InterviewSolution
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InterviewSolution
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A neutron makes is a head - on elastic collision with a stationary deuteron The fraction energy loss of the neutron in the collision isA. `16//81`B. `8//9`C. `8//27`D. `2//3` |
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Answer» Correct Answer - b Apply law of coservation of momentumm and coservation of energy Let the two balls of mass `m_(1)` and `m_(2)` collide each other electrically with volicities `u_(1)` and `u_(2)`. Their velocities become `v_(1)` and `v _(2)` after the collision. Applying conservation of linear momentum, we get `m_(1)u_(1)+m_(2)u_(2)= m_(1)v_(1) +m_(2)v_(2)` ...(1) Also from conservation of kinetic energy `(1)/(2)m_(1)u_(1)^(2) + (1)/(2)m_(2)u_(2)^(2) = (1)/(2) m_(1)v_(1)^(2) + (1)/(2) m_(2)v_(2)^(2)`...(2) Solving Eqs `(1)` and `(2)` we get `v_(1) = ((m_(1) - m_(2))/(m_(1) + m_(2))) u_(1) + ((2m_(2))/(m_(1) + m_(2))) u_(2)`...(3) `and v_(2) = ((m_(2) - m_(1))/(m_(1) + m_(2))) u_(2) + ((2m_(2))/(m_(1) +(m_(2))) u_(1))`...(4) On taking apporaximate value the mass deterons is twice the mass of neutron Given `u_(1) = u , u_(2) = 0 , m_(1)=m,m_(2)= 2m` Velocity of neutron, `v_(1) = ((m - 2m)/(m + 2m)) u = -(u)/(3)` Velocity of deuteron, `v_(2) = ("2mu")/(m + 2m) = (2)/(3) u` Fractional energy loss `= ((1)/(2)m u^(2) - (1)/(2)m( -(u)/(3))^(2))/((1)/(2)m u^(2))` `= 1 - (1)/(9) = (8)/(9)` |
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