Kinetic Energy loss in a completely inelastic collision

  • Thread starter aaj92
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  • #1
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Homework Statement



A particle of mass m[itex]_{1}[/itex] and speed v[itex]_{1}[/itex] collides with a second particle of mass m[itex]_{2}[/itex] at rest. If the collision is perfectly inelastic what fraction of the kinetic energy is lost in the collision? Comment on your answer for the casses that m1 is much much smaller than m2 and vice versa.

Homework Equations



KE = [itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex]


The Attempt at a Solution



m[itex]_{1}[/itex]v[itex]^{2}_{1}[/itex] = (m[itex]_{1}[/itex]+m[itex]_{2}[/itex])v[itex]^{2}_{f}[/itex]

if this is right... not really sure how to show as fraction of lost kinetic energy :/
 

Answers and Replies

  • #2
gneill
Mentor
20,925
2,867

Homework Statement



A particle of mass m[itex]_{1}[/itex] and speed v[itex]_{1}[/itex] collides with a second particle of mass m[itex]_{2}[/itex] at rest. If the collision is perfectly inelastic what fraction of the kinetic energy is lost in the collision? Comment on your answer for the casses that m1 is much much smaller than m2 and vice versa.

Homework Equations



KE = [itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex]


The Attempt at a Solution



m[itex]_{1}[/itex]v[itex]^{2}_{1}[/itex] = (m[itex]_{1}[/itex]+m[itex]_{2}[/itex])v[itex]^{2}_{f}[/itex]

if this is right... not really sure how to show as fraction of lost kinetic energy :/

Keep in mind that momentum is ALWAYS conserved. So you should be able to find an expression for Vf in terms of the masses and V1. Then you'll be able to directly compare the initial and final kinetic energies.
 
  • #3
25
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Thank you :) i think i got it
 

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