### Author Topic: Impact force  (Read 14725 times)

#### Fu-Kwun Hwang

• Hero Member
• Posts: 3062
##### Impact force
« on: May 27, 2009, 10:47:18 am »
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= ?T1
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\$.

Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list