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Kinetic Energy lost during elastic collision

  • Thread starter vetteLT4
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  • #1
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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?


2. Homework Equations
Elastic Collision:

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

3. 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.
 

Answers and Replies

  • #2
SammyS
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Are you confident of your result for the final velocity of the moving particle?

Simplify (1u-197u)/198u

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

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