Can you deduce the coefficient of restitution from impace force plot?

AI Thread Summary
The discussion revolves around whether the coefficient of restitution can be deduced from an impact force plot, where force is plotted against time. Participants explore the relationship between the areas under the curve and the slopes of the force graph, suggesting that these may relate to the coefficient of restitution. It is noted that the area under the curve represents impulse, but this alone does not provide information about energy loss or velocity changes. The conversation emphasizes the need to analyze the compression and decompression phases separately to derive equations that relate momentum and the coefficient of restitution. Ultimately, while the coefficient may be approximated, a definitive answer requires further analysis beyond the plot itself.
mokwana1221
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If you have the impact force on the y-axis (in Newtons) and the time on the x-axis (in milliseconds), can you get the coefficient of restitution from just this? The plot is for the force of impact exchanged between two balls as a function of time.

I know the coefficient of restitution is the change of velocities of the balls just after impact divided by the change just before.

Thanks
 
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welcome to pf!

hi mokwana1221! welcome to pf! :smile:

hint: what does the area under the curve represent? :wink:

(in the centre-of-mass reference frame, the coefficient of restitution is also the square-root of the ratio of kinetic energy after to kinetic energy before)
 
hmm it gives us the impulse (Newton*Seconds) but can the coeff. of restitution be deduced from that? i am still kind of lost unfortunately
 
also the area under the graph is triangular
 
hopefully somebody can help me out before my final tomorrow, thanks guys
 
I gather that the graph shows the force increasing linearly to a max then decreasing linearly to zero, the two slopes being different.
Intuitively, with restitution = 1 the two slopes would be the same. For less than 1, the second slope would be steeper. So it seems reasonable that the restitution is related to the ratio of the two areas.
Try considering the impact in those two phases. Let the two slopes be s1, s2, lasting t1, t2. The compression would peak, would it not, when the two bodies are at the same velocity? That should enable you to get enough equations.
 
have you done integrals? area represents an integral
 
tiny-tim said:
have you done integrals? area represents an integral
Hi tiny-tim.
I believe mokwana1221 established at post #3 that this much is understood. But that only gives the momentum exchange. It will not in itself say anything about the energy loss. See my previous post.
 
i see now i am kind of getting it. yes the two slopes are different (and naturally the two areas under the graph as well, divided by the peak).

1) can we get a definitive quantitative answer using this graph for the coefficient of restitution?

2) if yes, would it be the ratio of the two slopes, or the ratio of the two areas under the peak?

this was under the professor's 'advance topics' for the exam as we did not cover it but i am putting in the extra mile just in case he throws it on the exam

thanks
 
  • #10
the area of the first triangle is 10 and the area of the second is 5, the ratio is .5.

the slope of the first triangle is 5 and the slope of the second triangle is 10. the ratio is also .5.

so my questions is, is the coefficient of restitution .5? or am i missing something? or is it not possible to get the coefficient of restitution from just this plot (if this is the case, what other data is needed to get the coefficient of restitution)?

thanks guys i feel like this is a simple problem and that i am just missing something.
 
  • #11
mokwana1221 said:
t
so my questions is, is the coefficient of restitution .5?
Perhaps - or it could be a more complicated relationship involving a square or square root. It cannot be guessed at. Try what I suggested: consider the compression and decompression phases separately so that the speeds are the same at peak compression. Write out equations relating speed, momentum, time, whatever for start and end of the two phases. See where it leads.
 
  • #12
i am truly at a dead end. i do not see how one can get energy or velocity from this graph since multiplying force (N) times time (milliseconds) equals impulse.

i have been thinking and googling this question since i first posted and joined the forum hours ago. i am stuck.
 
  • #13
can anyone help me with a last effort?

thanks, kwan
 
  • #14
Have you tried what I suggested? Consider the compression and decompression phases separately so that the speeds are the same at peak compression.
Let the masses m1, m2, start at speeds u1, u2 (same straight line), reach a common speed of v at peak compression, and finish at speeds w1, w2.
Let the impulse during compression (i.e. area under first triangle) be I and that during decompression be J.
Let the coefficient of restitution be R.
Write out equations relating the above:
- four for momentum
- one from the definition of coefficient of restitution
If you do all that you'll be surprised how easy it is to get the answer.
 
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