Slipping block on a slipping ramp: final speed

[DavideGenoa]

Hi, friends! A block having mass $m$ frictionlessly slips from height $h$ on a ramp of mass $M$, which has an angle $\theta$ with the floor, where it slips with no friction.

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I would like to prove that, as my book says, the ramp's speed when the block leaves it is

$V=\sqrt{\frac{2m^2gh\cos^2\theta}{(m+M)(M+m\sin^2\theta)}}.$​

But I cannot reach this result. Because of the conservation of momentum in the direction parallel to the floor, which is a direction where no external force act on the system block-ramp, I would say that the speed $v$ of the block when it leaves the ramp is such that $mv\cos\theta=M V$ and, because of the conservation of mechanical energy, I would think that $mgh=\frac{1}{2}mv^2+\frac{1}{2} MV^2=\frac{1}{2}\big(\frac{M^2V^2}{m\cos^2\theta} +MV^2\big)$, and therefore $V=\sqrt{\frac{2m^2gh\cos^2\theta}{(m+M)(M^2+Mm\cos^2\theta)}}$
which is a wrong result. Where am I wrong?

I $\infty$-ly thank you for any help!

 

[paisiello2]

Where do you get the (m+M) term?

 

[haruspex]

Relative to the ground, what direction will the block be moving?

 

[haruspex]

Where do you get the term (m+M) term?

I assumed that was a typo.

 

[ehild]

See thread https://www.physicsforums.com/threads/wedge-and-block-initial-momentum.806992/. It is the same problem as yours.
Decide what you denote by v. The block slides on the ramp which is moving horizontally with velocity -V. If the relative velocity of the block with respect to the ramp is u, the horizontal component is ucos(θ) and in the rest frame of reference it is ucos(θ)-V.

 

[DavideGenoa]

Thank you all, friends! ehild's linked post answers the issue.