# Force to deflect ball colliding with wall

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.

• Since only the component of the velocity vector which is normal to the wall is changed, the force vector must also act normal to the wall, but with an opposite sign to the incident normal component. Jul 19 '15 at 22:07
• What formula does actually describe this force, though? Could you write it up in an answer? Jul 19 '15 at 22:21

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.

• Thanks. What does the time of impact depend on though? I have all properties of wall and ball, so there should be a formula telling me "the ball will deform for X ms and then bounce back". This bit is always elusive for me. Also, with v you mean the normal componenent of the velocity vector at impact time? Jul 20 '15 at 8:25
• Yes, v is the normal velocity at impact. Duration depends on the rigidity (not the elasticity, much) of the ball. A steel ball cannot deform during impact, so duration is very small, and peak force is very high. Soft rubber does deform, so the ball decelerates fairly slowly as it squishes, then regains its shape. A rough upper limit on duration would be impact velocity divided by ball diameter. Jul 20 '15 at 12:55
• Accepted as the most informative. Jul 21 '15 at 21:19

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.

• 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.

• The case is not an elastic collision. If it were, the coefficient of restitution would be one. Jul 20 '15 at 2:40
• Problem is Dt is to be found, as I don't know in advance the impact profile of the ball, nor the change in momentum. In fact I'm interested in a more low-level analysis: given the ball (elastic) and wall (rock solid) properties, and the ball impact angle and velocity, how long does the impact last and what force does the ball acquire? Jul 20 '15 at 8:27
• @Fabio It then depends on the elasticity of the ball. To say "the ball is elastic" is not enough; you need to know how elastic. You cannot avoid the time (or something in that relation, like the elasticity you are talking about) to find the force, because the force done depends on the duration. Jul 20 '15 at 8:32
• That seems like a dog-biting-its-own-leg kind of thing.. there must be a way around it, for example through Kinetic Energy manipulation? I mean, what do you do if you have a meteorite falling down on earth, and want to know: with what force it will hit the ground, if it will bounce back - and if yes, where and how many times? You only have, at most, mass, elasticity and limit velocity of the meteorite, as well as elasticity of the ground. Nothing else. Jul 20 '15 at 8:37
• Well, there are other expression of force - but all include the time in some way or another, because force fundamentally depends on time. I wouldn't know about meteorites. It is interesting, though - do you have an example of an article that actually gives the force done on Earth at impact? Jul 20 '15 at 8:48

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.

• Jul 20 '15 at 13:33
• Note that $\delta$, if the wall is undeformable and non-perforable, is the amount of compression the ball has experienced. If both ball and wall get compressed, you sum the total compression (I think..) Jul 20 '15 at 13:36
• If anybody is interested, this is not correct. There is a lot I don't understand here, but a bit more than before. I will updated my answer shortly with a better explanation for the future reader. Jul 21 '15 at 20:44

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

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