# The time-average force of ball undergoing collision between walls

Why do we take the time between collisions e.g time of ball in air during collision between two walls to find time-average force like in the following question:

A “superball” of mass $m$ bounces back and forth with speed v between two parallel walls, as shown. The walls are initially separated by distance $l$. Gravity is neglected and the collisions are perfectly elastic. Find the time-average force $F$ on each wall.

Why do we take the time between collisions i.e time taken by the ball to cover $l$ distance find time-average force, when actually in momentum and impulse related question we are more concerned with time of collision which could be the time of interaction between the wall and the ball. But instead we choose to take time of the ball in air. Why do we do it?

The ball gets its impact velocity only due to the momentary impulse it receives during the collision. The interaction is only for a very short period of time. That's true but the ball gets the velocity for quite a longer time. Now I want the wall removed and try to exert a constant force to have the same impact as before. For that average force we have to include the time the ball retains the collision's effect.

Mathematically, we will have

$$F_{avg}\Delta t = \int {Fdt}$$

• So based on the analogy you used, we will be required to exert an average force for time 't' (which is the time between the collisions) in order to maintain the velocity 'v'. Hence, we take the time between collisions. Did I get it right? – suiz Apr 20 '18 at 12:35
• Yeah , you got it.. – Jnan Apr 20 '18 at 13:52