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Momentum The momentum of a body is its mass times its velocity: p = mv . Suppose there is more than one body on our track. We then have a many-body system. 1 2 3 4 In such systems, the concept of momentum becomes particularly useful. Let’s see why. In general, two kinds of forces can act on the bodies in the system. There are internal forces between the bodies, and forces due to external agents acting on individual bodies. In what follows, we will make a specific assumption about the nature of the forces acting between bodies. We will assume that they satisfy Newton’s “third law”: The force of body j on body i is equal in magnitude and opposite in direction to the force of body i on body j. (Here, i and j stand for numbers which label the bodies.) This means that if the sub-system containing only those two bodies were placed in isolation (no external forces acting), then no net force would act on it. Although this may seem obvious, it does not hold for all types of force. Newton’s third law is not a law at all; it is just a description of a “nice” kind of force. We’ll symbolize the force on body i due to body j by Fij. In pictures, the third law looks like this: F 13 1 2 3 F 31 4 and in symbols, like this: F ij = − F ji . We’ll assume that a body exerts no force on itself, so F ii =0. (This is contained in the above equation.) Let’s also let F i be the external force acting on body i. We make no assumptions about its nature. Then Newton’s second law applied to body i reads m i ai = F i + 3 Fij . j system is zero. Then we have a conservation law: The right-hand side is the sum of the external force and all the internal forces acting on body i. Now let’s consider the total momentum of the system; it is the sum of the individual momenta: p tot = 3 m iv i . i How does the total momentum change in time? d 3 m v = 3 m a =3 F + 3 F dt i i i i i i i i i,j ij . The second term on the right-hand side is zero because the forces, taken in pairs, cancel out by Newton’s “third law”. Hence, we find dptot dt = Fext , where the sum of the external forces acting on the system is F ext = 3 F i . dp tot dt =0. This is not as trivial as it might seem, because it is true even when the internal forces are nonzero. The only thing that can change the total momentum is a nonzero total external force on the system. Example: collision of two bodies that stick together As our first example, suppose we have two bodies on the track. They move freely, collide and stick together, and the resulting body moves freely once again. The question is: what is the velocity of the resulting body? Let’s draw two diagrams showing the velocities before and after the collision: m1 v1 m2 M v2 w i The internal forces have no effect on the rate of change of the total momentum. Now let’s suppose that the total external force acting on the We’ll let w be the velocity of the final body, and M=m1+m2 be its mass. For consistency, we draw all velocities in the positive direction. If a velocity happens to turn out to be negative, then it points in the direction opposite to that shown in the diagrams. The statement that the bodies move freely means that no external forces act. Hence, the total momentum is conserved. m 1 v 1 + m2 v 2 = M w . Solving for the final velocity w, we find w= m1 v 1 + m2 v 2 . M You might be wondering: the law of conservation of momentum is all we needed, in order to solve for w. But what about conservation of energy? The answer is that kinetic energy appears not to be conserved, if we analyze the process at this level. To show this, simply calculate the final kinetic energy minus the initial kinetic energy. You will find m m ä1 ë 1 1 M w 2 − åååå m 1 v 21 + m 2 v 22 ìììì = − 1 2 (v 1 − v 2 )2 , 2 2M 2 ã2 í which is negative. Kinetic energy is lost. Where does it go? It must go into heat energy in the composite final system. The bodies making up the final system are macroscopic - they are made up of many atoms held together by interatomic forces. When the bodies collide and stick together, the atoms are jostled about and begin to move faster than before. This kind of energy is called heat. Experimentally, we would detect a rise in the temperatures of the bodies. If we looked on this finer scale, considering the kinetic energies of all the atoms making up the composite final system, we would find that the total kinetic energy is indeed conserved. Example: elastic collision of two free bodies Now let’s suppose that the two bodies, instead of sticking together, bounce off one another. We’ll assume that the collision is an ideal, elastic collision. This means that no energy is lost during the collision. (Our previous