What is the term for “shock” absorbed by car suspension? I’m looking for the correct term to find equaltions for car shock absorber physics. Searching for “shock” returns a lot of irrelevant results.
What is the correct term for the force exerted by a road bump on a car wheel when traveling at some velocity?
For example: a 30cm radius wheel traveling at 10m/s hits a 3 cm radius  semi circle bump. What is the force exerted on the wheel? (The wheel’s shock absorber then responds to this force)
 A: Assuming that the car is going at a constant speed, you can determine the impulse delivered to the car which is given by:
$$\vec{F}_\text{avg} t = \Delta \vec{p}$$
While the magnitude of momentum will be constant, the direction will change. There are two locations in the motion that we can look at.
Firstly, we need to look at the moment when the wheel goes onto the speed bump. We can assume an inelastic collision (so the tire stays in contact with the speed bump) where in an infinitesimal time interval, the velocity of the center of the tire is suddenly directed up the speed bump instead of along the flat horizontal road.

We have:
$$\sin\theta = \frac{R}{R+r}$$
so we can now calculate the change in momentum:

Therefore:
\begin{align*}
|\Delta \vec{v}| &= \sqrt{v^2+v^2-2v^2\cos(90-\theta)} \\ 
&= \sqrt{2v^2(1-\sin\theta)} \\
&= v\sqrt{\frac{2r}{R+r}} \\
\end{align*}
so $$|J| = mv\sqrt{\frac{2r}{R+r}}$$
This is known as an impulsive force.
Next, we can consider the impulse delivered to the wheel by the speed bump as it goes over it. I don't think there's a special name for it but we can calculate it in a similar way.
Let's consider the change in momentum from when the tire just got on the bump and when the tire is at the top of the bump (where it is going at the same speed and direction as before it got on the bump). As a result, the change in momentum would be the same but unlike the previous situation, we can calculate the time.
The angle the center of the wheel travels is
$$\alpha = \frac{\pi}{2} - \theta$$
Setting it equal to $\omega t = \frac{v}{R+r} t$, we can solve for the time:
$$t = \frac{R+r}{v}\left(\frac{\pi}{2} - \sin^{-1}\left(\frac{R}{R+r}\right)\right)$$
Therefore, the average force is given by:
$$|\vec{F}_\text{avg}| = \frac{\Delta p}{t} = \frac{mv^2\sqrt{\frac{2r}{R+r}}}{(R+r)\left(\frac{\pi}{2} - \sin^{-1}\left(\frac{R}{R+r}\right)\right)}$$
Interesting result is that:
$$F \propto v^2$$
So if you triple your speed, the force you experience from the speed bump increases by a factor of nine.
