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I see many people try to go over speed bump with one side of the car only. I’ve always thought that was weird, as the car gets shaken up either way. Intuitively speaking, crossing the speed bump on one side will reduce the amount of impact a car receives compared to crossing both sides, but it is not easy to prove at my physical knowledge level.

The variables that affect the amount of impact a car receives from a speed bump can be various, such as the shape of the car, the mass of the car, the shape of the speed bump, and the speed at which the car crosses the speed bump. Assuming that all of these variables are equal, which side of the impact is greater, one side or both sides?

P.S : I'm very sorry I didn't draw a diagram.

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When driving on one side, the center of mass of the car is raised to a lower height.

This means the (one front and one rear tire) bumps you experience should be smaller when driving on one side, as opposed to that experienced when both wheels go over$^1$ (two front and two rear tires), which raises the car to a greater height. This is because an object raised to a greater height has a greater potential energy at that height, so as this potential energy is converted into kinetic energy, it "hits" the ground at a greater speed.

$^1$ This may be different in the case of some cars which have more elaborate suspension systems, but the explanation above is basically what's happening. In older model cars this does make a difference and the amount of "bump" is less when going in on one side.

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  • $\begingroup$ thx for answering my question. Let me ask you one more question. Can you explain this in terms of momentum and impact? $\endgroup$
    – NOH WHIREA
    Mar 2 at 3:33
  • $\begingroup$ Certainly. As per the answer above, the "impact" and momentum is greater when (the COM of) the car is raised to a greater height. We know that when something is dropped from a certain height $h$, the kinetic energy it hits the ground with, is equal to $mgh$ so a greater height $h$ means a greater kinetic energy (and momentum) upon impact. So going in with both wheels, raises the car to a greater height than one wheel, meaning a greater impact or change in momentum when hitting the ground (note that this is referred to as "impulse" instead of "impact"). Cheers. $\endgroup$
    – joseph h
    Mar 2 at 3:47
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If both wheels hit the bump, the center line of the front of the car is raised to the height of the bump. If one wheel hits, the center line is raised half as much. The time it takes for one or two wheels to reach the top does not change.

I will make some very crude approximations. The bump accelerates the car upward. To make the problem simple, assume a is uniform. $d = 1/2at^2$. If you cut $d$ in half, you cut $a$ in half. You cut the force in half. You are only jostled half as much.

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