The maximum deformation of the ball?

AI Thread Summary
The discussion focuses on calculating the maximum deformation of a ball during a collision with a concrete floor. The initial calculations provided by the user yield an incorrect deformation value due to a sign error in the acceleration and initial speed directions. Participants emphasize the need to consider the total time of collision, which includes both compression and decompression phases, leading to conflicting interpretations of the problem. It is suggested that while acceleration may be constant during each phase, it can differ between compression and decompression. The complexity of the problem highlights the importance of correctly interpreting the conditions of the collision.
blueray101
Messages
6
Reaction score
0

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.
 
Physics news on Phys.org
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:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top