# Question regarding inelastic collisions

Concerning inelastic collisions: Is it true the kinetic energy after the collision is equal to 1/2 the total momentum?

## Answers and Replies

CORRECTION:
Concerning inelastic collisions: Is it true the kinetic energy after the collision is equal to 1/2 the total momentum squared divide by total mass?

jfizzix
Science Advisor
Gold Member
The kinetic energy has a range of values after the collision depending on just how inelastic the collision is.

The largest value the final kinetic energy could have would be equal to the initial kinetic energy, but this only happens in the limit that the collision becomes elastic.

The smallest value the final kinetic energy could have would be the kinetic energy if after the collision the particles stick together. This value depends on the ratio of the two masses, and the ratio of the two initial velocities. It is not simply half the initial kinetic energy (though I'd have to work it out to see if that's indeed the smallest possible value over all initial masses and velocities).

Nugatory
Mentor
The smallest value the final kinetic energy could have would be the kinetic energy if after the collision the particles stick together. This value depends on the ratio of the two masses, and the ratio of the two initial velocities. It is not simply half the initial kinetic energy (though I'd have to work it out to see if that's indeed the smallest possible value over all initial masses and velocities).

Two equal masses with equal and opposite velocities colliding and sticking together will yield zero kinetic energy.
(Think head-on SPLAT!).

jfizzix
jfizzix
Science Advisor
Gold Member
point taken!

jbriggs444
Science Advisor
Homework Helper
CORRECTION:
Concerning inelastic collisions: Is it true the kinetic energy after the collision is equal to 1/2 the total momentum squared divide by total mass?
In a perfectly inelastic collision, all the participants will stick together after the collision. The resulting momentum will be ##m_{tot}v_{cm}##. Yes, the resulting kinetic energy is ##\frac{1}{2}m_{tot}{v_{cm}}^2 = \frac{1}{2}\frac{(m_{tot}v_{cm})^2}{m_{tot}}##

excellent. thanks for all the reply's.