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ehild
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No. The wedge must be higher than that height h.gracy said:Is this maximum height equal to height of the wedge?
No. The wedge must be higher than that height h.gracy said:Is this maximum height equal to height of the wedge?
That's what I want to know,why block stops at height h?why not it continues climbing upto height of the wedge?ehild said:The wedge must be higher than that height h.
How did you figure it out?ehild said:The wedge must be higher than that height h
ehild said:No. The wedge must be higher than that height h.
gracy said:How did you figure it out?
No.But I am saying why h can not be equal to H.I know h can not be >H.ehild said:if it raised higher than the height of the wedge?
So,how should I interpret this?ehild said:No. The wedge must be higher than that height h.
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.gracy said:Thanks for all your answers.You really helped me.Have I calculated h correctly in my post #59?
Yoonique said:Isn't the maximum height of the block of initial velocity u is h_{max} = u^{2}(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.
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.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.
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.
Oh.Typo There.I meant h=1/2 MV^2(M+m)/m^2 g.ehild said:h=1/2 MV^2(M+m)/m^g. Do you really have m on the power of g?
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 said:Oh.Typo There.I meant h=1/2 MV^2(M+m)/m^2 g.
NO, you made it more clear. The problem was poorly worded.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
h=1/2 MV^2(M+m)÷m^2 g.ehild said:If g is in the denominator, use parentheses
I have given my best.ehild said:The problem was poorly worded.
Then write it as h=1/2 MV^2(M+m)/(m^2 g)gracy said:h=1/2 MV^2(M+m)÷m^2 g.
g is in multiplication with m^2.
gracy said:I have given my best.