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The $F$ in $$\mathrm{Impulse} = F\Delta t$$ is said to be the average force. For a ball dropped vertically onto a horizontal surface, the average force, F, on the ball from the floor is: $$F = \frac{\Delta{p}}{\Delta t}$$ $$\Delta{p} = p_f - p_i$$ $$\Delta{p} = mv_2 - (-mv_1)$$ $$\Delta{p} = mv_1 + mv_2$$ $$\Delta{p} = m(v_1 + v_2)$$ Therefore, the average force become, $$F= \frac{m(v_1 + v_2)}{\Delta t}$$

On the other hand, we know from Newton's second law, we know that:

$$F = ma$$ And therefore, in the case of the dropped ball, $$F = mg$$ Both are of the form "$F$ equals...", but are obviously different - What is the relation between the two? Is it correct to say that the equation derived from Newton's second law is the net force, as opposed to the former (the one derived from impulse) average force?

Would the average net force be

$$F= \frac{m(v_1 + v_2)}{\Delta t} + mg$$

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    $\begingroup$ I'm a bit confused. Aren't you comparing apples with oranges? In the first example involving impulse the force you are considering is the force that comes about by the collision of the ball with the floor. In the second example you are expressing the force on the ball (at any height) above the floor due to gravitational force. In the second example no collision is involved. $\endgroup$ – docscience Sep 21 '15 at 1:54
  • $\begingroup$ Also $\Delta t \ll 1$ means that $ g \ll \frac{v}{\Delta t} $ $\endgroup$ – John Alexiou Jul 27 '16 at 18:31
  • $\begingroup$ You are also confusing the concept of a net force and a contact force. $\endgroup$ – John Alexiou Jul 27 '16 at 18:34
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There are indeed two different forces: the force of gravity, working on the ball for as long as it is on Earth, and equal to $m\cdot g$. And the force due to the impact with the surface, which on average is indeed $\frac{\Delta p}{\Delta t}$.

If you consider a perfectly elastic collision, and the time interval from releasing the ball from height $h$ until it is once again back at height $h$, then the average net force must have been zero (because the ball is once again not moving).

To figure this out properly you need to make sure that you normalize things correctly. If you are only interested in the average force during the impact, you have a very short time $\Delta t$ corresponding to the impact. During that time, which is much less than the time of the drop from $h$, you can neglect the force of gravity - the impact force will be much, much larger (depending on the rigidity of the ball and surface, 100x or even more). If you consider the longer time of the drop, you need to take both into account - and can find a net force of zero averaged over the drop, impact, and rebound.

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Let's take an example of a ball falling from a height of $8\,\mathrm{m}$. $F=mg$ is same near the surface of the earth. The impulse experienced by the ball from the floor equals $m\frac{v_{final}-v_{initial}}{t}$, where $t$ is the time of contact. The latter is the average force and former is the instantaneous force with which it hits the floor. As per Newton's third law these got to be equal and opposite!

Does Newton's 2nd law depend on the contact time? I don't think it does.

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First you need to understand how impulse and Newton's second law differ in definition. Newton's second law is defined such that the net force on an object at any moment is equal to the product of its mass and acceleration, or $\vec{F}_{net} = m\vec{a}$. This gives the vector sum of all other forces acting on an object in an instant. Impulse, on the other hand, is defined using calculus. Specifically, $\displaystyle Impulse = \int_{t_1}^{t_2}\vec{F}dt$, where $\vec{F}$ is taken to be a force that varies over time. This expression devolves to $Impulse = F*t$ whenever F is a constant. Since average force over a period of time is a constant, we are allowed to use the latter expression in either case (whether it be a constant force or an average one). Therefore, $\vec{F} = m\vec{a}$ and $\displaystyle F= \frac{m(v_1 + v_2)}{t}$ are not the same thing; you are right to say that the former is the net force while the latter is the average force (when there is a collision, as that is how you derived the expression). Now, for your final question, there is not really such a thing as "average net force". There is an average force over a given period of time, and there is a net force on an object in an instant. What you're describing is really just an average forces, which you could obtain either from using the impulse-momentum theorem or the average of several net forces over time (assuming the changes in the net force are discrete).

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  • $\begingroup$ If there are several force on an object, and they vary with time, you will have a varying net force. You can average that net force if you want. So there really is such a thing as average net force. $\endgroup$ – garyp Jul 27 '16 at 16:13

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