I’ve just taught a year of physics (using Serway’s Physics for Scientists and Engineers, 5th ed., and Bigtrout is correct.
What I can add to the discussion is an excellent conceptual explanation supplied in the book used by our “low-track” physics course: Hewett’s Conceptual Physics, 8th ed. In Hewett’s “Next Time Questions” ancillary material, he poses the question:
"Which would be more damaging; driving into a massive concrete wall, or driving at the same speed into a head-on collision with an identical car traveling toward you at the same speed?
Answer: Both cases are equivalent, because either way, your car rapidly decelerates to a dead stop. The dead stop is easy to see when hitting the wall, and a little thought will show the same is true when hitting the car. If the oncoming car were traveling slower, with less momentum, you’d keep going after collision with more “give,” and less damage (to you!). But if the oncoming car had more momentum than you, it would keep going and you’d snap into a sudden reverse with greater damage [again, to you]. Identical cars at equal speeds means equal momenta–zero before, zero after collision."
Okay, this is my first post ever and I am about to immortalize myself as the member who knows the least about physics. However, here is my take, if anyone cares to see me give the horse another whack:
I am surprised that the best “real-world” demonstration that you all have come up with is the hand-clapping. Does anyone play pool? I do, and pretty well. Pool is one of the best ways to learn a little about physics. Take the cue ball and roll it into a wall (let us presume it is the magic immovable wall). Watch what the ball does after impact. Now take the cue ball and hit it into a stationary ball (the object ball in poolspeak). Observe how both balls react to the impact. Now take the cue ball in one hand, the object ball in the other, and observe again the reaction of the balls after the collision.
I know we don’t know the speed of the cue ball, but at least let’s assume that in demo three that both balls are travelling at the same speed. I have no mathmatical revelation to make from this. In fact, I studied Literature, so I am way out of my league. But it seems to me that there is only x amount of energy involved in each impact, and that the impact itself cannot generate energy, but yet the initial energy from all objects involved has to be accounted for. What was that I learned in high school about how energy never disappears, just transfers? In example two, the cue ball not only moves the object ball, but itself continues to move if a normal stroke is used. Only a type of spin put on the ball can cause it to stop on impact and remain stationary. In this circumstance, the object ball moves away from the collision at a higher speed than if the cue ball were allowed to continue on its original path.
Okay, I have exhausted my physics capabilities. Someone who knows what they are talking about should pick this up from here. Now I wonder how many of you are going to seek out a pool hall?
Pool balls closely simulate an elastic collision, in which kinetic energy is conserved. Cars colliding head on are much closer to an inelastic collision (especially if they are stuck together after the collision). In an inelastic collison, kinetic energy is NOT conserved.
For what it’s worth, I don’t think hand-clapping is the best analogy either. A better analogy is air-hocky pucks wrapped in a band of velcro such that the pucks stick together on collision. If two pucks, initially traveling at the same speed in the opposite direction, collide and stick together, they come to a dead stop. Now put velcro on a wall. If a puck collides with a wall and sticks to the wall, it (obviously) also comes to a dead stop. In each case, looking at one of the pucks, it undergoes the same change in momentum, and thus, the same “damage.”
Introduction to the Physics Principles of Collisions
The physics of collisions are governed by the laws of momentum.
One of these important laws can be expressed this way:
F x t = m x ∆v
The equation is known as the impulse-momentum change equation.
In a collision, an object experiences a force for a specific amount of time that results in a change in momentum. The result of the force acting for the given amount of time is that the object’s mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. In equation form, F x t = m x Δ v.
In a collision, objects experience an impulse; the impulse causes and is equal to the change in momentum. Consider a football halfback running down the football field and encountering a collision with a defensive back. The collision would change the halfback’s speed and thus his momentum.
If there are only two objects involved in the collision, then the momentum change of the individual objects are equal in magnitude and opposite in direction.
In a collision, the impulse experienced by an object is always equal to the momentum change.
One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value.
