Momentum question -- average over time or average over distance?

[imsmooth]

I saw this answer to the problem:
a 0.14kg ball traveling at 50 m/s is stopped by a glove. The glove moves 0.5m. What is the average force exerted by the ball?
https://answers.yahoo.com/question/...H2&p=baseball of mass 0.14kg moving at 50 m/s

How come you can't do this (which give a close answer of 350N)
kinetic energy of ball = work done by ball on glove
1/2 * 0.14kg * 50 (m/s) ^2 = Favg * 0.5m
Favg = 350N

 

[nasu]

You can. The problem here is that you can have average over time or average or distance and they are not equal, in general.
By your method you have an average over distance. Your average force times the distance will give the actual work.
In their solution the average is over time. Their average force times the time will give the actual impulse (equal to change in momentum).

The question does not specify what average. There was a disscution on the forum some time ago about what would be the "most common" meaning of the average force, when the problem does not specify.

 

[Herman Trivilino]

I saw this answer to the problem:
How come you can't do this (which give a close answer of 350N)

There's an error in the solution presented in that link. Fix that error and you get an answer that matches yours!

(Fnet)t = m(Vf - Vi)
(Fnet)(t) = 0.14 kg(0 - 50.0 m/s)
(Fnet)(t) = -7.5 kg.m/s

$(0.14)(50) = 7.0 \neq 7.5$

 

[nasu]

Their answer is conceptually flawed. They assume a constant acceleration which is not given and not necessary.
As they give the distance, your method is more reasonable because does not introduce artificial assumptions.

 

[Herman Trivilino]

The two solutions are equivalent. The phrase "find the average force" means the same thing as "find the force assuming it's constant". The relation

$\frac{1}{2}mv^2=Fd$

is valid only when the force is constant. That's equivalent to constant acceleration.

 

[nasu]

The relation $1/2 mv^2 =W_{net}$ is valid only when the initial velocity is zero.
If the net force is constant then we have $W=F d$
If the force is not constant over the distance d, we can define an average force by
$W=F_{average} d$
If we use this expression in the work-energy theorem, we have (with no initial velocity)
$1/2 mv^2 =F_{ave} d$ and we can use this relationship to find the average force.
No extra assumption is necessary (like the force being constant).
If the force were constant, what would be the point to even mention "average force" in the statement of the problem?

This is the average over distance. You can define average over time. If the force is constant the two averages are equal.

 

[Herman Trivilino]

If the net force is constant then we have $W=F d$
If the force is not constant over the distance d, we can define an average force by
$W=F_{average} d$

So using the constant force $F$ gives you the same result as using $F_{average}$ when the force is not constant. That is my point exactly.

The two solutions are equivalent. The only reason the answers came out different is because of the arithmetic error I pointed out. Fix that, and both solutions give the same answer, that's not a coincidence! The difference between a time average and a distance average is not relevant. In both solutions the same average is being used because $a_{ave}=F_{ave}/m$.

 

[imsmooth]

Thanks for the replies. I never checked the math on the link. A bad assumption on my part. This makes sense that both are equal because both methods seem correct, and I would think that both should yield the same answer.

 

[nasu]

So using the constant force $F$ gives you the same result as using $F_{average}$ when the force is not constant. That is my point exactly.

The two solutions are equivalent. The only reason the answers came out different is because of the arithmetic error I pointed out. Fix that, and both solutions give the same answer, that's not a coincidence! The difference between a time average and a distance average is not relevant. In both solutions the same average is being used because $a_{ave}=F_{ave}/m$.

Of course there is a difference and is relevant. The reason you get the same answer is that you make a conceptual mistake in assuming constant force where there is not the case.

There is nothing to dispute. You can easily see the difference taking a simple case with variable force, like harmonic oscillator with spring constant k and mass m.
Assume we release the mass at maximum displacement,d, with zero velocity.
At the equilibrium position it will have velocity $v=\omega d$.
The position average of the force over this interval will be
$<F>_{d}=\frac{1}{d}\int_0^d F(x)dx =\frac{1}{d}\int_0^d k x dx = \frac{1}{2} kd$
The time averaged force over the same interval will be
$<F>_{t}=\frac{1}{T/4}\int_0^{T/4} F(t)dt =\frac{1}{T/4}\int_0^{T/4} m \omega^2 x(t) dt =\frac{4 m \omega}{T} \int_0^{T/4} x(t) dt =\frac{2}{\pi}kd$
Clearly not the same thing.

