Read my post again, and realize that momentum is a vector.
For case 1, the initial momentum of car A is m[sub]A[/sub]v[sub]A[/sub] = (1000 kg)(50 km/hr) = 50,000 kg km/hr. The initial momentum of car B is m[sub]B[/sub]v[sub]B[/sub] = (1000 kg)(-50 km/hr) = -50,000 kg km/hr.
If you add these two momentum vectors together, you get a total momentum for the entire two-car system that is zero.
This does not change. After the collision, the total momentum is also zero.
You can’t just look at one vehicle to apply conservation of momentum. This law only applies if no outside force is applied. If you consider only one vehicle at a time, then certainly a force is applied–that of the impact from the other vehicle.
Instead, you must consider the two-car system as a whole.
This is so wrong, and represents such a gross misunderstanding of the physics in question, that I don’t even know where to begin.
See above comment.
Well, you need to compare the change in kinetic energy in each case.
You are apparently assuming that kinetic energy is conserved. It is not.
In an inelastic collision, kinetic energy is not conserved, so your analysis is flawed. In addition, you have come to the erroneous conclusion that conservation of momentum is “bogus,” which should be a clue that you have made a serious error.
Again, total energy and momentum are conserved. Kinetic energy is NOT conserved.
Again, KINETIC ENERGY IS NOT CONSERVED.
This is a joke, right?
No, seriously, where are the cameras?
:rolleyes:
Say, westom, why don’t you come on down here to the Pit?
My mistake.
Initially posted was 250,000 /.5 / 1000
It should have read
250,000 / .5 / 2000 = 250 = velocity squared
of 15.8 km/hr
BTW, I am trying to get you to deal with energy numbers. Once you apply those numbers properly, then conservation of momentum applies. None of this was possible until your every post including the underlying numbers. Without numbers, only lies, propaganda, anger, miscommunication, myths, and even pagan religions occur.
When done, the single vehicle at 100 km/hr is twice as destructive (dissipates twice the energy) of two 50 km/hr vehicles.
The balance between momentum and energy can define how energy dissipates But the bottom line remains - the 100 km/hr vehicle dissipates twice as much energy destructively as two 50 km/hr vehicles. That answers to the OP’s question - assuming energy defines what he has called detrimental.
Using numbers I have provided, you can better define how collision energy gets applied to two vehicles that must absorb or dissipate that energy detrimentally.
Remember, any and all vectors apply. The final answer must even assume collisions from all directions. Even that one vehicle is rolling after the initial contact. And that the crash occurs on vehicles with structually different crumple zones. For all we know, the 100 km/hr car snow plows throughthe parked car. The only definitive numbers are obtained from numbers the OP provided – 100 km/hr, 50 km/hr, and 0 km/hr. Limits all answers to ballpark conclusions. Only fact we can say with certainty is energy and speed in each car before the crash. And that, like all crashes (except hit and run), means vehicles end at 0 km/hr. Therefore all energy dissipates in the crash.
Any decent and honest person post corrections with numbers and equations. Such subjective sentences exist only from extremists, liars, junk scientists, and the least educated. You know the answer has errors? Good. So do I as I was originally hinting at in an earliest post. Now finish what was necessary and not found in your first posts - the numbers and equations. Stop posting subjectively. Stop posting emotions. Post the corrected equations. Explain what happens with all 2,500,000 and 5,000,000 kg km2/hr2. Explain it with numbers; not subjective (junk science) attacks and denials.
westom, I teach physics. We do instructional labs where we create collisions like this on a smaller scale, using carts on flat tracks or gliders on very low-friction air tracks. Our data always confirms that momentum is conserved during both elastic and inelastic collisions. It also confirms that kinetic energy is NEVER conserved during an inelastic collision. (Kinetic energy is only conserved in perfectly elastic collisions).
Forget the fucking cars–I’m going to use smaller numbers to try to make this easier to write and easier to understand. But first, you MUST concede that MOMENTUM IS CONSERVED DURING INELASTIC COLLISIONS. Seriously, dude, do 20 seconds of research online and you will find that any reliable source will confirm this. If you can’t get your head straight about that fact, nothing anybody says is going to help you and you’re going to remain ignorant. [The quick justification is that during either type of collision, the forces the objects exert on each other are the same magnitude but in opposite directions. These “equal and opposite” forces also must act on the objects for exactly the same amount of time. This leads to equal and opposite impulses on the system during the collision, resulting in no net change in momentum of the system.]
