# *2 trucks each going 50km/m hit head on with a force of still 50km/h ?

1. Jun 19, 2011

### nukeman

*2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Hey all, I am a first year physics student, and want some more info on this one here. I heard it in a lecture.

Everyone thinks that if 2 trucks are each going 50km/h and hit head on, it would be the same as a single truck hitting a solid wall at 100 km/h.

To be honest, that is what I always thought lol, but I was told that was wrong.

Can someone explain the physics of this for me?

Thanks

2. Jun 19, 2011

### DaveC426913

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

By whom?

Note that there's some error in the logic of your conclusion. Just because it may not be equivalent to 100km/h into a brick wall does not mean it is automatically equivalent to only 50km/h.

3. Jun 19, 2011

### nukeman

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Who was I told by? Someone in my physics class whom I was trying to explain something too.

So how come if a 50kg object was going 50km/h, which it then hit head on a exact same object, at the exact same speed, why would it not create a force = to this object hitting a solid wall at 100km/h ?

Sorry if I am really not explaining my self properly!

4. Jun 19, 2011

### AlephZero

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

By any engineer who is working on REAL impact modelling problems (not the collisions of point masses you study in Mechanics 101).

Trucks are flexible and plastically deformable. "Solid walls" (as an idealized concept) are not.

Agreed!

5. Jun 19, 2011

### DaveC426913

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

I do not actually know the answer. I can actually see both sides of the argument, so I do not know which one is correct.

Here is the alternate argument:
If a vehicle collides with a solid wall at point X, it sustains damage as the impact crushes it down to, say, a length of Y. The frontmost part of the car never passes point X.

If, instead of something unmoving like a wall, we substitute something else that can bring the car to a halt at point X, say, a counter-moving object, the exact same thing happens to the vehicle - it collides at point X and gets crushed down the same length Y. The frontmost part of the vehicle never passes point X.

In both cases the vehicle began decelerating at the same point (when its bumper reaches point X) and crumpled the same amount.

In short, it doesn't matter by what method it was brought it to a halt at X. All that matters is that it was brought to a halt at X.

6. Jun 19, 2011

### DickL

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

I think that part of the answer is that for in-elastic collisions the amount of crush is the a given. In a single car crash into the in-movable wall all of the crush will happen in the single car. In the 2 car crash the crush takes place equally in both cars, each car crushing only half the amount.

7. Jun 19, 2011

### DaveC426913

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Except that doesn't account for everything, like the doubled kinetic energy, so even if it's shared, they're sharing twice as much.

Again, I do not know the answer. (As a Science Advisor I don't want anyone to think I am making a claim.) I am simply pointing out aspects of the conundrum while waiting for someone more familiar with this thought experiment to weigh in.

8. Jun 19, 2011

### DickL

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

The 2 cases are: 1) 2 trucks (mass = m each) each going the same speed (v), 2) 1 truck (mass = m) going at double the speed (speed = 2v).

For these 2 cases, the KE is equal KE = 2m * v^2

Case 1: KE = m(truck 1) * v^/2 + m(truck 2) * v^2/2 = 2m*v^2
Case 2: KE = m(truck 1) * (2*v)^2/2 = 2m*v^2/2

9. Jun 19, 2011

### mender

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Without clarification, there are two possible answers:

1. The amount of damaged experienced by each of the vehicles will be the same when the initial speed of 50 km/h is reduced to 0 km/h either by hitting another identical vehicle moving in the opposite direction (head-on collision) or by hitting a non-movable wall.

2. To duplicate the amount of energy expended in the collision of the two vehicles, a single vehicle would need to hit the wall at 70.7 km/h.

Typically, the first scenario is the one being considered.

10. Jun 19, 2011

### Antiphon

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

If a truck going 50 hits another much much heavier and harder truck going 50, then the statement starts to become true for the smaller truck.

It's also the reason that driving a very heavy car is usually better in a collision than driving a lighter car.

11. Jun 19, 2011

### cjl

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Two identical vehicles hitting head on at some speed is indeed exactly identical to hitting a perfectly immovable wall at the same speed (rather than twice that speed, as a lot of people think). IIRC, this was tested by mythbusters recently too, with some very interesting results.

