1. three conserved quantities-energy, momentum, and angular momentum,
2. how those conserved quantities are transferred between objects,
3. the relationships between kinetic and potential energies,
4. equilibrium in general and stable equilibrium in particular,
5. restoring forces in general and spring forces in particular.

We are going to examine your motion in this half pipe as you zip up and down the walls and across the bottom. For simplicity, we will neglect both friction and air resistance in the questions that follow. We will also assume that, apart from the curved regions that connect each wall to the bottom, the half pipe's walls are perfectly vertical and its bottom is perfectly horizontal. We will also neglect any details associated with your size and shape-in effect,you're just a single, small object moving around in a fancy bowl.

1. After a minute or two of effort, you let yourself coast back and forth between the two sides of the half pipe. You keeping rising to the same height on each wall as you shuttle from one side to the other. At one moment, (a), you are coasting toward the left across the flat, horizontal bottom of the pipe and at another moment, (b), you are coasting toward the right across the bottom. Compare

1. the net force on you,
2. your acceleration,
3. your speed,
4. your velocity,
5. your momentum,
6. and your total energy

at those two times, (a) and (b). For example, you might answer two of the six lettered parts:

X. Equal in amount (or magnitude) but opposite in direction.
Y. This quantity is zero at both times.

Answer:

1. The net force on you is zero at both times.
2. Your acceleration is zero at both times.
3. Your speed is the same at both times.
4. Your velocity is equal in amount, but opposite in direction.
5. Your momentum is equal in amount, but opposite in direction.
6. Your total energy is the same at both times.

Why: At the bottom of the half pipe, you are simply coasting. With an upward support force balancing your downward weight, you experience zero net force. You are therefore not accelerating and move at constant speed and velocity. After rebounding from one of the walls, you coast backward with the same speed, but in the opposite direction. This direction reversal makes your velocity and momentum opposite what they were originally. However, your total energy (which incidentally consists only of kinetic energy at this point) is the same regardless of which way you are heading. You haven't done any work on anything so you retain all of your total energy.

2. You get back to business and, after some effort, you find yourself rising twice as high up each vertical wall as you did during question 1. How does your new speed along the half pipe's horizontal bottom, as you head toward the left, compare with your speed at that point during question 1?

Answer: Your new speed is greater than before by a factor of the square root of two (approximately 1.414).

Why: To rise twice as high as before on the vertical walls of the half pipe, you need twice as much total energy. That's because at the top of your travels, all of your energy is in the form of gravitational potential energy (GPE) and GPE is proportional to height. Doubling your maximum height above the pipe's bottom doubles your maximum GPE and therefore your maximum total energy. When you travel across the flat bottom of the half pipe, all your energy is in the form of kinetic energy. To double your kinetic energy, you must travel not twice as fast as before, but a factor of square root of 2 faster (about 1.414 times as fast). That's because your kinetic energy is proportional to the square of your speed.

3. A friend joins you in the half pipe. The two of you are exactly the same size and weight. Your friend is motionless in the middle of the pipe and you are coasting leftward when the two of you accidently collide. You push against one another with your arms for 1 second and avoid injury. As the result of this pushing, you come to a complete stop and your friend is now moving with exactly the direction and speed of your motion before the collision.

But suppose that the two of you had pushed against one another for only 0.5 seconds, with the same resulting motions (your friend assumes your motion and you stop). How would this shorter time of pushing affected (A) the forces the two of you exerted on one another and (B) the impulse you give to your friend?

Answer: (A) The forces you exert on one another are larger than before by a factor of 2. (B) The impulse you give to your friend is the same as before.

Why: Since your friend starts at rest and you end at rest, you are clearly transferring all of your momentum to your friend. Since that momentum is transferred via an impulse, the impulse you give to your friend is the same, regardless of how quickly it occurs. However, to give the impulse in half the time, it must involve larger forces. Since impulse is equal to force times time, halving the time of the impulse requires that the forces involved increase by a factor of 2.

4. After the collision in question 3, your friend travels up and down the opposite wall and then bumps into you again. You were still motionless when your friend reached you. This time, however, you hold onto one another when you collide and begin moving together instead of separately. (A) How fast do the two of you move and (B) why?

Answer: (A) The two of you will move half as fast as your friend did alone. (B) That is because the two of your are now sharing your friend's initial momentum. Since your total mass is twice that of your friend alone and since momentum is velocity times mass, your velocity as a pair must be half that of your friend alone.

Why: When you grab onto one another, you share the total momentum that the two of you had to start with. Since your friend initially had momentum and you had none, the two of you must share your friend's momentum. Because momentum is equal to mass times velocity, sharing that momentum between twice as much mass requires the velocity to drop in half.

