Find the average force at distance x

In summary: It probably amounts to the same thing, but I would take an elastic collision in the reference frame of the moving wall.
  • #1
TachyonLord
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Homework Statement



A “superball” of mass m bounces back and forth with speed v between two parallel walls, as shown. The walls are initially separated by distance l. Gravity is neglected and the collisions are perfectly elastic.
If one surface is slowly moved toward the other with speed V, the bounce rate will increase due to the shorter distance between collisions, and because the ball’s speed increases when it bounces from the moving surface.
Find the average force at distance x.
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Homework Equations


$$\int\vec F \cdot \vec dx= ΔK$$
where ΔK is the change in kinetic energy
$$K = \frac {1} {2}mv^2$$

The Attempt at a Solution


I tried using momentum conservation and the above equation, but I seem to get stuck, and my brain goes blank.
 

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  • #2
Its a problem from Kleppner.
 
  • #3
Start by deriving (or writing down) an expression for the change in the ball's speed after the collision assuming that the moving wall's mass is much much larger that the ball's mass ##m##.
 
  • #4
TachyonLord said:
I tried using momentum conservation and the above equation, but I seem to get stuck, and my brain goes blank.

Note that there is an external force moving the wall so momentum of the ball may not be conserved.
 
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  • #5
I am a bit confused on how the problem defines the average force at distance x:
1) The wall as it moves from distance l to distance x, will be hit by the ball let's say N(x) times so the average force we want is ##F=\frac{\Delta P}{\Delta t}## where ##\Delta P## the total change in the momentum of the ball due to the N(x) collisions and ##\Delta t=\frac{l-x}{V}## where ##V## the speed of the wall.
2) The wall moves at distance x and then stops. The ball will have a different speed let's call it v(x). We calculate the average force from that moment and afterwards which will be ##F=\frac{2mv(x)}{\frac{2x}{v(x)}}=\frac{mv^2(x)}{x}##

So is it 1) or 2) we want to find?
 
  • #6
Delta2 said:
I am a bit confused on how the problem defines the average force at distance x:
1) The wall as it moves from distance l to distance x, will be hit by the ball let's say N(x) times so the average force we want is ##F=\frac{\Delta P}{\Delta t}## where ##\Delta P## the total change in the momentum of the ball due to the N(x) collisions and ##\Delta t=\frac{l-x}{V}## where ##V## the speed of the wall.
2) The wall moves at distance x and then stops. The ball will have a different speed let's call it v(x). We calculate the average force from that moment and afterwards which will be ##F=\frac{2mv(x)}{\frac{2x}{v(x)}}=\frac{mv^2(x)}{x}##

So is it 1) or 2) we want to find?
I think it's 1. The wall keeps moving until the ball is squashed when ##x=0##. We are looking for the average force from the beginning until when the wall is at an arbitrary value of ##x##.
 
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  • #7
PeroK said:
Note that there is an external force moving the wall so momentum of the ball may not be conserved.
For OP's clarification, you mean that if the ball's momentum is ##mv_0## when ##x=L##, then it is ##mv## when ##x<L## in which case ##v>v_0##. This is stated in the problem. However, during any given collision with the wall momentum is conserved under the assumption that the collision takes place instantaneously.
 
  • #8
kuruman said:
For OP's clarification, you mean that if the ball's momentum is ##mv_0## when ##x=L##, then it is ##mv## when ##x<L## in which case ##v>v_0##. This is stated in the problem. However, during any given collision with the wall momentum is conserved under the assumption that the collision takes place instantaneously.

It probably amounts to the same thing, but I would take an elastic collision in the reference frame of the moving wall.
 
  • #9
PeroK said:
It probably amounts to the same thing, but I would take an elastic collision in the reference frame of the moving wall.
Why not the "lab" frame? The expressions for the final velocities in terms of the initial velocities are standard and simplify considerably when ##m_2>>>m_1##. Maybe you are thinking of different solution from mine.
 
  • #10
kuruman said:
I think it's 1. The wall keeps moving until the ball is squashed when ##x=0##. We are looking for the average force from the beginning until the wall is at an arbitrary value of ##x##.

I interpreted this slightly differently. When the wall reaches ##x## you have a number of collisions round about that point. The frequency of the collisions depends on the speed of the ball at that point. That gives approx an average change in momentum per unit time at that point.

This assumes an appropriate approx for ##v >> V##.
 
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  • #11
PeroK said:
I interpreted this slightly differently. When the wall reaches ##x## you have a number of collisions round about that point. The frequency of the collisions depends on the speed of the ball at that point. That gives approx an average change in momentum per unit time at that point.

This assumes an appropriate approx for ##v >> V##.
You know, I initially had that interpretation, but I changed it (please don't ask why). I edited the post and put it back the way I initially had it.
 
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  • #12
PeroK said:
It probably amounts to the same thing, but I would take an elastic collision in the reference frame of the moving wall.
Yes, it will help to take it in the reference frame of the wall, but no, it does not amount to the same thing. There will be a gain in KE at each bounce, as we are given:
TachyonLord said:
and because the ball’s speed increases when it bounces from the moving surface

kuruman said:
during any given collision with the wall momentum is conserved
Even if the wall were stationary momentum would not be conserved; the direction has changed. Did you mean KE?
 
  • #13
haruspex said:
Even if the wall were stationary momentum would not be conserved; the direction has changed. Did you mean KE?
I think we are talking past each other. I consider a given ball-wall collision as the standard problem in which a wall of mass ##m_2## moving with velocity ##V## to the right collides with a ball of mass ##m_1## moving to the left with velocity ##v_i##. I use energy and momentum conservation to find the final velocity ##v_f## of the ball after the collision in terms of ##v_i## and ##V## and then consider what ##v_f## becomes when ##m_2>>>m_1##. In that limit, the velocity of the wall does not change but the velocity of the ball is what it is until the next collision. What possessed me to do this is that I can use ##a=\frac{\Delta v}{\Delta t}## to find an acceleration for the ball. That's the path I followed to get a solution. It would be interesting to compare paths and solutions in the end.
 
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  • #14
kuruman said:
I think we are talking past each other. I consider a given ball-wall collision as the standard problem in which a wall of mass ##m_2## moving with velocity ##V## to the right collides with a ball of mass ##m_1## moving to the left with velocity ##v_i##. I use energy and momentum conservation to find the final velocity ##v_f## of the ball after the collision in terms of ##v_i## and ##V## and then consider what ##v_f## becomes when ##m_2>>>m_1##. In that limit, the velocity of the wall does not change but the velocity of the ball is what it is until the next collision. What possessed me to do this is that I can use ##a=\frac{\Delta v}{\Delta t}## to find an acceleration for the ball. That's the path I followed to get a solution. It would be interesting to compare paths and solutions in the end.
Ah, ok. In that case it does come to the same as working in the frame of the wall.
 
  • #15
kuruman said:
I think we are talking past each other. I consider a given ball-wall collision as the standard problem in which a wall of mass ##m_2## moving with velocity ##V## to the right collides with a ball of mass ##m_1## moving to the left with velocity ##v_i##. I use energy and momentum conservation to find the final velocity ##v_f## of the ball after the collision in terms of ##v_i## and ##V## and then consider what ##v_f## becomes when ##m_2>>>m_1##. In that limit, the velocity of the wall does not change but the velocity of the ball is what it is until the next collision. What possessed me to do this is that I can use ##a=\frac{\Delta v}{\Delta t}## to find an acceleration for the ball. That's the path I followed to get a solution. It would be interesting to compare paths and solutions in the end.
Using energy and momentum conservation and in the limit M>>m I get that V'=V and v'=-v (where M,m the mass of the wall and the ball, V',V the velocity of the wall after and before collision and v',v the velocity of the ball, after and before collision. So the magnitude of the velocity of the ball doesn't change before and after each collision which contradicts the problem statement. Where do I go wrong?
 
  • #16
Delta2 said:
Using energy and momentum conservation and in the limit M>>m I get that V'=V and v'=-v (where M,m the mass of the wall and the ball, V',V the velocity of the wall after and before collision and v',v the velocity of the ball, after and before collision. So the magnitude of the velocity of the ball doesn't change before and after each collision which contradicts the problem statement. Where do I go wrong?
I will send you a PM. We should wait to hear from OP.
 
  • #17
kuruman said:
I will send you a PM. We should wait to hear from OP.

Yes ok I think I see now, let's just say that it is ##V'\approx V## and ##v'\approx-v## but there can be more accurate approximation for v'.
 
  • #18
Delta2 said:
Yes ok I think I see now, let's just say that it is ##V'\approx. V## and ##v'\approx.-v## but there can be more accurate approximation for v'.
Yes, it's not an approximation and the result is not that the velocity just reverses direction. It's the same principle as the slingshot effect, only smaller scale.
 
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  • #19
kuruman said:
I think we are talking past each other. I consider a given ball-wall collision as the standard problem in which a wall of mass ##m_2## moving with velocity ##V## to the right collides with a ball of mass ##m_1## moving to the left with velocity ##v_i##. I use energy and momentum conservation to find the final velocity ##v_f## of the ball after the collision in terms of ##v_i## and ##V## and then consider what ##v_f## becomes when ##m_2>>>m_1##. In that limit, the velocity of the wall does not change but the velocity of the ball is what it is until the next collision. What possessed me to do this is that I can use ##a=\frac{\Delta v}{\Delta t}## to find an acceleration for the ball. That's the path I followed to get a solution. It would be interesting to compare paths and solutions in the end.

... and I considered the wall to be moving inexorably and used separation velocities for an elastic collusion. Either way it does amount to the same thing as far as the change in KE of the ball is concerned.
 
  • #20
PeroK said:
... for an elastic collusion.
There is no collusion. I repeat, no collusion! :smile:
 
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  • #21
haruspex said:
Ah, ok. In that case it does come to the same as working in the frame of the wall.
What's the difference between working in the lab frame and working in the frame of the moving wall? In the frame of the moving wall an observer at rest sees the far wall moving towards him with speed ##V##. It's the same situation except for the initial velocity of the ball. Am I missing something?
 
  • #22
kuruman said:
What's the difference between working in the lab frame and working in the frame of the moving wall? In the frame of the moving wall an observer at rest sees the far wall moving towards him with speed ##V##. It's the same situation except for the initial velocity of the ball. Am I missing something?
It's the same when both are done correctly, of course, but working in the wall's frame avoids the temptation of treating the wall as stationary during the bounce.
 
  • #23
Thanks for clarifying.
 

Related to Find the average force at distance x

1. What is the formula for calculating average force at a given distance?

The formula for calculating average force at a given distance is F = m*a, where F is the average force, m is the mass of the object, and a is the acceleration at that distance.

2. How do you determine the distance in this equation?

The distance in this equation is typically given or can be measured using a ruler or other measuring tool. It represents the distance between the object and the point at which the average force is being calculated.

3. What units are used for average force and distance?

The SI unit for average force is newtons (N), and the unit for distance is meters (m).

4. Why is calculating average force at a specific distance important?

Calculating average force at a specific distance is important because it helps us understand the amount of force an object experiences at a particular point, which can be crucial in determining the object's motion, stability, and potential impact on its surroundings.

5. Can average force at a distance be negative?

Yes, average force at a distance can be negative. This typically occurs when the object is experiencing a deceleration or is moving in the opposite direction of the applied force.

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