Finding distance that a pad is compressed after being dropped

AI Thread Summary
To determine how much the foam pad compresses when an egg is dropped from a height of 11.9 m, the egg's velocity just before impact is calculated to be 15.28 m/s. The discussion revolves around using equations of motion and impulse to find the compression distance, but there is confusion regarding the correct formulas and constants needed, particularly the spring constant. An attempt to solve for the distance resulted in a value of 0.144854 m, which was marked incorrect on the assignment page. Participants are encouraged to clarify the equations and approach to accurately compute the compression of the pad. The conversation highlights the challenges of applying physics concepts to real-world scenarios.
Greywolfe1982
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Homework Statement



An egg is dropped from a third-floor win-
dow and lands on a foam-rubber pad without
breaking.
The acceleration of gravity is 9.81 m/s2 .
If a 56.3 g egg falls 11.9 m from rest and the
7.29 cm thick foam pad stops it in 6.32 ms,
by how much is the pad compressed? Assume
constant upward acceleration as the egg com-
presses the foam-rubber pad. (Assume that
the potential energy that the egg gains while
the pad is being compressed is negligible.)
Answer in units of m.

Homework Equations



I'm not quite sure. I was thinking EP (mgh) = 1/2 kx2, but you aren't given the spring constant.

The Attempt at a Solution



I calculated that just before the egg hit the pad, its velocity was 15.28m/s. After this...I have no clue where to go.
 
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I think you probably have to do something with the impulse that the pad delivers to the egg:
The egg hits the pad with some initial velocity and then is brought to a complete stop before most likely bouncing up a little bit.
This is essentially an acceleration in the upward direction. Then you can use the equations:
d(x)=v(int)t=1/2at^2
v(final)=v(int) + at
to solve for your two unknowns.
Hope this helps, sorry my equations are not that easy to read (d=delta)
 
I think my brain is fried after doing physics for a few hours straight...but your first equation doesn't seem to make sense to me.

d(x)=v(int)t=1/2at^2

Two equals signs? How would you solve for x?
 
haha sorry the second equals was supposed to be a plus. my bad. Yea solving for x would have been interesting.
 
Tried that, got an answer of .144854 m, but according to my online assignment page, that's wrong. Is there a problem in my math, or is that the wrong formula to use?
 
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