In this question, we're presented with a scenario in which a block is dropped but is stopped at some point due to a rubber band stretching. We're then asked to determine the correct sequence of energy transfers in this process.

At first, the block and rubber band are held up and thus contain gravitational potential energy. Once the block is released, it falls straight down. As it falls, it loses gravitational potential energy and gains kinetic energy. At some point, however, the rubber band is stretched to a point where it stops the block from falling. Thus, all the kinetic energy of the block has now been transferred to the elastic potential energy of the rubber band.

For the unfamiliar, let's quickly define our rotational variables:

Moment of inertia for a point mass:

Rotational velocity:

This problem deals with the conservation of energy:

Before we begin manipulating this equation, let's make a few clarifications:

1. Let's make our initial condition when mass B is at its highest point and final condition when mass A is at its highest point

2. Since the masses are attached to a rigid rod and spinning in a uniform circle, we can see that 

3. Assume that a mass at its lowest point has a height 

4. We are discussing circular motion, so we will be using rotational velocities and rotational kinetic energies (In this problem, you will get to same answer using linear velocities. That is not always the though).

Moving on, let's define potential and kinetic energies:

Moving from left to right in our original equation for conservation of energy, let's expand each term:

Substituting all of these back into our original equation:

Before we move on, let's cancel out 2 terms; initial potential energy of mass A and final potential energy of mass B (see clarifications #1 and #3). Thus, we get:

Using clarification #2, we can reduce our equation further. Since we know  and are trying to determine , let's replace  and  to get:

Also, we know that , since those are both heights at the top of the circle. Using clarification #3, we can say that the height of the top of the circle is . So let's replace the heights in our equation to reduce it further:

Now let's substitute equations in for moment of inertia and rotational velocity (beginning of solution):

Note how all radii can be canceled out to obtain:

Factoring what's inside the parentheses, we get:

Rearranging, we get:

Rearranging to get velocity on its own side:

Rearranging for final velocity of mass A:

This is the final equation. Before we start plugging in values, let's check and make sure the units work out. We want the right side to have units of .

Time to plug and chug (we'll leave the units out to decrease cluster since we just proved that the units work):

This could be a lengthy and complex problem. However, both masses are the same. This simplifies the problem immensely since the total potential energy between the two masses never changes (they both change height at the same rate but in different directions, thus are always the same vertical distance away from point p). Therefore, kinetic energy is never going to change either and the velocity remains constant.

The question is essentially asking us if the system has enough kinetic energy in its horizontal position to make up for the difference in potential energy of the two masses when the rod is in its vertical position. In other words, is the initial kinetic energy equal to or greater than the final potential energy. Let's first calculate final potential energy (i.e. how much initial kinetic energy we need):

If we assume that point p is at a height of 0, we can remove initial potential energy and substitute the following expressions in for the height of each mass:

Substituting these into our original equation:

Factoring the expression:

We know each of these values, so time to plug and chug:

So that's how much energy we need to start with, let's see if we have it:

Factoring, we get:

Plugging in values:

We have more initial kinetic energy than we need, so yes, the rod will reach it's vertical position

By definition, the total mechanical energy of a system is always be conserved except in the presence of non-conservative forces, namely friction and air resistance.

Using conservation of momentum:

Plugging in values:

Initial kinetic energy:

Final kinetic energy:

Difference:

Plugging in values:

When the bowstring is drawn back, it acts like a spring and has potential energy:

When the string is released, the arrow turns all of that potential energy into kinetic energy:

Using conservation of energy to solve for velocity:

One thing to remember is that the mass of the arrow was given in grams, but kilograms are needed:

The first step is to figure out how much gravitational potential energy the wrecking ball has when it's lifted to it's maximum point.  To do this, we use trigonometry combined with the formula for gravitational potential energy:

Because the wrecking ball is at it's maximum height, this is also the wrecking balls total energy.  We then need to find the gravitational potential energy the wrecking ball would be at upon impact.  The difference between the beginning gravitational potential energy (total energy) and the end gravitational potential energy is the amount of kinetic energy the wrecking ball has at impact.

After we know the kinetic energy of the wrecking ball, we can figure out the velocity it has at impact.

Then we can multiply the now known velocity by the wrecking ball's mass to find its momentum. 

This problem begins out simply but will get increasingly complex since we need to account for the extension of the spring as it rotates. This problem will involve a wide variety of concepts. Let's begin with the expression for conservation of energy:

Let's designate the initial state as when the spring is released from horizontal and the final state is when the mass reaches its fastest velocity (lowest point). We can eliminate initial kinetic energy since the mass beings at rest. Let's also designate a reference height of 0 at 1 meter below the ceiling (this is where the mass's lowest point would be if the spring did not stretch). Now beginning with initial potential energy:

Where initial height 

 (1)

Moving on to final potential energy:

Where the final height is the displacement of the spring:

A negative sign is used with the displacement if we assume that stretching of the spring is a positive displacement. Now we need to develop an expression for the displacement of the spring. We can start with the expression for spring force:

Now we need an expression for force, using Newton's 2nd law:

Where the acceleration is the acceleration due to gravity plus centripetal acceleration:

Where the expression for centripetal acceleration is:

The problem statement tells us to assume that . Otherwise, we would have to say  and the problem would get very, very, very messy.

Plugging this into our expression for force:

Now rearranging for x:

Now plugging this back into our potential energy expression:

Let's clean this up a bit and multiply both sides by 2:

 (2)

We'll leave it like this for now and move on to final kinetic energy:

 (3)

Plugging equations 1, 2, and 3 back into our original expression for conservation of energy, we get:

Now we will spend the time solving this problem by rearranging for velocity. Let's begin by eliminating all fractions by multiplying both sides of the problem by :

Rearranging and factoring out velocity:

We know all of the values in the parenthesis so let's begin to plug them in. To keep things somewhat organized, let's do each term one at a time, beginning with :

Now :

Now the constant:

Putting these three terms together, we get:

Now using the quadratic formula to solve for , we get:

 or 

Squared values can't be negative so we will go with the first one. Therefore:

Note how this velocity is significantly lower than if this were a pendulum since the spring absorbed a large amount of energy from stretching.

Using conservation of energy:

Initially, all of the energy is gravitational potential energy, at the end, all of the energy is spring potential energy.

Solving for the spring constant

The height will be equal to the height above the spring when the object was dropped plus the distance the object compressed the spring.

Plugging in values: