How is the Average Force Calculated for a Superball Bouncing Between Two Walls?

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SUMMARY

The average force exerted by a superball of mass m bouncing between two walls with initial speed Vo is calculated based on the principles of elastic collisions. The change in velocity after each collision is definitively 2V, where V represents the speed of the wall. When one wall moves toward the other at speed V, the average force can be expressed in terms of the separation distance x between the walls. The discussion clarifies the definitions of variables and emphasizes the importance of understanding collision dynamics in this context.

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NOTE: I AM NOT POSTING A HW QUESTION.

It is the last problem, 4.29.

4.29 A "superball" of mass m bounces back and forth between two surfaces with speed Vo. Gravity is neglected a nd the collisions are perfectly elastic.
a. Find the average force F on each wall.
b. If one surface is slowly moved toward the other with speed V «v,
the bounce rate will increase due to the shorter distance between colli-
sions, and because the ball's speed increases when it bounces from the
moving surface. Find F in terms of the separation of the surfaces, x.
(Hint: Find the average rate at which the ball's speed increases as the
surface moves.)

Here is the link to the answer : http://physics141.uchicago.edu/2002/hw4.pdf

This is a problem from Kleppner and Kolenkow. I have a problem with the working. It is claimed that after every collision, the velocity changes by 2V.
My point is that initial velocity before striking the wall was Vo towards the left and after the collision, the velocity is Vo + 2V towards the right, thus making the change of velocity 2(Vo + V) and not just 2V. Can someone justify how it is that one can solve the problem as done above?
 
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For the moving wall, if the ball speed is Vb and the wall speed is Vw then the collision speed is (Vb + Vw) and the rebound speed is - (Vb + 2 Vw) (using a fixed frame of reference).

In the pdf file, V is used for the wall speed, and v is used for the ball speed, and it's noted that with each bounce cycle, the ball speed increases by 2 V, which seems to be the same as what you're asking. The distance between the walls is defined as x = L - Vt. (I swapped uppercase L for the lower case l in the pdf file). I think once you understand the variable names, then the rest of the problem solution makes sense.
 
Last edited:
Jeff Reid said:
For the moving wall, if the ball speed is Vb and the wall speed is Vw then the collision speed is (Vb + Vw) and the rebound speed is - (Vb + 2 Vw) (using a fixed frame of reference).

You mean velocity. Collision velocity, i.e. velocity of ball before impact, is only Vb, not Vb + Vw.
 

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