Kinetic & Potential Energy / Momentum Problem

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SUMMARY

The problem involves a small cube of mass m sliding down a circular path of radius R on a large block of mass M, both initially at rest and moving without friction. The velocity v of the cube as it leaves the block is derived using energy conservation principles. The correct expression for v when m equals M is v=√(gR), while the initial energy of the small cube is given by potential energy PEm=mgR, leading to the kinetic energy equation KEm=(1/2)mv². The final velocity is calculated as v=√(2gR), but the conservation of momentum must also account for the kinetic energy of the large block M.

PREREQUISITES
  • Understanding of kinetic energy (KE) and potential energy (PE) concepts.
  • Familiarity with conservation of momentum principles.
  • Basic algebra skills for solving equations.
  • Knowledge of gravitational acceleration (g) and its application in physics problems.
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  • Study the principles of energy conservation in mechanical systems.
  • Learn about momentum conservation in collisions and interactions.
  • Explore the effects of frictionless surfaces on motion dynamics.
  • Review advanced algebra techniques for solving physics equations.
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Students studying classical mechanics, physics educators, and anyone interested in solving problems related to kinetic and potential energy, as well as momentum in frictionless systems.

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Homework Statement


From Kleppner and Kolenkow:
A small cube of mass m slides down a circular path of radius R cut into a large block of mass M, as shown. M rests on a table, and both blocks move without friction. The blocks are initially at rest, and m starts from the top of the path.
Find the velocity v of the cube as it leaves the block.

Ans. clue. If m=M, v=√(gR)

http://img390.imageshack.us/img390/5575/blocksbi2.jpg

Homework Equations


KE=(1/2)mv2
PE=mgh
mivi=mfvf
KEi=KEf


The Attempt at a Solution


For m, initial energy PEm=mgR
Just before m leaves block M, KEm=(1/2)mv2=mgR
Thus, v=√(2gR).
From here, I set up equations for conservation of momentum as well as kinetic energy.
I then used substitution and tried to solve for the velocity of m as it leaves the block, M, but the algebra just isn't working out. Did I make a mistake?
 
Last edited by a moderator:
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Do not ignore the KE of the large block. ehild
 

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