# Block Wedge problem

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1. Nov 30, 2015

### AnwaarKhalid

1. The problem statement, all variables and given/known data
A block of mass m is placed on a triangular block of mass M, which in turn is placed on a horizontal surface as shown. Assume frictionless surfaces. Find the speed of triangular block when the small block reaches the bottom end,

2. Relevant equations

3. The attempt at a solution
I've uploaded my attempt at a solution. Please, tell me where I did wrong.

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2. Dec 1, 2015

### endoftimes

I think you might actually have the right answer, but you need to rearrange algebraically to solve for whatever it is you are supposed to have. I'm sorry it is hard to make out the photograph, but do you know what it is supposed to be?

3. Dec 1, 2015

### haruspex

You've lost me with your very first equation. You need to define your variables if others are to follow your logic.
Is that Acom? Does that stand for the acceleration of the common mass centre of triangle+block? As a vector or in a specific direction?

4. Dec 1, 2015

### AnwaarKhalid

Yes, that is acc. of the triangle as well as the block.

5. Dec 1, 2015

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6. Dec 1, 2015

### haruspex

That only answered one of my questions. Anyway, I agree with the 'book' answer.
Reading an image of your handwritten working is too much of a struggle. Please take the trouble to type it in, preferably using LaTeX.

7. Dec 1, 2015

### AnwaarKhalid

Common horizontal acc. of the blocks= (M+m)a= ma_x

8. Dec 1, 2015

### AnwaarKhalid

I'm sorry I do not know how to type using latex. Hope , these two images are clear enough.

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9. Dec 1, 2015

### haruspex

I said preferably with LaTeX. It is not mandatory. Typing using subscripts and superscripts (see X2 and X2 in the toolbar) is good enough.

10. Dec 1, 2015

### AnwaarKhalid

Common acc. of the masses = (M+m)acom=max
ax= gsinxcosx
Therefore, acom=mgsinxcosx\(M+m)
Now,
Tanx= h\BC
BC=h\tanx
Let the triangular block be displaced by x when the smaller block reaches the bottom end
Since there is no external force
Therefore,
Rcomi=Rcomf
M(BC/2) = M(BC/2 - x) + m(BC -x)
which gives,
x= mhcosx/(M+m)sinx
Now, using
v2 - u2= 2as
v={ (2m2ghcos2x)/(M+m)2 }1/2

#### Attached Files:

• ###### Snapshot.jpg
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11. Dec 2, 2015