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If we have a ball with a mass of 1kg, to keep it simple, and we drop it from a 9.8m height, it will travel 9.8m/s at its fastest, and its entire journey would take 1 second. In a perfectly closed system with no friction, air resistance and a perfectly flat ground to bounce it on, the ball should bounce back to its 9.8m height. The ball will exert a 9.8N force onto the ground, and the floor will exert a 9.8kg force back on the ball taking it back to the height it was originally.

I don't understand this, if the ground exerts a 9.8N force on the ball, shouldn't the ball seize motion as gravity is exerting a 9.8N force on the ball as well in the opposite direction resulting in a Net Force of 0?

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    $\begingroup$ This is not part of your question, but about your easy assumptions: With $g=9.8\,m/s^2$, a ball that you let got at $9.8\,m$ height does not only need $1s$ to fall down, and its final speed will not be $9.8\,m/s$. How could it? If it starts with $v_0=0$, and you assume it's final speed is $9.8\,m/s$, then it cannot make $9.8\,m$ in one second, since it would need your value of final speed as constant speed to make this. :-) You missed a factor of $\sqrt{2}$ somewhere. $\endgroup$
    – Koschi
    May 12 at 10:27
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Is the ball exerting 9.8N force on the ground?

When the body hits the ground with speed v it rebounds with the same speed v if the collision is elastic, so change in momentum is 2mv.

Now the key is the time interval. How long does the object stay in contact with the wall during collision? It could even be in the order of microseconds.

So force becomes 2mv/t (On average, in reality it varies continuously) which brings it to the order of mega newtons(considering v and m are close to 1 SI unit)

The weight here is just mg which is about 10$^5$ times lesser and barely creates a change in momentum in the microsecond time interval.

Might wanna read up on Impulses:https://en.wikipedia.org/wiki/Impulse_(physics)

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  • $\begingroup$ I’m only in my preliminary year of physics so I didn’t have such an understanding of impulse. I know now that the formula is F=mv/t, does that mean when 0<t<1 the force will be greater than the force applied to the original object? Where does the extra force come from in the physical world? $\endgroup$
    – LAMAR__44
    May 15 at 21:17
  • $\begingroup$ The force is electromagnetic if you see on a micro scale, cuz you are pushing the ball (and so its electrons) into the ground which has electrons too. This force is wayy stronger than gravitational force at such a short distance because or else the ball should just sink through $\endgroup$ May 16 at 3:26
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I've realised where I went wrong, the collision being less than a second causes time to be a value greater than zero but less than 1, since dividing by a decimal is basically multiplying, the change in momentum is greater. This can be explained in terms of Newtonian physics by taking the entire momentum of the system of the Earth into account which needs to be conserved. Since the ball exerts a slight change of momentum on the Earth it needs to be repelled back at equal height to conserve the momentum of the system if the collision was elastic.

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