0
$\begingroup$

It is said that force is an interaction that causes an object to accelerate. It is measured in newtons. $1$ N of force causes an object with a mass of $1$ kg to accelerate at $1$ m/s². Now let's imagine a box sliding on frictionless surface and hitting a standing box. Standing box will accelerate. But their interaction will last extremely short amount of time. Isn't the length of time of them touching each other approaching zero? If it is, isn't the acceleration approaching zero?

Intuitively we know it will move, but mathematically it confuses me.

Thanks for answers!

$\endgroup$
1
  • $\begingroup$ @Aleksander Why approaching zero? I would expect the acceleration to increase if time decreases. $\endgroup$ – Charles Tucker 3 Jun 6 at 16:02
1
$\begingroup$

You're correct that the acceleration only lasts as long as the force is applied. If there's a collision between two boxes as described and there's significant acceleration in a short amount of time, then the force between them must be very large.

This is where it's better to use $\Delta p = F\Delta t$, often refered to as "impulse". If the second box weighs 1kg and accelerates to 10 m/s, that's a change in momentum of $\Delta p = 10 kg \cdot m/s$. If you know they were in contact for only 0.01 seconds, then the force must be $F = \frac{\Delta p}{\Delta t} = {10}{0.01} = 1000N.$

Realistically the two boxes would be in contact longer than that - maybe 0.1-0.3 or so, resulting in a much lower force. However if the boxes are made out of very hard incompressible material - say blocks of diamond - then the $\Delta t$ would be very small and the force very high.

This is why it hurts more to run into a brick wall ($\Delta t$ small) than a mattress ($\Delta t$ large) even though the total integrated acceleration ends up being the same.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.