Consider a collision between two objects - object 1 and object 2. For such a collision, the forces acting between the two objects are equal in magnitude and opposite in direction (Newton’s third law). This statement can be expressed in equation form as follows.
F1 = - F2 (the forces are equal in magnitude and opposite in direction)
The forces act between the two objects for a given amount of time. In some cases, the time is long; in other cases the time is short. Regardless of how long the time is, it can be said that the time that the force acts upon object 1 is equal to the time that the force acts upon object 2. This is merely logical. Forces result from interactions (or contact) between two objects. If object 1 contacts object 2 for 0.050 seconds, then object 2 must be contacting object 1 for the same amount of time (0.050 seconds). As an equation, this can be stated as:
t1 = t2
Since the forces between the two objects are equal in magnitude and opposite in direction, and since the times for which these forces act are equal in magnitude, it follows that the impulses experienced by the two objects are also equal in magnitude and opposite in direction. As an equation, this can be stated as
F1 x t1 =-F2 x t2 (the impulses are equal in magnitude and opposite in direction)
But the impulse experienced by an object is equal to the change in momentum of that object (the impulse-momentum change theorem). Thus, since each object experiences equal and opposite impulses, it follows logically that they must also experience equal and opposite momentum changes. As an equation, this can be stated as
m1 Δv1 = - m2 x Δv2 (the momentum changes are equal in magnitude and opposite in direction)
The above equation is one statement of the law of momentum conservation.
In a collision, the momentum change of object 1 is equal to and opposite of the momentum change of object 2. That is, the momentum lost by object 1 is equal to the momentum gained by object 2. In most collisions between two objects, one object slows down and loses momentum while the other object speeds up and gains momentum. If object 1 loses 75 units of momentum, then object 2 gains 75 units of momentum. Yet, the total momentum of the two objects (object 1 plus object 2) is the same before the collision as it is after the collision. The total momentum of the system (the collection of two objects) is conserved.
A useful analogy for understanding momentum conservation involves a money transaction between two people. Let’s refer to the two people as Jack and Jill. Suppose that we were to check the pockets of Jack and Jill before and after the money transaction in order to determine the amount of money that each possesses. Prior to the transaction, Jack possesses $100 and Jill possesses $100. The total amount of money of the two people before the transaction is $200. During the transaction, Jack pays Jill $50 for the given item being bought. There is a transfer of $50 from Jack’s pocket to Jill’s pocket. Jack has lost $50 and Jill has gained $50. The money lost by Jack is equal to the money gained by Jill. After the transaction, Jack now has $50 in his pocket and Jill has $150 in her pocket. Yet, the total amount of money of the two people after the transaction is $200. The total amount of money (Jack’s money plus Jill’s money) before the transaction is equal to the total amount of money after the transaction. It could be said that the total amount of money of the system (the collection of two people) is conserved. It is the same before as it is after the transaction.
For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision. Momentum is conserved for any interaction between two objects occurring in an isolated system. This conservation of momentum can be observed by a total system momentum analysis or by a momentum change analysis.
Scenario Presented by Questioner - Head-on Collision Between Two Vehicles of Identical Masses and Velocities
My understanding of the original question is essentially that if a car traveling at for example, 50 mph collides with an immovable concrete wall, the car will suffer a certain amount of damage. If, however, the same car were to collide with another identical car moving in the opposite direction also at 50 mph, then the resulting head-on collision between the two vehicles would be equivalent to a single vehicle traveling at 50 mph colliding with an immovable cement wall creating what would be equivalent to a collision at 100 mph.
In terms of the physics involved, collisions between objects such as cars are governed by the laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision as here, then the momentum change of the individual objects are equal in magnitude and opposite in direction.
Certain collisions are referred to as elastic collisions. Elastic collisions are collisions in which both momentum and kinetic energy are conserved. The total system kinetic energy before the collision equals the total system kinetic energy after the collision. If total kinetic energy is not conserved, then the collision is referred to as an inelastic collision. The situation described here – a head-on collision between cars, is an inelastic collision.
