"In the other extreme where the collision is perfectly inelastic, such as when two balls of clay hit each other and stick, there is no difference between two balls of clay moving at 1 m/s each, colliding, and coming to rest, and one ball hitting a solid wall at 1 m/s and coming to rest. "
Explain how momentum is conserved then? Is it the same in both cases?
“Looking at it from an energy point of view: The crumple zones in cars are ment to absorb energy, right? What energy are we talking about? It’s the 60 MPH that the car is traveling times the weight of the car.”
60 MPH * mass is the momentum of the car. Both energy and momentum are conserved.
Agreed, yes don’t be hasty and don’t write replys while trying to listen to the rest of the show. 
Thanks. Ditto on the hasty.
You guys make it all too complicated. Lets just put up a brick wall and have two cars hit it on opposite sides at the same time. The is they would have hit head on if the wall wasn’t there. The wall is immovable whether hit by one car on one side only or hit on both sides at the same time. Now what is the difference to the vehicles if they hit one at a time or both at once? What would be the difference if the wall is removed and the two vehicles hit head on. The simple answer is that the caller was right and Tom and Ray were wrong.
YES,
ALWAYS.
BUT THE NEIGHBOR IS ALWAYS CORRECT.
“Now, point X itself would be a different matter. If it had some kind of impact meter, it would register twice the impact from the two cars as it would from the car vs wall because both cars would impart their force to it.”
Q.E.D. the cases are not the same.
The mass, momentum and energy of the system are conserved. The two systems that you describe have different total momentum and energy. Thus they are different. Momentum is described as mass (m) times () velocity (V) or mV. The kinetic energy of a particle is described as 1/2mV*V (The cars are on flat ground so there is no potential energy). The laws of conservation of momentum and energy state that the sum of the initial momentum of the system equals the sum of the final momentum of the system and the sum of the initial energy of the SYSTEM equals the sum of the final energy of the system. The terms that the first year physics books do not give you are the dissipation terms. The terms that convert the energy from one form to another. Without the dissipation, you can’t have an inelastic colision. Energy and momentum must be lost. In the case considered, all is lost at point X.
In your first case, two particles enter the infinitesimal region surrounding point X with V and -V. An instant before impact, the net momentum of the system is mV + mV. After impact the momentum of the system is m0 + m0. To conserve momentum 2(mV) has to be dissipated. With respect to energy, 0.5mvv + 0.5mvv exists prior to the collision and 0 exists after indicating that mv*v amount of energy is lost to the system.
In your second case, one particle is moving at the same V as in case 1 and the wall is not. Also note that you are assuming that the wall has large mass compared to the particle mass (i.e. it does not move). The initial momentum and energy are mV + 0 and 0.5mVV. Since V = 0 after the colision, the momentum and energy lost are mV and 0.5mVV.
Case two is clearly not the same as case 1 for conservative properties regarding the total quantities in the systems.
Momentum is always conserved, whether the collision is elastic or inelastic. That’s a Law of physics. Energy is conserved ONLY in the elastic case. In the perfectly inelastic case, all energy is converted into heat and sound. In the partially elastic case, some of the energy is converted to heat and sound, the rest is conserved as motion.
From a momentum standpoint, two identical cars going 60 mph and hitting each other = one of those cars hitting a wall at 60 mph. The caller was right. Tom and Ray were incorrect.
You’re analysis is correct. This would have made a good puzzler, since the “common sense” answer is not the correct one. It can’t be understood by just looking at the relative velocities.
I agree, the problem is completely symmetrical (just like there is a solid wall between the cars) if the cars are exactly the same and traveling at the same speed. Obviously, that is not true is there is a difference in speed and/or mass of the cars.
Thought experiment: Hang a scale between the two cars and hang one on the stone wall.
Divide the experiment into two phases: impact and dissipation.
If two cars hit the scale at the same time, the scale will add the forces. But those forces immediately dissipate into the two cars.
On the wall, (if the car is going 60), the scale will read 1/2 of the other scale, but all the energy will dissipate into the one car.
The scale readings are different. The impact on the cars are all the same.
The caller is correct the 60 mile an hour head-on collision with an identical car produces the same damage to my car as the 60 mile an hour collision with the wall.
The change in momentum argument:
From the point of view of the work done on the car to bring it to a stop, the work is force times distance. In both cases the car goes from 60 mph to 0 in the same distance, the crumple distance of one car. To bring the car to rest from 60 mph it takes a force to change the momentum of one car p = mv to zero. This force will be the same force over the same distance in both cases and over the same time. And so the work done on the car will be the same for the head-on and the wall collision. The wall does not move (immovable object) and so no work is done on the wall. The same amount of work is done on the other car.
Energy argument
The car has an initial kinetic energy KE = 1/2 mv^2 the final kinetic energy of one car is 0. The energy is dissipated by deforming the car.
For one car into the wall all the energy goes in to one car.
