Understanding Kinetic Energy and Momentum Conservation in Colliding Masses

AI Thread Summary
In a theoretical mechanics discussion, the conservation of momentum and kinetic energy during collisions of two masses, m1 and m2, was analyzed. When m1, moving towards m2, collides and recoils with the same speed but opposite direction, m2 must move to the right with a speed v' to conserve momentum, described by the equation v' = 2m1v / m2. Although m1 retains its kinetic energy post-collision, m2 gains kinetic energy, leading to a total kinetic energy that exceeds the initial energy. This discrepancy arises because kinetic energy is not always conserved in collisions, particularly when m2 is stationary and has a finite mass. The discussion highlights that for the final velocities to equal -v for m1, the collision must be elastic and m2 must have an infinitely large mass, or the mass ratio must approach zero.
Yosty22
Messages
182
Reaction score
4
In my theoretical mechanics class, we went over some very basic conservation laws (namely momentum) and talked about colliding masses.

Say you have 2 masses, m1 and m2. m1 is moving to the right (towards m2) with velocity v. m2 is stationary. After the two masses collide, m1 recoils with velocity -v (same speed, just the direction changed). In order for momentum to be conserved, mass m2 must move to the right after the collision with speed v'. In terms of v, m1, and m2, v' can be described as:
v' = 2m1v / m2.

This is all well and good, but some classmates and I were thinking about the kinetic energy here. If m1 is moving to begin with, it has some initial kinetic energy equal to .5m1v2. However, since it recoils at the same exact speed (just different direction), it's kinetic energy is the same after the collision as it was before. However, m2 must move to the right after the collision for momentum to be conserved, so after the collision, m2 also has kinetic energy equal to .5m2v2. That is, the kinetic energy after the collision is greater than the kinetic energy before. (I know kinetic energy doesn't always have to be conserved, but why would the total kinetic energy be greater after this collision?)
 
Physics news on Phys.org
Your observation is correct. If colliding with a stationary object, a mass with initial velocity v can only have final velocity -v if the collision is elastic *and* the object it collides with has infinite mass (or rather, the velocity will go to -v as the mass ratio goes to 0). In this limit, the heavy mass can absorb the additional momentum without changing its kinetic energy significantly.

If the objects have the same mass and the collision is elastic, the originally moving object will be at rest after the collision.
 
  • Like
Likes 1 person
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