Wedge and block initial momentum

[gracy]

Homework Statement

:block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom.
(image is attached)

Homework Equations

:Let the velocity of the triangular wedge be V.And v be the velocity of block m.
[/B]
applying conservation of linear momentum in the horizontal direction we get

MV +m(v cos theta -V)=0
And because of conservation of energy
1/2 MV^2+1/2 m[(v cos theta - V)^2 +v sin theta]^2
The answer comes out to be

The Attempt at a Solution

:I understood all these except conservation of linear momentum part.

MV +m(v cos theta -V)=0
here
MV +m(v cos theta -V) is final momentum
and 0 is taken as initial momentum.
why their initial momentum and velocity is taken as zero?I have learned in constrains, force responsible for motion of wedge is Nsin theta which is always there as long as block and wedge are in contact.And similarly motion of block is because of mg sin theta which should be always there if block is on the wedge.Then which initial condition are you referring to when these forces are not there so that they have zero velocity?I know we are taking blocks and wedge as our system now so normal forces become internal and cancels out.This problem must be a part of kinematics,Kinematics is the branch of classical mechanics which describes the motion of points, bodies (objects) and systems of bodies (groups of objects) without consideration of the causes of motion.

But I am confused why initial momentum is zero?
[/B]

 

[gracy]

The answer comes out to be

The answer comes out to be sorry I am not able to use latex.

 

[gracy]

Above is the answer

 

[gracy]

applying conservation of linear momentum in the horizontal direction we get

MV +m(v cos theta -V)=0

 

[AlephNumbers]

Nothing is moving initially, right?

 

[gracy]

initially,

That's what I want to know.What really is initial condition?When block was not placed on wedge?

 

[ehild]

That's what I want to know.What really is initial condition?When block was not placed on wedge?

The problem text is not clear. Did you copy it correctly?
The initial condition should be both velocities (that of the wedge and the block) and the position of the block. As the velocity of the block is asked when it reaches the bottom of the wedge, it must start from some height h. The simplest initial condition is that the whole system is in rest when the block is at height h.

 

[gracy]

The problem text is not clear

block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom.

Are you talking about this ?

 

[gracy]

The simplest initial condition is that the whole system is in rest when the block is at height h.

But if the block is at height "h"i.e on the block.Then they will not be at rest.Force responsible for motion of wedge is Nsin theta which is always there as long as block and wedge are in contact.And similarly motion of block is because of mg sin theta which should be always there if block is on the wedge.

 

[ehild]

Yes, I talk about your problem. The whole system can be initially in rest if you keep it in rest, and it starts to move when you release it.
By the way, the solution you cited is valid for the initial condition that the block was at height h initially, and the system was in rest.

 

[gracy]

height h initially,

Is this height of wedge?

 

[gracy]

The whole system can be initially in rest if you keep it in rest, and it starts to move when you release it.

What does it mean to keep something in rest and to release it?I often see these terms inh physics problems but I really don't understand how to interpret it?

 

[AlephNumbers]

Imagine that nothing is moving because you are holding the block in place at the top of the wedge. Then imagine letting go of the block, and allowing it to move. That is what it means.

 

[ehild]

Is this height of wedge?

h is the initial height of the block. The wedge can be of any height, but you put the block on the wedge at a point that is at height h above the ground.

 

[AlephNumbers]

Yes, or like ehild says, h can be any arbitrary height on the wedge.

 

[ehild]

What does it mean to keep something in rest and to release it?I often see these terms inh physics problems but I really don't understand how to interpret it?

Imagine you hold a pebble in your hand. It is in rest. It starts to fall when you release it, you let it fall out from your hand.

 

[gracy]

force responsible for motion of wedge is Nsin theta

Am I correct?

motion of block is because of mg sin theta

Am I correct?

 

[gracy]

Imagine that nothing is moving because you are holding the block in place at the top of the wedge. Then imagine letting go of the block, and allowing it to move. That is what it means.

But in the problem nothing such is mentioned that block is holded and then released.

 

[AlephNumbers]

Exactly. That is why I asked for you to elaborate upon whether or not the system is initially at rest. The solution that you provided is only valid under the assumption that the system is initially at rest.

 

[ehild]

What was the original text of the problem?

 

[gracy]

The original text is as follows
block of mass m slides from height h on smooth triangular wedge of mass M kept on smooth floor making theta angle with the horizontal.Find velocity of wedge when block reaches bottom.
Actually this problem was given by my teacher she must have forgot to say that block was released from rest.