For the head-on there are 2 cars with twice the energy, each gets its 1/2 share of the energy so the energy of the 60 mile an hour head-on dissipated by 1 car is the same as the energy of running into the wall.
Paul D (who went to MIT with Ray but who went on to become a physics professor, and who looks forward to hearing Wolfgang weigh in on this next week)
We are comparing collisions, and collisions are about rapid deceleration. Specifically, change in speed and the elapsed time the change in speed occurred over. A car hitting an immovable wall will decelerate from impact speed (let’s say 60 mph) to zero in time T. (the actual time will depend on how the car is built). So we have a damage profile that can be described as 60mph(T), ie a 60 mph deceleration in the time T.
In the a head-on collision of two identical cars, both cars have the same momentum (one in the opposite direction of course) so when they collide, they will both come to a stop about the plane where their bumpers first meet. The deceleration for each car will be 60 mph to zero, and the duration of deceleration will be T. (This is obvious because the momentum of each car exactly cancels out the momentum of the other car, just as the immovable wall completely cancelled out the momentum of the car that hit it). The damage profile for either car in the head-on will be 60mph(T), which is identical to the damage profile for the wall collision at 60 mph.
I think we have a consensus. Tom and Ray will have to barbecue some crow this weekend.
The kenetic energy of a body is 1/2 * m* V(**2)
If you hit the wall at 60 mph then you have ONE QUARTER of the energy that you would have if you had hit the wall at 120 mph.
One quarter.
Not one half.
It ain’t linear.
Even if the energy did scale linearly; well, it would be LINEAR!!!
So two cars hitting the wall have twice as much energy to dissipate as one car hitting the wall.
How could both of those cars possibly sustain more damage?
You need more energy in order to create more damage. But there ain’t no more energy - so there ain’t mo damage.
Yeash.
Running into an immovable brick wall at 60 mph is exactly like colliding head-on with a vehicle also moving at 60 mph, assuming the vehicle masses are equal. Two thought experiments can prove this:
Suppose the vehicle coming towards you were moving at 30 mph, not 60 mph. Then you would slam into it and continue moving forward (conservation of momemtum would show you would continue movng at 15 mph, assuming no friction). This would clearly not be like running into a brick wall at 60+30 = 90 mph. This is what Tom and Ray imply, as they say the collision of two 60 mph vehicles would be like running into a brick wall at 120 mph.
Suppose the collision were elastic, not inelastic. That is, suppose the two cars bounce off each other like billiard balls instead of sticking together. A billiard ball running into a brick wall at 60 mph would bounce back at 60 mph. Similarly, two billiard balls approaching each other at 60 mph would bounce back at 60 mph. If it were more, like 120 mph as Tom and Ray would imply, then you would violate the conservation of energy. You could increase speeds up to the speed of light just by utilizing head-on collisions.
All the discussion about crumple zones and whatnot is moot. The caller is not talking about a minor difference between two cars at 60mph and one car and wall at 120mph, but in something much more significant.
His mental error is that he somehow thinks that the immovable wall is subject to no effects from the impact of a 120mph, unlike a moving 60mph car. This is the fatal flaw, for in both cases the energy is the same. The easiest way to see this is for him to imagine his car being motionless, then being hit by a wall moving at 120mph. It makes no difference whether the car is moving or the wall is moving: you’re in trouble both ways.
That is not correct, the wall i this example is considered to be rigid and is not assumed to dissipate any of the energy from the crash. What you are saying would be true if we were talking about a highway barrier with an energy absorbing design (similar to a car crumple zone).
Heh, just had to log in and razz Tom & Ray over goofing this one up. It partially gets back for all those blasted lightbulbs and prisoners they’ve made me think about over the years.
Keith’s answer is the best so far. Along those lines: imagine a car driving at 60 mph into a wall with mirror on it. What you “see” is two cars approaching each other at a combined speed of 120 mph, and having a perfectly symmetrical head-on collision. But obviously the final result is exactly the same as one car hitting a wall at 60 mph.
It reminds me of a puzzler we were given in freshman physics. Centuries ago, a scientist perfected the first working vacuum pump. He arranged a demonstration for the king (since, of course, the king was his funding agency, and we all know you have to keep the funding agencies happy). The scientist had two large metal hemispheres. He pressed them together into a ball, and evacuated the air from the inside. Air pressure then kept the two halves joined. The scientist tied a team of horses to one hemisphere, and another team of horses to the opposite hemisphere, and showed how the horses could pull in opposite directions without being able to pull the ball apart.
Now here is the question (and it relates directly to this car-smash problem): Since the scientist was a good physicist, he knew it would be the exact same force pulling on the ball if he had used just ONE team of horses, and tied the opposite hemisphere to a tree. So why did he use two teams of horses instead of just one?
Answer below…
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Seriously, think about it, it’s fun …
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…the scientist used two teams of horses because he realized the king, knowing no physics, would be more impressed seeing TWO teams of horses unable to pull the ball apart than by seeing ONE team.
So, based on that, I hereby nominate Tom and Ray king for the day.