A question regarding energy and momentum conservation

[jehwig0107]

I have a problem in mechanics.

On the wedge and block only the gravisational force (mg) is exerted (and there is no friction in this system).

What is asked in the question is the final velocities of the wedge and the block (vB, vK). The velocity of the block is conserved when it reaches at the ground.

Do I need to solve this question with the conservation of momentum and conservation of energy? (this is what the solution tolds me) It seems to me that it is absurd, because if the momentum is conserved, then the total energy is encreased.

When we compare two energies:

1) E just before the separation : 0.5*(mB * vB^2 + mK * vK'^2) = mB * g * h

And when the momentum for horizontal direction is conserved : mB * vB'x - mK * vK' = mB * vB - mK * vK = 0

(vB' is the velocity of Block just before the seperation, and its magnitude is same with vB, the final velocity. vB'x is the horizontal component of vB')

vK is thus greater than vK' (because vB is greater than vB'x)

2) E after the separation : 0.5*(mB * vB^2 + mK * vK^2) > mB * g * h (1)

Therefore the total energy is not conserved if the horizontal momentum is conserved.

Am I wrong?

 

[Orodruin]

if the momentum is conserved, then the total energy is encreased.

No, this is not correct. If energy conservation is one of your criteria and momentum conservation the other, obviously the solution must reflect this.

vB' is the velocity of Block just before the seperation, and its magnitude is same with vB

This is not correct. There is no reason to assume this.

 

[jehwig0107]

No, this is not correct. If energy conservation is one of your criteria and momentum conservation the other, obviously the solution must reflect this.

This is not correct. There is no reason to assume this.

This condision is given by the question.
"The velocity of the block is conserved when it reaches at the ground."
That means, the magnitude of vB' (the velocity just before the seperation) is same with vB (the velocity after saperation).

 

[PeroK]

This condision is given by the question.
"The velocity of the block is conserved when it reaches at the ground."
That means, the magnitude of vB' (the velocity just before the seperation) is same with vB (the velocity after saperation).

All that means is that the block doesn't lose energy hitting the ground. The slope tapers off so there is a smooth transition to horizontal motion at the end of the slope.

 

[PeroK]

Do I need to solve this question with the conservation of momentum and conservation of energy? (this is what the solution tolds me) It seems to me that it is absurd, because if the momentum is conserved, then the total energy is encreased.

Kinetic energy is not conserved. The initial KE is zero. But, you have initial potential energy in the system.

 

[jehwig0107]

All that means is that the block doesn't lose energy hitting the ground. The slope tapers off so there is a smooth transition to horizontal motion at the end of the slope.

You mean that magnitude of vB' is differ from magnitude of vB?

And you mean that the velocity of slope is encreased because of the potential energy (mgh)?
That means, the velocity of slope is faster, compair to the case, if the velocity of block is not conserved?

 

[PeroK]

You mean that magnitude of vB' is differ from magnitude of vB?

And you mean that the velocity of slope is encreased because of the potential energy (mgh)?
That means, the velocity of slope is faster, compair to the case, if the velocity of block is not conserved?

No. I mean that in this problem there is no difference between $vB$ and $vB'$. They are both the speed of the block when all the potential energy has been converted to Kinetic Energy.

You are making this problem too complicated. You don't have to worry about any complications when the block slides off the wedge.

You can use conservation of energy and (horizontal) momentum to solve the problem.

 

[jehwig0107]

No. I mean that in this problem there is no difference between $vB$ and $vB'$. They are both the speed of the block when all the potential energy has been converted to Kinetic Energy.

You are making this problem too complicated. You don't have to worry about any complications when the block slides off the wedge.

You can use conservation of energy and (horizontal) momentum to solve the problem.

Thank you for your reply.
I am just wonderning now: Is there any difference between the kinetic energies just before the separation (1) in the case of simple inclined plane without curved edge (unlike the question above, there is no velocity conservation by collision with the ground) and (2) in curved plane?

 

[PeroK]

Thank you for your reply.
I am just wonderning now: Is there any difference between the kinetic energies just before the separation (1) in the case of simple inclined plane without curved edge (unlike the question above, there is no velocity conservation by collision with the ground) and (2) in curved plane?

If it was a simple wedge, then its speed would be the same as in your problem, but it would have a vertical component of velocity on impact with the ground. The horizontal component of its velocity would be different, so that would be a different problem.

 

[neilparker62]

Alternative approach:

"Reverse time" event is a perfectly inelastic collision in which two objects with masses m_B and m_k having approach velocities v_B and v_k respectively come to a simultaneous standstill. Collision energy ΔE is not lost but converted to PE. Hence equations pertaining to perfectly inelastic collisions applicable:

$$ΔE = ½μΔv^2=m_Bgh...1$$ and collision impulse $$Δp=μΔv...2$$ where μ is the reduced mass of the colliding objects $$μ=\frac{m_Bm_K}{m_B+m_K}$$ and Δv their relative velocity.

Obtain Δv from ...1 and simply divide Δp from ...2 by the respective masses to obtain respective velocities. The problem is a variant of Problem 4 in the following problem set which can be solved by similar means.

http://web.mit.edu/8.01t/www/materials/ProblemSets/Raw/f12/ps07sol.pdf