Trouble understanding collisions

  • Thread starter Thread starter Opus_723
  • Start date Start date
  • Tags Tags
    Collisions
Click For Summary
The discussion centers on understanding the dynamics of a collision between a light mass and a heavy mass, particularly regarding momentum and kinetic energy conservation. Initially, the light mass has momentum mv and kinetic energy (mv^2)/2, while the heavy mass is at rest. After the collision, the light mass reverses direction with the same speed, leading to a new momentum of -mv, which raises questions about energy conservation since the total kinetic energy seems unchanged. The user realizes that for energy conservation to hold, the light mass's post-collision velocity must be adjusted by a factor k, which is derived from the mass ratio. Ultimately, the user concludes that while the scenario initially seemed unphysical, a deeper analysis clarified the principles of momentum and energy conservation in collisions.
Opus_723
Messages
175
Reaction score
3
So we had some questions in lab today that left me confused. Someone please tell me what I'm not understanding here?

We start with a light mass moving at velocity v, having linear momentum p=mv and kinetic energy (mv^2)/2, and a heavy mass at rest, with linear momentum p=M(0)= 0 and kinetic energy (m(0^2))/2= 0. The total momentum of the system is 0 + mv = mv, and the total kinetic energy is 0 + (mv^2)/2 = (mv^2)/2.

Suppose the light mass collides with the heavy mass in such a way that it has a new velocity nearly equal to its original velocity but opposite in direction. It's new momentum is m(-v)= -mv, and its new kinetic energy is (m(-v)^2)/2= (mv^2), which is the same kinetic energy it began with. Since the initial momentum of the system was p = mv, we have
mv = x -mv, where x is the final momentum of the heavy mass. So x = 2mv. Since the heavy mass has a momentum, it must have kinetic energy. But the light mass has the same kinetic energy that it had at the start, so adding any energy to that would give us more kinetic energy than we started with.

What did I miss?
 
Physics news on Phys.org
Maybe I spoke too soon. I think I figured it out myself.

My first suspicion was that the situation the lab gave me was unphysical. That an object could never bounce back with the same speed it started with.

So I decided to work out the problem in more general terms so that if you start with the situation above, the light mass has a velocity -kv after the collision for some value of k. Then I applied conservation of kinetic energy and found that in order for energy to be conserved, k must be equal to (M-m)/(M+m). This is zero for equal masses and approaches one as the difference between the masses increases, but can never equal one.

So yay!

EDIT: For anyone wondering why I posted for help when I had an idea of where to go next, it's because the first time I tried the above, I lost a negative somewhere and got a silly result that suggested that objects ALWAYS bounce back with equal speed. After posting, I decided to double check my work. I'll just let that be a lesson to myself.
 
Last edited:

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 83 ·
3
Replies
83
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
12
Views
2K
Replies
12
Views
2K