cq_1_211

Your 'cq_1_21.1' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.

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Good answers. For much more detail see below. No need to revise; this info is included simply FYI.

&#Let me know if you have questions. &#

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

• 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?

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.