Energy and momentum conservation

AI Thread Summary
The discussion revolves around the conservation of energy and momentum in a physics problem involving a ball and a block. The key equations used include those for energy conservation and momentum conservation, but there is confusion about their applicability in the context of an inelastic versus elastic collision. It is clarified that while energy is conserved throughout the motion, momentum conservation only applies after the ball reaches point B due to an external force acting on the system before that point. The participants emphasize that since the ball cannot move with the block from point A to B, momentum is not conserved during that segment. The conversation concludes with a clear understanding of the conditions under which each conservation law applies.
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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