cq1_21_1 solution and discussion

The following is a solution to the given problem.

Please compare the given solution with your solution and submit a self-critique of any error(s) in your solutions, and/or and additional questions or comments you might have.

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####. Submit using the Submit Work Form.

A ball is tossed vertically upward and caught at the position from which it was released.

Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?

GOOD STUDENT ANSWER: The ball would be traveling at the same speed when it was released because gravity is a conservative force. This is the case when there are no factors, just the acceleration of gravity.

What, if anything, is different in your answer if air resistance is present? Give your best explanation.

GOOD STUDENT ANSWER: The ball would be traveling at a lower speed than when it was released because the ball would have lost energy during the toss due to air resistance.

A ball is tossed vertically upward and caught at the position from which it was released.

Ignoring air resistance will the ball at the instant it reaches its original position be traveling faster, slower, or at the same speed as it was when released?

The above answers are brief but to the point. More detailed analysis is given below:

We can answer this question in a number of ways.

Based on the equations of uniformly accelerated motion:

As the ball travels upward then downward, the acceleration acts through equal and opposite displacements, so the term 2 a `ds of the fourth equation is the same in both cases, except for the sign.

Going up, a and `ds are in opposite directions to 2 a `ds is negative, whereas coming down a and `ds are in the same direction so 2 a `ds is positive. Thus

2 a `ds_up = -2 a `ds_down,

with 2 a `ds_down being positive.

The fourth equation tells us that

vf^2 = v0^2 + 2 a `ds.

Going up we know that vf = 0, so

v0_up^2 = - 2 a `ds_up.

Going down we know that v0 = 0 so that

vf_down^2 = 2 a `ds_down.

Since

2 a `ds_up = -2 a `ds_down,

we conclude that

v0_up^2 = vf_down^2

so that

| v0_up | = | vf_down |;

i.e., the initial upward speed is the same as the final downward speed.

There's an even simpler way to answer based on the equations of motion.

The displacement `ds from start to end is zero, and the acceleration is constant throughout.
Since acceleration is constant we can treat the entire up-down motion as a single uniform-acceleration interval.
Since `ds for this interval is zero, we have 2 a `ds = 0.

Since vf^2 = v0^2 + 2 a `ds, we have vf^2 = v0^2 + 0, or vf^2 = v0^2 and | vf | = | v0 |.

The question can also be answered from the point of view of energy which bypasses the details of the equations of motion:

The net force up is the same as the net force down, and the displacement up is equal and opposite to the displacement down.

The work done by the net force on the way up is therefore equal and opposite to the work done on the way down, and the net work is zero.

Change in KE is equal to work done by the net force, so KE change must be zero, which implies that the speed of the object must be the same at both points.

In more detail the argument goes like this:

The net force up is the same as the net force down, and the displacement up is equal and opposite to the displacement down.

Thus the work done by the net force on the way up is equal and opposite to the work done by the net force coming down, so that the work done by the net force for the entire motion is zero.

Since KE change is equal to the work done by the net force, we conclude that KE change is zero.

KE change is the change in the quantity 1/2 m v^2. If there is no change in KE, there is no change in 1/2 m v^2. Since the mass of the object doesn't change, we conclude that v^2 must be the same at the beginning as at the end.

If v^2 is the same at two points, then we know that | v | is the same. Thus the speed has to be the same at both points.

Finally we can reason in terms of potential and kinetic energies. The change in gravitational PE going up is equal and opposite to the change going down, since the vertical displacements are equal and opposite. So the change in gravitational PE is zero. Only the gravitational force acts on the object, so nonconservative forces are not present and `dKE = - `dPE. Since `dPE = 0, `dKE = 0 and 1/2 m v^2, and therefore | v |, is the same at the beginning as at the end. [This argument can be further simplified; we don't have to worry about separating the up and down motions. All we need to know is that the altitude at the end is the same as at the beginning, so that `dPE from beginning to end is zero. The same conclusion about velocities follows immediately.

What, if anything, is different in your answer if air resistance is present? Give your best explanation.

One explanation:

Air resistance will enhance the slowing effect of gravity on the rising ball, which will as a result not rise as far. As a result of the decreased maximum height, the falling ball won't drop as far as it would if air resistance was not present, resulting in a lesser final speed.

Air resistance will then oppose the speeding up of the falling ball, so that the final velocity will be even less.

For a dense ball tossed gently upward, the effect of air resistance will be small, perhaps negligible with respect to the accuracy of our instruments. If the ball is less dense, and/or speeds are greater, air resistance will have a greater effect.

Explanation in terms of energy conservation:

In terms of energy conservation, assuming still air (i.e., no wind, no rising and falling of air), air resistance acts always in the direction opposite motion and therefore does negative work on the object.

