Two cars driving 30 and 40 mi/hr is like hitting a brick wall at 70 mi/hr

[Pupil]

I hear this from people sometimes. Upon hearing about a bad head on collision, I will hear a person promptly add the two speeds together and claim the wreck for the drivers is as bad as slamming into a brick wall with that new speed (S = s1+s2). Is this really a good analogy? I'm not so sure it is.

When two cars collide head on mv = F$$\Delta$$t applies, and the time it takes for the two cars to stop when they hit might be longer than slamming into a brick wall, so the force (which is what I think people mean by "is the same as running into...") could be smaller. I realize I'm not being extremely specific about the crash so there are a lot of things to nitpick over, but let's say we're talking about an inelastic accordion-like wreck on a frictioney (made up word) surface.

Am I wrong or does this analogy to a brick wall fall apart?

 

[rcgldr]

Only if there is a huge difference in mass, like a car going 30 mph head on into a train coming at 40 mph. If the masses are the same, then it's like hitting a solid wall at 35 mph, assuming it's a direct head on and not a partial head on where the cars spin.

 

[tiny-tim]

Upon hearing about a bad head on collision, I will hear a person promptly add the two speeds together and claim the wreck for the drivers is as bad as slamming into a brick wall with that new speed (S = s1+s2). Is this really a good analogy? I'm not so sure it is.

Hi Pupil!

A car "gives" a lot more than a brick wall does.

So it would be better to say it's the same as slamming into the front half of a stationary car up against a brick wall.

EDIT: mgb_phys has changed my mind … see below. ​

 

[Pupil]

Hi Pupil!

A car "gives" a lot more than a brick wall does.

So it would be better to say it's the same as slamming into the front half of a stationary car up against a brick wall.

LOL! Well, yeah, I suppose that would be a more apt analogy. With the impulse equation that was what I was getting at when you say it 'gives' more. The brick wall also accelerate much less than the car when hit by another car, so won't that slow the car's acceleration as well?

 

[mgb_phys]

Also imagine two identical cars hitting each other head on, both going at 30mph.
Would that be any different from one car hitting a wall at 30mph?

 

[Pupil]

Also imagine two identical cars hitting each other head on, both going at 30mph.
Would that be any different from one car hitting a wall at 30mph?

I would think so, since the $$\Delta$$t in mv = F($$\Delta$$t) is probably larger for the car collision than for the wall, which would result in a smaller average force.

However both cars would stop dead as if they ran into a brick wall, since their total momentum would be 0 (assuming perfect in-elasticity).

Right? :shy:

 

[Tony71502]

So if one car travels at 30 and one at 40, they both hit head on, and one car takes "70 mph" of force. What does the other take... 0? lol..

 

[tiny-tim]

However both cars would stop dead as if they ran into a brick wall, since their total momentum would be 0 (assuming perfect in-elasticity).

Yes, if both cars have the same speed, then it makes no difference whether the wall is there or not.

I prefer mgb_phys's reasoning to mine …

ignore my previous post!

 

[TurtleMeister]

Since I've been following this thread, I just have to ask. What is it with the brick wall thing? I know that it's common for people to say this, but It's a bad analogy any way you look at it. As others have already pointed out, there's a lot of difference between a brick wall and a car. If you're going to make analogies why not say that two cars, one traveling 30mph and the other traveling 40mph and colliding with each other head on, is similar to one car traveling 70mph and colliding head on with a parked car?

Also imagine two identical cars hitting each other head on, both going at 30mph. Would that be any different from one car hitting a wall at 30mph?

Yes, because a brick wall is not similar to a car. Let's phrase the last sentence a different way. "Would that be any different from one car hitting a beach ball at 30mph?"

So if one car travels at 30 and one at 40, they both hit head on, and one car takes "70 mph" of force. What does the other take... 0? lol..

No, they both experience the same force.

 

[mgb_phys]

If two identical cars hit each other head on they are going to both stop exactly over the midpoint of the collision. Therefore it is equivalent to hitting an immovable object at that same point.

 

[TurtleMeister]

If two identical cars hit each other head on they are going to both stop exactly over the midpoint of the collision. Therefore it is equivalent to hitting an immovable object at that same point.

Yes, that is true. I was not considering the brick wall to be immovable.

