Energy and momentum conservation

Click For Summary

Homework Help Overview

The problem involves an object with a mass of 5kg on a horizontal surface and a ball with a mass of 1kg released from a point in a semi-circular orbit. The task is to calculate the maximum height the ball reaches, considering conservation of energy and momentum principles.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of momentum and energy conservation equations in the context of the problem, questioning whether the collision is elastic or inelastic.

Discussion Status

Some participants have provided clarifications regarding the conservation laws applicable to the scenario, noting that momentum is not conserved during certain phases due to external forces. The discussion is exploring the conditions under which energy and momentum conservation apply.

Contextual Notes

Participants are examining the implications of the fixed baffle on momentum conservation and the conditions for applying conservation laws in this specific setup.

Cc518
Messages
23
Reaction score
0

Homework Statement


An object with a mass of 5kg is placed on a horizontal surface and it has a semi-circular orbit with radius 1m. Its left end is close to a baffle fixed on the ground. A ball with a mass of 1kg is released from the point A by static. The surface and the groove are both smooth. Ignore the air resistance. Calculate the maximum height at which the ball rises.

Homework Equations


xkCsabn3H61juPIvVbwmM35eEzbbrFfKDhMfIda4vBSLGnEx_zyntDFzWsektfgoHOpazA=s170.jpg

From A to B:
mgR=1/2 (VB)2
VB = √2gR
From B to C:
mVB = (M + m) VC
mgR=mgh+1/2(M+m) * (VC)² where h is the height

The Attempt at a Solution


I can use these relationships to get the answer, but I don understand why I can use these 2 equations in the same problem:
mVB = (M + m) VC which is an equation for inelastic collision,
mgR=mgh+1/2(M+m) * (VC)² which represents the energy is conserved

If I assume this is an inelastic collision, then the energy will not be conserved. mgR=mgh+1/2(M+m) * (VC)² This equation will not be true if it's an inelastic collision.

I also thought this is an elastic collision, and the two objects just move at the same final velocity. But how can I tell this is an inelastic or elastic collision?
 

Attachments

  • xkCsabn3H61juPIvVbwmM35eEzbbrFfKDhMfIda4vBSLGnEx_zyntDFzWsektfgoHOpazA=s170.jpg
    xkCsabn3H61juPIvVbwmM35eEzbbrFfKDhMfIda4vBSLGnEx_zyntDFzWsektfgoHOpazA=s170.jpg
    1.8 KB · Views: 323
Physics news on Phys.org
Cc518 said:
mVB = (M + m) VC which is an equation for inelastic collision,
That equation is a statement of momentum conservation. There's no inelastic collision here. (Energy is conserved.)
 
Doc Al said:
That equation is a statement of momentum conservation. There's no inelastic collision here. (Energy is conserved.)
Oh, I see :smile:
Thank you for reply.
 
To elaborate on @Doc Al 's reply, all conservation laws come with conditions attached. The conditions which are satisfied in a given situation can be any combination.
In the present problem, conditions for work conservation are satisfied throughout, whereas conditions for linear momentum conservation of the particle+block system are only satisfied after the particle reaches B. Why not before that?
 
haruspex said:
To elaborate on @Doc Al 's reply, all conservation laws come with conditions attached. The conditions which are satisfied in a given situation can be any combination.
In the present problem, conditions for work conservation are satisfied throughout, whereas conditions for linear momentum conservation of the particle+block system are only satisfied after the particle reaches B. Why not before that?

Because the object cannot move with the ball from point A to B, the momentum is not conserved?
 
Cc518 said:
Because the object cannot move with the ball from point A to B, the momentum is not conserved?
Yes. While the ball moves from point A to B there is an external force on the system (from the fixed baffle), so momentum is not conserved.
 
Thank you so much!
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
21
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
10
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
4K