Okay. Case 1
Two balls of soft clay, each with mass 1 kg are moving towards each other, each with a speed of 10 m/s. They undergo a head-on perfectly inelastic collision (they stick together). Friction is negligible for the small time interval just before to just after the collision.
Momentum
Before collison: net momentum = 1(10) + 1(-10) = 0
Immediately after the collision, the net momentum must be 0, because MOMENTUM IS ALWAYS CONSERVED, EVEN IN INELASTIC COLLISONS!
net momentum = (1+1)v = 0, so the final velocity of the stuck-together clay balls is 0.
Individually, each clay ball had a change in momentum of 10 kg m/s. But because they changed in opposite directions, the net change is 0.
Kinetic Energy
Before the collision: kinetic energy = .5(1)10^2 +.5(1)(-10)^2 = 100 Joules
After the collision: since final v = 0, the final kinetic energy = 0.
Individually, each clay ball “lost” 50 Joules of kinetic energy.
Where did the 100 Joules of kinetic energy go? Most of it becomes thermal energy during the collision. Energy is conserved! It’s just not *kinetic *energy anymore.
Case 2: Now lets say that one of the clay balls was sitting still and the other was moving towards it with an initial speed of 20 m/s.
Okay. Perfectly inelastic collision, same as before:
Momentum
Before collision: net momentum = 1(0) + 1(20) = 20 kg m/s
After collision: net momentum = same as before collision = 20 = (1+1)v; therefore v = 10 m/s (this is the speed of the stuck-together clay balls immediately after the collision)
Individually, each clay ball had a change in momentum of 10 kg m/s. But because they changed in opposite directions, the net change is 0.
Kinetic Energy
Before collision: total kinetic energy = .5(1)0^2 + .5(1)20^2 = 200 Joules
After collision: total kinetic energy = .5(1+1)10^2 = 100 Joules
Individually, each clay ball “lost” 50 Joules of kinetic energy.
Where did this 100 J of kinetic energy go? Most of it becomes thermal energy during the collision. Energy is conserved! And so is momentum!
The change in momentum and the change in kinetic energy for each object is the same in either case.
Yes, the cars will eventually come to a stop after the “Case 2” collision. But that’s most likely to be an insignificant event compared to the collision itself, unless something unpredictable and unfortunate happens–like the combined objects going over a cliff, or falling into a river, or being hit by a meteorite, or being stepped on by Godzilla.
This is a bizarre thread…
Simple thought experiment: you sit on a motorcycle. First case, you are at rest, and a car (car 1) going 100mph slams into a stationary one (car 2). Second case, you move at 50mph in the same direction as car 1. From your point of view, this car now moves at 50mph, while car 2 also moves at 50mph towards you. Both cars crash into each other. The only thing that has changed is your state of motion. Does this affect the crash in any way? No, of course not. Both situations are the same.
In case that’s not enough, let’s look at the math for general collisions. First, we’ll treat the elastic case – which doesn’t apply in this situation, but will help us in getting a general handle on the problem. In this case, momentum and (kinetic) energy are both conserved, so let car 1 have a mass of m[sub]1[/sub] and an initial velocity of u[sub]1[/sub], while car 2 has a mass of m[sub]2[/sub], and a velocity of u[sub]2[/sub]. Conservation of momentum gives us:
m[sub]1[/sub]u[sub]1[/sub] + m[sub]2[/sub]u[sub]2[/sub] = m[sub]1[/sub]v[sub]1[/sub] + m[sub]2[/sub]v[sub]2[/sub],
where the v[sub]i[/sub] are the velocities of the cars after the collision. Similarly, conservation of energy yields:
m[sub]1[/sub]u[sub]1[/sub][sup]2[/sup]/2 + m[sub]2[/sub]u[sub]2[/sub][sup]2[/sup]/2 = m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup]/2 + m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup]/2.