A force acting for a given amount of time will change an object’s momentum. Put another way, an unbalanced force always accelerates an object - either speeding it up or slowing it down. If the force acts opposite the object’s motion, it slows the object down. If a force acts in the same direction as the object’s motion, then the force speeds the object up. Either way, a force will change the velocity of an object. And if the velocity of the object is changed, then the momentum of the object is changed.
These concepts are merely an outgrowth of Newton’s second law. Newton’s second law (Fnet = m x a) stated that the acceleration of an object is directly proportional to the net force acting upon the object and inversely proportional to the mass of the object. When combined with the definition of acceleration (a = change in velocity / time), the following equalities result.
F = m x a
or
F = m x ∆v / t
If both sides of the above equation are multiplied by the quantity t, a new equation results.
F x t = m x ∆v
This equation represents one of two primary principles to be used in the analysis of collisions in physics.
To truly understand the equation, it is important to understand its meaning in words. In words, it could be said that the force times the time equals the mass times the change in velocity. In physics, the quantity Force x time is known as impulse. And since the quantity m•v is the momentum, the quantity mxΔv must be the change in momentum. The equation really says that the:
Impulse = Change in momentum
Let’s put some numbers to the described scenario to better understand the physics of a head-on inelastic collision.
Example 1: Inelastic Collision Between Two Identical 1000 kg Cars
Mass of car 1: 1000 kg
Velocity of car: 20 m/s
Momentum of car: 20000 kg*m/s
Mass of car 2: 1000 kg
Velocity of truck: -20 m/s
Momentum of truck: -20000kg*m/s
Note: the velocity and momentum of car 2 are negative because both variables are vector quantities.
In the above example of a collision between two identical car,s the total system momentum is conserved. Before the collision, the momentum of car 1 is +20000 kgm/s and the momentum of car 2 is -20000 kgm/s; the total system momentum is 0 kgm/s. After the collision, the momentum of car 1 is -10000 kgm/s and the momentum of car 2 is +10000 kgm/s; the total system momentum is 0 kgm/s. The total system momentum is conserved. The momentum change of car 1 (-10000 kgm/s) is equal in magnitude and opposite in direction to the momentum change of car 2 (+10000 kgm/s).
An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 400000 Joules (200000 J for car 1 plus 200000 J for car 2). After the collision, the total system kinetic energy is 100000 Joules (50000 J for car 1 and 50000 J for car 2). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.
In this scenario, what each car independently experiences in the two car collision is exactly the same as what each car would experience in a head-on with an immovable object such as a cement wall or a mountainside.
So, if you now think that two cars possessing similar masses and velocities colliding with one another is equivalent to a single car colliding with fixed wall or the side of a mountain at a comparable velocity, then you are in error. In the original scenario, there are two wrecked cars, not one wrecked car, and the kinetic energy utilized to wreck each of those cars in a collision requires twice the energy required to wreck only one car. (Kinetic energy = 1/2 mass X velocity^2).
Its true that two identical cars in a head-on collision traveling at identical velocities would each experience an equivalent amount of damage to each colliding with a cement wall at the same velocity. This might lull a person into a false sense of security concerning head-on collisions. Let’s compare a collision between a truck and a car in the same scenario to dispel that feeling.
Example 2: Inelastic Collision Between a 1000 kg Car and a 3000 kg Truck
The before- and after-collision velocities and momentum are listed below
Mass of car: 1000 kg
Velocity of car: 20 m/s
Momentum of car: 20000 kg*m/s
Mass of truck: 3000 kg
Velocity of truck: -20 m/s
Momentum of truck: -60000kg*m/s
Note: the velocity and momentum of the truck are negative because both variables are vector quantities.
In the above example of a collision between a truck and a car, the total system momentum is conserved. Before the collision, the momentum of the car is +20000 kgm/s and the momentum of the truck is -60000 kgm/s; the total system momentum is -40000 kgm/s. After the collision, the momentum of the car is -10000 kgm/s and the momentum of the truck is -30 000 kgm/s; the total system momentum is -40000 kgm/s. The total system momentum is conserved. The momentum change of the car (-30000 kgm/s) is equal in magnitude and opposite in direction to the momentum change of the truck (+30000 kgm/s).