5. Your friend heads home and you are alone again in the half pipe. You get yourself going so hard that you begin to pop up above the top of each vertical wall and are briefly in free fall. While you are airborne above one of the walls, which of the following quantities remain constant?

1. momentum,
2. angular momentum,
3. total energy,
4. kinetic energy,
5. potential energy

   Answer: The only two quantities that remain constant are angular momentum and total energy.

   Why: In free fall, gravity is pulling on you unopposed and changing your momentum, so your momentum is not constant. However, gravity exerts no torque on you about your center of mass, so your angular momentum is constant. You do retain all your energy, but it changes from kinetic energy on your way up to gravitational potential energy at the top, to kinetic energy on the way down. Thus only your total energy remains constant.

6. The half pipe becomes crowded again, so you move along to a somewhat different structure. It is also U-shaped, but its bottom never flattens out completely: it curves continuously from its left vertical wall to its right vertical wall. As a result, you find yourself tending to settle into the low point in the middle of the structure. (A) Why is that lowest point a stable equilibrium and what are (B) the net forces and (C) the accelerations that you experience like when you are in or near that lowest point?

Answer: (A) The lowest point is an equilibrium position because you experience zero net force when you are there. This equilibrium is stable because any shift away from it gives rise to restoring forces that push you back toward the equilibrium position. (B) The net force on you is zero at the equilibrium position, but points toward the equilibrium position when you are shifted away from it. (C) Although you do not accelerate when you are at the equilibrium position, you accelerate toward it when you are shifted away from it.

Why: The lowest point in this structure is essentially at the botom of two ramps, one to the left and one to the right. When you ride up on either ramp, there is a downhill force that pushes you back toward the bottom. That returning push is a restoring force and makes the bottom point a stable equilibrium. Only at that bottom do you experience zero net force and zero acceleration.

7. Ever a generous soul, you decide to use your tremendous skills and experience to teach local school children how to skateboard. Although you rarely fall anymore, you remember how unpleasant it is to bump your knees on the bottom of the half pipe. To soften the blow for one of your tiny students, you rig up a giant overhead spring to act as a safety system.

The top end of the spring connects to a tower high above the half pipe and the bottom end of the spring connects to a harness worn by the pupil. You have selected the spring length and stiffness so that it exerts zero force on your pupil when your pupil is even with the top of the half pipe's vertical walls and exerts an upward force equal to your pupil's weight when your pupil is standing on the horizontal bottom of the half pipe.

One of the first things that your pupil does is to dangle about from the spring and find equilibrium. After a minute or two, you find your pupil hanging motionlessly from the spring, without touching the half pipe at all. (A) At what height is your pupil located (for example: one-third of the way up the half pipe from its bottom or two feet above the top of the half pipe) and (B) why?

Answer: (A) The pupil's equilibrium height will be at the bottom of the half pipe, where the pupil is just able to stand on the horizontal bottom surface. (The pupil will need slightly bent knees to avoid touching the bottom of the pipe.) (B) At that height, the upward spring force on the pupil is exactly equal in amount to the pupil's downward weight. These two forces sum to zero and yield a net force of zero on the pupil.

Why: At any height above the bottom of the half pipe, the spring will fail to fully support the pupil's weight. At any height below the bottom of the half pipe, the spring will more than fully support the pupil's weight. Only right at the bottom will the upward spring force and the downward weight balance perfectly so thatthe pupil is at equilibrium.

8. While this spring arrangement certainly lessens the pain of ordinary falls, your pupil eventually makes the mistake of jumping off the top of a vertical wall toward the middle of the half pipe and hits the bottom surface hard. Use the concepts of (A) net force, (B) acceleration, (C) momentum, and (D) energy to explain why this impact occurs and does not conflict with your answer to question 7.

Answer: The bottom of the half pipe is the pupil's equilibrium height, meaning that at that height the pupil experiences (A) zero net force and (B) zero acceleration. But having already dropped a substantial distance after jumping off the top of the wall, the pupil reaches this equilibrium height with a considerable amount of (C) downward momentum and (D) kinetic energy. The pupil coasts right through the equilibrium height and collides hard with the bottom of the half pipe. Just because the pupil is in equilibrium at the bottom of the half pipe doesn't mean that the pupil will stop there if that pupil has considerable momentum when equilbrium is reached. The pupil's momentum will propel the pupil right on past equilibrium and, in this case, into the bottom of the half pipe.

Why: Inertia can keep things going even if they are in equilibrium. While the net force on an object may vanish at equilibrium, the object's momentum remains.