How long will it take for a projectile to be knocked off a wall? (See details)

Question

I am trying to calculate the impulse of an object that is thrown against a wall. Problem is, to calculate the impulse one needs to know not just the force, but also the time it takes for the momentum to change.

Is there a way to precisely calculate this? Like, if I know the dimensions of the object, and its speed and its angle when it hits the wall (and we’re assuming that the wall is even and at 90° exactly), could I find out how long the contact between the projectile and the wall will last?

I know it must be a very short time (am I right if I say that it could be around 0.1 second?) but I have realized that it makes a real difference in my calculations.

Answer 1

It's easy to obtain the minimum and maximum values:

The inelasitic limit is where the projectile will stick to or in the wall. Then simply all momentum by the projectile is transferred.

The absolutely elastic limit with an infinitely heavy wall means that the projectile will bounce off the wall and travel back at the same speed it hit the wall. In that case twice the momentum of the projectile is transferred.

Reality often lays somewhere between these two cases - and cannot be easily expressed as it depends heavily on the microphysics of the contact area. That in turn depends on the materials and the surface roughness up and down to microscopic level. You can obtain values for the contact time and penetration depth and rebound behaviour experimentally - but that's already challenging. Similating the behaviour in a computer to a level that you get useful results is at least as challenging. 100ms is a very long contact time - but depends on circumstance. In experiments with glass spheres bouncing off a floor I measured contact times more in the range of 1...10ms; it's very material dependent, though, as the elasticity plays a major role.

Answer 2

Impulse is defined as $F\Delta t$, but there is an equation that states that $\Delta p = F\Delta t$. This means that you can assign a positive direction for an object's velocity, calculate it's momentum before it hits a wall, and calculate it's momentum after it bounces off the wall. Accordingly, given the equal sign in the equation, impulse also equals $\Delta p$, so you can calculate impulse from change in momentum without having to know the $\Delta t$ term.

Answer 3

So you have that

$F\Delta t = m \Delta v$

so that

$\Delta t = \frac {m \Delta v} {F}$

Because you know what the velocity before and after the collision will be (and you know what $m$ is), you can calculate $m \Delta v$ (the change in momentum). For $F$ you need a device that can measure the force of the impact between the ball and wall. You could set up a spring behind the wall, so the when the ball hits the wall the spring will be compressed by a distance, say $\Delta x$. Knowing what the spring constant $k$ is, you could then calculate the force using Hookes Law $F = k \Delta x$ so that

$\Delta t = \frac {m \Delta v} {k \Delta x}$