Linear Momentum: Homework for Block on Triangular Block

[zorro]

Homework Statement

A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface. Assuming frictionless surfaces, calculate the velocity of the triangular block when the smaller block reaches the bottom end. The angle of inclination of triangular block is theta.

Homework Equations

The Attempt at a Solution

When the small block reaches bottom of the surface, linear momentum in x-direction is conserved.

Mv=m(u-v) where v is the absolute velocity of triangular block
i.e. v=mu/(M+m) and u is the relative velocity of small block

by applying conservation of energy,
u=(2gh)^2

but the answer is not v=[m(2gh)^2]/(M+m)

It is very complex in terms of theta.
Where am I wrong?

 

[Fightfish]

You have failed to consider that the system is not an isolated one; there is a net gravitational force acting on the small block.

 

[zorro]

I considered the Linear momentum only in x-direction. There is no net external force acting in x-direction.

 

[Fightfish]

A key thing is that your answer does not appear to involve theta - this implies that you have mixed up your 'u's. Your first 'u' would be the relative velocity of the small block in the x-direction, but in your COE equation, you used 'u' again to denote the velocity of the small block - this time not just in the x-direction.

 

[zorro]

A key thing is that your answer does not appear to involve theta - this implies that you have mixed up your 'u's. Your first 'u' would be the relative velocity of the small block in the x-direction, but in your COE equation, you used 'u' again to denote the velocity of the small block - this time not just in the x-direction.

Then please give me some hints on solving this.