Now trying to calculate the time average by assuming a constant force and kinematic equations for constant acceleration (as in the solution linked) will give you the the value of the position average. But this does not mean the "method" is justified. You can get the same expression by many nonsensical series of manipulations.

 

[Herman Trivilino]

The position average of the force over this interval will be
$<F>_{d}=\frac{1}{d}\int_0^d F(x)dx =\frac{1}{d}\int_0^d k x dx = \frac{1}{2} kd$

Note that nothing in that analysis involves the amount of time taken for the process to occur. You are therefore free to choose an amount of time over which to time-integrate that force such that the resulting impulse equals the product of the values for that same average force and that chosen time interval.

In other words, integrate with respect to distance and get an amount of work that equals the position-averaged force of 350 N times the given distance. Integrate with respect to time and get an impulse that equals the time-averaged force of 350 N times the chosen time interval.

There is nothing whatever conceptually wrong with using an average value of the acceleration in the kinematics equations for motion with constant acceleration. It makes no difference if the average value is a distance-average or a time-average. Your argument that they are in general different is not relevant because values of distance and time can be chosen so that they are in particular equal.

 

[PeroK]

Note that nothing in that analysis involves the amount of time taken for the process to occur. You are therefore free to choose an amount of time over which to time-integrate that force such that the resulting impulse equals the product of the values for that same average force and that chosen time interval.

In other words, integrate with respect to distance and get an amount of work that equals the position-averaged force of 350 N times the given distance. Integrate with respect to time and get an impulse that equals the time-averaged force of 350 N times the chosen time interval.

There is nothing whatever conceptually wrong with using an average value of the acceleration in the kinematics equations for motion with constant acceleration. It makes no difference if the average value is a distance-average or a time-average. Your argument that they are in general different is not relevant because values of distance and time can be chosen so that they are in particular equal.

In this particular question, if we do not assume a constant force (wrt time), then $F_{t-ave} = \frac{mv}{t}$ where $t$ can be as large as you wish. The problem, therefore, has no single solution (in terms of time average of force) unless you assume constant force.

The distance average force is, however, always $F_{d-ave} = \frac{mv^2}{2d}$.

So, I'm not sure what you're saying. Yes, you can always choose $t$ such that $F_{t-ave} = F_{d-ave}$, but that's not the only option for $t$ unless the force is constant.

 

[nasu]

@ Mister T
Yes, and you can choose your condition in such a way that you are always right, no matter what statement you make initially.

I can always find some way to get the same numerical value of the position average force for motion from A to B.
But this does not mean that the time average for the motion from A to B is the same as the position average from A to B.

If you adjust the time to get the same value, you simply find something else, not the time average.
Can you take that value, multiply by the time to go from A to B and get the momentum at B? This is the meaning of the time average.
You cannot the time and distance arbitrary, they are related by the kinematic equations of the specific motion.

 

[Herman Trivilino]

So, I'm not sure what you're saying. Yes, you can always choose $t$ such that $F_{t-ave} = F_{d-ave}$, but that's not the only option for $t$ unless the force is constant.

What I'm saying is that it is an option. That even in cases where the force is not constant it's possible to choose a value of $t$ such that $F_{t-ave} = F_{d-ave}$ will be true for any given value of $d$.

 

[Herman Trivilino]

Yes, and you can choose your condition in such a way that you are always right, no matter what statement you make initially.

I can always find some way to get the same numerical value of the position average force for motion from A to B.
But this does not mean that the time average for the motion from A to B is the same as the position average from A to B.

If you adjust the time to get the same value, you simply find something else, not the time average.
Can you take that value, multiply by the time to go from A to B and get the momentum at B? This is the meaning of the time average.
You cannot the time and distance arbitrary, they are related by the kinematic equations of the specific motion.

None of that is relevant to the two points you made. The first was that the two results were different because $F_{t-ave} \neq F_{d-ave}$. When I pointed out that the solutions were not different when the arithmetic error is avoided, you then claimed that the poster's solution was not valid because he assumed the acceleration is constant. My point was, and still is, that it's perfectly valid to use a value for $a_{ave}$ in that situation. It is in fact done repeatedly and it always gives the right answer because it's a valid method.

It is not a coincidence that you get the same answer both ways. The two solutions are logically equivalent. I can start with either as a premise and show that the other is a consequence.

 

[PeroK]

What I'm saying is that it is an option. That even in cases where the force is not constant it's possible to choose a value of $t$ such that $F_{t-ave} = F_{d-ave}$ will be true for any given value of $d$.

You just need the average velocity (wrt time) to be $v/2$.