Kinetic Energy lost during elastic collision

AI Thread Summary
In an elastic collision between a neutron and a gold nucleus, the final velocity of the neutron is calculated using the formula V1f = (m1 - m2)/(m1 + m2) * V1i. The neutron, with a mass of 1 u, collides with a stationary gold nucleus of mass 197 u. To determine the percentage of kinetic energy lost, the kinetic energy before and after the collision needs to be compared. The calculation involves simplifying the expression for the final velocity and then finding the ratio of the final kinetic energy to the initial kinetic energy. The discussion emphasizes the need for careful simplification to accurately determine the percentage of kinetic energy lost during the collision.
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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