Understanding the Wedge Problem in Conservation of Momentum and Energy

[manenbu]

Homework Statement

Attached in the files, scan from Resnick and Halliday. Also the answers (notice that I use a different meaning for u, but it shouldn't matter.

Homework Equations

Conservation of momentum, and conservation of energy.

The Attempt at a Solution

Initial Energy of the system is mgh, initial momentum is 0.
I'll call the velocity of the wedge u and the velocity of the block v.
The equations:
2mgh = mv^2 + Mu^2 - all potential energy goes to the velocities of both objects.
Mu = -mv \cos \alpha - conservation of momentum - only the horizontal component of v is taken into consideration.

I end up having this as an answer:
u = \sqrt{\frac{2gh}{M^2 + Mm \cos ^2 \alpha}}m \cos \alpha
Which is, not correct according to the answer.
What did I do wrong?

 

[kuruman]

I can't see the pictures pending approval, so I guessed at what the question is asking. I assumed that you are looking for the speed of the wedge and I got the same answer as you. What is the answer in the back of the book?

 

[manenbu]

I'll type it in:
A block of mass m rests on a wedge of mass M which, in turn, rests on a horizontal table, as shown in Fig. All surfaces are frictionless. If the system starts at rest with point P of the block a distance h above the table, find the velocity of the wedge the instant point P touches the table.

Answer is:
(\frac{u \cos \alpha}{\sqrt{1-u \cos ^2 \alpha}})\sqrt{2gh}

u = \frac{m}{m+M}

It has some similarities to my answer, so I guess the general idea I had in mind was correct. My guess is my conservation equations are not accurate (because I don't feel sure about them).

 

[kuruman]

The problem with your expression is that the x component of the block's velocity is v_block*cosα relative to the wedge, not relative to the table. Relative to the table it is v_block*cosα - V_wedge.

 

[manenbu]

so I need to replace it only in the momenton equation? What about the energy equation?

 

[Doc Al]

so I need to replace it only in the momenton equation? What about the energy equation?

You'll need to modify both equations.

 

[manenbu]

Meaning in the energy equation the vertical component of the block's velocity stays the same while the horizontal is relative?
I'll try.

 

[manenbu]

So here is yet another attempt.
I'm out of ideas.
Where did I go wrong?

 

[Doc Al]

I haven't had a chance to look at your attempt in detail (I will a bit later), but at first glance:
(1) Your energy equation is OK.
(2) Correct your momentum equation: The velocity of the wedge is -u (u is positive).

 

[kuruman]

It works. You should correct your momentum equation as Doc Al has suggested. From the momentum equation you should get

v cos\alpha = \frac{M+m}{m}u

In order not to confuse your u with the book's u, let

\frac{M+m}{m} = \frac{1}{\beta} so that

v cos\alpha = \frac{u}{\beta}

Put that in your energy equation and it should come out.

 

[manenbu]

Ok, I got it. Thanks!