Why does spring compression differ if falling balls experience the same acceleration?

[mprm86]

I don´t understand this: Accordingo to the second Newton´s Law, F=ma. You have a spring on the floor with constant k with an horizontal base over it. You drop a ball from different heights, h. Clearly, when h is great, then the string will be compressed a long distance. If you drop the ball from a not too big height, so it won't be compressed that much. This means that the force changes depending on the height, right? But, the acceleration for the two balls is the same, g, so the forces acting on them must be the same. So, what happens?

 

[dextercioby]

Heh,energy is conserved,both for the falling ball,and for the horizontal plate & spring.So a greater velocity on the impact (from a greater "h") implies a bigger amplitude of oscillation for the the body-spring system...Which also means that the max acc. & force (elastic) will be bigger,if "h" is big...

Daniel.

 

[mprm86]

Yes, i know that, but i still don´t get why the spring compresses more, because this would implies that the force that the ball dropped from a bigger height had is bigger than the one dopped from a low height. But this is not possoble, because the acceleration of both is g, and their masses are the same.

 

[cAm]

collisional forces work differently. I can't remember the equations exactly, but I am sure someone else here does.

 

[jdavel]

mprm86,

"the acceleration of both is g, and their masses are the same."

That's true until they hit the spring. But what does F=ma say about the acceleration of the masses after they hit the spring?

 

[dextercioby]

Collision means transfer of momentum & energy and the presence of contact forces...These forces cannot be calculated,really...

Daniel.

 

[Huckleberry]

mprm86, you are right to say that both of the balls would have the same acceleration, but they would have different velocities because the ball that was dropped from a higher starting position has more time to accelerate. The higher ball produces more force because it has more velocity. Try doing a google search on the formula f=ma and you will see some interesting things. Try these formulas for your problem instead f=d(mv)/dt or f=d(v1-v0)/(t1-t0). Hope those are right. Better check them first ;)

What was the question?
Huck