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Consider the configuration below shown in the attached picture!

The wedge can slide on the inclined plane and the cube on the wedge.Their motion is described by [itex] x_1 [/itex] and [itex] x_2 [/itex] respectively. There is no friction and the inclined plane doesn't move.

Here's the Lagrangian of the system:

[itex]

\mathcal L=\frac{1}{2} M \dot{x}_1^2+\frac{1}{2}m(\dot{x}_1^2+\dot{x}_2^2-2 \dot{x}_1\dot{x}_2 \cos\theta)+(M+m)gx_1 \sin\theta

[/itex]

And the Euler-Lagrange equations are:

[itex]

M\ddot{x}_1+m(\ddot{x}_1-\ddot{x}_2\cos\theta)=(M+m)g\sin\theta\\

m(\dot{x}_2-\dot{x}_1\cos\theta)=const

[/itex]

I want to know,what is the constant of motion appearing above?

One may expect it to be the component of total linear momentum parallel to the ground but it can't be that because it doesn't involve M.So what it is?

Thanks

The wedge can slide on the inclined plane and the cube on the wedge.Their motion is described by [itex] x_1 [/itex] and [itex] x_2 [/itex] respectively. There is no friction and the inclined plane doesn't move.

Here's the Lagrangian of the system:

[itex]

\mathcal L=\frac{1}{2} M \dot{x}_1^2+\frac{1}{2}m(\dot{x}_1^2+\dot{x}_2^2-2 \dot{x}_1\dot{x}_2 \cos\theta)+(M+m)gx_1 \sin\theta

[/itex]

And the Euler-Lagrange equations are:

[itex]

M\ddot{x}_1+m(\ddot{x}_1-\ddot{x}_2\cos\theta)=(M+m)g\sin\theta\\

m(\dot{x}_2-\dot{x}_1\cos\theta)=const

[/itex]

I want to know,what is the constant of motion appearing above?

One may expect it to be the component of total linear momentum parallel to the ground but it can't be that because it doesn't involve M.So what it is?

Thanks