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