# Block Slides Down a Movable Ramp...

1. Jun 30, 2015

### entropist1

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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?

Last edited: Jun 30, 2015
2. Jun 30, 2015

### Nathanael

[edit: sorry this post is wrong ]
This is the answer you said is correct? Are you sure it's correct? The first conservation of energy attempt you did looks good to me.

If you're not sure it's correct (I don't think it is) I would like to see the details of how you got it.

Last edited: Jun 30, 2015
3. Jun 30, 2015

### entropist1

I am sure it is correct from my analysis and also from solution manual . Notice also that considering conservation of energy and momentum using block velocity relative to incline yields the exact same answer .

4. Jun 30, 2015

### Nathanael

Ah sorry for being naive I should've thought about it more The mistake is that vb is not traveling at the angle of the incline with respect to the ground (because the incline is also moving) thus the equation mvbcos(α)=Mvi would not be true. (In other words that equation assume the direction of vb is at an angle α from the horizontal.)

5. Jun 30, 2015

### entropist1

I see ....so from the ground the block appears moving with less steep angle since wedge is moving to left right ?
so how can I calculate the angle with respect to ground ?

6. Jun 30, 2015

### Nathanael

Well the angle ought to be more steep (that is, closer to vertical).
Imagine if the mass of the wedge went to zero: it would be pushed out of the way without affecting the block very much, and so the block would fall nearly vertically.

There is the constraint that the relative y-displacement over the relative x-displacement (between block and wedge) must equal the tan(α)
(In words, this is simply the constraint that the block stays in contact with the wedge.)

Differentiating this constraint (w.r.t. time) gives you a relationship between Vbx Vby Vi and α
(Then of course we also have Vb^2=Vbx^2+Vby^2)
I haven't done it but this should make the problem solvable by conservation of energy.

(I think there are other ways but this is the first that comes to my mind.)

7. Jun 30, 2015

### entropist1

Thank you very much :D