Finding distance that a pad is compressed after being dropped

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SUMMARY

The discussion revolves around calculating the compression of a foam-rubber pad after an egg, weighing 56.3 g, is dropped from a height of 11.9 m. The egg reaches a velocity of 15.28 m/s just before impact and is brought to a stop by the pad in 6.32 ms. The user initially attempts to apply the energy conservation equation EP (mgh) = 1/2 kx² but struggles due to the absence of the spring constant. The correct approach involves using kinematic equations to determine the compression distance of the pad.

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This discussion is beneficial for physics students, educators, and anyone interested in understanding the dynamics of impact and material compression in 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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