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The softest thing overcomes the hardest thing in the universe. ...Lao Tzu (570-490 BC)
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Author Topic: Impact force  (Read 13405 times)
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Fu-Kwun Hwang
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on: May 27, 2009, 09:47:18 am » posted from:Taipei,T\'ai-pei,Taiwan

Two objects with m1 and m2 respectively, released from the same height h to the ground.
Under the condition: mass m2=2*m1

The time for object to reach the ground are T1 and T2 respectively,
Normally, the air reststance can be ignored for heavy objects, and we know that
T2=T1=\sqrt{2h/g}, where g is the gravity.

If the impact forces are F1 and F2 respectively.
 What is the relation between F1 and F2?
A. F2>2*F1
B. F2=2*F1
C. F2<2*F1

The impact force F=△P/△T, where △P is the momentum change during the impact and △T is the impact time.
Both object drop from the same height, so the velocity when impact start is v=\sqrt{2gh},
So △P2=2*△P1.
We need to know the relation between △T2, △T1 in order to compare F2/F1.
If △T2= △T1, then F2=2*F1.
However, is it true that △T2= △T1Huh
If △T2> △T1 , how to find out relation between F1 and F2?
You can check out the answer with the following simulation:

The impact time △T and Faverege will be shown. F(t) is available, too!

The impact force is modeled with F(y)=-k*(y-y_0)-b*v_y.

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