Inelastic Collisions and ratio of their masses

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SUMMARY

The discussion focuses on solving the problem of finding the mass ratio (m1/m2) of two objects undergoing a totally inelastic collision, where half of the initial kinetic energy is lost. The relevant equations include momentum conservation (m1v1 + m2v2 = (m1 + m2)vf) and kinetic energy considerations. The user attempts to incorporate the loss of kinetic energy into their calculations but struggles to derive the correct mass ratio. The correct approach involves setting up the kinetic energy equation to reflect the energy loss and solving for the mass ratio algebraically.

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  • Understanding of momentum conservation in collisions
  • Knowledge of kinetic energy and its equations
  • Familiarity with algebraic manipulation for solving equations
  • Concept of inelastic collisions and energy loss
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  • Learn how to derive mass ratios from conservation laws
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Students studying physics, particularly those focusing on mechanics and collision theory, as well as educators looking for examples of inelastic collision problems.

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Homework Statement



Two objects moving in opposite directions with the same speed undergo a totally inelastic collision, and half the initial kinetic energy is lost. Find the ratio of their masses, m1/m2.

Homework Equations



m1v1+m2v2=(m1+m2)vf
The objects are moving in opposite directions with the same speed, so v1=-v2=v

The Attempt at a Solution



The part that gets me is the part that says half the kinetic energy is lost. Could someone explain how I factor that into the equations? If I just solve for m1/m2, plugging in 1/2v for vf(I don't know if that's right) I get 3, which is not the correct answer.
 
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If the collision were perfectly elastic, kinetic energy is conserved. But it is totally inelastic and 1/2 of the energy is lost. Therefore you can say:

.5 * m1 * v1^2 + .5 * m2 * v2^2 = (.5 * (m1 + m2)* vf^2) * .5

Then start cranking on the algebra to get m1/m2.
 
That makes sense. However, when I solve for m1/m2 I get (.5vf^2 + v^20/(v^2-.5vf^2) when I need an actual ratio.
 

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