Car crash: Can it be an elastic collision?

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SUMMARY

The discussion clarifies that car crashes are inherently inelastic collisions, as they involve the conversion of kinetic energy into other forms, such as heat, due to deformation. The participants confirm that while textbook problems may present scenarios with elastic collisions, real-life car accidents do not exhibit this behavior. The conversation also emphasizes the importance of momentum conservation in collision calculations, particularly in scenarios involving different masses and velocities, as illustrated by the example of two cars colliding with specified masses and speeds.

PREREQUISITES
  • Understanding of elastic and inelastic collisions
  • Knowledge of momentum conservation principles
  • Familiarity with kinetic energy concepts
  • Basic algebra for solving collision problems
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  • Study the differences between elastic and inelastic collisions in physics
  • Explore real-world applications of crash dynamics and energy absorption
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Physics students, automotive engineers, safety analysts, and anyone interested in understanding the dynamics of car collisions and their implications on vehicle design and safety.

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If two cars are going towards each other, collide, and then are separated (have separate masses and separate velocities)... that is an elastic collision, correct?
 
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yup i think so, inelastic is when the objects hit and then stop dead
 
An elastic collision requires conservation of kinetic energy and momentum. If there is so much as a single dent caused by the collision some kinetic energy will have changed in for example heat. In reality car crashes are always inelastic.
 
There are two extremes.
Perfectly inelastic when the objects stop dead and there is a total conversion of K.E.
Perfectly elastic when there is zero loss of K.E.These collisions can happen with atomic scale objects such as gas molecules.
The collision referred to lies between the extremes but is more inelastic.
 
That's why I am confused... let me paraphrase the problem (I don't remember it verbatim- these numbers won't work but you'll get the idea):

Car 1 is 1.45kg and is going 4 m/s in the positive x direction. Car 2 is 0.65kg and is going -0.9 m/s. They collide, and after car 2 is going 2 m/s... find the new velocity of car 1.

There's where I got confused. I thought car crashes were inelastic, and momentum would be conserved by:
MiVi (car 1) + MiVi (car 2) = (Mcar1+car2)(Vf)

Since it gave a new velocity for car 2, I assumed it meant that they separated:
MiVi (car 1) + MiVi (car 2) = MfVf (car 1) + MfVf (car 2)

I plugged in the numbers and found car 1's new Vf.

Does that sound right to you?
 
correct
 
Your intuition is correct, you won't see two cars bounce off each other during a head on collision in real life. However with these kind of textbook problems it's important to just look at the data given regardless of how realistic it is. Your answer is correct.
 
Yeah, impossible to have elastic collision. Elastic collision = bad because KE is conserved and we would be bouncing back pretty fast!

That's why car manufacturers have crush zones to help reduce KE into other forms of energy including heat.
 
Cyosis said:
Your intuition is correct, you won't see two cars bounce off each other during a head on collision in real life. However with these kind of textbook problems it's important to just look at the data given regardless of how realistic it is. Your answer is correct.


Thank you.
 

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