Thus, we have two equations with two unknowns, v[sub]1[/sub] and v[sub]2[/sub]. So this can be solved to yield:
v[sub]1[/sub] = (m[sub]1[/sub]u[sub]1[/sub] - m[sub]2[/sub]u[sub]1[/sub] + 2m[sub]2[/sub]u[sub]2[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
v[sub]2[/sub] = (m[sub]2[/sub]u[sub]2[/sub] - m[sub]1[/sub]u[sub]2[/sub] + 2m[sub]1[/sub]u[sub]1[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
Now let’s treat the other extreme, the perfectly inelastic collision. In this case, kinetic energy is not conserved – a lot of it goes into the deformation of the cars, and into all the sound and fury associated with that. In the perfectly inelastic case, both cars will ‘stick together’ after the collision. Thus, conservation of momentum yields:
m[sub]1[/sub]u[sub]1[/sub] + m[sub]2[/sub]u[sub]2[/sub] = (m[sub]1[/sub] + m[sub]2[/sub])v,
because now, both cars form one object (i.e. a twisted mass of broken metal) of mass m[sub]1[/sub] + m[sub]2[/sub], going at some velocity v. This is elementary to solve:
v = (m[sub]1[/sub]u[sub]1[/sub] + m[sub]2[/sub]u[sub]2[/sub])/(m[sub]1[/sub] + m[sub]2[/sub]).
In reality, no collision is ever perfectly elastic or inelastic; so we’ll need a way to interpolate between both solutions, such that for perfect elasticity, we get the first one, and for perfect inelasticity, we get v[sub]1[/sub] = v[sub]2[/sub] = v. This is done by introducing the coefficient of restitution, c:
v[sub]1[/sub] = (cm[sub]2[/sub](u[sub]2[/sub] - u[sub]1[/sub]) + m[sub]2[/sub]u[sub]2[/sub] + m[sub]1[/sub]u[sub]1[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
v[sub]2[/sub] = (cm[sub]1[/sub](u[sub]1[/sub] - u[sub]2[/sub]) + m[sub]2[/sub]u[sub]2[/sub] + m[sub]1[/sub]u[sub]1[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
As you can see, for c = 0, this reduces to the inelastic case, and for c = 1, this covers the elastic one. Intermediate values describe realistic cases, which are partly elastic, and partly inelastic.
Now, how do we see that both collisions are equivalent? We look at the velocity change of the second car, |u[sub]2[/sub] - v[sub]2[/sub]|. If car 2 is stationary before the collision, then v[sub]2[/sub] = (cm[sub]1[/sub]u[sub]1[/sub] + m[sub]1[/sub]u[sub]1[/sub])/(m[sub]1[/sub] + m[sub]2[/sub]) = (c1000kg * 100mph + 1000kg * 100mph)/2000kg = c50mph + 50mph. For car 2 moving at 50mph, we have v[sub]2[/sub] = (c1000kg*(50mph - (-50mph)) - 1000kg50mph + 1000kg50mph)/2000kg = (c1000kg100mph)/2000kg = -c*50mph, where the minus signs come from the fact that the cars move in opposite directions.
In the first case, then, |u[sub]2[/sub] - v[sub]2[/sub]| = |0 - (c50mph + 50mph)| = c50mph + 50mph; this is easily seen to be the right answer, because in the elastic case, all the energy gets transferred to the second car, thus causing it to move at 100mph, while in the inelastic case, the combined object made from both cars moves (immediately after the impact) at 50mph. In the second case, |u[sub]2[/sub] - v[sub]2[/sub]| = |-50mph - c50mph| = c50mph + 50mph. Thus, both collisions are equivalent, regardless of how elastic or inelastic they are – as of course must be the case.
Good job folks, especially Robby.
FWIW I also teach physics and, yep, both the math and the concepts stand up properly.
Westom, I and others have posted numbers and equations repeatedly, but as the saying goes, there are “none so blind as those that will not see.”
First, you have posted what everyone had to do to have a correct answer. Others assumed your same conclusion. But posted no numbers and other reasons why. So they were insulting an educated reader. In the real world, a correct answer also provides reasons why with numbers - as you have done and robby eventually did. A “scammer or politician” (a term used repeatedly to say this) could have posted the same answer - and is still 100% wrong. Because he did what makes him a “scammer or corrupt politician”. He did not post the numbers, in equations, with underlying facts. That defines scammers and politicians. They tell everyone what to know rather than what is necessary to be informed.
Second, after collision, the total kinetic energy is 100 joules. So where are those 100 joules? The cars are now rolling in a jumbled heap or ‘out of control’ manner; maybe even killing pedestrians. A crash is not over until the second 100 joules dissipates. Parked cars are not clay balls. A crash is not done until everything stops moving.
