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

[captainjack2000]

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!

 

[Hootenanny]

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.

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

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

 

[captainjack2000]

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?

 

[Hootenanny]

Looks ok to me.

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

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.

 

[captainjack2000]

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?

 

[Hootenanny]

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