MythBusters: Does two cars crashing at 50 mph each = a 100 mph crash into wall?
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MythBusters: Does two cars crashing at 50 mph each = a 100 mph crash into wall?
Many familiar with physics will probably know the result. This is a long time belief based on intuition.
What do you think?
Part 1 http://www.youtube.com/watch?v=q_vCuE4KaUU
Part 2 http://www.youtube.com/watch?v=c-JGIYLZZUg
^ two parts at almost 11 minutes each but you really only need to watch the second one as the first one is preliminary testing. Worth the 11 minutes!!
The result actually proved Jammie wrong!
What do you think?
Part 1 http://www.youtube.com/watch?v=q_vCuE4KaUU
Part 2 http://www.youtube.com/watch?v=c-JGIYLZZUg
^ two parts at almost 11 minutes each but you really only need to watch the second one as the first one is preliminary testing. Worth the 11 minutes!!
The result actually proved Jammie wrong!
Last edited by -J-P-L-; 05-15-10 at 06:26 PM.
#2
I would think that it's equivalent to hitting a wall at 50 mph. When two identical cars going 50 hit each other, they will both "stop" since each has the same momentum (mass times velocity). The same happens when the car hits the solid wall.
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I think 2 cars hitting head on at 100mph would be different from one car hitting the wall at 100mph.
#14
2 cars each of mass 1M hitting each other at 50 MPH is, indeed, equivalent to 1 car of mass 2M hitting a wall at 100 MPH.
That's the trick here, it's simple conservation of energy. You have a certain amount of kinetic energy being transferred as wreckage into both cars (total mass 2M) when 2 cars of mass 1 M hit each other at 50 MPH.
From a conservation of energy perspective, this is equivalent to 1 car of mass 2M hitting a stationary wall at 100 MPH.
It is not equivalent to 1 car of mass 1M hitting a 100 MPH wall, as was indicated by the video.
#15
2 cars of mass M hitting each other at 50 MPH is, indeed, not the same as 1 car of mass M hitting a stationary wall at 50 MPH. That's the simple part.
(1/2)(m)(v)^2 + (1/2)(m)(v)^2 = total energy
mv^2 = (1/2)(M)(2v)^2
where m = mass of car 1 and car 2
M = equivalent mass of a single car necessary to simulate the same collision for a one car-one wall set up
2mv^2 = 4Mv^2
M = 2m
Edit: That's totally wrong. That's yielding a result of M = (1/2)m
Interesting. I think this result is implying that 2 cars of mass m hitting each other at velocity v is equivalent to 1 car of mass (1/2)m (not 2m like I stated previously) hitting a single stationary wall at velocity 2v
Someone help out and clarify the math please.
(1/2)(m)(v)^2 + (1/2)(m)(v)^2 = total energy
mv^2 = (1/2)(M)(2v)^2
where m = mass of car 1 and car 2
M = equivalent mass of a single car necessary to simulate the same collision for a one car-one wall set up
2mv^2 = 4Mv^2
M = 2m
Edit: That's totally wrong. That's yielding a result of M = (1/2)m
Interesting. I think this result is implying that 2 cars of mass m hitting each other at velocity v is equivalent to 1 car of mass (1/2)m (not 2m like I stated previously) hitting a single stationary wall at velocity 2v
Someone help out and clarify the math please.
Last edited by syzygy; 05-16-10 at 01:22 AM.