Energy and momentum conservation

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SUMMARY

The discussion centers on the conservation of energy and momentum in a physics problem involving a 5kg object and a 1kg ball in a semi-circular orbit. The equations used include mgR = 1/2 (VB)² for energy conservation and mVB = (M + m) VC for momentum conservation. Participants clarify that while energy is conserved throughout the motion, momentum is not conserved from point A to B due to the external force from the fixed baffle. The conditions for applying conservation laws vary depending on the specific segment of the motion being analyzed.

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  • Understanding of Newton's laws of motion
  • Familiarity with conservation of energy principles
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Cc518
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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?
 

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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!
 

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