Collision physics explanation needed

Discussion in 'Physics & Math' started by MacGyver1968, Oct 5, 2010.

  1. MacGyver1968 Fixin' Shit that Ain't Broke Valued Senior Member

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    I recently watched an episode of "Mythbusters" and some of the physics in the episode confused me. I was hoping someone could explain it to me in layman terms.

    In a previous mythbusters episode, the MB team had crashed two cars into each other head-on, each traveling at 50 mph. One of the mythbusters had stated that this was the same as a single car hitting a wall at 100 mph. Many viewers wrote in to the show, claiming this was incorrect...that the impact would be equal to a single car hitting a wall at 50 mph...so the MB's decided to put it to the test.

    They crashed 2 cars into a wall, one at 50 mph and one at 100 mph and recorded the damage to the vehicle and the g-forces involved. Then they crashed two cars head-on at 50 mph and compared the damage. The viewers were correct. The damage and g-forces were consistent with the 50 mph wall crash. This totally surprised me, and I didn't understand how this was possible. I would appreciate it if someone could explain it in simple terms.
     
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  3. adoucette Caca Occurs Valued Senior Member

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    Visualize it this way.

    A car hits an immovable wall at 50 mph, call the location of the wall X.
    The rate of decleration of the passenger from the point the car's bumper hits X until the car stops is Y, and from this one can calculate the force on the person.

    Now instead of the wall have the two cars hit at point X.
    Since the other car has equal mass and velocity, at the end of the collision, the mass of both cars will still be on their respective sides of X, thus the impacting car had the exact same effect as an immovable wall.

    Now cut the weight of one car and you will see that X gets displaced in the direction of the lighter car, thus slower deceleration on the heavier car.

    Arthur
     
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  5. MacGyver1968 Fixin' Shit that Ain't Broke Valued Senior Member

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    Ok...that makes sense...it just seems so counter-intuitive.

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    Would it be true that the impact would equal a 100 mph wall crash if one of the cars had twice the mass of the other?

    Thanks for the quick response.
     
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  7. przyk squishy Valued Senior Member

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    Walls don't move or crush. Cars do. Two 50 mph cars crashing into one another is equivalent to a 100 mph car crashing into a stationary car, not a stationary wall.
     
  8. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Arthur's post 2 is 100% correct, but it may help to note that car going 100mph into solid wall has four times the kinetic energy to dissipate in crunching up the front of the car.

    When two equal mass car with their speedometer showing 50mph collide head on each has only 1/4 the energy of the 100 mph car, but as there are two of them together they have half a much energy but they come to sudden stop like the 100mph car does.

    That fact alone (half the total energy) tells you there will be less crunching of metal. Now when you add fact there is twice as much metal to crunch, each car is going to dissipate in crunching metal only 1/4 as much energy in crunching its metal as the 100 mph car did.
     
  9. John Connellan Valued Senior Member

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    Exactly, and what would you prefer to hit at 100 mph, a solid wall or a stationary car with no handbrake on?
     
  10. adoucette Caca Occurs Valued Senior Member

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    No.

    Into an immovable wall, the mass of the car is not an issue.

    If deceleration time remains constant, then the force on the person in the car would be determined by the velocity.

    If you increase the speed of the collision, the force increases at V^2

    In other words double the speed, quadruple the force, triple the speed, nine times the force.

    Arthur
     
  11. James R Just this guy, you know? Staff Member

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    Probably the best way to look at the problem is in terms of energy dissipated, as Billy T said.

    A car hitting a wall at 100 mph has 4 times the kinetic energy of a car hitting the wall at 50 mph. Two cars hitting each other at 50 mph together have twice the energy of a car hitting a wall at 50 mph.

    So, you'd expect the most damage for the 100 mph car hitting the wall. For two 50 mph cars colliding head on the damage will be shared equally between the two cars, giving the same damage to each car as would be incurred by that single car hitting a wall at 50 mph.

    The issue of the force on each car during the collision depends on the mass of the car, its initial speed and the time taken for the collision to occur (from initial contact to the time when the car's centre-of-mass comes to rest).

    If the collision of a 50 mph car with a wall takes the same time as a head-on collision of a 50 mph car with another identical car, then the forces felt by a passenger in the car will be the same in each case.

    Does the above match what the show found?
     
  12. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I of course agree.

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    But I will explain why: It is essentially impossible to say any thing quantative about the forces the driver will experience in say a 50 mph collision with a brick wall.

    At first they will be very small until the seat belt starts to restrain his forward motion. If he has no seat belt on (shame, shame!) the force for a somewhat longer period on him will be zero, but then when he collides with something the force will be much higher. If this stupid driver at least has an air bag or a steering wheel that buckles these force will not become as large as the force peak on the seat-belt-less passenger. He is likely to use his face to make a large deep dent in the dashboard or shatter the window going partially thru it.

    When I was about 12, the boyfriend of an older girl I was interest in (with no luck), who was older than her, had a car in which he died.
    No one had heard of seat belts nor steering wheels that collapsed back then. The stiff shaft of his went completely thru his chest.
     
    Last edited by a moderator: Oct 6, 2010
  13. MacGyver1968 Fixin' Shit that Ain't Broke Valued Senior Member

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    Yes..that is what they found. The 2 head-on cars had the same amount of damage and g-loads as the car that hit the wall at 50 mph.
     
  14. Fraggle Rocker Staff Member

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    You can simplify the problem by just looking at the g-force, which is strictly a matter of the rate of deceleration. Whether the car hits a brick wall, or runs head-on into another car of exactly the same mass moving at exactly the same speed, its front bumper will for all practical purposes decelerate to zero mph at the moment of impact. The occupants will decelerate from 50mph to zero mph in the time it takes their bodies to move from their place in a fully operational car to their new place a few feet forward in a crumpled car.

    They will decelerate from 70 feet per second to zero feet per second in about one tenth of one second. This is more than twenty gees. Even with an air bag that might well be fatal, but it will certainly cause calamitous injury. Today's cars are built to get you through a collision with a brick wall at 25-30mph without serious injury. This goes way beyond that; your body (and brain!) are absorbing 3-4 times as much energy.

    The details of the collision don't really matter. All that matters is the rate of deceleration. That's the key to solving any problem of this type.
     
  15. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    That is true but the peak deceleration of your body can be, and usually is, very much greater than the average, especially if you slam into something stiff. See post 9 for more details.
     
  16. Tach Banned Banned

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    You need to write the equations of conservation of momentum and energy.

    1. Car vs. car collision

    1.1. Momentum conservation

    mv+m(-v)=0

    1.2. Energy conservation

    E_deformation_1=mv^2/2+mv^2/2=mv^2

    2. Car vs. wall collision

    2.1. Momentum conservation

    mv+M*0=(M+m)*v_small (the wall recoils VERY little because M>>m)

    v_small=mv/(M+m) <<v

    2.2. Energy conservation:

    E_deformation_2+(M+m)v_small^2/2=mv^2/2=E_deformation_1/2

    So,

    E_deformation_2<E_deformation_1/2
     
    Last edited: Oct 24, 2010

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