Jump to content

Car Crash forces


Kethevin

Recommended Posts

I have a silly question, but I can’t wrap my head around the whys and I know you all are math wizards.



It’s the age-old question.  A car going 50mph has a head-on collision with another car going 50mph in the opposite direction…

Now I understand that both cars receive an impact force equal to 50mph because it’s spread across both cars.

What I’m wondering is, if one car was going 30mph and the other 70mph, would it still be 50mph to both or would there be a noticeable difference?

Lastly, if the impact is divided between the two objects, why does a car going 100mph into a wall not distribute a portion of that to the wall but instead seems to take the full force?

Link to comment
Share on other sites

I'm going to break up your question:

1.) I think that the 30 mph car would get hit worse than the 70 mph car

2.) Cars =/= walls. Many cars have crumple zones, they're designed to break like that to spread out the force. The wall does receive a portion of the force, but walls are sturdier, and so it's less noticeable.

Link to comment
Share on other sites

A head-on at 50 for each car is 100 closing velocity, and aside from a crumple zone for each car, it's closer to each hitting a wall at 100. Or better, it's the same as a stationary car being hit by another at 100---just set the frame of reference with one of the cars, and it's "not moving" and the other is closing at 100.

 

Link to comment
Share on other sites

We have a 1 t  car going 20 m/s. It's mechanical energy is 1000 * 202 / 2 = 200 kJ.
So, we have burnt 200 kJ of fuel and converted its chemical energy to the kinetic energy of car.

We have another car - motionless.
When they collide, these 200 kJ release and distribute between both cars.

We have another car, going 20 m/s too.
So, totally we have 400 kJ of the fuel energy deposited in these cars.
When they collide, we get back 400 kJ of the post-fuel energy.

So, if there is no difference, then it's no difference between pouring 1 or 2 litres of petrol onto the car and lighting it.

Link to comment
Share on other sites

50mph vs 50mph - both cars experience an impact equivalent to hitting a stationary wall car at 100mph.

30mph vs 70mph - both cars experience an impact equivalent to hitting a stationary wall car at 100mph.

There is a little difference between the two scenarios - there is a kinetic energy difference between the two combinations, with the 30-70 combination having a slightly higher total kinetic energy than the 50-50 combo. However, the relative speed is the same and total momentum is the same 

 

(Another slight correction - total momentum is not the same, as momentum is a vector with a direction [see further below], total momentum change is the same in the 2 scenarios however. The total of the *modulus* ["sign-free"] of the momenta will be the same :) )

 

I think that the excess kinetic energy in the 30-70 crash will be present in the wreckage as motion (ie: the combined wreck, if we say they clump together) in the direction of travel of the 70mph car.

With the 50-50 impact, the wreckage will be motionless, or its CoM will be anyway.

All 4 cars see the same impact. All cars decelerate just as rapidly and for the same amount of time, just that in the 30-70 combo, the 70 car doesn't quite get to 0mph and the 30mph car is decelerated to a negative figure (ie: it is propelled backwards somewhat), meaning that in the 50-50 combo, there is no kinetic energy leftover, and in the 30-70 combo there is. If all 4 cars are identical, all 4 will be deformed the same amount and all passengers will be subjected to the same forces. The difference is in the energy leftover in the wreck (which of course, could result in further forces on the car and occupants, but the initial impact is the same).

***edit***

Remember, momentum has direction (ie: if car 1 has momentum in the direction it is travelling, car 2 has negative momentum in that direction), there appears to be excess momentum in the wreck of the 30-70 crash, but the momentum change is the same.

***edit #2***

Hitting a stationary wall - well it strongly depends on the wall. In the real world yes, a portion of the impact will be distributed to the wall and its components.

If we assume an infinitely strong and rigid wall:

When the car hits the wall, force is transmitted into the wall which wants to make it move, this applies a force between the wall and whatever is holding it up - this force is essentially applied between this support and the car itself with the wall as a neutral go-between, meaning that with an infinitely strong, rigid wall, the full 100mph is felt by the car.

If the wall is a normal, ordinary wall, some of the force will be absorbed by deformation of the wall and putting parts of it into motion, meaning that the car experiences slightly less than the maximum force. Effectively the wall "cushions" the blow a little.

Edited by p1t1o
Tex pointed out an error
Link to comment
Share on other sites

23 minutes ago, p1t1o said:

50mph vs 50mph - both cars experience an impact equivalent to hitting a stationary wall at 100mph.

WRONG! A classic and very common mistake but that doesn't change the simple fact that it is WRONG!

Both cars will experience an impact equivalent of hitting a stationary wall at 50mph! You're forgetting the total amount of energy is divided between the two cars.

But why not let Adam and Jamie explain.

The full experiment (including the full scale crashes) can be found here: http://www.dailymotion.com/video/x2n9j62

Edited by Tex_NL
Link to comment
Share on other sites

 

5 minutes ago, Tex_NL said:

WRONG! A classic and very common mistake but that doesn't change the simple fact that it is WRONG!

Ah ok, not wrong, I just misspoke :( - the impact is the same as hitting a stationary *car* at 100mph, so the stationary car will absorb its fair share of the impact, you are right there.

If I make that change, the picture should be correct, both scenarios are still equivalent. (the correction also makes the last question makes a bit more sense)

Edited by p1t1o
Link to comment
Share on other sites

1 minute ago, p1t1o said:

the impact is the same as hitting a stationary *car* at 100mph

This would indeed be correct. When hitting a stationary car the 'load' will be more or less equally distributed between the two.

Link to comment
Share on other sites

I'd think as long as the closing velocity is identical, the effects would be nearly so, frankly. Conservation of energy and momentum still apply. The wall analogy never made sense, because walls don't have crumple zones, though as I said, you can treat the crash as a stationary car vs a moving car at the closing velocity.

BTW, mythbusters is wrong so often that I'd not use them as a source, even when they are right. (I'm still POed at the shooting fish in a barrel episode, lol).

Link to comment
Share on other sites

This thread is quite old. Please consider starting a new thread rather than reviving this one.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.
Note: Your post will require moderator approval before it will be visible.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...