To move a block up to the top of the wedge

[rudransh verma]

Homework Statement

A block of mass m is pushed towards a movable wedge of mass $\eta m$ and height h, with a velocity u. All surfaces are smooth. The minimum u for which the block will reach the top of the wedge is

Relevant Equations

KE=1/2mv^2

I think on top of the wedge the KE of both the wedge and block will be same but this fact doesn't take me anywhere. The base length of the wedge is not given. Maybe that would have helped.

 

[haruspex]

I think on top of the wedge the KE of both the wedge and block will be same

Do you have some basis for thinking that? Wild guesses waste time and get you no credit.
Draw a diagram, define some variables, write some equations. You should no longer need to be told.

 

[rudransh verma]

Do you have some basis for thinking that? Wild guesses waste time and get you no credit.
Draw a diagram, define some variables, write some equations. You should no longer need to be told.

I did. Here it is.
$F_a-mg\sin \theta=ma$
$F_a$ as applied force
What is slant distance of the wedge s?
If we know that then
$(F_a-mg\sin \theta)s=\frac12mv^2-\frac12mu^2$
v as final velocity.

 

[Steve4Physics]

Sorry, I hit the 'Post reply' button by accident before I'd finished my reply. So I deleted the post. I'll try again...

You haven't defined your variables. For example you use ‘$F_a$’ in your post but the diagram doesn’t show it. Your diagram shows ‘$F_{applied}$’. Are these meant to be the same? If so, you shouldn’t use different names for the same quantity. It causes confusion.

But, more importantly, I think the question (and therefore your diagram) is wrong. And using $\eta$ creates a readability problem. I’ll use $\alpha$ instead of $\eta$.

Is the question word-for-word as printed in your book? Or have your reworded it?

I’m guessing the real question is this:
1. A block, mass m, moves at speed u on a horizontal frictionless surface.
2. The block meets the thin end of a stationary frictionless wedge resting on the surface. The wedge’s mass is $\alpha$m and its height is h.
3. The block slides up the wedge, simultaneously making the wedge move.
4. Find an expression for the minimum value of u for the block to reach the top of the wedge.

Hints (assuming the above is the correct question): Don’t try to solve this using forces. Use conservation of momentum and conservation of energy.

 

[russ_watters]

I think on top of the wedge the KE of both the wedge and block will be same but this fact doesn't take me anywhere.

What advice did I give you in your thread yesterday? That several people gave you and then got annoyed when you ignored us...?

At this point you should be able to do better than this. You should be using an energy method here.

 

[russ_watters]

I would use the following equation to solve this problem.

$W_{net}= \Delta KE + \Delta PE$, where $W_{net}$ is work done by applied forces (excluding force of gravity)

The equation that I gave is a general form of work energy theorem.

 

[rudransh verma]

Is the question word-for-word as printed in your book?

Yes. I too think what is the meaning of $\eta$ here. Seems strange.
I don't really understand the question but if the block and the wedge both are moving initially with u then it would require more velocity v than u to reach the top and so requires a force.

At this point you should be able to do better than this. You should be using an energy method here.

I was going to ask about it since I don't know where it came from. I have not read about it in my book resnik. And so I will not use it since then you will say "don't do guess work".

 

[rudransh verma]

3. The block slides up the wedge, simultaneously making the wedge move.

"A block of mass m is pushed towards a movable wedge of mass ηm and height h, with a velocity u."
I think it says the wedge is moving with u velocity. There is a "comma" after "height h"

 

[Steve4Physics]

"A block of mass m is pushed towards a movable wedge of mass ηm and height h, with a velocity u."
I think it says the wedge is moving with u velocity. There is a "comma" after "height h"

I read it differently because of the comma! But it's ambiguous.

However, whichever object has velocity u, the overall method is the same.

Also, I don't like the phrase 'pushed towards'. It could imply that there is an external force continuously acting on the block. That would make the problem impossible as we are not told the force's magnitude or direction.

I believe the intended meaning is that the block is given an initial push and then continues sliding (with no force) at constant velocity (which I believe is u) until it meets the wedge.

 

[rudransh verma]

I believe the intended meaning is that the block is given an initial push and then continues sliding (with no force) at constant velocity (which I believe is u) until it meets the wedge.

Whatever you say. $u=\sqrt{2gh}$
But I don't think this is right. I have not read about conservation of momentum. I believe it says Total Momentum before and after the collision is constant. What do you say?

 

[Steve4Physics]

$u=\sqrt{2gh}$

That formula is only valid if ALL is the moving object's kinetic energy is converted to potential energy at the highest point (height h). It can be derived from $\frac 1 2 mu^2 = mgh$.

