Why Is My Velocity Calculated Differently than the Solutions?

[member 731016]

Hi!

For this part (e) of this problem,

Source: https://ocw.mit.edu/courses/8-01sc-...647ea989a352a972dc4b3dfe_MIT8_01F16_pset7.pdf

The solutions are,

However, I don't understand why they only used a component of the initial velocity as it comes off the incline. I used 4.38 m/s because I thought that once the block reaches the horizontal surface, the vertical and horizontal components of the velocity from the incline would combine to give 4.38 m/s. Do you please know why or whether the solutions are wrong?

Many thanks!

 

[PeterDonis]

Moderator's note: Thread moved to intro physics homework forum.

 

[PeterDonis]

once the block reaches the horizontal surface, the vertical and horizontal components of the velocity from the incline would combine

I'm not sure what you mean by "combine". You might try thinking in terms of what property of the block is unchanged across the transition from the inclined plane to the horizontal segment.

The solutions

Where are you getting these from?

 

[PeterDonis]

You might try thinking in terms of what property of the block is unchanged across the transition from the inclined plane to the horizontal segment.

A hint to this is that the problem statement never asks you for the speed of the block. It does ask for something else at the transition. What?

 

[member 731016]

I'm not sure what you mean by "combine". You might try thinking in terms of what property of the block is unchanged across the transition from the inclined plane to the horizontal segment.Where are you getting these from?

Hi Peter,

Thank you for your replies. I got the solutions from this guy's Github - .

Many thanks,
Callum

 

[PeterDonis]

I got the solutions from this guy's Github

You might want to try solving the problems yourself to check whether that source is reliable. (Hint: I don't think it is.)

 

[member 731016]

You might want to try solving the problems yourself to check whether that source is reliable. (Hint: I don't think it is.)

Hi Peter,

Yeah, I'm not sure, I did look at some of the other solutions which did seemed correct to me. The distance I calculated was 5 m which was using the velocity of the block at the bottom of the incline to be 4.38 m/s. Would you please know whether my solution is correct?

Many thanks,
Callum

 

[PeterDonis]

using the velocity of the block at the bottom of the incline to be 4.38 m/s

Why do you think that's correct? Not "because that's what the solution I read says". How would you do the calculation?

(Hint: you don't even need to calculate the block's velocity at all to solve the entire problem, all five parts. All you need is the fact that the work done by a force over a distance is the force times the distance.)

 

[PeterDonis]

(Hint: I don't think it is.)

On looking a little deeper I might have been too quick to dismiss this source as giving wrong numerical answers. However, I still don't think it works very well as a study guide.

 

[PeterDonis]

The distance I calculated was 5 m which was using the velocity of the block at the bottom of the incline to be 4.38 m/s.

Assuming this velocity is correct, how are you getting 5 meters?

 

[member 731016]

Thanks for your reply. I have written up how I got 5 meters:

 

[PeterDonis]

I have written up how I got 5 meters:

Equations in images are not permitted. Please use the PF LaTeX feature to post equations directly. You will see a "LaTeX Guide" link at the bottom left of each post window.

 

[PeterDonis]

I have written up how I got 5 meters

Your write up tells me nothing since you have given no equation that includes the distance $d$. What is the equation for $d$?

 

[PeterDonis]

I have written up how I got 5 meters

Note also that $\mu_k = 0.3$ for the horizontal part of the problem, not $0.2$.

 

[member 731016]

Your write up tells me nothing since you have given no equation that includes the distance $d$. What is the equation for $d$?

Thanks for your reply. I solved it the same way as the 'solutions' apart from the fact that I used the initial velocity to be 4.38 m/s. Many thanks

 

[haruspex]

why they only used a component of the initial velocity as it comes off the incline

The question should have made it clear exactly what happens at the bottom of the ramp.