12. Jun 19, 2011

### DaveC426913

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

As I concluded:

http://warp.povusers.org/grrr/collisionmath.html

"From the point of view of one of the vehicles it makes absolutely no difference whether it hits a rock wall at 50 km/h or another identical vehicle which was traveling at the same speed in the opposite direction. The amount of force applied to the vehicle is the same in both situations. "

13. Jun 19, 2011

### DaveC426913

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

But let me point out one thing: read the original poster's question:

emphasis on "it would be the same"

What, exactly, would be the same? Damage to one car? No! Total energy expended? Yes.

Two vehicles collidng at 50km/h each would expend the same total energy as one vehicle colliding with a wall at 100km/h.

You were correct, if you plead ambiguity in the question!

14. Jun 19, 2011

### Fredrik

Staff Emeritus
Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

It would be (roughly) the same as a single truck hitting a solid wall at 50 km/h. This is pretty obvious if you realize that the front of the car will have the same acceleration in these two scenarios, i.e. it will go from 50 to 0 in essentially no time at all. Note that it's the motion of the front relative to the back that will deform the car. And in these two scenarios, that motion is the same.

There was another thread about this not too long ago. Someone recently bumped it to post a link to the video of the Mythbusters experiment. I'll see if i can find it.

Edit: I didn't find the post, but this is the video:

(You can skip the first minute and start at 1:00).

Last edited by a moderator: Sep 25, 2014
15. Jun 19, 2011

### cjl

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Nope. Two identical trucks at 50 km/h would have a total of twice as much energy as one truck at 50 km/h, while a single truck at 100 km/h would have four times as much as a single truck at 50 km/h (since K.E. scales as v2). So, the single truck at 100 km/h has twice as much energy as the head on case.

16. Jun 20, 2011

### mender

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Which is why you take the square root of the number of vehicles and multiply that by the initial speed of each vehicle.

2^.5 =1.414, x 50 km/h = 70.7 km/h

17. Jun 20, 2011

### cabraham

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

The masses of the 2 colliding objects is important. A pair of identical truckes, say 2 tons each, colliding at 50 mph is equivalent in damage to either truck hitting an immobile wall at 100 mph. The 2 masses in this case are equal.

With unequal masses, things change. If I'm cruising down the road at 30 mph, & a mosquito is flying straight towards my windshield at 30 mph, what is the equivalent impact speed? For me in my 1.5 ton car, I don't even realize that I hit something. The mosquito splatters on my windshield, & I cannot even feel it.

To the mosquito, it's as if it flew at 60 mph straight into a windshield of a parked car. Not good for the poor mosquito. So if the masses are orders of magnitude apart, the impact equivalent speed could be the sum total of the 2 speeds for the lighter object, but not for the more massive object. The more massive object incurs an equivalent impact speed near zero as far as damage goes.

Make sense?

Claude

Last edited: Jun 20, 2011
18. Jun 20, 2011

### cjl

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

I'm not sure if this was a typo or a conceptual mistake on your part, but two identical trucks colliding at 50mph is equivalent in damage, energy, and every other way to hitting an immobile wall at 50mph. Not 100mph.

This part is correct though. For the case of two identical objects at some speed V, a head on impact is exactly like an impact with an immovable object at speed V. However, as the mass becomes biased towards one object, the heavier object will have the same damage as if it hit an immobile object at <V (with the extreme case being effectively no impact, as with a car and a mosquito), and the lighter object will act as though it hit an immobile object at >V (with the limiting case being 2V).

19. Jun 20, 2011

### cabraham

Re: *2 trucks each going 50km/m hit head on with a force of still 50km/h ????

Oh my gosh, of course. Sorry, my bad. I've posted on this issue before, & somehow I meant 50 mph, but my fingers typed 100 mph. I was thinking ahead. Since the masses are identical, the equivalent impact is the same for each vehicle, equal to the speed of either one, namely 50 mph.

In the case where the masses greatly differ, things change. We seem to agree. Thanks, & again, my apologies to all.

Claude