example was a completely inelastic collision.) We’ll also assume that the masses don’t change during the collision. We first draw the appropriate diagrams: m1 v1 m1 m2 w1 m2 v2 w2 Velocities after the collision are denoted by w’s. Suppose the initial velocities are known, and we want to solve for the final velocities. No external forces act, so we may invoke conservation of total momentum and of total energy. Conservation of momentum: before = after m 1 v 1 + m2 v 2 = m 1 w 1 + m2 w 2 Conservation of energy: before = after m1 2 m m m v 1 + 2 v 22 = 1 w 21 + 2 w 22 2 2 2 2 These are two equations in two unknowns, the w’s, so we have enough information to find a solution. The equations are quadratic, so there will be two solutions. Some very tedious algebra reveals them to be w 1 = v 1 and w1 = (m 1 − m 2 )v 1 + 2m 2 v 2 , m1 + m2 with w2 given by the same equations with 1 and 2 interchanged everywhere. What is the physical significance of these two solutions? Well, the first one just says that body 1 sails straight through without changing velocities at all. This is a valid solution, representing no collision. But it is not very interest- ing. The second solution is the one we are after. How can we check it? There are two ways. First, suppose the masses were the same. Then it’s more or less obvious that the two bodies just trade velocities. Setting m1=m 2 in the second solution, we indeed find w1=v2. Second, suppose that one of the bodies, say number 2, had zero velocity and were very massive. Then it’s as though the first body were hitting a brick wall. It would just bounce back with final velocity equal and opposite to its initial velocity. Setting v2=0 in the second solution and then taking the limit as m2 becomes infinite, we indeed find w1 approaches –v1. So, we are convinced that the solution makes physical sense. But if you tried to reproduce the solution for w 1 , you will know that it was obtained only after a lot of tedious algebra. In physics, that sometimes means that we have not found the best way of analyzing the problem. In fact, there does turn out to be a much better point of view from which to look at systems containing many particles. Let’s see what it is. • which is acted on by a force equal to the sum of the external forces acting on the bodies of the system. The center of mass frame We return to our equation for the rate of change of the total momentum of the system. Written in terms of the positions xi, it reads d2 3 m x = Fext . dt2 i i i v cm = We can make this look like Newton’s law for a single body whose total mass is the sum of all the masses in the system, by multiplying and dividing by the total mass M: M d 2 x cm dt2 = F ext 1 where x cm = 3m x M i i i In general, the center of mass moves as time goes on. The velocity of the center of mass is just the time derivative of xcm, or . This new quantity x c m is called the center of mass. It is the average of the positions of all the bodies, weighted by their masses. The above two equations say that the motion of the center of mass of the system is the same as the motion of a single body • whose mass is equal to the total mass of the system; • which is located at the center of mass of the system; and 1 3m v M i i i Now, let’s do a trick. We will view the whole system from a frame of reference which is moving along with the velocity of the center of mass. Let’s say that a body is moving at 10 m s-1 in the original reference frame, and that the center of mass is moving in the same direction at 6 m s-1. Then the velocity of the body when viewed in the center of mass frame is 4 m s- 1. To get the velocity in the center of mass frame, you just subtract vcm. We will symbolize quantities in the center of mass frame by putting primes on them. Then v iN = v i − v cm . The momentum of a body in the center of mass frame is just its mass times its velocity in the center of mass frame: p iN = m iv iN = m i(v i − v cm ) . Now for the punch line. We can show that the sum of the momenta in the center of mass frame is zero. 3 p iN = 3 m i(v i − v cm) i i = 3 m iv i − M vcm i =0 . The last line follows directly from the definition of the velocity of the center of mass. Application to previous example How does this help us in our previous collision example? Well, there are only two momenta before the collision, so they are equal and opposite in the center of mass frame: p 2N = − p 1N . There are only two momenta after the collision (we’ll call them q1 and q2), which are also equal and opposite to one another in the center of mass frame: q 2N = − q 1N . We now apply conservation of energy in the center of mass frame: p 1N 2 p 2N 2 q 1N 2 q 2N 2 + = + . 