If it is a smooth change, with a small arc connecting the two flat surfaces, at least some of the vertical component of velocity will get converted to horizontal. However, the object is shown as a block, so its moment of inertia becomes relevant. To avoid its getting quite messy, take it as a point mass at the centre of a light block.
If the radius of curvature of the arc is R its length is Rθ and the normal includes a centripetal component mu²/R, which could be a lot more than mg. This gives a frictional force μmu²/R acting over that distance and taking away μmu²θ of the KE. Note that this cannot be magicked away by making R small.

The solution you post assumes a sudden change, and that the vertical component of momentum is simply lost. This immediately takes away $\frac 12mu^2\sin(\theta)$ of the KE. But the friction complicates matters. There will be an impulse at each end of the block, leading to impulsive friction at each, also instantly reducing the speed.

In short, a problem like this with friction is fiendishly difficult.

 

[PeterDonis]

I solved it the same way as the 'solutions' apart from the fact that I used the initial velocity to be 4.38 m/s.

If you aren't going to respond to my questions, we can just close this thread. Either you're here to get help or you're not.

As I've already said, you should not assume that the "solutions" are correct. You should solve the problem for yourself. You're not even following the problem statement since part d) of that asks you for a symbolic equation that relates $d$ to other parameters. Can you give such an equation?

 

[member 731016]

Note also that $\mu_k = 0.3$ for the horizontal part of the problem, not $0.2$.

Sorry I am quite bad at explaining myself. I happen to be dyslexic so apologies if I am a little slow.

 

[PeterDonis]

I happen to be dyslexic so apologies if I am a little slow.

It's fine to take your time. The key thing is to think through the solution for yourself, without referring to anyone else's attempt at a solution.

 

[PeterDonis]

The question should have made it clear exactly what happens at the bottom of the ramp.

I agree this is a key missing piece in the question. I have been assuming that there would be something in the course notes or lectures that would clarify this (since it's an introductory physics course I would not expect it to go into all of the complexities you describe), but so far I have not found anything.

 

[member 731016]

The question should have made it clear exactly what happens at the bottom of the ramp.

If it is a smooth change, with a small arc connecting the two flat surfaces, at least some of the vertical component of velocity will get converted to horizontal. However, the object is shown as a block, so its moment of inertia becomes relevant. To avoid its getting quite messy, take it as a point mass.
If the radius of curvature of the arc is R its length is Rθ and the normal includes a centripetal component mu²/R, which could be a lot more than mg. This gives a frictional force μmu²/R acting over that distance and taking away μmu²θ of the KE. Note that this cannot be magicked away by making R small.

The solution you post assumes a sudden change, and that the vertical component of momentum is simply lost. This immediately takes away $\frac 12mu^2\sin(\theta)$ of the KE. But the friction complicates matters. There will be an impulse at each end of the block, leading to impulsive friction at each, also instantly reducing the speed.

In short, a problem like this with friction is fiendishly difficult.

Thank you for your reply, yeah it seems a bit dodge to me that that vertical component of the momentum is suddenly loss. If I consider a ball rolling down the incline, I don't think it's vertical component would be lost like that.

Many thanks,
Callum

 

[member 731016]

Note also that $\mu_k = 0.3$ for the horizontal part of the problem, not $0.2$.

Yes sorry, my mistake.

 

[member 731016]

It's fine to take your time. The key thing is to think through the solution for yourself, without referring to anyone else's attempt at a solution.

Ok thank you!

 

[member 731016]

If you aren't going to respond to my questions, we can just close this thread. Either you're here to get help or you're not.

As I've already said, you should not assume that the "solutions" are correct. You should solve the problem for yourself. You're not even following the problem statement since part d) of that asks you for a symbolic equation that relates $d$ to other parameters. Can you give such an equation?

I agree, I cannot assume the "solutions" are correct since they have not been credited by MIT. I did solve part (d) to get the same symbolic equation as the "solutions" which I integrated the force with respect to position using the limits, zero and d.

 

[PeterDonis]

If I consider a ball rolling down the incline, I don't think it's vertical component would be lost like that.

