Max Velocity of Wedge: When Block Reaches Height h

[gracy]

Homework Statement

:There is a wedge of mass kept on smooth surface. M and block of mass m is sliding in the depression of the wedge.And there is wall in left of the wedge.At which instant velocity of the wedge will be maximum?(surface of contact between wedge and block is smooth)
[/B]

Homework Equations

But the horizontal component of normal force is balanced by the wall reaction and vertical force is balanced by normal reaction on wedge .So no motion of wedge.So
conservation of energy as no dissipative force is there.
blocks potential energy is converted into it's kinetic energy completely.

The Attempt at a Solution

:
But when it reaches at this point

After this point the

When the block reaches at height h and the block height is achieved at this point,Vertical force is still balanced but horizontal force is not balanced by any force as wall is not on right hand side.
So block accelerate.Now the block kinetic energy at mid point let's say 1/2 mv^2 of the depression is not completely converted to it's potential energy.As some goes in kinetic energy of the wedge.

My teacher says velocity of the wedge will be maximum at the below instant

I am not getting why?As can be seen from the image no net force is acting on thee wedge at this instant.Because the only force acting is in vertical direction which is balanced by normal force.I think velocity of the wedge should be maximum at

As in this the only force acting is in horizontal so maximum force ,maximum acceleration hence maximum velocity.

[/B]

 

[Yoonique]

You meant the maximum velocity of wedge or the block?

 

[gracy]

You meant the maximum velocity of wedge or the block?

maximum velocity of wedge

 

[gracy]

Some errors were there in my original post.I am correcting them
1. Homework Statement :There is a wedge of mass kept on smooth surface. M and block of mass m is sliding in the depression of the wedge.And there is wall in left of the wedge.At which instant velocity of the wedge will be maximum?(surface of contact between wedge and block is smooth)
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64652-5de495f229ba4bd42a727d2570e0f6fc.jpg

2. Homework Equations
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64654-a429551536e8d18cafe546da0c38a3dd.jpg
But the horizontal component of normal force is balanced by the wall reaction and vertical force is balanced by normal reaction on wedge .So no motion of wedge.So
conservation of energy as no dissipative force is there.
blocks potential energy is converted into it's kinetic energy completely.
3. The Attempt at a Solution :

https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64659-5168d6ca9d612915662a0111d957a41e.jpg
When the block reaches at height h
as in the below image(I forgot to specify height "h"in my picture.) the block maximum height in right hand side is "h" ,Vertical force is still balanced but horizontal force is not balanced by any force as wall is not on right hand side.
So block accelerate.Now the block kinetic energy at mid point let's say 1/2 mv^2 of the depression is not completely converted to it's potential energy.As some goes in kinetic energy of the wedge.

My teacher says velocity of the wedge will be maximum at the below instant
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64656-7d13d77e1bad91857f1a6f7eca50d4bb.jpg
I am not getting why?As can be seen from the image no net force is acting on thee wedge at this instant.Because the only force acting is in vertical direction which is balanced by normal force.

 

[Yoonique]

Using conservation of momentum, ΣP_i = ΣP_f.
ΣP_i = 0
0= MV + mv
V= -mv/M
If V = V_max, then V_max = -mv_max/M
So now the question is when is v=v_max. The maximum velocity of the block is when the block has maximum kinetic energy and minimum potential energy by conservation of energy. So that is the point where the block is at the lowest point in the depression.
Since there is a wall left side of the wedge, when the wedge has a maximum velocity to the left (block has maximum velocity to the right), there is a normal force by the wall on wedge to decelerate it to 0. So the wedge has V=0. But when the wedge has a maximum velocity to the right, there is no wall on the right side of the wedge thus it will move to the right at maximum velocity which happens when the block has a maximum velocity to the left.

Or you can see it in this way: V = V_max, then V_max = -mv_max/M
Taking right as positive. Since the wedge has to move to the right, V_max must be positive so v_max must be negative which means the block has to move to the left with v_max.

 

[haruspex]

My teacher says velocity of the wedge will be maximum at the below instant
https://physicsforums-bernhardtmediall.netdna-ssl.com/data/attachments/64/64656-7d13d77e1bad91857f1a6f7eca50d4bb.jpg
I am not getting why?As can be seen from the image no net force is acting on thee wedge at this instant.Because the only force acting is in vertical direction which is balanced by normal force.

A moment before the situation shown, where was the block in relation to the wedge? What direction would have been its force on the wedge? Which way would the wedge have been accelerating?
Same questions for a moment after the position shown.

 

[gracy]

A moment before the situation shown, where was the block in relation to the wedge?

What direction would have been its force on the wedge?

Which way would the wedge have been accelarting?

In the right direction.

 

[haruspex]

In the right direction.

Right. So had it reached max speed then?
What about the situation a moment after that in the diagram?

 

[gracy]

in the diagram?

Which diagram?This one?

 

[haruspex]

Which diagram?This one?

Yes.

 

[gracy]

What about the situation a moment after that in the diagram?