In the ball world, car 2 keeps moving because of that 100 joules. In the real world, the car is either a rolling ball of scrap metal, is tearing tires down the road, is plowing through a nearby store, rolling with parts still flying off in many directions, etc. I even suggested car one might be snow plowing through car 2. And still that is only a second collision? Of course not. A car crash is over when everything stops moving. A crash is never done until all kinetic energy (the second 100 joules) dissipates. If car 2 is moving (with brakes applied), then that is not “detrimental”? Of course it is “detrimental”. Detriment continues until a car finally rolls (as in overturned) to a stop.
Third, the OP did not say balls or sliding blocks. And he used a vague term “detrimental”. If a stopped car 2 is moving due to an inelastic collision, then it is still crashing. Because a car (ie brakes applied) must still dissipate the second 100 joules. The second 100 joules may be an “insignificant event” or it may be the worst part of a crash. As repeatedly noted in earlier posts, too many hypotheticals.
Need for a definition of ‘detrimental’ was necessary.
Many assumed the entire collision was an instantaneous event when two cars (or clay balls) were in contact. Something ‘detrimental’ continues until everything stops moving - until all kinetic energies are zero. Until smashed cars stop moving as a threat to everything and one around them.
Thank you for doing what most before you refused to do and what robby only eventually did with prodding.
Momentum was conserved. But that was irrelevant. A car crash is not two clay balls. A parked car or one waiting at a stop light is not “detrimental” when moving or rolling at 50 km/hr due to a crash? Of course it is. If the intent was to demonstrate conservation of momentum, the original question was vague and misleading.
[Moderator Warning]
westom, insults, even oblique ones, are not permitted in General Questions. This is an official warning. Don’t do this again.
Colibri
General Questions Moderator
Depending on the coefficient of friction due to the surface of the road, the deformation of their wheels, etc… it’s being burned off as waste heat, sonic energy and probably some light if metal is sparking against the pavement at a certain rate. And yes, the collision is over once the object collide and either stick together or do not stick together. After that we are no longer dealing with the mathematics of collision, but the mathematics of a new object, Car1+Car2, and it is no longer a collision problem.
P[sub]net start[/sub] = P[sub]net finish[/sub]. Momentum would not be conserved if P[sub]net start[/sub] = P[sub]net finish[/sub] + or - X kg m/s. Adding an external force like friction no longer has it being a closed system. Momentum is always conserved in a closed system, it’s one of the fundamental laws of reality. When adding friction over time it creates impulse. Impulse is, in turn, a vector quantity. As friction always works against the direction of motion, it will provide a vector which reduces and eventually cancels out P. P + -P = 0. Once we include μ[sub]k[/sub] in our system, P[sub]net start[/sub] =P [sub]net finish[/sub] due to vector addition, and momentum is still conserved.
The OP also did not give us full information in the beginning. As P is conserved, the after effects of the 50 mph crash will likewise depend on the surroundings if the cars are still moving after the collision is resolved, just as the 100 mph crash will. If you don’t believe me, picture the results of a Mack truck traveling at 50 m/s having a head on collision with a carbon fiber bike traveling at -50 m/s.
Is it fair to say that if we consider the Earth to be part of the system, that friction is transferring momentum from the cars to the Earth - but that due to the huge mass difference between the Earth and the cars, the additional velocity of the Earth after the relative motion has disappeared is essentially undetectable?
I haven’t the energy to read through all the posts, but look at it this way:
Case one: we watch the cars hit each other each doing 50 km/h. Our perspective is such that we look at the cars from the side.
Case two: let’s do the EXACT SAME THING except that we are moving the background at 50 km/h towards the left. This makes it appear like one car is standing still and the other is going twice as fast.
The cars will still hit and crumple the same way regardless how we move the background, since their speed relative to each other is the same.
That’s about as simple as I can make it.
Let me see if I’m getting the arguments straight.
It seems that everyone is in agreement that the collisions themselves between two cars going at 50 k/h on the one hand and one car going 100 k/h and another stopped on the other hand are essentially identical. Except for the very slight differences arising from relativity, there is a pair* of inertial reference frames where the one event is the same as the other in the sense that you could superimpose them on each other completely.
