# A question regarding energy and momentum conservation

• jehwig0107
In summary, the conversation discusses a problem in mechanics involving a wedge and a block with only gravitational force acting on them. The question asks for the final velocities of the wedge and the block, and the solution involves using conservation of momentum and energy. However, there is a disagreement about whether the velocity of the block is conserved when it reaches the ground. The summary concludes that the solution can be solved using conservation of energy and momentum, and there may be a difference in the kinetic energies before the separation in a curved plane compared to a simple inclined plane.
jehwig0107
Moved from a technical forum, no template.

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?

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jehwig0107 said:
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.
jehwig0107 said:
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.

Orodruin said:
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).

jehwig0107 said:
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.

jehwig0107 said:
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.

PeroK said:
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?

jehwig0107 said:
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.

PeroK said:
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.
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?

jehwig0107 said:
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.

Alternative approach:

"Reverse time" event is a perfectly inelastic collision in which two objects with masses mB and mk having approach velocities vB and vk 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

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## 1. What is energy conservation?

Energy conservation refers to the principle that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that in any physical process, the total amount of energy remains constant.

## 2. How is energy conservation related to momentum conservation?

Energy conservation and momentum conservation are closely related because in a closed system, the total energy and momentum must remain constant. This means that any changes in one quantity must be balanced by changes in the other.

## 3. What is momentum conservation?

Momentum conservation refers to the principle that in a closed system, the total momentum remains constant. This means that the total mass times velocity before a collision or interaction must equal the total mass times velocity after the collision or interaction.

## 4. How do energy and momentum conservation affect everyday life?

Energy and momentum conservation are fundamental principles that govern the behavior of all physical systems, including everyday objects and processes. They are important in understanding and predicting the motion and interactions of objects, from a simple game of billiards to the movement of planets in our solar system.

## 5. What happens if energy or momentum is not conserved?

If energy or momentum is not conserved, it means that there is an external force or influence acting on the system. This can result in unexpected or unpredictable behavior, and is often a sign of an incomplete understanding of the system or a missing piece of information.

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