But in the current problem, some of the moving object's kinetic energy is transferred to another object. So the above formula is not useful here.

Don’t use a formula unless you understand the context in which it is valid.

I have not read about conservation of momentum. I believe it says Total Momentum before and after the collision is constant.

That's right.

Let’s assume the question is as described in Post #4.

This is what happens:

When the moving block (initial velocity u) and stationary wedge meet, the block will slide up the wedge and also the wedge will accelerate.

For some value of u, the block will just reach the top of the wedge. At this instant, the block and wedge will have the have same velocity, say v.

Here is an outline of what you must do.

Using the conservation of momentum. calculate v in terms of u, m and $\alpha$.

If you haven’t learned about conservation of momentum for simple ‘collisions’ you need to read-ahead and/or watch a YouTube video or two.

After you have an expression for v, you can use conservation of energy to find u in terms of m, $\alpha$, h and g.

Post your attempt, showing all your working.

Edit. This Post overlaps @vcsharp2003's Post #11, but since I've already written it, here it is.

 

[rudransh verma]

At this instant, the block and wedge will have the have same velocity, say v.

How do you know this? Did you do experiment?
$mu=mu'+\eta mv$
$v=(u-u')/\eta$ where u' is the velocity of the block after collision.
Now conservation of energy
$(1/2)mu^2=1/2\eta mv^2+mgh$
$(1/2)u^2=(1/2)\eta (u-u')^2/\eta^2+gh$
$(2\eta u^2-2(u-u')^2)/4\eta =gh$
$(\eta u^2-(u-u')^2)/2\eta=gh$
$(u^2-(u-u')^2/\eta=2gh$
$(u^2-\eta v^2)=2gh$
$u^2=(-\eta 2gh+2gh)$
$u=\sqrt{2gh(1-\eta)}$

 

[vcsharp2003]

How do you know this? Did you do experiment?

There is no need of an experiment. The question itself is saying that at the instant block reaches the top, it stops it's upward motion. It therefore means that block + wedge are now moving together with same velocity.

 

[vela]

I was going to ask about it since I don't know where it came from. I have not read about it in my book resnik. And so I will not use it since then you will say "don't do guess work".

What is "it"?

I suggest you stick to problems in the book so you don't waste time trying to solve problems you're not ready for. If this problem had been in the book, it would have been in the chapter on momentum, so presumably you'd have already learned about momentum before attempting the problem.

 

[rudransh verma]

suggest you stick to problems in the book so you don't waste time trying to solve problems you're not ready for.

Yeah.

If this problem had been in the book, it would have been in the chapter on momentum, so presumably you'd have already learned about momentum before attempting the problem.

actually in resnik momentum and it’s conservation are in different chapters. But in my problem book, question are from both under one chapter.

 

[haruspex]

$mu=mu'+\eta mv$
$v=(u-u')/\eta$ where u' is the velocity of the block after collision.

When after the collision? You only care about the point in time at which the block reaches the top. Both velocities will be changing until then.

Now conservation of energy
$(1/2)mu^2=1/2\eta mv^2+mgh$

What happened to the remaining KE of the block?

$u=\sqrt{2gh(1-\eta)}$

Always try to check your answers by considering a special case. What does that result say if the block and wedge have the same mass?

I did. Here it is.

So why didn’t you post it initially?

 

[rudransh verma]

What happened to the remaining KE of the block?

Hmm.

You only care about the point in time at which the block reaches the top. Both velocities will be changing until then.

According to momentum conservation,
$$mu=mv+\eta mv$$
$$u=v(1+\eta)$$
According to conservation of energy,
$$(1/2)mu^2=(1/2)\eta mv^2+(1/2)mv^2+mgh$$
$$u^2=\eta v^2+v^2+2gh$$
After grinding for a while
$$u=\sqrt{2gh(1+\frac1\eta)}$$

 

[rudransh verma]

Both velocities will be changing until then.

Why?
Is it because both bodies will collide. Due to Newtons third law both will exert forces on each other. The block will deaccelerate from u to v while the wedge will accelerate from 0 to v. Both bodies will attain same velocity (given in question). The initial KE will get converted to KEs of the block and the wedge + PE of the block.

 

[haruspex]

Why?
Is it because both bodies will collide. Due to Newtons third law both will exert forces on each other. The block will deaccelerate from u to v while the wedge will accelerate from 0 to v. Both bodies will attain same velocity (given in question). The initial KE will get converted to KEs of the block and the wedge + PE of the block.

Because while the block is sliding up the wedge the normal force between them has a horizontal component.

