How Do Energy and Momentum Conservation Apply to a Mass Sliding Down a Wedge?

In summary, the conversation is discussing a scenario where a mass slides down a wedge on a frictionless table, causing the wedge to slide left. The question is asking about the application of linear momentum and energy conservation in this situation. The final equation expressed is mgh = (1/2)mv^2 + (1/2)Mu^2, where v is the velocity of the mass directed parallel to the slope. The conversation also clarifies that both the horizontal and vertical components of the mass and wedge need to be considered for the equation.
  • #1
captainjack2000
99
0
1.
A wedge mass M is at rest on horizontal frictionless table. Mass m is on wedge. There is no fricton between parrticle and wedge. Height of wedge = h and angle of incline= theta. The massm slides down wedge from rest and wedge slides left on table. How does linear momentum conservation apply?
The question also asks Hows does the energy conservation apply in this situation. Write down an equation expressing the consequence of energy conservation.(Hint the square of a velocity is equal to the sum of the squares of its components)

3. Can you just say that the horizontal components of the momentum must be constant before and after the mass m begins to slide? Therefore mvcos(theta) = Mu where u is the horizontal velocity of the wedge wrt the table and vcos(theta)is the horizontal component of the mass m wrt to the table?

Does the potential energy mgh = (Mu^2)/2 + (m(vcos(theat))^2)/2

Bit confused!
 
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  • #2
captainjack2000 said:
Can you just say that the horizontal components of the momentum must be constant before and after the mass m begins to slide? Therefore mvcos(theta) = Mu where u is the horizontal velocity of the wedge wrt the table and vcos(theta)is the horizontal component of the mass m wrt to the table?
Looks ok to me.
captainjack2000 said:
Does the potential energy mgh = (Mu^2)/2 + (m(vcos(theat))^2)/2
Why are you only considering the horizontal velocity of the mass?
 
  • #3
Should I be equating the vertical components to get

mgh = m(vsin(theta))^2/2

What about the kinetic energy of the wedge traveling horizontally?
 
  • #4
Hootenanny said:
Looks ok to me.

Why are you only considering the horizontal velocity of the mass?

captainjack2000 said:
Should I be equating the vertical components to get

mgh = m(vsin(theta))^2/2

What about the kinetic energy of the wedge traveling horizontally?
You need to consider both the horizontal and vertical components of the mass and the wedge, remember that speed is a scalar quantity and direction doesn't matter.
 
  • #5
So just to clarify:

You could say mgh = (1/2)mv^2 + (1/2)Mu^2 where v is the velocity of the mass m directed parallel to the slope?
 
  • #6
captainjack2000 said:
So just to clarify:

You could say mgh = (1/2)mv^2 + (1/2)Mu^2 where v is the velocity of the mass m directed parallel to the slope?
Looks good to me :approve:
 

FAQ: How Do Energy and Momentum Conservation Apply to a Mass Sliding Down a Wedge?

What is energy conservation?

Energy conservation is the principle that energy cannot be created or destroyed, but can only be transformed from one form to another. This means that the total amount of energy in a closed system remains constant over time.

How does momentum conservation work?

Momentum conservation is the principle that the total momentum of a closed system remains constant over time. This means that the sum of all the individual momenta of the objects in the system remains the same, regardless of any internal or external forces acting on the system.

What types of systems does energy and momentum conservation apply to?

Energy and momentum conservation apply to closed systems, which are systems that do not exchange matter or energy with their surroundings. This can include objects on a frictionless surface, certain chemical reactions, and even the entire universe.

How are energy and momentum related?

Energy and momentum are related through the concept of work. Work is done when a force is applied to an object and causes it to move. The amount of work done is equal to the change in the object's energy, which in turn affects its momentum.

Are there any exceptions to energy and momentum conservation?

While energy and momentum conservation are fundamental principles of physics, there are a few situations where they may not hold true. These exceptions include situations involving strong gravitational fields, nuclear reactions, and certain quantum phenomena. However, these exceptions are rare and do not invalidate the overall validity of energy and momentum conservation.

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