Car crash: Can it be an elastic collision?

AI Thread Summary
In a car crash scenario, the collision is classified as inelastic due to the loss of kinetic energy, evidenced by damage like dents. Elastic collisions require conservation of both kinetic energy and momentum, which is not the case in real-life car accidents. The discussion highlights a specific problem involving two cars with different masses and velocities, where the conservation of momentum is applied to find the new velocity of one car post-collision. Despite the theoretical framework, real-world collisions do not exhibit elastic behavior, as energy is converted into other forms like heat. The conversation emphasizes the importance of understanding textbook physics while acknowledging the practical implications of car safety design.
kikidoll
Messages
9
Reaction score
0
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?
 
Physics news on Phys.org
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.
 

Similar threads

Replies
41
Views
2K
Replies
6
Views
2K
Replies
3
Views
713
Replies
2
Views
2K
Replies
13
Views
1K
Replies
9
Views
4K
Replies
5
Views
2K
Replies
18
Views
2K
Replies
16
Views
3K
Back
Top