 

[rudransh verma]

Because while the block is sliding up the wedge the normal force between them has a horizontal component.

I think it should be the component of weight that will slow down the block. The normal component is not in opposite direction to motion of the block.

Due to Newtons third law both will exert forces on each other.

Also is there any action reaction(Newton's third law). I think there is at the instant the block touches the wedge.

 

[Steve4Physics]

I think it should be the component of weight that will slow down the block. The normal component is not in opposite direction to motion of the block.

For a block (B) sliding up a non-accelerating (e.g. fixed) wedge, that’s a sensible way to think about it.

But here the wedge is accelerating. That complicates the problem.

It is simpler to think about the separate vertical and horizontal components of B’s velocity.

B’s weight ($\vec W_B$) acts vertically downwards so it has zero horizontal component.

Therefore B's horizontal acceleration cannot be produced by a component of $\vec W_B$.

B's horizontal acceleration is produced by the horizontal component of the wedge's normal reaction on B.

The net vertical force on B is the vector sum of $\vec W_B$ and the vertical component of the normal reaction of the wedge on B. B's vertical acceleration is produced by this net vertical force.

It should be clear that solving this problem using momentum and energy is simpler than using forces. Part of the skill of physics problem-solving is recognising which approach to take.

Also is there any action reaction(Newton's third law). I think there is at the instant the block touches the wedge.

Yes of course. It is impossible for there not to be. They are:
- the normal reaction of the wedge on B and
- the normal reaction of B on the wedge.

Edit - typo's.

 

[rudransh verma]

That complicates the problem.

It is tough

B’s weight (W→B) acts vertically downwards so it has zero horizontal component.

Therefore B's horizontal acceleration cannot be produced by a component of W→B.

B's horizontal acceleration is produced by the horizontal component of the wedge's normal reaction on B.

The net vertical force on B is the vector sum of W→B and the vertical component of the normal reaction of the wedge on B. B's vertical acceleration is produced by this net vertical force.

By taking the x and y-axis parallel to the slope and perpendicular to the slope of the wedge, I have drawn the fig. Please see. Clearly its the horizontal component of the weight of the block that is slowing down the block.

 

[Steve4Physics]

Clearly its the horizontal component of the weight of the block that is slowing down the block.

The weight acts vertically downwards. Its horizontal component is $mgcos(90º) = 0$! [typo' corrected]

Maybe the problem is that you are thinking about ‘a component of a component’. mgsinθ is the downhill component of weight and contributes to the slowing down. But mgsinθ has no horizontal component.

For example, if an object were moving along the x-axis, then by considering ‘a component of a component’ you could show it also had a component of velocity along the y-axis which is silly!

 

[rudransh verma]

Maybe the problem is that you are thinking about ‘a component of a component’. mgsinθ is the downhill component of weight and contributes to the slowing down. But mgsinθ has no horizontal component.

I think you are taking the standard orientation of coordinate axises x and y. My x-axis is making $\theta$ angle with horizontal assuming the surface of the wedge is also making $\theta$ angle with horizontal . I have drawn the orientation in top right corner. Please look at it.

 

[kuruman]

I think you are taking the standard coordinate axises x and y. My x-axis is making $\phi$ angle with horizontal. I have drawn the orientation in top right corner. Please look at it.

In that case you cannot conserve momentum along the x-direction because the component of the weight in your chosen x-direction is an external force acting on the block.

 

[rudransh verma]

In that case you cannot conserve momentum along the x-direction because the component of the weight in your chosen x-direction is an external force acting on the block.

Wait! There was server issue while I was writing. Now look.

 

[Steve4Physics]

I think you are taking the standard orientation of coordinate axises x and y. My x-axis is making $\theta$ angle with horizontal assuming the surface of the wedge is also making $\theta$ angle with horizontal . I have drawn the orientation in top right corner. Please look at it.

I agree that you can choose any convenient orientation for the xy axes. That's not the issue.

Suppose the block's weight = 100N and θ (angle of slope to horizontal) =30º.

Can you tell us the magnitude of the weight's horizontal component? And how you calculated it?

 

[rudransh verma]

In that case you cannot conserve momentum along the x-direction because the component of the weight in your chosen x-direction is an external force acting on the block.

Do you mean when the block is moving up the wedge.
I have two question.
1) Does the momentum conservation takes place from the moment the block touches the wedge to the top because I have written the eqn like that? It’s quite different from the conservation of the two balls colliding.
2) The downward mg does no effect on the motion of the block but the horizontal normal force does. Why?

