Gravitation & Conservation of Momentum and Energy

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SUMMARY

The discussion focuses on calculating the velocities of two spheres with masses M and 2M, and radii R and 3R, released from a distance of 12R. The gravitational force is described by the equation Fg = Gm1m2/r^2, and the conservation of energy is applied to derive the equation for the change in energy. The final energy equation is set to zero due to the absence of non-conservative forces. The participants seek clarification on expressing the velocity of the mass M (v1) in terms of the velocity of the mass 2M (v2) without including the radius R in the final expression.

PREREQUISITES
  • Understanding of gravitational force equations (Fg = Gm1m2/r^2)
  • Knowledge of conservation of energy principles (E = K + U)
  • Familiarity with momentum conservation equations (Mv1i + 2Mv2i = Mv1f + 2Mv2f)
  • Basic algebraic manipulation skills to solve equations for variables
NEXT STEPS
  • Study the derivation of gravitational force equations in classical mechanics
  • Explore detailed examples of conservation of energy in collision problems
  • Learn how to manipulate momentum conservation equations for multi-body systems
  • Investigate the relationship between kinetic energy and potential energy in gravitational systems
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics, as well as anyone interested in understanding the principles of gravitation and momentum conservation in collision scenarios.

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1. Two spheres having masses M and 2M and radii R and 3R, respectively, are simultaneously released from rest when the distance between their centers is 12R. Assume the two spheres interact only with each other and we wish to find the speeds with which they collide. Write an equation from one of the models and solve it for v1, the velocity of the sphere of mass M at any time after release in terms of v2, the velocity of 2M.

2. Fg = Gm1m2/r^2

Change in E = E

3. Do I just use conservation of energy?

E = K + U

Ei = 0 + [G(M)(2M)]/(R^2)

Ef = 1/2Mv1^2 + 1/2(2M)v2^2 + 0

change in E = [1/2Mv1^2 + 1/2(2M)v2^2] - [G(M)(2M)]/(R^2)
This equals 0 since there are no nonconservation forces doing work.
change in E= Wnc = 0

My answer is sqrt(M/36R - 2v2^2)

This is confusing me because it asks for v1 in terms of v2 and I have R in the answer. How do I solve it without R?
 
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I also have to solve this using conservation of momentum.

Mv1i+2Mv2i = Mv1f + 2Mv2f

I don't get how to solve this for v1 in terms of v2 when I don't know initial or final velocities.
 

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