 

[TurtleMeister]

So I've changed my mind. You can make an analogy with the brick wall if you consider the brick wall to be immovable. But the OPs example is still incorrect when saying that two cars, one traveling 30mph and the other traveling 40mph and hitting each other head on, is similar to one car hitting a brick wall at 70mph. It is actually similar to one car hitting a brick wall at 35mph.

 

[Pupil]

But the OPs example is still incorrect when saying that two cars, one traveling 30mph and the other traveling 40mph and hitting each other head on, is similar to one car hitting a brick wall at 70mph. It is actually similar to one car hitting a brick wall at 35mph.

How so?

 

[Cyrus]

I would suggest someone calculate or reference something. This is getting to be a bit of a ridiculousness thread...

 

[TurtleMeister]

How so?

By calculating the average speed of the two cars. As "mgb phys" has already pointed out:

If two identical cars hit each other head on they are going to both stop exactly over the midpoint of the collision. Therefore it is equivalent to hitting an immovable object at that same point.

However, he neglected to mention that both cars must be traveling at the same speed for this to be true. If one car is traveling at 30mph and the other at 40mph it would be the same if both cars were traveling at 35mph. 30+40=70, 35+35=70. If one car was traveling at 40mph and the other car at 60mph then it would be the same as hitting an "immovable" brick wall at 50mph.

 

[TurtleMeister]

I would suggest someone calculate or reference something. This is getting to be a bit of a ridiculousness thread...

Well, I'm not very good with math but I'm going to give this a try. The solution stated in my previous post works just fine as long as both cars have the same mass. But what if they have different mass and are traveling at different speeds?

For example, suppose we have Car1 which weighs 2500 lbs and is traveling at 30mph, and we have Car2 which weighs 3500 lbs and is traveling at 40mph. Car1 and Car2 crash head on.

Now, I want to crash Car1 into an immovable brick wall with the same force as when it collided with Car2. How fast does Car1 have to be traveling? If I do the same with Car2, how fast does it have to be traveling?

Here is my solution and I need the much wiser PF members to tell me if I'm on the right track or not. Since I know the following to be true when both cars have the same mass and same speed:

(M1 * V1²) / 2 = (M2 * V2²) / 2

All I have to do is solve for V1 or V2.

Speed of Car1 = sqrt((M2 * V2²) / M1) = 47.33 mph
Speed of Car2 = sqrt((M1 * V1²) / M2) = 25.35 mph

The numbers look about right but I'm not sure if it's correct or not.

 

[rcgldr]

(M1 * V1²) / 2 = (M2 * V2²) / 2

This is only true in an elastic collision where kinetic energy is conserved. I would assume an inelastic collision where the cars front ends collapse and the cars remain attached to each other. In this case you just need to conserve momentum so that the momentum of the combined cars moving as the same speed is the same as the momentum was before the collision to determine the final speed:

(m₁ + m₂) v = m₁ v₁ + m₂ v₂

v = (m₁ v₁ + m₂ v₂) / (m₁ + m₂)

Then the impact for car₁ is related to v₁-v and for car₂ it's related to v₂-v.

 

[Cyrus]

Well, I'm not very good with math but I'm going to give this a try. The solution stated in my previous post works just fine as long as both cars have the same mass. But what if they have different mass and are traveling at different speeds?

For example, suppose we have Car1 which weighs 2500 lbs and is traveling at 30mph, and we have Car2 which weighs 3500 lbs and is traveling at 40mph. Car1 and Car2 crash head on.

Now, I want to crash Car1 into an immovable brick wall with the same force as when it collided with Car2. How fast does Car1 have to be traveling? If I do the same with Car2, how fast does it have to be traveling?

Here is my solution and I need the much wiser PF members to tell me if I'm on the right track or not. Since I know the following to be true when both cars have the same mass and same speed:

(M1 * V1²) / 2 = (M2 * V2²) / 2

All I have to do is solve for V1 or V2.

Speed of Car1 = sqrt((M2 * V2²) / M1) = 47.33 mph
Speed of Car2 = sqrt((M1 * V1²) / M2) = 25.35 mph

The numbers look about right but I'm not sure if it's correct or not.

There we go. That's what I'm talking about. Now instead of debating what if's, we can talk in absolutes.

