Chapter 7: Impulse and Momentum Tuesday, September 17, 2013

Transcription

1 Chapter 7: Impulse and Momentum Tuesday, September 17, :00 PM In the previous chapter we discussed energy, and in this chapter we discuss momentum. The concepts of momentum and energy provide alternative perspectives to Newton's laws of motion and pathways into deeper understandings of Newtonian mechanics. The basic understanding of Newton's second law that we have discussed so far is that the acceleration of an object is caused by the net force acting on it. However, there are situations where it is difficult to measure the force acting on an object, and so therefore it's difficult to apply Newton's second law. For example, the force may act for only a very short time, or over a very short distance, or both. This is the case in collisions, for example; kicking a ball, hitting a baseball with a bat, a slap shot in hockey, hitting a tennis ball with a racquet, a car collision, etc. In such situations, it's typical that the force varies dramatically over a short time; this is much more complicated than the situations we've dealt with so far, where we often assumed that the force acting is constant (which is a reasonable approximation in some situations). Ch7Preliminary Page 1

2 How then do we analyze such situations? Well, we can think about averaging the force to make things simple. But should we average the force over time or over distance? Each of the ideas has advantages and useful properties; averaging the force over time leads to the concepts of impulse and momentum, whereas averaging the force over distance leads to the concepts of work and energy, as we studied in the previous chapter. Newton's second law of motion can be expressed in terms of these new concepts in the following ways: Impulse = change in momentum (impulse-momentum theorem) Work = change in energy (work-energy theorem) We'll see that there is a conservation principle for momentum, just as there is a conservation principle for energy. Conservation principles are very useful in physics; the world is a very complicated place, and so if you can identify some quantities that are constant throughout the complicated processes that you are analyzing, then it gives you something to hang on to. Conservation principles are useful in solving physics problems, and often allow one to solve problems more simply and more directly than using Newton's laws of motion by themselves. If you dig deeper into things, you'll find that conservation principles are a consequence of certain somewhat abstract symmetry principles. For example, the fact that momentum is conserved is a consequence of the fact that the fundamental laws of physics are invariant with respect to spatial translations. In other words, if you do an experiment here in St. Catharines, and then slid your apparatus over to Buffalo, or Toronto, or anywhere, then Ch7Preliminary Page 2

3 you will find that the experimental results are the same. This is a very deep connection between underlying symmetries of the universe and conservation principles; our deepest understandings of the universe are currently expressed in these terms of symmetry and invariance. Other examples: Conservation of energy is a consequence of invariance with respect to time translations, and conservation of angular momentum is a consequence of invariance with respect to spatial rotations. This is the story in Newtonian mechanics, but similar, and similarly deep (some would say more fundamental) symmetry principles apply in quantum mechanics, too, and you'll encounter them eventually if you continue your studies in this direction. Back to the story of how to cope with forces when they vary wildly in magnitude over very short times or over very short distances. Ok, let's now explore the effect of impulse on motion. Recall from above that the definition of impulse is the average force acting on an object multiplied by the time interval over which the force acts: How does impulse affect motion? Consider the following calculation: This calculation suggests that the quantity mv may be important, and so it's worthwhile giving it a name; we call it momentum. The relation that we've just derived above, which describes the effect of impulse on motion, is called the impulse-momentum theorem. Ch7Preliminary Page 3

4 examples: tennis, baseball, hockey, catching an egg or water-balloon or a hard-thrown ball, air-bag in a car Example: A baseball of mass 150 g is thrown towards home plate with a speed of 100 km/h. The batter hits the ball with an impulse of 10 N s so that it reverses its motion. Determine the speed at which the ball leaves the bat. Ch7Preliminary Page 4

5 Example: A tennis ball of mass 90 g arrives at your racquet with a speed of 80 km/h and you hit it directly back at a speed of 60 km/h. (a) Determine the impulse that you exert on the ball. (b) Determine the magnitude of the average force that you exert on the ball if it is in contact with your racquet for 14 ms. Solution: Choose "towards the right" to be the positive direction, and therefore "towards the left" is the negative direction. Ch7Preliminary Page 5

6 Principle of conservation of momentum Ch7Preliminary Page 6

7 Consider the impulse-momentum theorem, If it happens that the impulse is zero, then the change in momentum will also be zero. In other words, if the impulse acting on a system is zero, then the momentum of the system is conserved. Yet another way to say this is that if the net external force on a system is zero, then the total momentum of the system is conserved. Examples: Person sitting on a chair; what are the external and internal forces acting on the person-chair system? Car accelerating forwards; what are the external and internal forces acting on the driver-car system? Hockey stick hitting a puck; what are the external and internal forces acting on the puck? The principle of conservation of momentum is a generalization of Newton's third law of motion. That is, the principle of conservation of momentum is considered to be more fundamental than Newton's third law of motion, and furthermore Newton's third law of motion can be derived from the principle of conservation of momentum. Furthermore, the principle of conservation of momentum is more general than Newton's third law of motion because the former applies to light and fields as well as particles, whereas the latter applies only to particles. Furthermore, because force is the derivative of momentum, one can get by without the force concept as long as one uses momentum. This is true to an even greater extent in more advanced approaches to physics (quantum mechanics, relativity, field theories) where force is not very useful, but momentum is extremely useful. The principle of conservation of momentum is useful in analyzing collisions, explosions, etc. Consider the following example. Example: A car of mass 1000 kg travelling east at 40 km/h collides with a car of mass 1500 kg going west at 50 km/h in a completely inelastic collision (i.e. they stick together). Determine the velocity of the car/car combination immediately after the collision. Ch7Preliminary Page 7

