Force to deflect ball colliding with wall

Question

Suppose I have a ball with a certain coefficient of restitution. The wall can be considered indeformable and with infinite mass. Everything's in 3 dimensions, and the ball can hit the wall at any angle.

The ball has a certain velocity at the time of contact, and we need to calculate the new velocity after the impact.

So far, I use simple linear algebra to get a reflected vector using the normal with the wall - scaling it by the restitution coefficient. This gives me a velocity "deflected" away from the wall with a smaller magnitude than before the impact.

However I can't figure out how to calculate a Force vector instead of a new velocity.

How do you get the force impressed by the wall upon the collision? Most formulas need a Dt parameter, the time the impact lasted: I don't know that, it should be obtainable with the ball characteristics.

Answer 1

Given the mass m of the ball, the incident normal speed v, and the coefficient of restitution $\rho$, Then the integral of F over the duration of the collision $\Delta t$ is $$\int_0^{\Delta t} F dt = \frac{m(1 + \rho)v }{\Delta t}$$ assuning no rotational effects are incurred.

This follows from the fact that at any instant the acceleration of the ball away from the wall is F/m, so the integrated acceleration over the duration of the collision is the total change in normal velocity of the ball. Since the post-bounce velocity is simply the approach velocity times the coefficient of restitution, the total velocity change is as indicated.

If you want to simplify the force profile to assume a constant force FB during the collision (which is clearly not accurate) then $$\frac{(FB)(\Delta t)}{m} = (1 + \rho)v $$

The collision clearly does not produce a uniform force level over the duration of the collision, because the elastic forces on the ball will vary with the amount of deformation of the ball.

Answer 2

Simple momentum change will do:

$$\vec F = \frac{d \vec p}{dt} $$

Your case is an elastic collision, so it simplifies to the difference between end states: $\vec F = \frac{\Delta \vec p}{\Delta t}$. Key thing: you need to know about the duration of the impact.

Think about it...

• If your wall is very elastic - a vertical trampoline - the ball might still reach the same speed after impact even though the collision takes much longer to finish before the ball is sent off again. Much less force is required, but over a longer period of time.

• Countrary, a very hard wall will apply much greater force to make the ball reach the same speed in shorter time.

Answer 3

Ok, I might have "solved it" although there are still grey areas.

First the equation for the force experienced by a body colliding against a body with infinite mass:

$$ \textbf{F} = \textbf{n} \delta e_r m k_0 $$

Explanation:

• $ \textbf{n} $: unit normal pointing outside the colliding wall. This is the direction of the force (ie, where we want the ball to be deflected)
• $ \delta $: length of penetration at the "peak" of the collision
• $ e_r $: coefficient of restitution defined between $ [0, 1] $, where 1 is total bounciness
• $ m $: mass of the ball
• $ k_0 $: a stiffness constant of the ball

This can be generalised for two spheres collisions but I needed a much simpler case. The difficulty here becomes mainly how to choose $ k_0 $, which I'm still investigating.

A few references for the curious:

http://baseball.physics.illinois.edu/CORNormalization.pdf

https://arxiv.org/abs/physics/0601168

If anybody wants to help finishing this off (essentially, finding a table of Ks I can use for different materials, and verifying the formula makes sense) I'll accept their answer.

Answer 4

If the collision was linear, and reaction force $F$ was proportional to deflection $x$ by some constant stiffness coefficient $k$ then the shape of the force over time curve would be that of a cosine. Peak deflection $\delta$ occurs at time $t=0$, contact frequency is $\omega$ and the time span is $\Delta t$.

$$ \begin{aligned} F & = k x \\ x & = \delta \cos \omega t \\ k & = m\,\omega^2 \\ \Delta t &= \frac{\pi}{\omega} \end{aligned}$$

In the context of a collision, if we know the time the collision occurs over $\Delta t$, then the following can be estimated

$$\begin{aligned} \omega & = \frac{\pi}{\Delta t} \\ k & = \frac{m \pi^2}{\Delta t^2} \\ F_{\rm max} &= \left(\frac{m \pi^2}{\Delta t^2} \right) \delta \end{aligned}$$

The total momentum $J$ transferred to the ball from the wall should equal in magnitude with twice the incoming momentum of the ball if the collision is perfectly elastic.

$$ J = \int F\,{\rm d}t = \int_{-\Delta t/2}^{\Delta t/2} k \delta \cos \omega t {\rm d}t = 2 \left( \frac{\pi}{\Delta t} \right) \delta = 2 m v $$

This allows the estimation of the peak deflection $\delta$ as well as the peak force

$$ \begin{aligned} \delta & = \frac{\Delta t}{\pi} v \\ F_{\rm max} & = \frac{\pi}{\Delta t} m v \end{aligned}$$

So peak force is a linear function of incoming momentum, and the contact time $\Delta t$ which is a function of the materials involved and the incoming velocity.

Also note that you might be interested in the average force over $\Delta t$ which is would you would apply in a simulation. This is found by the ratio $F_{\rm ave} = J/\Delta t$ or

$$ F_{\rm ave} = \frac{2}{\pi} F_{\rm max} $$