The wedge doesn't accelerate any more as force is no longer there so it moves with constant velocity.Right?

 

[haruspex]

The wedge doesn't accelerate any more as force is no longer there so it moves with constant velocity.Right?

In your teacher's diagram, the block is at the midpoint of the depression and moving left relative to the wedge. Where will it be relative to the wedge a moment later? What direction will its force on the wedge be?

 

[gracy]

Where will it be relative to the wedge a moment later

What direction will its force on the wedge be

One Component of normal force on wedge by block is In the left direction but it will be balanced by normal force by wall and the ground balances the other component of normal force on wedge by block.

 

[haruspex]

View attachment 81621

One Component of normal force on wedge by block is In the left direction but it will be balanced by normal force by wall and the ground balances the other component of normal force on wedge by block.

In the position in your teacher's diagram, it is no longer in contact with the wall.

 

[gracy]

it is no longer in contact with the wall.

Yes. I got your point .Wedge is no longer in contact with the wall,because the wedge has moved further.

What direction will its force on the wedge be?

In the left direction, so it will decelerate the wedge.

 

[gracy]

But why not the wedge has maximum velocity at this position

This is the moment just before the instant when block comes to mid point of depression with velocity in left direction.i.e

 

[gracy]

Actually I am not getting what this theta (angle) is?Is horizontal component Nsin theta or N cos theta?

 

[haruspex]

But why not the wedge has maximum velocity at this position
View attachment 81674
This is the moment just before the instant when block comes to mid point of depression with velocity in left direction.i.e

As long as the block is to the right of the midpoint of the depression, it is exerting a force to the right on the wedge. So the wedge is accelerating to the right, and is at less than its maximum speed. Maximum speed occurs when acceleration is zero.

 

[gracy]

Maximum speed occurs when acceleration is zero.

As long as the block is to the right of the midpoint of the depression, it is exerting a force to the right on the wedge.I wanted to ask is the force which causes acceleration of wedge varies as the block moves ?

 

[gracy]

here will it be relative to the wedge a moment later? What direction will its force on the wedge be?

In the left direction, so it will decelerate the wedge.

Is it right?

 

[haruspex]

Is it right?

Yes.

 

[gracy]

Please answer my post 319.It will help me to solve problem.

 

[haruspex]

I wanted to ask is the force which causes acceleration of wedge varies as the block moves ?

Not sure I understand the question. The force causing the acceleration of the wedge is the horizontal component of the normal force between block and wedge. That varies, of course.

 

[gracy]

That varies, of course

Why?Because theta changes?

 

[haruspex]

Why?Because theta changes?

Yes, and also because the normal force changes.

 

[gracy]

because the normal force changes.

Why normal force changes?

 

[haruspex]

Why normal force changes?

Because the accelerations of the wedge and block change, and because the angle changes.
A full analysis of the dynamics would be a bit of a challenge, and I suspect wll beyond what you are expected to do here. The question as posed is answered by the simple logic I've outlined. The wedge will be at max and min speeds when there is no horizontal force on it, and that will be when the block is at the lowest point. By momentum conservation, if the block is moving to the left relative to the wedge it will be the max, if to the right it will be the min.

 

[gracy]

The wedge will be at max and min speeds when there is no horizontal force on it,

Because as long force is there ,acceleration will also be there and so velocity will go on increasing and there will not be any instant when we can say that it is maximum velocity that's why when there is no horizontal force no acceleration as well as deceleration ,at that moment velocity will be maximum and that is certainly the midpoint of depression.Right?One more thing
Velocity of wedge at mid point of depression is equal to velocity of wedge just before coming at mid point of depression when there was horizontal force acting on wedge.

 

[gracy]

Please tell me my post #28 is correct or not?

 

[haruspex]

Velocity of wedge at mid point of depression is equal to velocity of wedge just before coming at mid point of depression when there was horizontal force acting on wedge.

I don't understand. If the block is not at the middle of the depression then the speed of the wedge is changing.

 

[gracy]

Because as long force is there ,acceleration will also be there and so velocity will go on increasing and there will not be any instant when we can say that it is maximum velocity that's why when there is no horizontal force no acceleration as well as deceleration ,at that moment velocity will be maximum and that is certainly the midpoint of depression.Right?

Is this correct?

 

[gracy]

If the block is not at the middle of the depression then the speed of the wedge is changing.

But it must be having some instantaneous velocity just before coming to mid point of depression.

 

[haruspex]

But it must be having some instantaneous velocity just before coming to mid point of depression.

Yes... so?

 

[gracy]

But it must be having some instantaneous velocity just before coming to mid point of depression.

But it must be having some instantaneous velocity just before the block comes to mid point of depression.[/QUOTE]By it I mean wedge.

 

[haruspex]

But it must be having some instantaneous velocity just before the block comes to mid point of depression.

By it I mean wedge.[/QUOTE]
I agreed the wedge would have some instantaneous velocity before the block reaches the mid point. But it will not be quite as much as when the block is at the mid point.