The maximum deformation of the ball?

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Homework Help Overview

The problem involves analyzing the collision of a ball with a concrete floor, focusing on the average acceleration during the collision and the maximum deformation of the ball. The scenario includes initial and final velocities, the time of collision, and the need to determine the distance traveled by the ball's center during deformation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations involving the time interval for deformation and whether the correct time value is being used. There is also a focus on the implications of assuming constant acceleration during the collision and how that affects the interpretation of the problem.

Discussion Status

Some participants have pointed out potential errors in the original poster's calculations, including a sign error and the interpretation of the time interval for deformation. There is an ongoing exploration of the assumptions regarding constant acceleration and how they may lead to conflicting conclusions.

Contextual Notes

Participants note that the distance traveled during compression must equal the distance traveled during decompression, raising questions about the validity of assuming constant acceleration throughout the entire collision.

blueray101
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Homework Statement


Part 1: I've worked out part 1 already. The answer is 4650m/s^2
A ball travels vertically downwards until it hits a concrete floor with speed 22.7-m/s. It then bounces vertically upwards at 10.8-m/s. Examination of a high speed video shows that the collision took 7.2-ms. Considering just the collision, what is the magnitude of the average acceleration?

Part 2:
For some balls, the acceleration of the center of the ball, in a collision like this, is fairly constant. So, assuming constant acceleration, what is the maximum deformation of the ball? (i.e. what is the maximum distance that the center of the ball travels downwards?)

Homework Equations



The 0 next to x and v are subscripts.
x=x0+v0*t+0.5*a*t^2

Rearranged into:

-(v0*t+0.5*a*t)=x0

The Attempt at a Solution


I basically just subbed in the values.
x0=-(-22.7*7.2*10^-3 +0.5*-4650*(7.2*10^-3)^2)
=0.283968
Aprox. 0.284

Apparently, my answer is wrong.
 
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blueray101 said:
I basically just subbed in the values.
x0=-(-22.7*7.2*10^-3 +0.5*-4650*(7.2*10^-3)^2)
=0.283968
Aprox. 0.284

Apparently, my answer is wrong.

Are you taking the correct time interval of travel of the ball in deformation?
 
drvrm said:
Are you taking the correct time interval of travel of the ball in deformation?
As far as I am aware, t=7.2 milliseconds which is 7.2*10^-3. Am I suppose to use a different value?
 
blueray101 said:
As far as I am aware, t=7.2 milliseconds which is 7.2*10^-3. Am I suppose to use a different value?

but , is total time of collision is time interval for deformation?
one can think that after deformation their is a recoil which may lead to velocity of return after collision- that is total time of collision has two prts. think about it.
 
dvrm has pointed out one key error, but there is a more egregious one. You have a sign error, leading to an answer an order of magnitude too large. The initial speed and the acceleration are in opposite directions, so the ut and at2 terms should partly cancel, not reinforce.
More subtly, there is a flaw in the question.
blueray101 said:
For some balls, the acceleration of the center of the ball, in a collision like this, is fairly constant. So, assuming constant acceleration, what is the maximum deformation of the ball?
It certainly reads as though you are to take acceleration as constant throughout the collision, but unfortunately that is effectively contradictory information. The distance traveled during compression must equal the distance traveled during decompression. If tne acceleration is constant throughout then the rebound speed must equal the impact speed. Given conflicting data, different solution methods, each valid, can lead to different answers.
Instead, you can take the acceleration to be constant during each phase, but different in each. There is enough information to solve on that basis.
 
Last edited:

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