/s. As discussed before, the man with the smaller mass has the larger speed, which means he had the greater acceleration.  It should be noted that when talking about the momentum equation with the rifle and bullet, I put a negative sign in the equation. If you notice the last problem, I did not do that. I solved for the final velocity and it ended up negative. When solving problems, do not put negatives in the equation. Keep everything positive because when you solve for the desired variable, the negative will come out in the answer. While we are at it, let's determine the meeting point of the two people. If the starting distance between them is 10.0 meters, we can express the distance traveled by the man on the left as x, and distance traveled by the man on the right as (10 - x).  So, the speed of the man on the left can be expressed as, Where 2 is his speed, x is his distance, and t is the time to get there. Similarily, the man on the right would be, Where 3 is his speed, (10 - x) is his distance, and t is the time to get there. Since t is the same for both men, we can solve each equation for t and set them equal to each other. and, Setting them equal gives, If we cross multiply the 2 and 3 we have, Solving this equation for x gives 4 meters. Therefore, the man on the left travels 4 meters, and the man on the right travels 6 meters. Before we work problems that deal with collisions, we need to examine the different types of collisions. Elastic Collisions There are three types of collisions: elastic, inelastic, and completely inelastic. Elastic collisions occur when there is no energy lost during the collision. In practice, there will always be a loss of energy when two objects collide. Those collisions are called inelastic. The energy is converted into heat and sound. As a result, the velocities of the objects will be less. But, for the purposes of beginning physics, we will consider collisions with no loss of energy. When two objects experience an elastic collision together, they will bounce off each other. Let's consider a collision when one object is moving, and the other is at rest.  If the two masses are equal, it would stand to reason that the moving object would give all of its energy to the object at rest, which means the object at rest would fly off with a speed equal to the moving object's speed. Consequently, mass one would be at rest after the collision. This makes sense since we said it gave all of its energy to mass two. (Notice the apostrophe after the V1 and V2. That is the symbol that denotes final velocities. It is read as V one prime and V two prime. These symbols are useful here to avoid having i's for initial and f's for final along with the numbers)  What happens when the masses are not equal? There are two equations that enable us to find the final speeds as long as one of the masses is at rest. The proof begins by taking a look at the conservation of energy and momentum. (If you are not interested in seeing the mathematical proof, skip down to the result found under the heading "Relative Velocity and Momentum Conservation." I have included a few key proofs in this book to show the beauty of starting with fundamental equations and ending with some very interesting results that predict with a great deal of accuracy the real world.)  We will first deal with the energy equation. Notice this equation does not include potential energy. We are not considering different heights so it is not necessary. This equation is dealing with the energy of each car before and after the collision. As a result, the closed system, comprised of the two cars, must conserve energy. Moving right along. The first step is to move the initial and final energy of mass one to the left, and the initial and final energy of mass two to the right. Also, we can cancel the 1/2 since it is shared by all four terms.  We can now factor mass one out of the left, and mass two out of the right.  Using the algebraic property: a^2-b^2 = (a+b)(a-b), we have,  We will leave that equation for now and come back to it after we mess around with the momentum equation. We want to do the same thing to the momentum equation: move the mass one velocities to the left and mass two velocities to the right.  We can again factor mass one and mass two and get,  Putting the last equation with the last energy equation we have,  Notice the left side of the momentum equation shows up in the left side of the energy equation.  Since m1(V1 - V1') equals m2(V2' - V2), we are going to substitute the last term into the energy equation, which gives,  Notice now that both sides have a mass two, and the quantity (V2'-V2). We can cancel those terms from the equation and have,  This result may not seem very significant, but a little algegraic manipulation will prove otherwise. First, get all of the initial velocities on the left and the final velocities on the right.  For the last step, factor a negative one out of the right side of the equation so that the V1' is first and the V2' is second.  