Block on Wedge problem

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Homework Statement



Attached in the files, scan from Resnick and Halliday. Also the answers (notice that I use a different meaning for [tex]u[/tex], 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 [tex]mgh[/tex], initial momentum is 0.
I'll call the velocity of the wedge [tex]u[/tex] and the velocity of the block [tex]v[/tex].
The equations:
[tex]2mgh = mv^2 + Mu^2[/tex] - all potential energy goes to the velocities of both objects.
[tex]Mu = -mv \cos \alpha[/tex] - conservation of momentum - only the horizontal component of [tex]v[/tex] is taken into consideration.

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

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Answers and Replies

  • #2
kuruman
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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
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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:
[tex](\frac{u \cos \alpha}{\sqrt{1-u \cos ^2 \alpha}})\sqrt{2gh}[/tex]

[tex]u = \frac{m}{m+M}[/tex]

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
kuruman
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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
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so I need to replace it only in the momenton equation? What about the energy equation?
 
  • #6
Doc Al
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so I need to replace it only in the momenton equation? What about the energy equation?
You'll need to modify both equations.
 
  • #7
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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
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So here is yet another attempt.
I'm out of ideas.
Where did I go wrong?
 

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  • #9
Doc Al
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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
kuruman
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It works. You should correct your momentum equation as Doc Al has suggested. From the momentum equation you should get

[tex]v cos\alpha = \frac{M+m}{m}u[/tex]

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

[tex]\frac{M+m}{m} = \frac{1}{\beta}[/tex] so that

[tex]v cos\alpha = \frac{u}{\beta}[/tex]

Put that in your energy equation and it should come out.
 
  • #11
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Ok, I got it. Thanks!
 

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