Force in Head-on Collisions: Equal Risk or More?

[Tweeter]

I know this is a really basic question but I'm a little rusty on my physics, and this is just something I have been curious about. Assume you have two cars of equal weight traveling toward each other and they meet in a head-on collision. However, one car is going twice as fast as the other (let's just say 30 mph vs 60 mph); would both cars absorb the same force, or would the car going faster absorb more of the force? In other words, all other factors being equal, would the drivers both be at equal risk of injury, or would the driver in the faster car be at more risk? Also, here is one other scenario: if the driver in one car was sitting still, and the other car hit him head on at 60 mph, would the force be the same in both cars for this situation? I am assuming that it would depend on whether or not the driver in the car sitting still had his foot on the break or if it was in neutral, as this would affect the rate of his backwards acceleration. Thanks for the help!

 

[gneill]

Hi Tweeter, welcome to Physics Forums.

What does Newton's Third Law of Motion have to say about such forces?

 

[Tweeter]

Hi Tweeter, welcome to Physics Forums.

What does Newton's Third Law of Motion have to say about such forces?

Ok, so I just refereshed my memory on Newton's 3rd Law. I guess the forces would therefore have to be the same on both cars. Let me ask this question in a different way now. If you still had the two cars of the same weight traveling toward each other (car A at 60 mph and car B at 30 mph), would car A deccelerate to 15 mph and car B deccelerate to 15 mph in the opposite direction after the collision (a 45 mph change for both cars)? The part that is confusing me is that if one of the cars held down the brake during the impact (as opposed to just being in neutral), it seems like that would slow down that car's rate of accelerate / deceration, and therefore lessen the force of the impact (according to the equation F = M*A?

 

[gneill]

If no brakes are applied then conservation of momentum holds sway. You can work out the final velocity of the amalgam of cars (assuming that they become stuck together) using that conservation law. You can then determine the change in velocity that each experiences.

If one of the cars is applying brakes then you have an external force to consider and the problem becomes a bit more complex. Then the details of the actual impact become important, such as exactly when the brakes are applied, how much braking force there is (do the wheel lock and friction between the tires and road take over?), the amount of time over which the actual impact takes place (time for all the metal to deform), and so forth. You can get an idea of the amount of energy "stolen" from the impact by braking from the work done by the frictional forces involved.

 

[Tweeter]

Thanks for taking the time to explain that. Much appreciated.