This last result points to a very interesting fact concerning collisions and relative velocity. Relative Velocity and Momentum Conservation Welcome back if you decided to skip the proof. The result of combining the conservation of energy and momentum equations gave,  Recall from chapter two, V1 - V2 is the equation for finding the relative velocity between two moving objects. In fact, that result would be the velocity of object one as seen by object two. Now notice the negative sign in front of the relative velocity equation on the right side, which represents the relative velocity after the collision. The negative sign means the opposite direction. So, this equation is saying that the relative velocity after the collision is equal to but in the opposite direction of the relative velocity before the equation! Let's consider the above example with the two balls to illustrate this point. Before the collision, mass one is traveling at a speed V1. Let's suppose that speed is 5.0 m/s. Mass two is at rest, so the relative velocity would be found by, this means that the velocity of mass one as seen by mass two is 5.0 m/s. After the collision, mass two flies off with the same speed that mass one had, 5.0 m/s, and mass one is at rest. So, the relative velocity is, This means the velocity of mass one as seen by mass two is negative 5.0 m/s, which means to mass two, it looks like mass one is traveling at 5.0 m/s to the left. Now, to show this satisfies the result we just found in the proof, we have, The two negatives on the right make a positive, so the two sides are equal. That was a very simple example, but it shows how the equation works. The beauty of this equation is that it applies to any elastic collision. So, it does not matter if both are moving before the collision, or if one is moving and the other is at rest. If you remember, we started this because we wanted to have two equations that would enable us to calculate the final speeds of the objects after the collision if the two masses were not equal. The proof was the starting point. I will not go into detail on how these two equations are derived, but they are a result of combining the conservation of momentum equation with the relative velocity equation while setting V2 equal to zero.  Combining these two equations and solving once for V1' and once for V2' we get,  These two equations will give the final velocities of both objects after an elastic collision as long as one is at rest before the collision. I REPEAT! These equations only work for an elastic collision when one object is at rest before the collision! Before we do another example with those equations, there are three scenarios to examine. 1. Mass one greater than Mass two If mass one is greater than mass two, V1' will be a positive number, because in the first equation it is M1 - M2. This means mass one will continue in the same direction after the collision. Also, V1' will be smaller than V1 because in the equation V1 is multiplied by a number that is less than one, due to the term (M1-M2)/(M1+M2). Simply stated: mass one will slow down after the collision. According to the second equation, V2' will be positive as well, which is pretty obvious. What is not so obvious is the fact V2' will actually be larger than V1. Mass two will be traveling faster than mass one was before the collision. This is a result of the term 2M1/(M1+M2). That term will always be greater than one as long as M1 is greater than M2.  2. Mass two greater than Mass one If mass two is greater than mass one, V1' will be a negative number, because the term M1 - M2 will be negative. This means mass one will be moving in the opposite direction after the collision. It bounces off in the opposite direction. V1' will also be smaller than V1 because the (M1-M2)/(M1+M2) term will always be less than one. V2' will be positive again, which I think is still obvious. However, V2' will be smaller than V1 because the 2M1/(M1+M2) term will also be less than one. But, V2' will be very close to V1 the closer M1 is to M2. V2' will be very small if M2 is a lot larger than M1.  3. Mass one equals Mass two We already know the results of this scenario, but let's show that the two equations do indeed give those results. When mass one equals mass two, the (M1-M2)/(M1+M2) term becomes zero; therefore, V1' is zero. The 2M1/(M1+M2) term reduces to one, because the bottom term would become 2M1 as well. (It doesn't matter if it is 2M1 or 2M2 because they are equal anyway). Therefore, V2' just equals V1.  For an example, let's say a 5.0-kg ball traveling at 30.0 m/s strikes another ball at rest with a mass of 10.0-kg. Find the final velocities for both balls, show that momentum and energy is conserved, and show that the relative velocity before equals the opposite relative velocity after. Before plugging in the numbers, we can already determine that ball one will bounce off of ball two and travel in the opposite direction. Plugging in the numbers for V1' and V2' gives and, Notice how since mass two is twice the size of mass one, the velocity of mass two is twice the size of mass one. It would seem that since mass two was twice as large it would be traveling at half the speed. But, remember mass one was the only object moving. When mass one hit mass two, mass two in a sense had to stop mass one and make it go in the opposite direction at 1/3 its original speed. Therefore, the change in mass one's velocity was actually 40, because Vf - Vi would be -10 minus 30 which equals - 40 m/s. So, mass one still had the overall greater acceleration. Plugging in the numbers into the conservation of momentum, both sides should be the same. Indeed, both sides equal 150 kgm/s. Plugging into the energy conservation equation should also show that both sides are equal. Both sides do equal 2250 J. Finally, plugging into the relative velocity equation yields, Both sides do indeed equal 30 m/s. This means that before the collision the velocity of mass one as seen by mass two was a positive 30 m/s. (Obviously since mass one