26 Impulse and Momentum

Transcription

1 6 Ipulse and Moentu First, a Few More Words on Work and Energy, for Coparison Purposes Iagine a gigantic air hockey table with a whole bunch of pucks of various asses, none of which experiences any friction with the horizontal surface of the table. Assue air resistance to be negligible. Now suppose that you coe up and give each puck a shove, where the kind of shove that you give the first one is special in that the whole tie you are pushing on that puck, the force has one and the sae value; and the shove that you give each of the other pucks is siilar in the following respect: To each puck you apply the sae force that you applied to the first puck, over the sae exact distance. Since you give each of the pucks a siilar shove, you ight expect the otion of the pucks (after the shove) to have soething in coon and indeed we find that, while the pucks (each of which, after the shove, oves at its own constant velocity) have speeds that differ fro one another (because they have different asses), they all have the sae value of the product v and indeed if you put a ½ in front of that product and call it kinetic energy K, the coon value of 1 v is identical to the product of the agnitude of the force used during the shove, and the distance over which the force is applied. This latter product is what we have defined to be the work W and we recognize that we are dealing with a special case of the work energy principle W = K, a case in which, for each of the pucks, the initial kinetic energy is zero. We can odify our experient to obtain ore general results, e.g. a saller constant force over a greater distance results in the sae kinetic energy as long as the product of the agnitude of the force and the distance over which it is applied is the sae as it was for the other pucks, but it is interesting to consider how different it would see to us, in the original experient, as we ove fro a high-ass puck to a low ass puck. Iagine doing that. You push on the high-ass puck with a certain force, for a certain distance. Now you ove on to a low-ass puck. As you push on it fro behind, with the sae force that you used on the high-ass puck, you notice that the low-ass puck speeds up uch ore rapidly. You probably find it uch ore difficult to aintain a steady force because it is siply ore difficult to keep up with the low ass puck. And of course, it covers the specified distance in a uch shorter aount of tie. So, although you push it for the sae distance, you ust push the low-ass puck for a shorter aount of tie in order to ake it so that both pucks have one and the sae kinetic energy. Pondering on it you recognize that if you were to push the low-ass puck for the sae aount of tie as you did the high-ass puck (with the sae force), that the low-ass puck would have a greater kinetic energy after the shove, because you would have to push on it over a greater distance, eaning you would have done ore work on it. Still, you iagine that if you were to push on each of the pucks for the sae aount of tie (rather than distance), that their respective otions would have to have soething in coon, because again, there is soething siilar about their respective shoves. Now we Move on to Ipulse and Moentu You decide to do the experient you have been thinking about. You place each of the pucks at rest on the frictionless surface. You apply one and the sae constant force to each of the pucks for one and the sae aount of tie. Once again, you find this ore difficult with the lower ass pucks. While you are pushing on it, a low-ass puck speeds up faster than a high-ass puck does. As a result you have to keep pushing on a low-ass puck over a greater distance and 180

2 it is going faster when you let it go. Having given all the pucks a siilar shove, you expect there to be soething about the otion of each of the pucks that is the sae as the corresponding characteristic of the otion of all the other pucks. We have already established that the saller the ass of the puck, the greater the speed, and the greater the kinetic energy of the puck. Experientally, we find that all the pucks have one and the sae value of the product v, where v is the post-shove puck velocity. Further, we find that the value of v is equal to the product of the constant force F and the tie interval t for which it was applied. That is, F t = v The product of the force and the tie interval for which it is applied is such an iportant quantity that we give it a nae, ipulse, and a sybol J. J = (6-1) Also, as you probably recall fro chapter 4, by definition, the product of the ass of an object, and its velocity, is the oentu p of the object. Thus, the results of the experient described above can be expressed as J = p The experient dealt with a special case, the case in which each object was initially at rest. If we do a siilar experient in which, rather than being initially at rest, each object has soe known initial velocity, we find, experientally, that the ipulse is actually equal to the change in oentu. J = p (6-) Of course if we start with zero oentu, then the change in oentu is the final oentu. Equation 6-, J = p, is referred to as the Ipulse-Moentu Relation. It is a cause and effect relationship. You apply soe ipulse (force ties tie) to an object, and the effect is a change in the oentu of the object. The result, which we have presented as an experiental result, can be derived fro Newton s second law of otion. Here we do so for the case in which the force acting on the object is constant during the tie interval under consideration. Note that the force which appears in the definition of ipulse is the net external force acting on the object. Consider the case of a particle, of ass, which has but one, constant force (which could actually be the vector su of all the forces) acting on it. 181

