Momentum and CollisionMedium
Question
A neutron makes a head-on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is
Options
A.16/81
B.8/9
C.8/27
D.2/3
Solution
Let m be the mass of neutron, mass of deuteron will be 4m. So applying conservation of momentum,
mv + 0 = mv1 + 2mv2 & v - 0 = (v2 + v1) [Relative velocity of approach = relative of separation]
v2 = v1 + v
mv = mv1 + 2m(v1 + v) = 3mv1 + 2mv - mv = 3mv1
mv + 0 = mv1 + 2mv2 & v - 0 = (v2 + v1) [Relative velocity of approach = relative of separation]
v2 = v1 + v
mv = mv1 + 2m(v1 + v) = 3mv1 + 2mv - mv = 3mv1
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