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.
  • #1
gracy
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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]
 

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  • #2
gracy said:
The answer comes out to be
The answer comes out to be sorry I am not able to use latex.
 
Last edited:
  • #3
upload_2015-4-6_11-51-29.png

Above is the answer
 
  • #4
gracy said:
applying conservation of linear momentum in the horizontal direction we get

MV +m(v cos theta -V)=0
WEDGE MOTION.jpg
 
  • #5
Nothing is moving initially, right?
 
  • #6
AlephNumbers said:
initially,
That's what I want to know.What really is initial condition?When block was not placed on wedge?
 
  • #7
gracy said:
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.
 
  • #8
ehild said:
The problem text is not clear
gracy said:
block slides on smooth triangular wedge kept on smooth floor.Find velocity of wedge when block reaches bottom.
Are you talking about this ?
 
  • #9
ehild said:
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.
 
  • #10
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.
 
  • #11
ehild said:
height h initially,
Is this height of wedge?
 
  • #12
ehild said:
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?
 
  • #13
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.
 
  • #14
gracy said:
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.
 
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  • #15
Yes, or like ehild says, h can be any arbitrary height on the wedge.
 
  • #16
gracy said:
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.
 
  • #17
gracy said:
force responsible for motion of wedge is Nsin theta
Am I correct?
gracy said:
motion of block is because of mg sin theta
Am I correct?
 
  • #18
AlephNumbers said:
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.
 
  • #19
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.
 
  • #20
What was the original text of the problem?
 
  • #21
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.
 
  • #22
gracy said:
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.
 
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  • #23
gracy said:
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.
 
  • #24
ehild said:
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?
 
  • #25
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.
 
  • #26
ehild said:
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.
 
  • #27
gracy said:
MV +m(v cos theta -V)=0
- MV +m(v cos theta -V)=0
is it correct now?
 
  • #28
gracy said:
- MV +m(v cos theta -V)=0
is it correct now?
Now it is correct.
 
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  • #29
gracy said:
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?
 
  • #30
gracy said:
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.
 
  • #31
gracy said:
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.
 
  • #32
ehild said:
Is the formula in Post#3 really for v?
No.Sorry.Thankfully Typo.The formula is fo V.I forgot to press caps lock.
 
  • #33
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?
 
  • #34
velocity.jpg
 
  • #35
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)
 

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