Why Is My Velocity Calculated Differently than the Solutions?

[ChiralSuperfields]

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.

 

[ChiralSuperfields]

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