Wedge and block initial momentum

[gracy]

please can you answer post 35.

 

[ehild]

My doubt is that now block is not on wedge,so force which was responsible for motion of block is no longer there.So block should have constant velocity that is same as it's velocity when the block was at bottom most part because after that force was removed (as block came off from wedge)

There is no force when the box has left the wedge, but there is interaction between the block and wedge during the transition , and that force is not the same as it was along the wedge.

 

[gracy]

And one more question.

What is the maximum height reached by small "m"(after climbing that much height it stops)

Block stops with respect to wedge but in the ground frame both blocks and wedge are moving with same velocity let's say V
So mu=(m+M)V
And conservation of energy
1/2 mv^2=mgh(maximum height)+1/2mV^2+1/2MV^2

 

[ehild]

Write the full question.

 

[gracy]

Is conservation of momentum applied in non inertial frame?As conservation of momentum is derived from Newton's second law and Newton's second law is only applicable in inertial frames ,conservation of momentum should only be applied in inertial frames,right?

 

[Yoonique]

Is conservation of momentum applied in non inertial frame?As conservation of momentum is derived from Newton's second law and Newton's second law is only applicable in inertial frames ,conservation of momentum should only be applied in inertial frames,right?

Yes you are right. But is you want the conservation of momentum to be applied in non inertial frame, you need to add in fictitious forces into your free body diagram.

 

[gracy]

And conservation of energy
1/2 mv^2=mgh(maximum height)+1/2mV^2+1/2MV^2

Hmmm...correction needed.
1/2 mu^2=mgh(maximum height)+1/2mV^2+1/2MV^2

 

[Yoonique]

So mu=(m+M)V

I think this is wrong. The use of conservation of momentum should be in the same direction. So it should be mucosθ=(M+m)V

 

[gracy]

mu=(m+M)V

1/2 mu^2=mgh(maximum height)+1/2mV^2+1/2MV^2

From these equations I solved for h(maximum height upto which block rises on wedge .
h=1/2 MV^2(M+m)/m^g
I want to know whether this h=Height of wedge(H)
Or h<H

 

[gracy]

I think this is wrong. The use of conservation of momentum should be in the same direction. So it should be mucosθ=(M+m)V

No.It is correct.

 

[AlephNumbers]

Write the full question.

Before this goes any farther, please do as ehild asks.

 

[gracy]

Write the full question.

A block is moving with velocity u as shown in my image attached.And there is a wedge of mass M.The block rises on wedge and attains maximum height ?(h)Find maximum height.

 

[AlephNumbers]

Are the wedge and block both on frictionless surfaces still?

 

[gracy]

Are the wedge and block both on frictionless surfaces still?

Yes.

 

[gracy]

One thing I would like to mention here is my teacher told me that when the block reaches it's maximum height it stops and then it's velocity is same as wedge .They move with common velocity V.

 

[ehild]

The wedge and the block collide. Was the wedge in rest before the collision? Is the collision elastic? The momentum is not conserved as the block gains vertical component of momentum.

 

[gracy]

. Was the wedge in rest before the collision?

Yes.

Is the collision elastic?

Yes.

 

[gracy]

Please answer.I have solved for h (in my post 59)and my teacher says it's correct.I just wanted to know why block stops at that height?

 

[ehild]

If both energy and horizontal momentum are conserved, and the block slides up the wedge, you can calculate the maximum height applying momentum and energy conservation between the stages 1) and 2). 1): wedge is in rest, block moves with speed u towards the wedge; 2) Both the wedge and the block move with the same velocity V, block is at height h.

After reaching maximum height, the block will slide backwards, and leave the wedge at the end. You can determine both velocities, that of the wedge and that of the block at this stage.

 

[gracy]

After reaching maximum height

Is this maximum height equal to height of the wedge?

 

[ehild]

Is this maximum height equal to height of the wedge?

No. The wedge must be higher than that height h.

 

[gracy]

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?

 

[gracy]

The wedge must be higher than that height h

How did you figure it out?

 

[Yoonique]

Isn't the maximum height of the block of initial velocity u is h_max = 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.

 

[gracy]

Please someone guide me.

No. The wedge must be higher than that height h.

How did you figure it out?

 

[ehild]

Would the block stayed on the wedge if it raised higher than the height of the wedge?

 

[gracy]

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.

 

[ehild]

You can make the wedge of any height. So it can be h.

 

[gracy]

No. The wedge must be higher than that height h.

So,how should I interpret this?

 

[ehild]

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.

 

[gracy]

Thanks for all your answers.You really helped me.Have I calculated h correctly in my post #59?

 

[ehild]

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.

 

[ehild]

Isn't the maximum height of the block of initial velocity u is h_max = 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.

 

[Yoonique]

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.

 

[AlephNumbers]

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

 

[gracy]

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.

 

[ehild]

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.

 

[ehild]

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.

 

[ehild]

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.

 

[gracy]

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.

 

[gracy]

The problem was poorly worded.

I have given my best.

 

[ehild]

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)

 

[ehild]

I have given my best.

Well, your teacher had to be more clear.