Elastic vs inelastic

1. Aug 11, 2014

KurtWagner

say I have two objects of the same mass. Is the collision always going to be elastic?
mv + mv = mv + mv

the masses cancel out and thus kinetic energy is conserved right?

does this mean that for two cars with exactly the same mass the collision is going to be elastic?

if so, would there be any damage to the cars? as the there is no change in kinetic energy then there is no energy to crumble the car body right?

2. Aug 11, 2014

KurtWagner

like for example when a moving car hits a resting car.

3. Aug 11, 2014

billy_joule

Whether a collision is elastic or not has nothing to do with the masses.

Imagine a head on collision at 100km/hr between:
A) two 1000kg vehicles
B) one 500kg and one 1500kg vehicle

Does it make any sense to think the cars in case A) will be undamaged and the cars in case B) will be damaged?

In the real world there are no perfectly elastic collisions. Some collisions can be closely approximated as elastic but a car crash is definitely not one of them.

4. Aug 11, 2014

KurtWagner

how about a rear ender between two cars of equal mass on ice.

first car is 10m/s second is at rest. after the collision the velocities are conserved right?

5. Aug 11, 2014

billy_joule

No, cars are specifically designed to be inelastic in collisions for safety reasons.

Collisions between Billiard balls are close to elastic.

6. Aug 11, 2014

Orodruin

Staff Emeritus
No, same mass does not imply elastic collision. The collision between two cars is typically quite inelastic.

Edit: Maybe I should stop opening several tabs and answering when I get to them without checking if it was already answered ... :)

7. Aug 11, 2014

KurtWagner

so how do you use the conservation of momentum for two cars of equal masses where the collision is inelastic.

why does the differences in speeds not break the conservation of momentum

8. Aug 11, 2014

Orodruin

Staff Emeritus
Conservation of momentum is one equation but you have two variables (the post-collision velocities of each car). As such, it does not uniquely determine the velocities and you need more input, such as the total kinetic energy being conserved (elastic) or the velocities being equal after collision (fully inelastic).

9. Aug 11, 2014

KurtWagner

am i right in assuming that most car collisions are completely inelastic?

10. Aug 11, 2014

KurtWagner

or should i say. most car collisions where the breaks are engaged in both cats

11. Aug 11, 2014

KurtWagner

*cars

12. Aug 11, 2014

SteamKing

Staff Emeritus
"brakes", not "breaks".

13. Aug 11, 2014

KurtWagner

thanks for that :p

14. Aug 11, 2014

haruspex

Any car collision more than a nudge is mostly inelastic.

15. Aug 11, 2014

billy_joule

The brakes have nothing to do with it. Prior deceleration, like mass, has nothing to do with whether a collision is elastic or not.

16. Aug 11, 2014

KurtWagner

so this brings me back to my misunderstanding involving the conservation of momentum.

for example a moving car hitting a stationary one.

initial
car a: 1000kg 10m/s
car b: 1000kg 0m/s

final
car a: ?
car b: 5m/s

using the conservation of momentum on this would leave car a going 5m/s right?

kinetic energy is conserved right?

what am I doing wrong?

17. Aug 11, 2014

Nathanael

Kinetic energy is not conserved in this case.
The initial KE was $\frac{1}{2}10^2M=50,000$ Joules.
Whereas the final KE was $\frac{1}{2}5^2M+\frac{1}{2}5^2M=25,000$ Joules.

Kinetic Energy was lost (50% of it).

...

When two objects collide, there are infinite possible combinations of final velocities that will satisfy conservation of momentum.
Many of these possible solutions violate conservation of energy (because energy comes from nowhere)
Some of these possible solutions result in a decreased amount of kinetic energy (so-called "inelastic collions")

But there is only 1 solution of conservation of momentum which leaves the kinetic energy unchanged.
This sitatuion is referred to as an "elastic collision"

...

In your example, the "elastic solution" would be $a_{final}=0$ and $b_{final}=10\frac{m}{s}$

18. Aug 11, 2014

KurtWagner

HA!

Thank you. I did not realize. The square of the velocity!!

That is exactly what I was missing.

5^2 plus 5^2 is not the same as 10^2. Lol