was moving at 30 m/s towards an at rest mass two). After the collision, the speed of mass one as see by mass two is a negative 30 m/s. From mass two's perspective, mass one is moving away at 30 m/s. It is interesting to note that the two equations found do not allow for a situation where mass one is twice the size of mass two, and mass two flies off with a speed that is twice the value of the original speed of mass one, and mass one stays at rest. For example, say mass one is 10-kg and mass two is 5-kg. If mass one hits mass two with a speed equal to 10 m/s, mass two cannot fly off at a speed equal to 20 m/s and mass one stays at rest after the collision. Even though momentum would be conserved because, Which shows that the intial and final momenta are equal to 100 kgm/s. However, the question of energy conservation remains. Is energy conserved with this situation? This initial energy is equal to 500 Joules where the final energy would be equal to 1000 Joules! Therefore, since energy is not conserved this situation can never happen. That is why the two equations derived above do not allow it either. What happens if both are moving? We cannot use the two equations above. But, we can come up with two new equations exactly the same way. We can start with the conservation of momentum and the relative velocity result, only this time V2 is not zero.  Without showing the steps, if you combine the two equations and solve once for V1' and once for V2' you will have,  These two equations are quite a bit more complicated than the other two. There are four variables to manipulate that result in different values for V1' and V2'. For example, if mass one is larger than mass two, and V2 is larger than V1, and heading towards mass one, mass one could be traveling in the negative direction after the collision. However, even if V2 is larger than V1, mass one could be big enough compared to mass two so that mass one still travels in the positive direction after the collision. Let's do an example and still show that momentum and energy is conserved, and the relative velocity equation still holds. Suppose mass one has a mass of 10.0-kg and is traveling at 10.0-m/s to the right. Mass two has a mass of 5.0-kg and is traveling to the left at 20.0-m/s. After the collision, the velocities are found by, and After the collision, the larger mass is sent in the opposite direction. But, the smaller mass is sent in the opposite direction as well. Again, the smaller mass definitely experienced the larger acceleration, because the change in velocity is greater. Mass one was traveling at a positive 10 m/s and ended up going a negative 10 m/s, thus resulting in an overall change of 20 m/s. Mass two was traveling at a negative 20 m/s and ended up going a positive 20 m/s, which results in an overall change of 40 m/s.  Showing that momentum is conserved, Both sides equal zero! And now for energy conservation. Again, both sides equal 1500 J. Finally, the relative velocity before and after can be found by, It turns out that the relative velocity is 30 m/s before, and a negative 30 m/s after. To finish this section, let's consider a two-dimensional collision. Say a ball with mass of 10.0-kg is traveling in the positive x-direction at 10.0 m/s. Another ball with mass of 5.0-kg is at rest. After the collision, ball one flies off at an angle of 30.0 degrees with a speed of 6.0 m/s. Find the speed and direction of ball two.  We cannot just use the above equations for this problem because not all of the velocities are in the same direction. Up until now, we have only considered objects traveling in the x-direction. In order to solve this problem, we have to break it into two parts: the x part and y part. We need to do this because momentum is conserved in the x-direction and the y-direction separately. This means any speed in the x-direction does not affect the momentum in the y-direction, and vice versa. So, let's start with the x-direction first. Momentum in the x-direction Since we are exclusively dealing with the x-direction, the momentum conservation equation becomes,  Before the collision ball one's velocity is all in the x-direction, and ball two isn't moving. After the collision, some of ball one's speed is in the x-direction, but some is also in the y-direction. Therefore, we need to find the x-component of the velocity by using the techniques learned in chapter two.  The diagram above shows the x and y-component of the final velocity. In order to find Vx, we need to use the cosine function, Knowing all of the given information, we can set up the conservation equation as, Solving for V2x yields 9.6 m/s. Now on to the y-direction. Momentum in the y-direction The momentum conservation equation now looks like this for the y-direction.  In this direction, neither balls are moving before the collision. We need to find ball one's speed in the y-direction by using the sine function, Now the equation becomes, Here, V2y is a negative 6.0 m/s. This means ball two is heading down. Putting it all together the diagram for each ball becomes,  The last thing needed is to find the resultant velocity of ball two, which is V2', and the angle of travel for ball two. The angle is found by using the inverse tangent function, And the resultant is found by, Therefore, ball two is has a velocity of 11.3 m/s 32 degrees below the horizontal. You can verify that energy is conserved by plugging in the initial and final velocities. You do not need to plug in the x and y-direction because energy is not a vector. |