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9 The velocity of the two cars together after the collision is 14 km/h to the West. I wonder if kinetic energy is conserved in the previous problem? Let's check this out: Kinetic energy is not conserved; since the gravitational potential energy does not change for the cars (the collision takes place on level ground) we conclude that mechanical energy is not conserved in the collision. What happened to the lost kinetic energy? Where did it go? Collisions in two dimensions We must be careful to recognize that momentum is a vector, not a scalar. In situations where momentum is conserved, carefully note that this means that each component of the momentum is separately conserved. This is illustrated in the following example. Example: A truck of mass 2000 kg travelling east at 60 km/h collides with a car of mass 1000 kg going north at 80 km/h in a completely inelastic collision (i.e. they stick together). Determine the velocity of the car/truck combination immediately after the collision. Ch7Preliminary Page 9

10 Solution: This is a two-dimensional problem, so we'll adopt the usual math-class conventions: Ch7Preliminary Page 10

11 This is the final velocity of the two cars together. If you prefer to express the velocity in terms of its magnitude and direction, then: Thus, the final velocity of the car-truck combination is 48.1 km/h at an angle of 33.7 degrees North of East. I wonder if kinetic energy is conserved in the previous problem? Check it out! Ch7Preliminary Page 11

12 Example: A rubber ball of mass 5 kg travelling to the right at 10 m/s collides with a ball of mass 10 kg going to the left at 4 m/s. After the collision, the first ball moves at a speed of 2 m/s to the left. Determine the velocity of the second ball immediately after the collision. Thus, the velocity of the second ball after the collision is 2 m/s to the right. Is kinetic energy conserved in the previous example? Check it out! Elastic and inelastic collisions Ch7Preliminary Page 12

13 In many every-day collision problems, kinetic energy is not conserved; you've seen this in the past few examples. However, there are some collisions where kinetic energy is nearly conserved; a collision of two billiard balls is an example. Even in this situation, you can understand that kinetic energy is not exactly conserved, because there is a little sound when the billiard balls collide, and that sound carries energy, which means that the final kinetic energy is very slightly less than the initial kinetic energy. In the ideal situation where the kinetic energy is exactly conserved in a collision, and we do know that physics textbooks often deal with ideal situations in order to keep things simple, the collision is called an elastic collision. A collision in which kinetic energy is not conserved is called inelastic. The following example illustrates an elastic collision. Example: A block of mass 2.4 kg moving to the right with a speed of 5.6 m/s collides head-on and elastically with a block of mass 1.6 kg that is initially at rest. Determine the velocities of the blocks immediately after the collision. Solution: The "head-on" phrase signifies that the motions take place in a single straight line, both before and after the collision. The fact that the collision is elastic means that kinetic energy is conserved. Choose "right" to be the positive direction. What do you think will happen after the collision, qualitatively? By conservation of momentum, By conservation of kinetic energy, Ch7Preliminary Page 13

14 By multiplying each term in the second equation by 2, and using the given value of the velocity of the second block before the collision, these two equations reduce to Ch7Preliminary Page 14

15 Do these expressions for the velocities of the blocks after the collision make sense? Can you check special cases? Do they correspond to what you've experienced at the billiards table, or at the bowling alley? To complete our problems, substitute the given values to obtain: Do these results make sense? Example: A block of mass m moving to the right with a speed of 4.3 m/s collides head-on and elastically with a block of mass 2m moving to the left with a speed of 2.9 m/s. Determine the velocities of the blocks immediately after the collision. Solution: Ch7Preliminary Page 15

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18 Another example of a collision in two dimensions Example: A billiards ball moves with a velocity of v B to the right and strikes a stationary billiards ball. After the elastic collision, one of the billiards balls moves with speed v 1 at an angle of 30 degrees from the horizontal line, and the second ball moves with speed v 2 at some other angle. Determine expressions for v 1, v 2, and the unknown angle in terms of v B. Solution: Remember that momentum is a vector quantity. In this situation, momentum is conserved; this means that each component of the total momentum is conserved. Because the collision is elastic, kinetic energy is also conserved. Also, we'll assume that both balls have the same mass. Conservation of momentum in the x-direction: Conservation of momentum in the y-direction: Conservation of kinetic energy: Ch7Preliminary Page 18

19 We now have three independent equations for the three unknowns (the speeds of the two balls after the collision and the unknown angle), so the problem is solvable. The angle seems to be the biggest pain, so let's eliminate the angle as a first step to solving the equations. A nice trick, worth remembering as a standard trick to try in these types of situations, is to solve Equation 1 for cosine theta, solve Equation 2 for sine theta, and then square the two expressions and add them to eliminate theta: Ch7Preliminary Page 19