A rolling ball introduces additional complexities, though, because some of the work done on the ball as it rolls down the incline goes into increasing its rate of rotation instead of its linear velocity. This is a fairly common "gotcha" for physics students analyzing experiments with inclined planes like those done by Galileo and getting stumped about the results they are seeing.

 

[PeterDonis]

I did solve part (d) to get the same symbolic equation as the "solutions"

What equation is that? Please post it explicitly here.

 

[member 731016]

Ok let me just take a photo of my working.

 

[member 731016]

 

[PeterDonis]

Ok let me just take a photo of my working.

Again, this is not allowed. Please use the PF LaTeX feature to post equations directly in the thread.

 

[member 731016]

Again, this is not allowed. Please use the PF LaTeX feature to post equations directly in the thread.

Ok thanks for letting me know.

 

[PeroK]

Hi!

For this part (e) of this problem,
View attachment 317728
Source: https://ocw.mit.edu/courses/8-01sc-...647ea989a352a972dc4b3dfe_MIT8_01F16_pset7.pdf

The solutions are,
View attachment 317729
However, I don't understand why they only used a component of the initial velocity as it comes off the incline. I used 4.38 m/s because I thought that once the block reaches the horizontal surface, the vertical and horizontal components of the velocity from the incline would combine to give 4.38 m/s. Do you please know why or whether the solutions are wrong?

Many thanks!

In the simplest terms, when the block reaches the end of the ramp it collides with the ground.

Imagine a car or bicycle coming to the end of a ramp like that. There would be a definite vertical bounce.

And with a block it's not clear what happens when that leading edge hits the ground.

Not a great question, IMO, as clearly you need the simplifying assumption of a simple totally inelastic collision at the bottom of the ramp.

 

[member 731016]

In the simplest terms, when the block reaches the end of the ramp it collides with the ground.

Imagine a car or bicycle coming to the end of a ramp like that. There would be a definite vertical bounce.

And with a block it's not clear what happens when that leading edge hits the ground.

Not a great question, IMO, as clearly you need the simplifying assumption of a simple totally inelastic collision at the bottom of the ramp.

Thanks for your reply, I see what you mean. If there was no friction my solution would be correct I think. Many thanks!

 

[PeroK]

Thanks for your reply, I see what you mean. If there was no friction my solution would be correct I think. Many thanks!

If there were no friction, the block would never stop. Even if the slope and ground were pure ice, there would still be a collision between the leading edge and the horizontal ice. In practical terms, the ice would be chipped and energy lost.

Even with a ball the vertical motion would be translated into a sequence of bounces, rather than horizontal velocity.

 

[haruspex]

Thanks for your reply, I see what you mean. If there was no friction my solution would be correct I think. Many thanks!

No, you are missing the point. The sudden loss of speed on hitting the horizontal surface in the given solution has nothing to do with friction. It is because of the inelastic impact in the vertical direction. And making it elastic doesn’t help; it would bounce, but the vertical momentum and energy it had on the ramp still does not contribute to the horizontal speed after impact.

 

[member 731016]

If there were no friction, the block would never stop. Even if the slope and ground were pure ice, there would still be a collision between the leading edge and the horizontal ice. In practical terms, the ice would be chipped and energy lost.

Even with a ball the vertical motion would be translated into a sequence of bounces, rather than horizontal velocity.

Ok thank you!

 

[member 731016]

No, you are missing the point. The sudden loss of speed on hitting the horizontal surface in the given solution has nothing to do with friction. It is because of the inelastic impact in the vertical direction. And making it elastic doesn’t help; it would bounce, but the vertical momentum and energy it had on the ramp still does not contribute to the horizontal speed after impact.

Thanks for your reply! One question: how is the inelastic impact in the vertical direction? Is it not an angle to vertical since the incline is slanted? Many thanks!

 

[PeroK]

Thanks for your reply! One question: how is the inelastic impact in the vertical direction? Is it not an angle to vertical since the incline is slanted? Many thanks!

The simplifying assumption for this problem is that (at the point where it reaches the horizontal surface) the block loses all its vertical momentum (hence does not bounce) and retains all its horizontal momentum. In other words, a totally inelastic collision in the vertical direction.