But it seems that one reference frame has the earth stopped and the other has the earth moving at 50 k/h. And the cars are ultimately going to come to rest in both reference frames. So, in one of the collisions, everything has to slow down another 50 k/h. And the question is whether or not that matters. There’s one side that says that that is all post-collision stuff and that it is consequently mostly irrelevant. Then there’s westom who thinks that that is part of the collision (which is reasonable)** but that somehow the law of conservation of momentum gets violated in the process, even though this has never been shown to ever happen (which is not reasonable).
Westom goes on and on about the need to define detrimental, but it seems that it would be helpful to define collision to avoid this.
** but what gets lost here is that the collisions themselves are exactly the same in the sense that the collisions are not the same if two cars are going 20 k/h and to other cars are going 200 k/h. The key really seems to be the post-collision effects.
Also some random comments:
There is no physics section on the SAT.
While charlatans and corrupt politicians often hide behind flowery rhetoric, an intuitive understanding of various concepts and how they relate to each other is invaluable. The problem is that in physics as well as in life, one’s first intuition may not be the proper one, and the calculations help home you in on the right one. When things can’t be quantified, this gets a lot harder, but it remains all the less still advantageous.
Hi, Opal.
It’s not always possible to define a proper form for the answers to various questions, especially if you don’t know where the questions will lead. I am reminded of a phrase that I can’t remember that encapsulates this apparently by Asimov: “The most exciting phrase to hear in science, the one that heralds new discoveries, is not ‘Eureka!’ but rather, ‘Hmmmm, that’s funny.’”
Not exactly.
The important thing to realize is that momentum is a vector quantity, not a scalar quantity. That is, depending on directionality, vectors can be negative. For example, speed is scalar. You cannot be going slower than 0 miles an hour. Velocity is a vector, if our coordinate system privileges right as the positive direction, then moving to the left at 30 m/s is -30 m/s. So when we calculate ΣP, we have to take the initial momentum of an object into account as well as the force of friction. Since the force of friction is a force (measured in Newtons) and it’s applied for a certain amount of time, it produces impulse. Impulse is equal to FT but also ΔP. ΔP is still, in essence, a form of P, it just denotes which direction the vector’s going.
So imagine we have a toy car with a mass of 1kg, moving at 5 m/s. Its P is 5 kg m/s. Acting against it is the force of friction. Assume 1N of force acting for 5 seconds. ΔP is -5 kg m/s.
ΣP[sub]start[/sub]: 5 kg m/s + -5 kg m/s = 0 kg m/s
ΣP[sub]finish[/sub] (the toy car comes to a stop): (1kg) (0 m/s) = 0 kg m/s
0 kg m/s = 0 kg m/s
Momentum is conserved.
Yes, this should be the correct summary.
Westom, I am a third year physics student. Although I doubt anything I can say or do will change your mind, there is still the possibility that something my professors can say will change your mind. Do you want me to ask them this question?
Actually, I agreed mostly with what was provided. Even suggested it in a first post. Conservation of momentum does exist. But momentum alone was insufficient. Where kinetic energy went was also relevant and ignored.
The original question was poor if its purpose was to explain physics that also explains clay ball collisions. Tangent and robby finally provided answers with numbers that were incomplete. A parked car suddenly at 50 km/hr sideways or into an intersection is not part of the crash? That energy also must be accounted for.
So many answers were devoid of numbers for both momentum and energy equations. An answer that only said (subjectively) that collisions were equivalent was misleading and incomplete. In this case, a parked or waiting car suddenly doing 50 km/hr is irrelevant to the crash? Energy numbers and the resulting velocity suggested otherwise.
Momentum says the 100 km/hr crash or two 50 km/hr vehicles crashing are similar at the initial impact. But undissipated energy after a 100 km/hr crash implies that higher energy crash has further consequences. That undissipated kinetic energy was being ignored. And explains why the two cases are not same.
Appreciate the problem with most answers. The always required numbers were missing until robby and Tangent provided some of them.
No, you still do not understand the basic math or concepts behind the problem.
I’ll use simpler numbers to help you out here.
1)ΔE[sub]k[/sub] = -50J
2)ΔE[sub]k[/sub] = -50J
The only difference is that collision 2 will still have a velocity vector after its initial impact. However that is no longer part of the collision, but part of the mechanics for the new object, Car[sub]1[/sub] + Car[sub]2[/sub]. Car[sub]1+2[/sub] may grind against the ground until it gently comes to rest, may smack into another car, may go off a cliff. But that is not part of the collision problem.