Kinetic Energy loss in a completely inelastic collision

AI Thread Summary
In a perfectly inelastic collision, a particle of mass m1 collides with a stationary particle of mass m2, resulting in both particles moving together post-collision. The conservation of momentum is key, allowing the final velocity (vf) to be expressed in terms of the initial velocity (v1) and the masses. The initial kinetic energy (KE) can be calculated using KE = 1/2 m v^2, and the lost kinetic energy can be determined by comparing initial and final kinetic energies. When m1 is much smaller than m2, most kinetic energy is lost, while if m1 is much larger, less energy is lost. Understanding these dynamics is crucial for analyzing energy loss in inelastic collisions.
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Homework Statement



A particle of mass m_{1} and speed v_{1} collides with a second particle of mass m_{2} 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 = \frac{1}{2}mv^{2}


The Attempt at a Solution



m_{1}v^{2}_{1} = (m_{1}+m_{2})v^{2}_{f}

if this is right... not really sure how to show as fraction of lost kinetic energy :/
 
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aaj92 said:

Homework Statement



A particle of mass m_{1} and speed v_{1} collides with a second particle of mass m_{2} 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 = \frac{1}{2}mv^{2}


The Attempt at a Solution



m_{1}v^{2}_{1} = (m_{1}+m_{2})v^{2}_{f}

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.
 
Thank you :) i think i got it
 
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