Does Kinetic Energy Loss Equation Hold When Both Objects Are Moving?

AI Thread Summary
In a perfectly inelastic collision, the kinetic energy loss equation can be expressed as KE final/KE initial = M2/(M1 + M2), but this is typically applied when one object is stationary. When both objects are moving prior to the collision, the situation becomes more complex, yet the ratio can still be calculated. The equation's simplicity arises from its specific case, making it easier to handle mathematically. While the equation holds in both scenarios, the calculations differ when both objects are in motion. Understanding these dynamics is crucial for accurate energy loss assessments in collisions.
MotoPayton
Messages
96
Reaction score
0
http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

When a block in motion(M1) collides with a a stationary object (M2) in a perfectly inelastic collision, the fraction of KE final/KE initial can be modeled as M2/M1+M2.

My question is if M2 has to be stationary for this equation to work.

If both objects are moving before the collide in a perfectly inelastic collision does the ratio of lost kinetic energy still hold?
 
Physics news on Phys.org
MotoPayton said:
My question is if M2 has to be stationary for this equation to work.
Yes.
If both objects are moving before the collide in a perfectly inelastic collision does the ratio of lost kinetic energy still hold?
It's a bit more complicated when both blocks are initially moving. But it's easy enough to work it out.
 
I ran the math and it makes sense now.

The only reason we use the M1/M1+M2 is because it is the only ratio of energies that has a simple mathematical form.
 
MotoPayton said:
The only reason we use the M1/M1+M2 is because it is the only ratio of energies that has a simple mathematical form.
Yes. It's a special case that is particularly easy to deal with.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top