 

[nooma]

Ok well I'm from Aus so i'll use km/h instead :P

if you think of it in terms of change in momentum (as has been previously stated), for the cars the change in momentum will be different for either car. Assuming that both cars weigh 1,000kg, car1 is traveling west at 80km/h and car2 is traveling east at 100km/h.

total initial momentum = mv
= (1,000 x 100/3.6) - (1,000 x 80/3.6)
= 27777.78 - 22222.22
= 5555.6 Ns^-1 east

initial momentum = final momentum
5555.6 = mv where m is now the combined mass of the cars
v= 2.7 ms^-1 East(10 km/h)

now we also know that $$\Delta$$p = $$\Delta$$Ft

assuming that the crash takes 0.5 sec to happen (including all crumpling etc)... i have just plucked this number out of thin air btw

$$\Delta$$p = 22222.2 Ns^-1 east

22222.2 Ns^-1 east = F x 0.5
therefore F_av = 44444.4 N

to find the force if the car had hit an imovable brick wall...

t = 0.5 sec
a = (v - u)/t
F = ma

from this we can solve to see what speed the car would have to be traveling at to experience the same force

44444.4 = 1000 x (-u)/0.5
-u = 22.22 ms^-1
= 80 km/h

thus is can be shown that two cars hitting head on can be shown to be the same as one car hitting a brick wall at the average of their speeds

 

[TurtleMeister]

(M1 * V1²) / 2 = (M2 * V2²) / 2

This is only true in an elastic collision where kinetic energy is conserved.

If M1 = M2 and V1 = V2 then it has to be true (before the collision). I was originally bebating whether I should use momentum or kinetic energy.

nooma, thanks for the detailed proof that if the cars are of the same mass we need only calculate their average speed.

So, I'm assuming that the current consensus is that my method is wrong? I guess I'll have to give it some more thought.

To refresh:
Car1 initial speed = 30mph weight = 2500lbs
Car2 initial speed = 40mph weight = 3500lbs
If the cars crash head on, at what speed must the cars travel in order to hit an immovable wall with the same force? Anyone have answers?

 

[Lok]

Relativity still applies.. so two cars colliding at 30 mph each will be the same as two cars colliding with one stationary and one at 60 mph ( relativity effects too small ).

But 2 cars will absorb through their chassis a certain amount of energy, not the case with a car colliding with a brick wall which by virtue will absorb very little energy.

 

[nooma]

If M1 = M2 and V1 = V2 then it has to be true (before the collision). I was originally bebating whether I should use momentum or kinetic energy.

nooma, thanks for the detailed proof that if the cars are of the same mass we need only calculate their average speed.

So, I'm assuming that the current consensus is that my method is wrong? I guess I'll have to give it some more thought.

To refresh:
Car1 initial speed = 30mph weight = 2500lbs
Car2 initial speed = 40mph weight = 3500lbs
If the cars crash head on, at what speed must the cars travel in order to hit an immovable wall with the same force? Anyone have answers?

when dealing with these sort of collisions you should always use momentum, as momentum is conserved whilst kinetic energy is often lost. If you look at the collision in the example I used;

total kinetic energy before = 0.5x1000x(100/3.6)² + 0.5x1000 x(80/3.6)²
=632716 J

kinetic energy after = 0.5x2000x2.7²
= 7290

As you can see there is considerable energy lost during the collision

If you substitute a different mass into the calculations that I used you can also find the speed needed for the same force to be exerted if one of the cars was in a collision with an imovable object


lets see if we change the mass of the car traveling west

total initial momentum = mv
= (1,000 x 100/3.6) - (2,000 x 80/3.6)
= 27777.78 - 44444.44
= 16666 Ns-1 west

initial momentum = final momentum
16666 = mv where m is now the combined mass of the cars
v= 5.5 ms-1 west(20 km/h)

now we also know that p = Ft

assuming that the crash takes 0.5 sec to happen (including all crumpling etc)... i have just plucked this number out of thin air btw

p = 16666 Ns-1 east

16666 Ns-1 west = F x 0.5
therefore Fav = 33333 N

to find the force if the car had hit an imovable brick wall...

t = 0.5 sec
a = (v - u)/t
F = ma

from this we can solve to see what speed the car would have to be traveling at to experience the same force

33333 = 1000 x (-u)/0.5
-u = 33.3 ms-1
= 120 km/h for the 1000 kg car

=60 kmh for the larger car

 

[J_Cervini]

My take: The hypothesis holds true if each car is identical with opposite and equal velocities, which causes a perfect head on collision; meaning all potential energy of both vehicles is exhausted equally, to zero at impact. No spin - just pancake.
The difference in the "30 mph and 40 mph" velocities is that at the end of impact there will be a resulting forward velocity in relation to the faster vehicle. He has more momentum thus he carries forward. This is lost energy from the standpoint of relating this impact to that of a single car hitting a wall. This lost energy results in less force at impact.

