Two objects with m

_{1} and m

_{2} respectively, released from the same height

**h** to the ground.

Under the condition: mass m

_{2}=2*m

_{1}The time for object to reach the ground are T

_{1} and T

_{2} respectively,

Normally, the air reststance can be ignored for heavy objects, and we know that

T

_{2}=T

_{1}=$sqrt{2h/g}$, where g is the gravity.

If the impact forces are F

_{1} and F

_{2} respectively.

What is the relation between F

_{1} and F

_{2}?

A. F

_{2}>2*F

_{1}B. F

_{2}=2*F

_{1}C. F

_{2}<2*F

_{1}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 ?P

_{2}=2*?P

_{1}.

We need to know the relation between ?T

_{2}, ?T

_{1} in order to compare F

_{2}/F

_{1}.

If ?T

_{2}= ?T

_{1}, then F

_{2}=2*F

_{1}.

However, is it true that ?T

_{2}= ?T

_{1} If ?T

_{2}> ?T

_{1} , how to find out relation between F

_{1} and F

_{2}?

You can check out the answer with the following simulation:

The impact time ?T and F

_{averege} 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

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Creative Commons Attribution 2.5 Taiwan License
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