Elastic or Inelastic Collision?

Question

Your car rolls slowly in a parking lot and bangs into the metal base of a light pole. In terms of safety, is it better for your collision with the light pole to be elastic or inelastic? Explain.

Here is what I have so far, please advise:

On one hand it is better to have an elastic collision, because then the body receives less of a jerk. If the collision is completely inelastic, then all the kinetic energy of the moving car converts into shock energy all at once which gives the body of the driver a more severe jerk.

IYet it could be better if it the collision is inelastic because then the light pole gives your car only enough impulse to bring it to rest. If the collision is elastic, the impulse given to the car is about twice as much. This additional impulse-which acts over a short period of time-could cause more injury.

Answer 1

For the sake of safety to the people inside, an inelastic collision is better. The best case is all of the energy is absorbed by deforming the lightpost or the car without collapsing the interior cabin space and the car decelerates to a complete stop (along with the person inside). This is why a car is designed with front end crumple zones.

If the collision is elastic, that would be bad news. Not only does the car and the body inside decelerate to a stop, but then they have to be accelerated in the opposite directions, prolonging the amount of time the occupant is subjected to stresses.

Answer 2

Inelastic. I'm 200% sure of this.

If I'm ever trying to figure something out intuitively, I always think of the extremes. So let's think of that in this case.

First extreme: Elastic collision.

You've found some crazy new alloy that is SUPER SUPER strong. It takes massive forces to bend it. If you were in a car made of this, when you struck the light pole the front of the car wouldn't bend barely at all. What do you think is going to happen to you? You're going to fly into the dashboard, no question. Just because the car looks fine doesn't mean you do! The car would bounce off the lightpole but you would keep going forward, with some nasty whiplash caused by the seatbelt.

Now for the other extreme: Inelastic.

The front of your car is made of jello. When you hit the light pole, the jello is going to deform like crazy. You'll still shoot forwards a bunch, but not nearly as much. Why? A huge amount of energy has gone into deforming the jello instead of deforming your face.

Here's a shorter, nicer answer. Think of it like this: this crash, no matter what your car is made of, is going to produce a certain amount of energy. Do you want that energy to go into destroying the car, or do you want it to go into destroying you? It has to go somewhere.

A saturn dealership in my city proudly displays a car who's front end is wrapped right around a flagpole. Sure the car is in bad shape, but since the crash was so inelastic the driver and passengers were totally fine.

Hope this helps.

Answer 3

this is probably not any help in any respect. the thank you to define, defining top as advantageous (+), given the values: m1 = 3 kg u1 = 5 m/s v1 = m2 = 5 kg u2 = -3 m/s (a) Inelastic - kinetic capability isn't conserved v * [m1 + m2] = m1u1 + m2u2 v * [ (3 kg) + (5 kg) ] = [ (3 kg) * (5 m/s) ] + [ (5 kg) * (-3 m/s) ] v * [ 8 kg ] = [ 15 kg-m/s ] + [ -15 kg-m/s ] v * [ 8 kg ] = [ 0.0 kg-m/s ] v = 0.0 kg-m/s (b) Elastic, kinetic capability is conserved [0.5*m*u^2] + [0.5*m*u^2] = [0.5*m*v^2] + [0.5*m*v^2] [0.5 * (3 kg) * (5 m/s)^2] + [0.5 * (5 kg) * (-3 m/s)^2 ] = [0.5 * (3 kg) * v^2] + [0.5 * (5 kg) * v^2] [ (a million.5 kg) * (25 m^2/s^2) ] + [ (2.5 kg) * (9 m^2/s^2) ] = [ (a million.5 kg) * v^2 ] + [ (2.5 kg) * v^2 ] [ 37.5 J ] + [ 22.5 J ] = [ (a million.5 kg) * v^2 ] + [ (2.5 kg) * v^2 ] [ 60 J ] = [ (a million.5 kg) * (v1)^2 ] + [ (2.5 kg) * (v2)^2 ] [ (v1)^2 * (a million.5 kg) ] = [ 60 J ] - [ (2.5 kg) * (v2)^2 ] (v1)^2 = [ 40 J ] - [ a million.sixty seven kg * (v2)^2 ] v1 = SQRT { [ 40 J ] - [ a million.sixty seven kg * (v2)^2 ] }