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).

Chapter 7: Impulse and MomentumTuesday, September 17, 2013 10:00 PM

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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 slide your apparatus over to Buffalo, or Toronto, or anywhere, then you will find that the experimental results are the same. This is a very deep

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How does impulse affect motion? Consider the following calculation:

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:

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, front-ends of cars that "crumple" during collisions

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 Ns 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

Consider the impulse-momentum theorem,

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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?

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Car accelerating forwards; what are the external and internal forces acting on the driver-car system?

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Hockey stick hitting a puck; what are the external and internal forces acting on the puck?

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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

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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.

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.

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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:

Collisions in two dimensions

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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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Thus, the final velocity of the car-truck combination just after the collision 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.

This is the final velocity of the two cars together, just after the collision. If you prefer to express the velocity in terms of its magnitude and direction, then:

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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!_____________________________________________________________

Example: A rubber ball of mass 3 kg moving to the right at 5 m/s collides with a ball of mass 4 kg moving to the left. Immediately after the collision, the first ball moves at a speed of 1 m/s to the left, and the second ball moves at a speed of 1 m/s to the right. Determine the velocity of the second ball before the collision.

Solution:

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The total momentum before the collision is:

The total momentum after the collision is:

By the principle of conservation of momentum,

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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 (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?

The speed of Ball 2 before the collision is 3.5 m/s, and its velocity is negative, meaning it moves to the left before the collision.___________________________________________________

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Solution 1: Let's substitute the given information immediately and proceed to solve equations (1) and (2).

Our strategy is to solve equation (1) for one of the velocities and then substitute the resulting expression into equation (2):

Now substitute this expression into equation (2) and solve for the other velocity.

By conservation of momentum,

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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Solution 2: For more general insights, let's solve equations (1) and (2) in

Now substitute this value into equation (3) to determine the value of the other velocity after the collision.

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Solution 2: For more general insights, let's solve equations (1) and (2) in general, substituting given information only at the very end. Recall that equations (1) and (2) are

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Do these results make sense?_____________________________________________________

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

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

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:

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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

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

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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:

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.________________________________________________________

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The next example is a special case of the situation described by the most recent boxed red equations.

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 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?

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

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:

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Example: A rubber ball of mass 3 kg moving to the right collides elastically and head-on with a ball of mass 4 kg moving to the left. Immediately after the collision, the first ball moves at a speed of 1 m/s to the left, and the second ball moves at a speed of 1 m/s to the right. Determine the velocities of the two balls before the collision.

Solution:

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The first solution amounts to no collision, because the velocity of

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The first solution amounts to no collision, because the velocity of the first ball is the same both before and after the collision. (Substituting the first solution into equation (3) shows that the second ball also has the same velocity before and after the collision.) Thus, this solution is not interesting, and we can disregard it.

The second solution is the one that corresponds to an actual collision. Substituting the second solution into equation (3), we obtain

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 kinetic energy:

Conservation of momentum in the x-direction:

Conservation of momentum in the y-direction:

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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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! (Physics majors should definitely do

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

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

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Give it a try; a little algebra is good for you! (Physics majors should definitely do this.)_______________________________________________________________

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.

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:

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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.

https://www.youtube.com/watch?v=qB0Fq3UXEqA

Formula for centre of mass of several "point" objects:

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Example: Determine the centre of mass of the two bars considered as one unit.

Example: Determine the centre of mass for the system of three identical coins.

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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.

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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-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):

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The following argument is meant to justify treating an extended object, such as a baseball, as a point particle. We have tacitly made this assumption throughout the course, and indeed it's an excellent model and approximation to reality in most cases, but it's nice if we can provide a clear justification for the validity of this model, the "particle model."

Strategy: Pretend that the mass of each bar is concentrated at its geometrical centre.

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Two fun devices/phenomena

Galileo's cannon

Try dropping a basketball with a tennis ball sitting on top of it. Can you analyze the situation 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.

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.) _________________________________________________________

http://ap.smu.ca/demos/index.php?option=com_content&view=article&id=78&Itemid=85

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This was a popular "executive" toy a few decades ago.

Here's a Newton's cradle in action:

http://www.youtube.com/watch?v=fWSgm5aMsbU

Also see

http://ap.smu.ca/demos/index.php?option=com_content&view=article&id=85&Itemid=85

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Additional problems and solutions:

Example: Bob, who has a mass of 75 kg, can throw a 500 g rock with a speed of 30 m/s. The distance through which his hand moves as he accelerates the rock forward from rest until he releases it is 1.0 m.Determine the constant force Bob exerts on the rock.a.If Bob is standing on frictionless ice, what is his recoil speed after releasing the rock?

b.

Solution: Bob clearly has a "rifle-arm."

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Example: A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular to each other, at 20 m/s. The third piece has twice the mass as the other two. Determine the speed and direction of the third piece.

Solution:

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Example: A 10 g bullet is fired into a 10 kg wood block that is at rest on a wood table. The block, with the bullet embedded, slides 5.0 cm across the table. Determine the speed of the bullet. (The coefficient of friction is 0.20.)

Solution:

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Example: A 1500 kg weather rocket accelerates upward at 10.0 m/s2. It explodes 2.00 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

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Example: The figure shows a collision between three balls of clay. The three hit simultaneously and stick together. Determine the speed and direction of the resulting blob of clay just after the collision.

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Example: A 20 g ball is fired horizontally toward a 100 g ball that is hanging motionless from a 1.0-m-long string. The balls undergo a head-on, elastic collision, after which the 100 g ball swings out to a maximum angle of 50 degrees. Determine the initial speed of the 20 g ball.

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