Here's a question which pretty much everyone gets wrong. But the readers of this blog aren't just a random sample, so I bet most of you will get it right:
Two identical vehicles A and B, both traveling at speed V directly toward the other vehicle, collide exactly head-on. At another test track, car C collides with an indestructible concrete barrier at speed V'. All other things being equal, the crash-test dummy occupants of vehicles A, B, and C will experience the same forces (and therefore injuries) if...
1. V' = V/2
2. V' = V
3. V' = 2V
4. Something else (explain)How might the answer change if the A and B do not have equal masses?
I'll let my astute readers answer, but I'll give a place to start for those who want to make sure they're thinking from a physics perspective. Momentum is mass times velocity. Force is the time rate of change of momentum. Go from there. Also, in this case I have to suggest you not try to find the answer experimentally!
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You have not given us all the information. Is this happening on a flat surface? Is there an atmosphere? Are any of the test dummies insured? Are the test dummies allowed to veer off at the last second?
2. V' = V
I'm no physicist, but it looks to me like there is no answer for the unequal masses case, as the dummies in the heavier car will experience lesser forces than those in the lighter car (so there's no way to make them both equal to the forces experienced by the dummies in car C, since they won't be equal to one another). My non-physicist brain also gives the answer "2" on the first question.
V' = V is normally the correct answer (in which case total energy release in the collision is twice as great, but is spread between two vehicles), though in the real world it isn't precisely true, since you'll get different collision mechanics.
Smash a car into a wall, smash a car into a mirror. (2), V'=V. In the unequal mass case the less massive car comes to a complete stop then accelerates in the 180-degree opposite direction. It gets creamed while the more massive car is cushioned for continuing on its trajectory, albeit slowed.
http://en.wikipedia.org/wiki/MythBusters_(2009_season)
Episode 115
What the posters who are choosing V' = V are overlooking is that the time over which the collision occurs matters. The two cars are closing with relative velocity 2V, so there is only half as much time before a hood ornament ends up in the driver's lap as for a car hitting a stationary barrier at V. To get the same force (rather than the same change in momentum), you would need (3) V' = 2V.
@Eric: no, the time is the same. In both cases there is a velocity change of V which takes place over a distance d determined by the structural characteristics of the cars. If the deceleration is constant the time is 2d/V - in practice it'll be something a bit different, but there is no difference in the two cases.
The relative velocity is only relevant if one of the vehicles is so massive that it suffers negligible deceleration - then the other one suffers in the way you describe.
What the posters who are choosing V' = V are overlooking is that the time over which the collision occurs matters. The two cars are closing with relative velocity 2V, so there is only half as much time before a hood ornament ends up in the driver's lap as for a car hitting a stationary barrier at V.
Incorrect -- bear in mind that both vehicles crumple, whereas the wall is specified as indestructible. If the wall were as destructible as the car, yes, the collision with the wall would be less harmful.
For the equal mass case, the answer is V' = V. If the 2 cars have unequal mass, then the answer is V' = V_cm, where V_cm is the velocity of the initial car wrt the center of mass of the two car system.
yes, but where will the hood ornament be when the driver's lap coincides with it? what is the cars' velocity relative to that point in each scenario?
3. It's only the relative velocity that matters. For the situation where there are 2 different cars of different masses colliding, the dummies will never all have the same average forces since the big car will always accelerate more slowly.
If both cars are of equal mass then the will both come to a complete stop as if they had hit an imovable object. Thus V' =V.
Besides the force question, what I want to know is, given that the kinetic energy involved is greater, how does that change the dynamics/results of the collision?
The KE is not any different. The change in KE of car A (or B) is the same as car C. It also occurs in the same amount of time so the power involved in the deformation of the vehicle is the same as well.
I've always wondered why the IIHS uses a somewhat crushable impact structure when doing their head-on offset tests, but I figure it is to simulate the crash dynamics that result when two vehicles do not hit squarely head on. As the real world tests by Mythbusters showed, repeatedly, it is very hard to get the absolutely perfect symmetry needed for to make the real world match up with this toy problem.
@14
Oh right, I was still thinking about the discussions I'm used to having about such collision puzzlers, where the velocities are usually argued as different (durr). I plead "lack of sleep".
Right answer is V'=sqrt(2)*V, i think.
Lets do the math:
When two bodies collide, the devastating effect in collision depends only on their relative velocity V1-V2.
Kinetic energy, which has the destructive effect is equal to
1/2*M1*M2/(M1+M2)*(V1-V2)^2
The rest of the kinetic energy is associated with the movement of the center of mass of the system.
This energy in the collision does not change, and has no effect of destruction .
In given case, if faced two identical cars moving toward each other with one and the same velocity V then the energy on which the destruction depends is
1/2*M*M/(M+M)*(2*V)^2 = M*V^2
So, all the kinetic energy is spent on destruction.
Now, consider the case of the car collides with an indestructible concrete barrier at speed V'.
To get the kinetic energy spent on destruction lets follow again the formula
1/2*M1*M2/(M1+M2)*(V1-V2)^2
Setting M1=M,V1=V'and M2=>infinity and V2=0 we obtain
1/2*M*V'^2
Of course, this result is obvious without any calculations.
Finally, equalizing destructing kinetic energies in both cases we have
1/2*M*V'^2 = M*V^2 =>
=> 1/2*V'^2 = V^2
=> V'=sqrt(2)*V
Nice calculation, Martin, but you forgot that the energy in the first case is spread between two vehicles, while in the latter case it is all in the one vehicle.
Plus noise, but that comes from the destruction of the vehicle.
CCPhysicist said:
"...you forgot that the energy in the first case is spread between two vehicles, while in the latter case it is all in the one vehicle. Plus noise, but that comes from the destruction of the vehicle."
Hmm... You may be right but i do not see any reason why all released at the collision energy should move in a specific direction,namely into the vehicle.
Energy is a scalar physical quantity not a vector.
Although the concrete barrier has been assumed to be indestructible this does not mean that it cannot absorbe energy.
It certainly can vibrate and is connected to the Earth where a part of the released energy can finally arrive and dissipate to heat.
I still think that the energy released at the collision between the car and the barrier is distributed equally between them.
But of course, i may be totally wrong :)
All things being equal then V' = V/2