Question

Determine the fraction of the magnitude of kinetic energy lost by a neutron (m1 = 1.01 u) when it collides head-on and elastically with a target particle at rest which is 11H (m = 1.01 u).

Answer 1

as mass of both particles are same

let initial velocity of neutron = u

mass of neutron = m1 = m

target particle mass = m2 = m

initial velocity of target particle = 0

as its elastic collision

let the final velocity of neutron = v1

final velocity of target particle = v2

so

mu + 0 = m v1 + m v2

v1 + v2 = u/2 .................(1)

we know that

for elastic collsion

e = 1 = (v1 - v2)/(u - 0)

v1 - v2 = u ...........(2)

from 1 and 2

v1 = 3u/4 m/s

KE intial = KEi = 1/2 m u²

KE final = KEf = 1/2 m (3u/4)² = 9/32  m u²

so fraction = (KEf - KEi ) / KEi   

= - 0.437

answer