Question: Newton's 2nd Law and Dropping a Ball from Different Heights

  • Context: Undergrad 
  • Thread starter Thread starter mprm86
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Discussion Overview

The discussion revolves around the implications of Newton's second law (F=ma) in the context of dropping a ball from different heights onto a spring. Participants explore the relationship between the height from which the ball is dropped, the resulting force on the spring, and the acceleration of the balls during the fall and upon impact.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how the force on the spring can change with height if the acceleration due to gravity (g) remains constant for both balls dropped from different heights.
  • Another participant suggests that energy conservation implies a greater velocity upon impact for the ball dropped from a greater height, leading to a larger amplitude of oscillation in the spring system.
  • A participant expresses confusion about the relationship between the spring's compression and the forces acting on the balls, asserting that the acceleration should remain constant.
  • One reply introduces the idea that collisional forces operate differently and that the equations governing them may not be straightforward.
  • Another participant emphasizes that while both balls have the same acceleration before hitting the spring, their velocities differ due to the time spent accelerating, which affects the force exerted upon impact.

Areas of Agreement / Disagreement

Participants generally agree that both balls experience the same acceleration due to gravity before impact, but there is disagreement regarding the implications of this on the forces experienced by the spring and the nature of the collision. The discussion remains unresolved with multiple competing views on the relationship between height, velocity, and force.

Contextual Notes

Participants note that the forces during the collision cannot be easily calculated and that the discussion involves assumptions about energy conservation and the behavior of collisional forces.

mprm86
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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?
 
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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.
 
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.
 
collisional forces work differently. I can't remember the equations exactly, but I am sure someone else here does.
 
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?
 
Collision means transfer of momentum & energy and the presence of contact forces...These forces cannot be calculated,really...

Daniel.
 
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
 

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