What is the speed of a wedge sliding down an inclined plane?

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SUMMARY

The discussion focuses on calculating the speed of a wedge sliding down an inclined plane when a mass slides down it. The correct approach involves applying the conservation of momentum due to the absence of external forces in the horizontal direction. The initial attempt using kinematics was incorrect, while the conservation of energy equation was also deemed complex for this scenario. Ultimately, the conservation of momentum provides the necessary framework to solve for the wedge's speed.

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Homework Statement



A point particle of mass m is sliding down a wedge inclined at an angle of [tex]\alpha[/tex] to the horizontal. The wedge has a mass m and is free to slide on a smooth horizontal surface. When the mass has fallen a height h, what will be the speed of the wedge?

Homework Equations





The Attempt at a Solution



I tried a kinematics approach with a lot of angle-bashing and eventually came up with:

[tex]v = \frac{h}{g^2} \cot \alpha[/tex]

But this is wrong, and I have a feeling this is too complex for kinematics

I've thought about a conservation of energy approach with

[tex]mgh = \frac{1}{2}mu^2+\frac{1}{2}mv^2[/tex]

where 'u' is the speed of the particle and 'v' is the speed of the block, but I don't know how to divde up the speeds!

thanks
 
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Have you tried the conservation of momentum? There's no external forces in the x direction, so horizontal momentum must be conserved.
 
Thanks, I hadn't considered conservation of momentum, I think I've got it now
 

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