# Block on Wedge problem

## 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?

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kuruman
Homework Helper
Gold Member
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?

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).

kuruman
Homework Helper
Gold Member
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.

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

Doc Al
Mentor
so I need to replace it only in the momenton equation? What about the energy equation?
You'll need to modify both equations.

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

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

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Doc Al
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).

kuruman
Homework Helper
Gold Member
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.

Ok, I got it. Thanks!