An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 800000 Joules (200000 J for the car plus 600000 J for the truck). After the collision, the total system kinetic energy is 200000 Joules (50000 J for the car and 150000 J for the truck). The total kinetic energy before the collision is not equal to the total kinetic energy after the collision. A large portion of the kinetic energy is converted to other forms of energy such as sound energy and thermal energy. A collision in which total system kinetic energy is not conserved is known as an inelastic collision.
The original question examined whether two identical cars possessing equivalent masses colliding head-on at identical velocities is equivalent to each car individually experiencing a collision in which their individual velocities are combined, thus doubling the velocity at which the collision occurs for each car. This does not occur. Rather, each car experiences the collision at the velocity at which the car is traveling at that instant of impact. The velocities of the cars do not combine. Each car would receive an equivalent amount of damage.
The outcome would be very different if a smaller compact car of less mass collides head-on with a fully loaded semi-truck possessing
a much greater mass traveling at the same velocity as the car but in the opposite direction. The two vehicles would share the kinetic energy involved in the collision evenly but due to their very different masses, the compact car would experience much greater damage, analogous to an insect with relatively low mass colliding with the windshield of a moving semi-truck of much greater mass.
Likewise, according to the equation for impulse (impulse = force X time), the greater the time over which a collision occurs, the correspondingly less peak force there will be upon impact. An example would be running into a haystack versus a fixed brick wall. The collision with the haystack occurs over a longer period of time, correspondingly reducing the force of the impact. In comparison, a collision with a fixed brick wall occurs over a brief time period with a correspondingly larger peak force resulting in s substantial increase in damage.
Any other interpretation is clever, but misleading, wordsmithing.
Relevant formulas for inelastic collisons
Before Collison
Momentum = mass car x velocity car
Kinetic energy = 1/2 mass car x velocity car ^2
Note: This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The kinetic energy is dependent upon the square of the speed.
After Collision
momentum = (mass car 1 + mass car 2) x velocity car 2
Kinetic energy = 1/2 (mass car 1 + mass car 2) x velocity^2 of car 2
Ratio of kinetic energies before and after collision:
Kinetic energy final / Kinetic energy initial = mass car 1 / mass car 1 + mass car 2
Fraction of Kinetic Energy Lost in Collision
Kinetic energy initial - kinetic energy final / kinetic energy initial = mass car 2 / mass car 1 + mass car 2
From conservation of momentum:
mass car 1 x velocity car 1 = (mass car 1 + mass car 2) x velocity car 2 yields velocity car 2 = mass car 1 / mass car 1 + mass car 2 x velocity car 1
Experiment
This link is to an animation that portrays the inelastic collision between two 1000-kg cars. The before- and after-collision velocities and momentum are shown in the animation’s data tables. Try it!
Right, IIRC what they actually found was that the force turned out to go down ever so slightly. I believe they calcuated it as 0.99… repeating.
I remember they did mention this unexpected finding could not be attributed to any limitations on the accuracy of their measurement devices or anything like that.
If memory serves, they weren’t able to offer any reasonable explanation for the deviation from theory.
Heh, I read the whole thing, and it seems to me, as is so often the case, that the problem arose out of the difficulty of communicating in seemingly “plain English” simple concepts of Physics. If two people discussing a scenario do not actually agree about what they are saying, it’s usually impossible to reach an understanding on the truth of the matter. And that proliferated through the thread, compounded by those who chimed in with an imprecise understanding of Physics.
Of course the force is doubled … of course there’s twice as much work performed … but in the second case, this work is divided between the two cars …
The catch in the OP is we are only looking at one car … and just looking at the one car, we cannot tell whether it hit a brick wall or on-coming traffic … the amount of work performed on the one car is the same … assuming we’re not using AMC Pacers …
