Conservation of Momentum/Mechanical Energy Question (Fairly Challenging)

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Homework Help Overview

The problem involves a small block of mass m moving on a horizontal table and transitioning onto a sloped big block of mass M, which can also move. The scenario assumes no friction and requires finding the speed of the small block after it leaves the slope, utilizing conservation of momentum and mechanical energy principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss setting up conservation of momentum and mechanical energy equations, questioning the correct form and variables involved. There are attempts to derive relationships between the velocities after separation and concerns about the complexity of the resulting equations.

Discussion Status

Some participants have provided guidance on setting up the equations and eliminating variables, while others express uncertainty about the applicability of conservation of kinetic energy due to the nature of the collision. The discussion reflects a collaborative effort to clarify the setup and approach without reaching a definitive conclusion.

Contextual Notes

There are indications of confusion regarding the assumptions of the problem, particularly whether the collision is elastic and how to properly apply the conservation laws. Participants are also navigating the challenge of expressing the final answer in terms of the initial speed v0.

Quotexon
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Homework Statement


A small block of mass m is moving on a horizontal table surface at initial speed v0. It then moves smoothly onto a sloped big block of mass M. The big block can also move on the table surface. Assume that everything moves without friction
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Find the speed v of the small block after it leaves the slope.

Homework Equations


Conservation of Momentum: mv0 = (M+m)v → v=(mv0)/(M+m)
Conservation of Mechanical Energy: fairly obvious
The height that it rises to is h = (1/2g)(Mv0^2/M+m), I derived this from conservation of momentum and conservation of mechanical energy

The Attempt at a Solution


I set up the conservation of momentum: (M + m)v = -mv1 + Mv2 Is this correct?
Now I ask, how should I set up the conservation of mechanical energy equation? Should I solve for one variable in terms of the other? The answer should be in terms of v0.
The answer is v1 = ((M-m)/(M+m))v0
 
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Quotexon said:
Conservation of Momentum: mv0 = (M+m)v → v=(mv0)/(M+m)
Don't assume that m & M have the same velocity.
Conservation of Mechanical Energy: fairly obvious
The height that it rises to is h = (1/2g)(Mv0^2/M+m), I derived this from conservation of momentum and conservation of mechanical energy
Don't worry about the height. That's an intermediate point that we don't care about.

The Attempt at a Solution


I set up the conservation of momentum: (M + m)v = -mv1 + Mv2 Is this correct?
Set initial momentum equal to final momentum. Let v1 and v2 be the final velocities after they separate.
Now I ask, how should I set up the conservation of mechanical energy equation?
Intial KE = final KE
Should I solve for one variable in terms of the other? The answer should be in terms of v0.
You'll have two equations. Eliminate v2 and solve for v1.
 
How would a conservation of kinetic energy equation look like?

1/2(M+m)v^2 = 1/2mv1^2 + 1/2Mv2^2 ?

Conservation of momentum: (M+m)v = -mv1 + Mv2 ?

If I solve for v1 using these equations, it turns out very convoluted and does not resemble the answer. I feel as though it should be cleaner than this. Am I doing something incorrectly?
 
Does this problem even meet the criteria to use conservation of kinetic energy? It doesn't seem to be elastic, does it?

Please help, this problem has been bugging me for awhile.
 
Quotexon said:
How would a conservation of kinetic energy equation look like?

1/2(M+m)v^2 = 1/2mv1^2 + 1/2Mv2^2 ?
OK, except for the left hand side. Initially, only m is moving.

Conservation of momentum: (M+m)v = -mv1 + Mv2 ?
Same issue as above.

If I solve for v1 using these equations, it turns out very convoluted and does not resemble the answer. I feel as though it should be cleaner than this. Am I doing something incorrectly?
Clean up the equations as I suggest and try again. A bit of a pain, but you'll get the required answer with a bit of work.
 
Wow, it worked! Thanks so much for your help. I greatly appreciate it.
 
Excellent! :approve: (And you are most welcome.)
 

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