2m 1 2m 2 2m 1 2m 2 There are thus only two possibilities; either q 1N = p 1N and q 2N = p 2N , or q 1N = − p 1N and q 2N = − p 2N . The first says that both bodies just sail through the collision without changing their momenta; this is not interesting. The other says that the momenta of both bodies just reverse their directions, when viewed in the center of mass frame. Since the masses of the bodies don’t change during the collision, we therefore conclude that the velocities just reverse their directions, when viewed in the center of mass frame. This fact simplifies the analysis considerably, as we will now see. Let’s use this fact to get our solution for w1 the easy way. The first step is to find the velocity of the center of mass. It is v cm = m1 v 1 + m 2 v 2 . m 1 + m2 The second step is to find the velocity of body 1 in the center of mass frame: v 1N = v 1 − m1 v 1 + m 2 v 2 m2 = (v − v ) . m 1 + m2 m 1 + m2 1 2 The third step is to use what we just discovered the final velocity in the center of mass frame is just the negative of this: w 1N = − m2 (v − v ) . m 1 + m2 1 2 The last step is to transform back to the original frame by adding vcm. We find w 1 = w 1N + vcm = (m 1 − m 2 )v 1 + 2m 2 v 2 , m 1 + m2 which is the result we found before. Example: explosive fragmentation of a body Here’s another example. Suppose a body of mass M is flying along with constant velocity v. The body has a charge contained within it, and it explodes, releasing some energy ∆E. The body breaks into two pieces with masses m 1 and m 2 (such that m1+m2=M), and the energy released during the explosion appears as kinetic energy of the fragments. What are the final velocities of the two bodies? Let’s let the final momenta of the two bodies be q1 and q2. We go into the center of mass frame, in which we know that q 2N = − q 1N . q N = − q1N 2 m 2 m1 q1N Before the collision, the total energy in this frame was zero (because the original body was at rest in the center of mass frame!) After the collision, the total energy in this frame is ∆E . Hence, q 1N 2 q 2N 2 ∆E = + . 2m 1 2m 2 Solving, we find q 1N 2 = 2m 1 m 2 ∆E . M This has two solutions, corresponding to body 1 going in the positive or negative direction. Both are physically equivalent, so we’ll choose the positive one. Then the final velocity of body 1 in the center of mass frame is w 1N = q 1N = m1 2m 2 ∆E m1M . The velocity of body 2 in this frame is obtained by switching 1 and 2 in this formula, and multiplying by –1. Transforming back to the original frame, we find the result 2m 2 ∆E w 1 = v cm + w 1N = v + . m1M It is instructive to calculate the final total energy of the system in the original frame. We find E final = = and the fact that the sum of the momenta in the center of mass frame is zero: 3 m iv iN = 0 , i we find 3 i 1 1 m iv 2i = 3 m i ãäv 2cm + 2v cm v iN + v iN 2 ëí 2 i 2 äå 3 m ëì v 2 + v 3 m v N + 3 1 m v N 2 å i ì cm cm i i i i ãi í i i 2 1 1 m 1 w 21 + m 2 w 22 2 2 = 1 2 1 M v2 + ∆E , 2 = 1 1 M v2cm + 3 m iv iN 2 . 2 i 2 as it should be. The total energy is increased by ∆E, no matter which frame we calculate it in. Energy and the center of mass frame This last fact is an example of an important property of the center of mass frame: the total kinetic energy of a system is equal to the kinetic energy of the center of mass plus the sum of the kinetic energies of the bodies of the sytem, as calculated in the center of mass frame. The proof is just a short calculation. Using the velocity transformation law v i = v cm + v iN , The first term on the last line is the kinetic energy of the center of mass. Another way to write the last equation is E = E cm + E N . An example with an internal force Consider a pair of masses joined by a spring. The whole thing can move freely along the track. How does it move? Well, we know that the center of mass will just move with constant velocity, since no external forces act. And we expect that the length of the spring will oscillate about its equilibrium value. What will be the frequency? L x2 − x1 − L x1 The complete motion of the system is an oscillation of the relative coordinate superimposed upon the uniform motion of the center of mass. x2 The equations of motion are m 1 x¨ 1 = + k (x 2 − x 1 − L) m 2 x¨ 2 = − k (x 2 − x 1 − L) Adding these, we find m 1 x¨ 1 + m 2 x¨ 2 = 0 , which is just the statement that the center of mass moves with constant velocity. We also find ä k k ìëì x¨ 2 − x¨ 1 = − åååå + ìì (x 2 − x 1 − L) , ã m1 m2 í which has the solution x 2 − x 1 = L + A cos (ωt − δ) , where the frequency is ω= lates about its equilibrium value L. The constants A and δ are determined by initial conditions. k k + . m 1 m2 Hence, the relative coordinate x 2 –x 1 just oscil-