20 There are two possible solutions here. The first solution, v 1 = 0, leads to theta = 0 and v 2 = v B. This solution represents the situation that the incident ball misses the stationary ball, and just continues on with its original speed. This is not interesting. (Although it is interesting that the solutions of the system of equations includes this non-interesting possibility.) The interesting solution is Substituting the expression for v 1 into Equation 3, we obtain an expression for v 2 : Substituting the expressions for v 1 and v 2 into Equation 2, we can solve for the angle theta: Ch7Preliminary Page 20

21 One can't help but notice that the sum of the two angles in the previous example is a right angle, a fact that will be familiar to all of you billiards players out there. It's an exercise in algebra to adapt the method of the previous example to prove that the two billiards balls always leave the collision so that the angle between them is a right angle. (Of course, we assume that the balls have the same mass and that the collision is elastic.) Give it a try; a little algebra is good for you! Centre of Mass The world is a very complicated place, as you know, with enormously complicated processes going on. In this course, when we analyze the motion of a tossed baseball, we don't consider the motion of each individual atom in the baseball, but we rather treat the baseball as a whole. This deserves a little more discussion. The motion of each individual atom in the baseball is quite complicated, and analyzing the motions of all of the atoms in a baseball is quite a long way beyond our capability. There are too many atoms, their motions are too complex, and we have no way of measuring all of them to see if our analysis would be correct, if it were even possible to carry out such an analysis. The atoms are vibrating, jiggling, etc., in an extremely complex way. Nevertheless, all of the complications are not very important, in a way. If all we want to do is to figure out where the baseball will land when we throw it, and that is the kind of thing that we want to do in this course, then we don't need to consider all the complicated individual motions of each atom in the baseball. We can quite successfully treat the motion of the baseball as a whole. This is what we did starting back in Chapters 2 and 3, and it's what we continue to do: we treat the extended object (baseball in this case) as if all of its mass were concentrated at some "average" location; this average location is called the centre of mass of the object. Ch7Preliminary Page 21

22 For a an object of uniform density, the centre of mass of the object is at its geometrical centre. If the density is not uniform, then the centre of mass will be offset from the geometrical centre towards the part of the object where the mass is more concentrated. The centre of mass of an object need not lie within the object itself, which leads to some interesting tricks. Consider a doughnut of uniform density; the centre of mass is at the geometrical centre, right in the centre of the hole: High-jumpers have figured out long ago how to make use of the fact that the centre of mass of a human body is not within the body when the body is contorted into a strange configuration. They have figured out how to jump over a bar while their centre of mass actually passes below the bar! Such efforts culminated in the invention of a technique called the Fosbury Flop after Dick Fosbury, who used it to win the gold medal at the 1968 Olympics. The following shows some pretty videos of a high-jumper doing the Fosbury Flop, over and over and over again; however, don't listen to everything they say about science (for example, no, power is not force divided by time), but what they do say about the rotation of the body at take-off is critical, and notice how the jumper arches his back and places his arms and legs. Formula for centre of mass of several "point" objects: Ch7Preliminary Page 22

23 Formula for centre of mass of several "point" objects: For continuous mass distributions, one can use a similar formula involving integration. You'll learn how to do this if you take MATH 1P02 or MATH 1P05. Example: Determine the centre of mass for the system of three identical coins. Ch7Preliminary Page 23

24 Example: Determine the centre of mass of the two bars considered as one unit. Strategy: Pretend that the mass of each bar is concentrated at its geometrical centre. If there are no external forces on a system, then momentum is conserved, which implies that the velocity of the centre of mass of the system is also conserved. You can see that this is true as follows; we'll derive the relation for the x-component, and the arguments for the y- Ch7Preliminary Page 24

25 component and z-component of momentum are the same. Start with the relation for the x-component of the centre of mass, assuming that there are three point-objects undergoing a collision (the argument is the same no matter how many point objects there are): Now divide each term of the previous equation by a suitable time interval, to obtain: The right side of the previous equation represents the x-component of the total momentum of the three particles. If there are no external forces acting on the system, then the quantity on the right side of the previous equation is conserved, because momentum is conserved. This means that the quantity on the left side of the equation is also conserved. If we extend the argument to the other two components of momentum, this means that if no external forces are acting on the system, the velocity of the centre of mass of the system is constant (i.e., conserved). You can also think of the quantity on the left side of the previous equation as the x-component of the centre of mass momentum of the system of three particles. This kind of reasoning justifies treating a baseball as if all of its mass is concentrated at its centre of mass when solving kinematics problems, doesn't it? (Well, the baseball is a continuous mass distribution, so we really should use calculus here, but you get the idea, I hope.) Ch7Preliminary Page 25

26 Two fun devices Galileo's cannon In class we dropped a basketball with a tennis ball sitting on top of it. Can you analyze the demonstration to explain why the tennis ball shoots up with such a high speed? What simplifying assumptions are reasonable in your calculation? Newton's cradle This was a popular "executive" toy a few decades ago. Here's a Newton's cradle in action: Ch7Preliminary Page 26