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Re: [Phys-L] real-world physics



You can accelerate ("decelerate") a falling object at all sorts of different rates (do it over a given time, or over a given distance), and have the resulting "final velocity" be in the same direction as the original direction, zero, or opposite to the original direction. Whether you count it as a bounce or not is probably a matter of defining the sign of the final velocity as being opposite to the original velocity.


I'm not sure saying the buffer is "long" has much meaning to it unless you say what the "length" of such a buffer is. This would require normalizing to an average acceleration of "g" or something like that.


 
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________________________________
From: brian whatcott <betwys1@sbcglobal.net>
To: Paul Nord <Paul.Nord@valpo.edu>
Cc: Phys-L@Phys-L.org
Sent: Thursday, June 7, 2012 10:35 AM
Subject: Re: [Phys-L] real-world physics

I fear that I shall need someone to draw a picture: take a case where an
object dropping near Earth is accelerating at -1g and via a very long
buffer is decelerated thru 0 to +1g. Does that object not bounce up?
Where is my misconception? Does -1g to  +1 g constitute this doubling? I
am think doubling -1 is -2 or doubling one is two. Would you straighten
me out please?

Brian


On 6/7/2012 9:23 AM, Paul Nord wrote:
If the bubble wrap acts like an ideal spring, completely reversing the motion does imply that a lower limit on the maximum acceleration of 2g.

In this case, it's simpler to consider the duration of the collision.  Starting at a speed of 120 mph (50 m/s) and changing that to zero implies an average speed during collision of 25 m/s.  Traveling a distance of 4 inches (0.1 m) at this speed means that the collision takes 0.004 seconds.  Knowing our delta-v and delta-t we can say that the average acceleration during that time is 12,500 m/s/s (1250 g).

That is the calculation that the Mythbusters should have done before attempting to build this.

Paul


On Jun 6, 2012, at 5:46 PM, brian whatcott wrote:

Hmmm..  with a compressing spring, an increasing force, providing an increasing retardation.
So certainly not constant acceleration.  I think I was responding to the idea of doubling the acceleration by the reversed direction, which seems also mistaken

Brian W


On 6/6/2012 11:54 AM, brian whatcott wrote:
Comparable speed at reciprocal heading, with constant acceleration possibly??

Brian W

On 6/6/2012 10:57 AM, Bernard Cleyet wrote:
On 2012, Jun 06, , at 08:19, John Clement wrote:

I wonder if the boxes were filled with bubble wrap.
A possible worsening, as if not popped, bouncing, doubling the acceleration.




  Since these are
professionals I suspect they first tried a dummy and that they know how to
calculate forces.

So testing an idea is not all that farfetched.
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Forum for Physics Educators
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