Wedge and block initial momentum

In summary, the problem involves a block of mass m sliding from a height h on a smooth triangular wedge of mass M that is placed on a smooth floor at an angle theta with the horizontal. The goal is to find the velocity of the wedge when the block reaches the bottom. The solution involves applying conservation of linear momentum and energy, and assumes that the system is initially at rest before the block slides down from height h.
  • #71
gracy said:
Is this maximum height equal to height of the wedge?
No. The wedge must be higher than that height h.
 
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  • #72
ehild said:
The wedge must be higher than that height h.
That's what I want to know,why block stops at height h?why not it continues climbing upto height of the wedge?
 
  • #73
ehild said:
The wedge must be higher than that height h
How did you figure it out?
 
  • #74
Isn't the maximum height of the block of initial velocity u is hmax = u2(M+msinθ)/(2g(M+m))? Can someone confirm this?

I got this answer even though the block is traveling towards the wedge at velocity u, but there is a smooth transition when the block collides with the wedge so it is traveling up the wedge at velocity u the moment when they collide. So I took the initial total momentum in the horizontal direction as mucosθ.
 
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  • #75
Please someone guide me.
ehild said:
No. The wedge must be higher than that height h.
gracy said:
How did you figure it out?
 
  • #76
Would the block stayed on the wedge if it raised higher than the height of the wedge?
 
  • #77
ehild said:
if it raised higher than the height of the wedge?
No.But I am saying why h can not be equal to H.I know h can not be >H.
 
  • #78
You can make the wedge of any height. So it can be h.
 
  • #79
ehild said:
No. The wedge must be higher than that height h.
So,how should I interpret this?
 
  • #80
I have answered it already. It can happen that the block stops just at the upper edge of the wedge. In that case h=H. But it is irrelevant.
 
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  • #81
Thanks for all your answers.You really helped me.Have I calculated h correctly in my post #59?
 
  • #82
gracy said:
Thanks for all your answers.You really helped me.Have I calculated h correctly in my post #59?
Oh, it was the answer to the last question? I think, the idea is good, but you have some mistakes in the formula h=1/2 MV^2(M+m)/m^g. Do you really have m on the power of g? Check the dimensions and place parentheses where needed.
 
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  • #83
Yoonique said:
Isn't the maximum height of the block of initial velocity u is hmax = u2(M+msinθ)/(2g(M+m))? Can someone confirm this?

I got this answer even though the block is traveling towards the wedge at velocity u, but there is a smooth transition when the block collides with the wedge so it is traveling up the wedge at velocity u the moment when they collide. So I took the initial total momentum in the horizontal direction as mucosθ.

We do not know anything about the transition from the horizontal motion of the block to its motion up the wedge. But the only external force acting on the block-wedge system is gravity; so the horizontal component of momentum is conserved.
The problem also states that the mechanical energy is conserved. You can consider process between the initial state (block moving horizontally with speed u), and the state when the block is in rest relative to the wedge, as inelastic collision. As the mechanical energy is conserved the deficiency in kinetic energy is equal to the potential energy gained by the block.
 
  • #84
ehild said:
We do not know anything about the transition from the horizontal motion of the block to its motion up the wedge. But the only external force acting on the block-wedge system is gravity; so the horizontal component of momentum is conserved.
The problem also states that the mechanical energy is conserved. You can consider process between the initial state (block moving horizontally with speed u), and the state when the block is in rest relative to the wedge, as inelastic collision. As the mechanical energy is conserved the deficiency in kinetic energy is equal to the potential energy gained by the block.
Does that mean the block doesn't go up the wedge at the velocity of u the moment it meets the wedge? So I can't use the moment of transition as an initial state? I got an exact question from my exam papers and the answer was the one I calculated.
 
  • #85
Yoonique said:
Does that mean the block doesn't go up the wedge at the velocity of u the moment it meets the wedge? So I can't use the moment of transition as an initial state? I got an exact question from my exam papers and the answer was the one I calculated.

I think ehild is right on this one. Why don't you make a separate thread for the question from your exam? This thread is already getting a bit bloated.

And of course I just contributed to the bloat
 
  • #86
ehild said:
h=1/2 MV^2(M+m)/m^g. Do you really have m on the power of g?
Oh.Typo There.I meant h=1/2 MV^2(M+m)/m^2 g.
 
  • #87
Yoonique said:
Does that mean the block doesn't go up the wedge at the velocity of u the moment it meets the wedge? So I can't use the moment of transition as an initial state? I got an exact question from my exam papers and the answer was the one I calculated.

Gracy's question was not really clear. Copy the question from your exam paper word by word and your derivation, please, in a separate thread as AlephNumbers suggested.

If we assume that the horizontal component of momentum is conserved during the whole process, and V is the final common velocity of the block and wedge, Gracy's equation (m+M)V=mu was correct, as both u and V are horizontal.
 
  • #88
gracy said:
Oh.Typo There.I meant h=1/2 MV^2(M+m)/m^2 g.

And do you really multiply by g? If g is in the denominator, use parentheses.
 
  • #89
AlephNumbers said:
I think ehild is right on this one. Why don't you make a separate thread for the question from your exam? This thread is already getting a bit bloated.

And of course I just contributed to the bloat
NO, you made it more clear. The problem was poorly worded.
 
  • #90
ehild said:
If g is in the denominator, use parentheses
h=1/2 MV^2(M+m)÷m^2 g.
g is in multiplication with m^2.
 
  • #91
ehild said:
The problem was poorly worded.
I have given my best.
 
  • #92
gracy said:
h=1/2 MV^2(M+m)÷m^2 g.
g is in multiplication with m^2.
Then write it as h=1/2 MV^2(M+m)/(m^2 g)
 
  • #93
gracy said:
I have given my best.

Well, your teacher had to be more clear.
 

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