3 As always, in applying Newton s second law of otion, we start by drawing a free body diagra: F a In order to keep track of the vector nature of the quantities involved we apply Newton s nd Law in vector for (equation 14-1): 1 = a F In the case at hand the su of the forces is just the one force F, so: Solving for F, we arrive at: a = 1 F F = a ultiplying both sides by we obtain F t = a Given that the force is constant, the resulting acceleration is constant. In the case of a constant acceleration, the acceleration can be written as the ratio of the change in v that occurs during the tie interval, to the tie interval itself. v a = Substituting this into the preceding expression yields: v = F t = The change in velocity can be expressed as the final velocity v (the velocity at the end of the tie interval during which the force acts) inus the initial velocityv (the velocity at the start of the tie interval): v = v v. Substituting this into F t = v yields which can be written as v = ( v v ) = v v 18

4 Recognizing that v is the final oentu and that v is the initial oentu we realize that we have F = p p On the left, we have what is defined to be the ipulse, and on the right we have the change in oentu (equation 6-): J = p This copletes our derivation of the ipulse oentu relation fro Newton s nd Law. Conservation of Moentu Revisited Regarding the conservation of oentu, we first note that, for a particle, if the net external force on the particle is zero, then the ipulse, defined by J =, delivered to that particle during any tie interval, is 0. If the ipulse is zero then fro J = p, the change in oentu ust be 0. This eans that the oentu p is a constant, and since p = v, if the oentu is constant, the velocity ust be constant. This result siply confirs that, in the absence of a force, our ipulse oentu relation is consistent with Newton s 1 st Law of Motion, the one that states that if there is no force on a particle, then the velocity of that particle does not change. Now consider the case of two particles in which no external forces are exerted on either of the particles. (For a syste of two particles, an internal force would be a force exerted by one particle on the other. An external force is a force exerted by soething outside the syste on soething inside the syste.) The total oentu of the pair of particles is the vector su of the oentu of one of the particles and the oentu of the other particle. Suppose that the particles are indeed exerting forces on each other during a tie interval. To keep things siple we will assue that the force that either exerts on the other is constant during the tie interval. Let s identify the two particles as particle #1 and particle # and designate the force exerted by 1 on as F. Because this force is exerted on particle #, it will affect the otion of particle # and we can write the ipulse oentu relation as F p = (6-3) Now particle #1 can t exert a force on particle without particle # exerting an equal and opposite force back on particle 1. That is, the force F1 exerted by particle # on particle #1 is the negative of F. F 1 = F Of course 1 F ( eff of on 1 ) affects the otion of particle 1 only, and the ipulse-oentu relation for particle 1 reads 183

5 Replacing F1 with F we obtain Chapter 6 Ipulse and Moentu F1 = p 1 = (6-4) p 1 Now add equation 6-3 ( F = p ) and equation 6-4 together. The result is: = p1 p + p1 + 0 = p On the right is the total change in oentu for the pair of particles ptotal = p1 + p so what we have found is that which can be written as 0 = p TOTAL p = 0 TOTAL (6-5) Recapping: If the net external force acting on a pair of particles is zero, the total oentu of the pair of particles does not change. Add a third particle to the ix and any oentu change that it ight experience because of forces exerted on it by the original two particles would be canceled by the oentu changes experienced by the other two particles as a result of the interaction forces exerted on the by the third particle. We can extend this to any nuber of particles, and since objects are ade of particles, the concept applies to objects. That is, if, during soe tie interval, the net external force exerted on a syste of objects is zero, then the oentu of that syste of objects will not change. As you should recall fro Chapter 4, the concept is referred to as conservation of oentu for the special case in which there is no net transfer of oentu to the syste fro the surroundings, and you apply it in the case of soe physical process such as a collision, by picking a before instant and an after instant, drawing a sketch of the situation at each instant, and writing the fact that, the oentu in the before picture has to be equal to the oentu in the after picture, in equation for: p = p. When you read this chapter, you should again consider yourself responsible for solving any of the probles, and answering any of the questions, that you were responsible for back in Chapter