 

[turbo]

There is a factor that nobody has taken into account. Cars have crumple zones to make them safer in crashes. I would rather have someone head-on me at 30 or 40 (additive) mph than slam into a concrete abutment at 70mph. Concrete and bricks are not designed with crumple zones. Depending on crash speed and design, your car's crumple zone might be used up too soon in a crash with an immovable object, while in a two-vehicle crash, both vehicles would be deforming, absorbing energy and reducing their momentums.

 

[turbo]

Also, when cars are crash-rated, they are not subject to head-ons over and over. They are slammed into immovable objects with only about half of the nose of the vehicle contacting the block. This takes out the absorptive effect of half of the front crumple zone, and probably allow the testers to point out deficiencies that would prompt designers to improve their vehicles' body/chassis to better re-distribute forces in asymmetrical crashes.

 

[nooma]

however the crumple times would remain the same for a head on between two cars and a stationary object

 

[tiny-tim]

however the crumple times would remain the same for a head on between two cars and a stationary object

Yes: if two equal cars with equal speed collide, it can make no difference to the crumpling whether a fixed wall is between them or not.

 

[TurtleMeister]

With the help of Wikipedia and previous posters, I think I've finally found a solution. Please disregard my previous solution. It's fubar. :)

I think we are focusing too much on how the cars absorb the energy of the impact. We can make things much simpler if we just ignore this and calculate how fast the car has to travel in order to hit an immovable brick wall with the same momentum as hitting another moving car. If we ignore the impact time, or crumpling time, we can simply assume that the cars are elastic. Since momentum is conserved in elastic collisions we can use:

m1v1 + m2v2 = m1u1 + m2u2

where

m1 = mass of car1
m2 = mass of car2
u1 = velocity of car1 before collision
u2 = velocity of car2 before collision
v1 = velocity of car1 after collision
v2 = velocity of car2 after collision

solving for v1

(u1 * (m1 - m2) + (2 * m2 * u2)) / (m1 + m2)

and solving for v2

(u2 * (m2 - m1) + (2 * m1 * u1)) / (m1 + m2)

Because momentum and kinetic energy are conserved in elastic collisions we can now use the before and after velocities to determine the velocities needed for each car to strike an immovable object with the same momentum.

V1 = (u1 - v1) / 2

and

V2 = (u2 - v2) / 2

If we plug in the values from the original problem:

m1 = 2500lbs (car1)
m2 = 3500lbs (car2)
u1 = 30mph (car1)
u2 = -40mph (car2)

we find that:

Velocity of car1 after impact = -51.67 mph
Velocity of car2 after impact = 18.33 mph

Velocity of car1 to strike an immovable wall with same momentum: 40.83 mph
Velocity of car2 to strike an immovable wall with same momentum: -29.17 mph



Edit:
Just a little more math to prove that calculations are correct.

1 Relative speed of car1 and car2 before impact = 30 + 40 = 70
2 Relative speed of car1 and car2 after impact = 51.67 + 18.33 = 70 (elastic rebound)
3 Relative speed of car1 and car2 with brick wall = 40.83 + 29.17 = 70

Notice that as previous posters have already pointed out, the wall is not even necessary in #3. The velocities of both vehicles has been adjusted (in #3) so that rebound velocity equals the initial velocity. Car1 has an initial velocity of 40.83 and a rebound velocity of -40.83. Car2 likewise -29.17 and 29.17. However, the change in momentum after impact for each car remains the same.

Additional proof that calculations are correct:

Car1 change in velocity after hitting car2 = 30 - (-51.67) = 81.67
Car1 change in velocity after hitting brick wall = 40.83 - (-40.83) = 81.66

Car2 change in velocity after hitting car1 = -40 - 18.33 = -58.33
Car2 change in velocity after hitting brick wall = -29.17 - 29.17 = -58.34

Of course the actual change in velocity would be half that indicated because real cars are not elastic and would absorb the impact instead of rebounding.

As you can see car1 hitting car2 at a relative speed of 70mph is the same as car1 hitting a brick wall at 40.83 mph. And car2 hitting car1 at a relative speed of 70mph is the same as car2 hitting a brick wall at 29.17mph. Proof that driving a heavy vehicle has it's advantages.