Kinetic Energy lost during elastic collision

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SUMMARY

The discussion focuses on calculating the percentage of kinetic energy lost by a neutron during an elastic collision with a gold nucleus. The relevant equations include the final velocity formula for elastic collisions, V1f = (m1 - m2) / (m1 + m2) * V1i, and the kinetic energy formula, K.E. = 1/2 m*v^2. The final velocity of the neutron is derived as (1u - 197u) / 198u * V1i. To find the percentage of kinetic energy lost, participants are guided to simplify the ratio of final to initial kinetic energy.

PREREQUISITES
  • Understanding of elastic collisions and momentum conservation
  • Familiarity with kinetic energy calculations
  • Basic algebra for simplifying expressions
  • Knowledge of particle physics, specifically neutron and gold nucleus properties
NEXT STEPS
  • Calculate the percentage of kinetic energy lost using the derived final velocity
  • Explore the implications of elastic collisions in nuclear physics
  • Review conservation laws in particle interactions
  • Study the differences between elastic and inelastic collisions
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of elastic collisions and kinetic energy in particle physics.

vetteLT4
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1. Homework Statement [/b]
A neutron (mass 1 u) experiences an elastic head-on collision with a gold nucleus (mass 197 u) that is originally at rest. What percentage of its original kinetic energy does the moving particle lose?


Homework Equations


Elastic Collision:

V1f= (m1-m2)/(m1+m2) * V1i
K.E. = 1/2 m*v^2

The Attempt at a Solution


The final Velocity of the moving particle is (1u-197u/198u) *V1i
But I don't know where to go from here in terms of solving the percentage of Kinetic Energy lost.
 
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Are you confident of your result for the final velocity of the moving particle?

Simplify (1u-197u)/198u

Find: \frac{(1/2)m{v_{1f}}^2}{(1/2)m{v_{1i}}^2} as a percentage. Lots of stuff should cancel.
 

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