# Block on Wedge problem

1. Aug 25, 2009

### manenbu

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

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.

2. Relevant equations

Conservation of momentum, and conservation of energy.

3. 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?

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Last edited: Aug 25, 2009
2. Aug 25, 2009

### 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?

3. Aug 25, 2009

### 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.

$$(\frac{u \cos \alpha}{\sqrt{1-u \cos ^2 \alpha}})\sqrt{2gh}$$

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

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).

4. Aug 25, 2009

### kuruman

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

5. Aug 25, 2009

### manenbu

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

6. Aug 25, 2009

### Staff: Mentor

You'll need to modify both equations.

7. Aug 25, 2009

### 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.

8. Aug 25, 2009

### manenbu

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

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9. Aug 26, 2009

### Staff: Mentor

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).

10. Aug 26, 2009

### 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.

11. Aug 28, 2009

### manenbu

Ok, I got it. Thanks!