PE doesn't change between the beginning and the end of the interval, so the result is a lesser KE at the end than at the beginning.

If we want to think in terms of the fourth equation of motion:

The direction of the air resistance (downward when the ball travels upward, upward when the ball travels downward) implies that the acceleration of the ball has greater magnitude on the way up than on the way down (downward air resistance reinforces the downward pull of gravity; upward air resistance works against the downward pull of gravity).

`ds has the same magnitude going up as coming down, so the 2 a `ds term has greater magnitude on the way up than on the way down.

Since 2 a `ds is negative on the way up and positive on the way down, the value of 2 a `ds for the entire motion is negative.

Since vf^2 - v0^2 = 2 a `ds, it follows that vf^2 - v0^2 is negative, i.e., v0^2 is greater than vf^2 so that | v0 | > | vf |.

Final speed is less than initial.

The following is a solution to the given problem.

Please compare the given solution with your solution and submit a self-critique of any error(s) in your solutions, and/or and additional questions or comments you might have.

Simply copy your posted document into a text editor and insert revisions, questions, and/or self-critiques, marking your insertions with ####. Submit using the Submit Work Form.

The above answers are brief but to the point. More detailed analysis is given below:

We can answer this question in a number of ways.

Based on the equations of uniformly accelerated motion:

As the ball travels upward then downward, the acceleration acts through equal and opposite displacements, so the term 2 a `ds of the fourth equation is the same in both cases, except for the sign.

Going up, a and `ds are in opposite directions to 2 a `ds is negative, whereas coming down a and `ds are in the same direction so 2 a `ds is positive. Thus

2 a `ds_up = -2 a `ds_down,

with 2 a `ds_down being positive.

The fourth equation tells us that

vf^2 = v0^2 + 2 a `ds.

Going up we know that vf = 0, so

v0_up^2 = - 2 a `ds_up.

Going down we know that v0 = 0 so that

vf_down^2 = 2 a `ds_down.

Since

2 a `ds_up = -2 a `ds_down,

we conclude that

v0_up^2 = vf_down^2

so that

| v0_up | = | vf_down |;

i.e., the initial upward speed is the same as the final downward speed.

There's an even simpler way to answer based on the equations of motion.

The displacement `ds from start to end is zero, and the acceleration is constant throughout.

Since acceleration is constant we can treat the entire up-down motion as a single uniform-acceleration interval.

Since `ds for this interval is zero, we have 2 a `ds = 0.

Since vf^2 = v0^2 + 2 a `ds, we have vf^2 = v0^2 + 0, or vf^2 = v0^2 and | vf | = | v0 |.

The question can also be answered from the point of view of energy which bypasses the details of the equations of motion:

The net force up is the same as the net force down, and the displacement up is equal and opposite to the displacement down.

The work done by the net force on the way up is therefore equal and opposite to the work done on the way down, and the net work is zero.

Change in KE is equal to work done by the net force, so KE change must be zero, which implies that the speed of the object must be the same at both points.

In more detail the argument goes like this:

The net force up is the same as the net force down, and the displacement up is equal and opposite to the displacement down.

Thus the work done by the net force on the way up is equal and opposite to the work done by the net force coming down, so that the work done by the net force for the entire motion is zero.

Since KE change is equal to the work done by the net force, we conclude that KE change is zero.

KE change is the change in the quantity 1/2 m v^2. If there is no change in KE, there is no change in 1/2 m v^2. Since the mass of the object doesn't change, we conclude that v^2 must be the same at the beginning as at the end.

One explanation:

Air resistance will enhance the slowing effect of gravity on the rising ball, which will as a result not rise as far. As a result of the decreased maximum height, the falling ball won't drop as far as it would if air resistance was not present, resulting in a lesser final speed.

Air resistance will then oppose the speeding up of the falling ball, so that the final velocity will be even less.

For a dense ball tossed gently upward, the effect of air resistance will be small, perhaps negligible with respect to the accuracy of our instruments. If the ball is less dense, and/or speeds are greater, air resistance will have a greater effect.

Explanation in terms of energy conservation:

In terms of energy conservation, assuming still air (i.e., no wind, no rising and falling of air), air resistance acts always in the direction opposite motion and therefore does negative work on the object.

PE doesn't change between the beginning and the end of the interval, so the result is a lesser KE at the end than at the beginning.

If we want to think in terms of the fourth equation of motion:

`ds has the same magnitude going up as coming down, so the 2 a `ds term has greater magnitude on the way up than on the way down.

Since 2 a `ds is negative on the way up and positive on the way down, the value of 2 a `ds for the entire motion is negative.

Since vf^2 - v0^2 = 2 a `ds, it follows that vf^2 - v0^2 is negative, i.e., v0^2 is greater than vf^2 so that | v0 | > | vf |.

Final speed is less than initial.