Kinetic Energy lost during elastic collision

  • Thread starter vetteLT4
  • Start date
  • #1
1
0
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.
 

Answers and Replies

  • #2
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,422
1,057
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.
 

Related Threads on Kinetic Energy lost during elastic collision

Replies
3
Views
2K
  • Last Post
Replies
2
Views
3K
Replies
2
Views
724
O
  • Last Post
Replies
5
Views
71K
Replies
3
Views
3K
Replies
9
Views
2K
Replies
1
Views
3K
Replies
9
Views
10K
Replies
4
Views
23K
Top