 

[kuruman]

Do you mean when the block is moving up the wedge.
I have two question.
1) Does the momentum conservation takes place from the moment the block touches the wedge to the top because I have written the eqn like that? It’s quite different from the conservation of the two balls colliding.
2) The downward mg does no effect on the motion of the block but the horizontal normal force does. Why?

1. Linear momentum of the two-mass system (block + wedge) is conserved in the horizontal direction as long as there is no horizontal force acting on this system. "Conserved" means that it is the same no matter when you calculate it. Its value before the block touches the wedge is the same as its value when the block is moving up the wedge. Before the block touches the wedge, all the horizontal momentum is carried by the block. When the block moves up the wedge, that initial value is divided between the block and the wedge but the sum of the two parts is the same.

2. I cannot fully understand guess what you are asking here. If you isolate the block and call that the system, then its momentum is not conserved because it is acted upon by the normal force exerted by the block wedge and the vertical force of gravity exerted by the Earth. Its acceleration is dictated by both these forces and I don't see why you exclude the weight.

On edit: Typo correction. Crossed out "block" and "added" wedge in item 2.

 

[rudransh verma]

1. Linear momentum of the two-mass system (block + wedge) is conserved in the horizontal direction as long as there is no horizontal force acting on this system. "Conserved" means that it is the same no matter when you calculate it. Its value before the block touches the wedge is the same as its value when the block is moving up the wedge. Before the block touches the wedge, all the horizontal momentum is carried by the block. When the block moves up the wedge, that initial value is divided between the block and the wedge but the sum of the two parts is the same.

Ok!

2. I cannot fully understand guess what you are asking here.

I am asking if the block is moving up the wedge then it is acted upon by the downward weight and the horizontal normal force. You said the blocks speed will slow down because of the horizontal normal force even though the direction of the force is not opposite to the motion of the block.
If you are right then why does mg have no effect on the motion of the block. Why doesn’t mg also slow down the block?

 

[haruspex]

The downward mg does no effect on the motion of the block but the horizontal normal force does. Why?

The "mg" of the block acts on the block, not on the wedge. The vertical force on the wedge is the vertical component of the normal force from the block plus the "mg" of the wedge, and this is exactly balanced by the normal force from the ground.

 

[rudransh verma]

The "mg" of the block acts on the block, not on the wedge. The vertical force on the wedge is the vertical component of the normal force from the block plus the "mg" of the wedge, and this is exactly balanced by the normal force from the ground.

I am asking shouldn’t the force be in the direction or opposite to the direction of the motion in order to work?

 

[haruspex]

I am asking shouldn’t the force be in the direction or opposite to the direction of the motion in order to work?

That's the horizontal component of the normal force, obviously.

 

[rudransh verma]

That's the horizontal component of the normal force, obviously.

I am not able to understand how can only the normal horizontal force can slow the block. The block is not just moving in -x direction but also in the +y direction.I have drawn the fig of the two instants of time 1 and 2. Clearly the block is also moving along y axis. Normal and weight mg ,both should slow down the block.

 

[haruspex]

I am not able to understand how can only the normal horizontal force can slow the block. The block is not just moving in -x direction but also in the +y direction.I have drawn the fig of the two instants of time 1 and 2. Clearly the block is also moving along y axis. Normal and weight mg ,both should slow down the block.

Sorry, I got confused. I thought you were asking about the acceleration of the wedge.
For the block, there is a horizontal component of the normal force slowing the horizontal component of the block's velocity, while in the vertical direction there is the block's weight and the vertical (upward) component of the normal force. The weight has the greater magnitude, so the vertical velocity also reduces.

 

[rudransh verma]

Sorry, I got confused. I thought you were asking about the acceleration of the wedge.
For the block, there is a horizontal component of the normal force slowing the horizontal component of the block's velocity, while in the vertical direction there is the block's weight and the vertical (upward) component of the normal force. The weight has the greater magnitude, so the vertical velocity also reduces.

Yeah! Thank you!

In that case you cannot conserve momentum along the x-direction because the component of the weight in your chosen x-direction is an external force acting on the block.

And there will be no violation of momentum conservation in x direction but in y direction we cannot apply this law because there is mg and by the way we don’t have to.

 

[kuruman]

I am asking if the block is moving up the wedge then it is acted upon by the downward weight and the horizontal normal force. You said the blocks speed will slow down because of the horizontal normal force even though the direction of the force is not opposite to the motion of the block.

Draw a correct free body diagram of the block and you will see that the horizontal component of the normal force on the block wedge is opposite to the motion of the block. While you'e at it, you might as well draw a free body diagram of the wedge.