 

[ehild]

But if the block is at height "h"i.e on the block.Then they will not be at rest.Force responsible for motion of wedge is Nsin theta which is always there as long as block and wedge are in contact.And similarly motion of block is because of mg sin theta which should be always there if block is on the wedge.

You should place that block on to the wedge at some time. Before that, the wedge is in rest. When you place the block on it, the block is also in rest. The whole system starts to move when you release the block.
Before you release the block, your force also acts, and the net force is zero.

 

[ehild]

The original text is as follows
block of mass m slides from height h on smooth triangular wedge of mass M kept on smooth floor making theta angle with the horizontal.Find velocity of wedge when block reaches bottom.

You see, height h is mentioned. The motion starts when the block is at height h.

 

[gracy]

You see, height h is mentioned. The motion starts when the block is at height h.

Hmmmm...you are right.Please can you help me the second pat of this problem?

 

[ehild]

You wrote the equation concerning the momentum of the system. Conservation of energy also can be used. You made little mistakes in both equations in the OP. Check the signs and parentheses.

 

[gracy]

You wrote the equation concerning the momentum of the system. Conservation of energy also can be used.

With the help of both of these principles I solved for v.And my answer is in post 3.

 

[gracy]

MV +m(v cos theta -V)=0

- MV +m(v cos theta -V)=0
is it correct now?

 

[ehild]

- MV +m(v cos theta -V)=0
is it correct now?

Now it is correct.

 

[gracy]

1/2 MV^2+1/2 m[(v cos theta - V)^2 +v sin theta]^2

mgh=1/2 MV^2+1/2 m[(v cos theta - V)^2 +(v sin theta)^2]
is it correct now?

 

[ehild]

With the help of both of these principles I solved for v.And my answer is in post 3.

You said v was the speed of the block with respect to the wedge. Is the formula in Post#3 really for v? The question is the speed of the wedge, V.

 

[ehild]

mgh=1/2 MV^2+1/2 m[(v cos theta - V)^2 +(v sin theta)^2]
is it correct now?

Yes, it is correct.

 

[gracy]

Is the formula in Post#3 really for v?

No.Sorry.Thankfully Typo.The formula is fo V.I forgot to press caps lock.

 

[gracy]

This question has one more part.
If the block comes down(i.e block is now no longer on wedge)what is the velocity of wedge?

 

[gracy]

 

[gracy]

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)

 

[AlephNumbers]

What forces act on the wedge after the block is no longer on it?

 

[AlephNumbers]

Yes, you are correct. The velocity should not only be constant, but it should be the same velocity that you solved for.

 

[gracy]

But as per my teacher I am not right.

 

[AlephNumbers]

Perhaps I am mistaken then. The surface that the wedge is on is frictionless, yes?

 

[gracy]

The surface that the wedge is on is frictionless, yes

Yes.

 

[AlephNumbers]

Oh, right. The linear momentum has to be equal to zero, right? The block will have a velocity, but it will be directed at an angle of 0 degrees.

 

[gracy]

The linear momentum has to be equal to zero, right?

Along which axis?

 

[AlephNumbers]

The x-axis is what we are worried about here. You do not know the exact details of the separation of the block and wedge, so consider what would happen if the block simply collided with a frictionless surface at an angle, with a velocity.

 

[gracy]

MY teacher gave me these two equations
- MV+mv=0
mgh=1/2 MV^2 +1/2 mv^2
Solving for these two I got an answer

 

[gracy]

Solving for these two I got an answer

 

[gracy]

collided with a frictionless surface

I didn't understand.

 

[AlephNumbers]

The motion of the block is directed down the incline before it reaches the bottom of the wedge. Then the block hits the frictionless surface at an angle θ, with a velocity v. What happens then?

 

[AlephNumbers]

You have the right equations, but solving the second part of this problem requires a conceptual understanding of frictionless surfaces.

 

[gracy]

You have the right equations

And what about answer.Is it right or wrong?

 

[ehild]

You can assume that the transition from the slope to the horizontal track is smooth, that is, the block will not bounce. This means that the angle theta gradually decreases to zero. That also means that the velocity of the wedge will change.
Conservation of momentum and energy applies, they are the same, as they were while the block moved on the wedge. If V is the backward speed of the wedge and v is the forward speed of the block after separation, the total momentum is still zero and the total energy is mgh, as it was at the instant when the motion started.
The equations in Post #44 are correct.