How realistic that assumption is is another matter.

 

[haruspex]

how is the inelastic impact in the vertical direction?

The level ground only prevents vertical motion, so only the vertical component of velocity is affected.

 

[PeterDonis]

The simplifying assumption for this problem is that (at the point where it reaches the horizontal surface) the block loses all its vertical momentum (hence does not bounce) and retains all its horizontal momentum. In other words, a totally inelastic collision in the vertical direction.

This is the assumption that is being implicitly made in the solution the OP refers to. However, I don't see it stated anywhere in the problem itself.

 

[kuruman]

Here is my personal take of this problem. One can use the work-energy theorem from the beginning of motion to the end. There are two entities that do work on the object: (a) Earth and (b) surface. The sum of the works is zero because the kinetic energy starts and ends at zero. Thus, one writes $$0=\Delta K=W^{\text{Earth}}+W^{inclined}+W^{flat}$$where the work done by the surface is separated into two pieces because the forces that do this work are different.

One can write expressions for all three forces that do work and assemble the answer. It is safe to assume that the displacement of the object, as it makes the transition from the inclined to the flat surface, is negligible and, therefore, there is no term other than the above three to consider. In my opinion, going through the intermediate step of finding the kinetic energy at the bottom of the incline is unneeded for getting the final answer unless it is meant as an exercise to the student.

 

[PeterDonis]

the work done by the surface is separated into two pieces because the forces that do this work are different

There are actually three "pieces" to the work done by the surface: on the incline, on the flat surface, and at the transition. The specific value of the transition work depends on what assumption is made about the transition, but (except with the highly unrealistic assumption that the transition occurs with no loss of kinetic energy whatever) it is not zero.

 

[PeroK]

Here is my personal take of this problem. One can use the work-energy theorem from the beginning of motion to the end. There are two entities that do work on the object: (a) Earth and (b) surface. The sum of the works is zero because the kinetic energy starts and ends at zero. Thus, one writes $$0=\Delta K=W^{\text{Earth}}+W^{inclined}+W^{flat}$$where the work done by the surface is separated into two pieces because the forces that do this work are different.

Is that not the mistake the OP made?

 

[kuruman]

There are actually three "pieces" to the work done by the surface: on the incline, on the flat surface, and at the transition. The specific value of the transition work depends on what assumption is made about the transition, but (except with the highly unrealistic assumption that the transition occurs with no loss of kinetic energy whatever) it is not zero.

It might be unrealistic, but I think it is the intent of the problem's author to ignore the "third piece" otherwise there would have been a statement about how to model the transition. It's no more unrealistic than having a car pass a truck where the velocities of the vehicles are given but not their lengths. Is a point car needing zero time to pass a point truck more realistic than a point mass losing zero energy during the transition from inclined to horizontal? I think the answer depends on one's tolerance of what is and is not realistic, so I will stop here.

 

[haruspex]

I think it is the intent of the problem's author to ignore the "third piece"

Clearly not, since the official answer considers energy to have been lost in an inelastic collision with the horizontal surface.
Beyond that, it is unacceptable for the student to be expected to guess the author's thinking. The issues I listed in post #16 - smooth or sudden transition to horizontal, impulsive friction if sudden, increase in normal force due to direction change if smooth - may well occur to a good student. And a good student should not be penalised for so being.

 

[PeterDonis]

the official answer

Um, the answer referenced by the OP is not "official". It's not a solution posted by MIT. It's just some random Internet person's solution.

 

[PeterDonis]

it is unacceptable for the student to be expected to guess the author's thinking. The issues I listed in post #16 - smooth or sudden transition to horizontal, impulsive friction if sudden, increase in normal force due to direction change if smooth - may well occur to a good student. And a good student should not be penalised for so being.

I agree, this is something that should have been mentioned either in the problem itself or somewhere in the class materials the student would be expected to use when working the problem.

 

[member 731016]

Thank you: haruspex, PeroK, kuruman and PeterDonis for helping with this problem!