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## Homework Statement

A block of mass m is held at rest at height h on an inclined plane of mass M, placed on a horizontal surface as shown. The block is then released.

Find the speed of the inclined plane v_{I} just as the block slides off of it. Neglect all frictional forces. Express your answer in terms of m, M, h and g.

The wedge angle with respect of horizontal is alpha

## Homework Equations

## The Attempt at a Solution

I have solved this using N2L and reached the expression

$$

v_{I}=\sqrt { \frac {2 m^2 gh \cos^2a}{(M+m)(M+ m\ sin^2 a)}}$$

However when I try solving it using conservation of momentum and energy approach .Where $$v_{b}$$ is block velocity relative to table and incline velocity $$v_{i}$$

block velocity x component $$v_{b}cos(\alpha)$$

$$\frac{1}{2} m (v_b)^2+\frac{1}{2} Mv_i^2=mgh$$

$$m (v_b \cos \alpha) + M v_I = 0$$

This system yields a different answer from the above correct answer .

Another one considers block relative velocity to incline $$v_{b|i}$$ so block X_velocity relative to table is $$v_{b|i} \cos \alpha + v_i$$

$$\frac{1}{2} m (v_{b|i} cos \alpha +v_i)^2+\frac{1}{2} Mv_i^2=mgh$$

$$m (v_{b|i} \cos \alpha + v_i) + M v_I = 0$$

Solving this system yields same answer as N2L analysis .

Why does first energy/momentum approach yields wrong answer while the second energy/momentum yields correct answer?

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