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.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top