KE Loss Qs 8: Elastic Collisions Explained

  • Thread starter Thread starter yolo123
  • Start date Start date
  • Tags Tags
    Loss
AI Thread Summary
The discussion revolves around the conditions under which all kinetic energy (KE) is lost during collisions. Participants debate the validity of statements regarding equal mass and opposite velocity collisions, with examples like billiard balls illustrating elastic collisions where KE is conserved. The consensus suggests that the original phrasing of "All KE is lost when" may be misleading, as it overlooks scenarios where KE can be conserved in elastic collisions. Clarification is sought on whether the statement should indicate that all KE "can" be lost rather than stating it as a definitive condition. Ultimately, the conversation highlights the complexity of collision types and energy conservation principles.
yolo123
Messages
63
Reaction score
0
8. All KE is lost when:
(a) two bodies of equal mass collide,
(b) two bodies of opposite initial velocity collide,
(c) two bodies of equal mass and opposite initial velocity collide,
(d) never.

None of these make sense to me.
(a) Billard balls have elastic collisions.
(b) Billard balls again disprove this...
(c) Billard balls...
(d) ...... Elastic collisions exist.
 
Physics news on Phys.org
Kinetic energy is lost in inelastic collisions. Does that help? Total energy is however conserved.
 
Yes. I don't see how that helps though. My counter-examples still hold true.
 
Your 'counterexample' on part (d) is incorrect; so what if elastic collisions exist? That tells us nothing about a case where *all* KE is lost, which is what the question is about.

As for the question itself, I'm not actually sure! :o. Are you completely sure it was "All KE is lost when" and not "All KE *can* be lost when"? That would make more sense, because in all the examples it can be conserved as well, depending on whether the collision is elastic or not. Or maybe I'm just delirious because it's midnight. Oh well.
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Collision between snooker balls
Replies
3
Views
710
Solving the Elastric 3-Body Collision Problem
Replies
2
Views
2K
Ambiguous question on momentum and how it works...
Replies
13
Views
1K
Find elastic energy in the compressed spring.
Replies
12
Views
3K
Physics Help with elastic collision
Replies
5
Views
2K
Two Particles Connected by Massless Rod: Dynamics Analysis
Replies
5
Views
2K
Can total KE ever increase for inelastic collisions?
Replies
5
Views
2K
Elastic Collision and Momentum of Ice Skaters
Replies
16
Views
3K
Complex collision with masses and Velcro
Replies
4
Views
2K
Elastic and inelstic collisions conceptual questions.
Replies
6
Views
3K

Hot Threads

Recent Insights

Back
Top