# Average force for repeated elastic collisions

I was asked to compute the average force exerted by a rigid spherical ball (mass $m$) on the floor over time $t \rightarrow \infty$, in a situation where the ball is dropped vertically from a height $h$ and makes repeated elastic collisions with the floor which is hard and smooth.

I reasoned that over infinite time, the normal force exerted at the instants of collision would occupy a small part of the time while the rest of the time there would be no normal force between the ball and the floor, so the answer would tend to zero.

However, all the options provided are non-zero, in terms of $mg$ and independent of $h$. Please help me understand the duration for which the force actually acts and the significance of averaging it over infinite time. I would also like to know what changes if the time was finite: either directly specified (say $t$), or if the number of collisions (say $n$) was specified.

• I'm not sure if I understand the question fully, so forgive me if I am incorrect. But doesn't the ball simply bounce up and down, going nowhere in the grand scheme of things, so I would assume that the average normal force is just combatting the force of gravity, making the answer simply "mg".
– Me2
Apr 25 '15 at 22:45

Here's how. The ball gains a velocity $v$ due to gravity before hitting the ground. So each time it hits the ground its velocity is changed from $v$ to $-v$ (taking down as positive) during the collision, then returning again with $v$.
The force $F_1= m\frac {dv}{dt}$ is experienced by the ball due to the collision, however this force is felt after the weight gets cancelled out by the normal contact force. So we have $$F_N=F_1+W$$ $$\Rightarrow F_N=m\frac{dv}{dt}+mg$$ If the total time taken was $t$, where $t$ is the sum of the duration of all collisions, then the sum of the forces experienced would be (assuming you know basic calculus) $$\int_0^t F_N\,dt=m\int_0^t \frac{dv}{dt}\,dt+\int_0^t mg\,dt$$ $$\Rightarrow \int_0^t F_N\,dt=m\int_v^{-v} dv+\int_0^t mg\,dt$$ $$\Rightarrow \int_0^t F_N\,dt=2mv+mgt$$ Hence the average force would be $$F_{avg}=\frac{\int_0^t F_N\,dt}{t}=\frac{2mv+mgt}{t}=\frac{2mv}{t}+mg$$ as $t \rightarrow \infty , \frac1t\rightarrow0,\;$ $$\therefore F_{avg}=mg$$ Notice, if this kind of bouncing happened horizontally (e.g. a ball colliding back and forth between two walls) then the $F_{avg}$ would be zero, just as you had thought. Hope this helps.
There are two external forces on the ball - the downward gravitational force (exerted by Earth if you are on Earth) and the normal contact force (exerted by the ground in the upward). The gravitational force is almost constant if the ball is dropped from a height which is much less than the radius of Earth (or any other celestial body). If we consider the mass of the ball to be $$m$$, then the gravitational force on the ball is $$mg$$. The average net downward force on the ball is $$mg$$. We have already said the net average force is zero in one full cycle. So the average net upward force is also $$mg$$. But wait, we have found the average force exerted by the ground on the ball, but we are asked to find the average force exerted by the ball on the ground. Seeking little help from Newton's third law, we can conclude that the average force exerted by the ball on the ground is $$mg$$ which is nothing but its weight.