If you are right then why does mg have no effect on the motion of the block. Why doesn’t mg also slow down the block?

Where did I say that mg has no effect on the motion of the block? In post #29 I wrote

If you isolate the block and call that the system, then its momentum is not conserved because it is acted upon by the normal force exerted by the block wedge and the vertical force of gravity exerted by the Earth[/color]. Its acceleration is dictated by both these forces and I don't see why you exclude the weight.

How does the "vertical force of gravity exerted by the Earth" differ from the weight also known as mg?

 

[kuruman]

And there will be no violation of momentum conservation in x direction but in y direction we cannot apply this law because there is mg and by the way we don’t have to.

That would be true if the x direction is the same as the horizontal direction. If that is what you think, you are contradicting yourself. In post #22 you wrote

By taking the x and y-axis parallel to the slope and perpendicular to the slope of the wedge, I have drawn the fig. Please see. Clearly its the horizontal component of the weight of the block that is slowing down the block.

Unless the angle of the incline is zero, the horizontal direction and the direction parallel to the slope cannot be the same. Furthermore, you mention "the horizontal component of the weigh". You seem to be unaware that the weight does not have a horizontal component and that's serious. The horizontal component of the weght is zero no matter how you choose to orient the x and y axes. That's because the direction of the weight serves as the definition of "vertical". Have you ever wondered how people, since ancient times, were able to build vertical walls on sloped hillsides? How did they figure out which way is vertical?

 

[rudransh verma]

That would be true if the x direction is the same as the horizontal direction. If that is what you think, you are contradicting yourself. In post #22 you wrote

Unless the angle of the incline is zero, the horizontal direction and the direction parallel to the slope cannot be the same. Furthermore, you mention "the horizontal component of the weigh". You seem to be unaware that the weight does not have a horizontal component and that's serious. The horizontal component of the weght is zero no matter how you choose to orient the x and y axes. That's because the direction of the weight serves as the definition of "vertical". Have you ever wondered how people, since ancient times, were able to build vertical walls on sloped hillsides? How did they figure out which way is vertical?

I got it. Thank you to you too.

 

[pbuk]

There has been a big diversion in this thread considering the angle of the block and the dynamics applying while the block is moving up the ramp. These are irrelevant and do not lead to an answer to the initial question so for posterity:

[Edited for clarity]
Homework Statement:: A block of mass $m$ moves with initial velocity $u$ towards a movable wedge of mass $\eta m$ and height $h$ that is initially stationary. All surfaces are smooth.

What is the minimum value of $u$ for which the block will reach the top of the wedge?

The key to answering this question lies in realising that if we are looking for the minimum initial velocity then we are looking for the value which means the block just reaches the top of the wedge i.e. it stops traveling up the wedge when it reaches the top. This also means that the block stops traveling along the wedge i.e. the velocity of the wedge and the block at the final position are equal; let's call this $v$.

We can now write two equations using (i) conservation of momentum and (ii) conservation of energy which we can solve to find the two unknowns $u$, which is the answer to the question, and $v$, which we don't need.

And that's it - no free body diagram, no inventing parameters that are not in the question such as the angle or length of the wedge, no components of forces, no SUVAT equations, none of these will get you any nearer the answer.

 

[nasu]

@pbuk Have you actually tried this method? Is the KE conserved?

 

[kuruman]

@pbuk Have you actually tried this method? Is the KE conserved?

The KE of the block + wedge system is not conserved but the KE + PE (mechanical energy) of the two-mass system is conserved as is the horizontal linear momentum.

 

[haruspex]

There has been a big diversion in this thread considering the angle of the block and the dynamics applying while the block is moving up the ramp. These are irrelevant and do not lead to an answer to the initial question

Quite so, but the OP successfully answered the question, using your approach, in post #17. The digression then occurred because it transpired (post #18) that the OP did not really understand the mechanical process by which the block and wedge arrive at the same speed.

 

[pbuk]

@pbuk Have you actually tried this method?

Yes of course. Have you tried any other method?

Is the KE conserved?

There is no law of conservation of kinetic energy. Total energy is of course conserved (because all surfaces are smooth and there are no external forces), so what other form of energy is involved here?

Quite so, but the OP successfully answered the question, using your approach, in post #17.

Ah yes, I didn't see that - useful then that we refer back to that and the successful method at the end of this thread.

 

[berkeman]

Thread is closed temporarily while the Mentors consider tying it off since the OP's questions have been answered well.

 

[berkeman]

Thread is closed temporarily while the Mentors consider tying it off since the OP's questions have been answered well.

After a Mentor discussion, the thread will remain closed. Thanks all for helping the OP.