Mechanics and Introduction to Relativity Lecture notes, lecture 7 PDF

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

10:00 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).

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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 slided your apparatus over to Buffalo, or Toronto, or anywhere, Ch7 Page 2

then 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.

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examples: tennis, baseball, hockey, catching an egg or water-balloon or a hard-thrown ball, air-bag in a car

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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.

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_________________________________________________________________ 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.

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Principle of conservation of momentum

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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.

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Solution: In the moment of the collision, the net external force acting on the car-car system is zero, so we can use the principle of conservation of momentum.

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:

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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?

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

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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:

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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! ________________________________________________________________ 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.

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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 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,

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By conservation of kinetic energy,

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

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Summarizing what we've derived, the velocities of the two balls after the collision are

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? _____________________________________________________

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Let's consider the equations we just derived above (in the red box) for a few more minutes, so we can bring hockey and baseball and bowling and billiards into the discussion.

You may have noticed that the faster you throw a tennis ball against a wall, the faster it bounces back at you. If the wall were moving towards you, instead of stationary, surely the same would be true, for the same reason. And indeed it is true, as you may have noticed from playing hockey or baseball. For what is a baseball bat or a hockey stick but a small moving wall? You can shoot a harder slap shot if a rebound comes straight at you than if you are shooting the puck from rest. It's easier to hit a home run off a fastball than it is off a knuckle ball or slow breaking pitch. We can understand this by examining the equations

Let m1 be the mass of the ball or puck and let m2 be the mass of the bat or stick. If you let the mass of the bat or stick be about 5 times (or so) the mass of the ball or puck, you'll get a sense for this effect by looking at the first equation: the greater you make the speed of the ball or puck before the collision, the greater is its speed after the collision. A complete analysis of this situation would be much more complex. The bat or stick is not motionless before the collision, and worse, it is moving in a very complicated way, including both translation and rotation. At impact, the player exerts complex forces on the bat or stick. But even ignoring all these complications, it's neat that we can get a sense for why this effect occurs by examining the equations for a much simpler phenomenon. ______________________________________________________ A good way to analyze the equations (in the previous red box) mathematically is to look at special or extreme situations. For example, if the two objects have the same mass, then to predict what happens we can let m2 = m1 in the boxed equations to obtain Ch7 Page 16

Thus, the incident object stops and all of its velocity is transferred to the second object. You can see this in action at the billiards table if you give the cue ball a little bit of back-spin, so that it slides without spinning. If the cue ball makes a direct hit on a second ball without spinning, then it does indeed stop dead and the second ball moves off in the same direction and with the same speed (nearly; the collision is not perfectly elastic) as the cue ball. If the cue ball spins, then our analysis is not good enough to predict what will happen, and we would have to take the cue ball's spin into account. As a second case, consider what happens if the second mass is much greater than the first mass, so that the ratio of the first mass to the second mass is close to zero:

This describes what happens when you throw a tennis ball against a wall. The tennis ball bounces back with the same speed, but opposite direction, that it had when it hit the wall. (The reality, of course, is that the collision is not perfectly elastic, so the speed of the ball after it bounces is always less than the in-going speed.) And the wall doesn't move at all after the collision. As a third case, consider what happens if the first mass is much greater than the second mass, so that the ratio of the second mass to the first mass is close to zero:

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This describes what happens when a bowling ball strikes a single pin. The mass of the bowling ball is much greater than the mass of the pin. When the bowling ball strikes the pin it just keeps going in the same direction at nearly the same speed, whereas the pin flies forward at nearly double the speed of the incoming bowling ball. ________________________________________________________ What if, instead of the incoming object striking a stationary object, we have two objects approaching each other in a head on elastic collision?

Can you set up the equations for conservation of momentum and conservation of kinetic energy and derive expressions for the final velocities of the two objects? Try it! (Physics majors should definitely try this as a test of their algebra skills.) Arrived at after a solid two pages of algebra (more if you include absolutely every step), here are my results:

There's a pleasing symmetry to the two expressions, isn't there? If you interchange the two masses, the results for the two final velocities should also be interchanged, and they are. Also, and this is very important, if you set the "before" velocity of the second object to zero, we should obtain, as a special case, the results of our

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previous big calculation (in the earlier red box), and we do. The result for the final velocity of the first object makes it clear that the faster the incoming pitch, the faster the outgoing batted ball. Is this clear? ________________________________________________________ The next example is a special case of the situation described by the most recent boxed red equations. 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 2 m moving to the left with a speed of 2.9 m/s. Determine the velocities of the blocks immediately after the collision. Solution:

Substituting the given values into equations (1) and (2), we obtain

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Another example of a collision in two dimensions Example: A billiards ball moves with a velocity of vB to the right and strikes a stationary billiards ball. After the elastic collision, one of the billiards balls moves with speed v1 at an angle of 30 degrees from the horizontal line, and the second ball moves with speed v2 at some other angle. Determine expressions for v1, v2, and the unknown angle in terms of vB. 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.

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Conservation of momentum in the x-direction:

Conservation of momentum in the y-direction:

Conservation of kinetic energy:

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:

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There are two possible solutions here. The first solution, v1 = 0, leads to theta = 0 and v2 = vB. 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

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Substituting the expression for v1 into Equation 3, we obtain an expression for v2:

Substituting the expressions for v1 and v2 into Equation 2, we can solve for the angle theta:

____________________________________________________________ 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 t...