It is found that if a neutron suffers an elastic collinear collision with deuerium at rest , fractional loss of energy is `P_(d)` while for its similar collision with carbon nucleous at rest , fractional loss of energy is `P_(c )` are respectively `.........`

Options

A. `(.89,.28)`
B. `(-28,.89)`
C. (0,0)
D. `(0,1)`

Answer

A

Explanation

From law of conservation of momentum
`mv_(0) +mv_(1)+2mv_(2)`
` :. V_(0) =v_(1)+2v_(2) " "…..(2)`
` :. V_(1)+2v_(2) =v_(2) -v_(1)`
` :. 2v_(1) =-v_(2)`
From equ . (2) `v_(0)= -2v_(1) -v_(1) =-3v_(1)`
` :. v_(1) = - (v_(0))/3 `
` :. v_(2) = (2v_(0))/3 `
Decrease in kinetic energy
`P_(d) =(1/2mv_(0)^(2)-1/2mv_(1)^(2))/(1/2mv_(0)^(2))`
`=(v_(0)^(2) -(v_(0)^(2))/9)/(v_(0)^(2)) " " [ :. v_(1)=-(v_(0))/3]`
`P_(d) =8/9 = 0.8888 approx 0.89`
Now is econd collision :

From law of conservation of momentum
`mv_(0) =mv_(1) +mv_(2)`
` :. v_(0) =v_(1) +12v_(2) " ".....(3)`
For elastic collision e = 1
`v_(2) -v_(1) = v_(0) " "....(4)`
From (3) and (4)
`v_(1)+12v_(2)=v_(2)-v_(1)`
` :. 2v_(1) = 011v_(2)`
` :. v_(1) -(11v_(2))/2 and v_(2) = -(2v)/13`
` :. v_(1)= -(11v_(2))/2 and v_(2) = - (2v_(1))/13 `
`Now , 13v_(1) = -11v_(0) +v_(1) and v_(2) = - (11v_(0))/13 + v_(0)`
` :. 2v_(1) =-11v_(0) " " :. v_(2) =(2v_(0))/13 `
` :. v_(1) = (11v_(0))/13`
Decrease in kinetic energy
`P_(c) = ((1/2mv_(0)^(2))-1/2m(-(11v_(0))/13)^(2))/(1/2mv_(0)^(2))`
`:. (v_(0)^(2)-(121v_(0)^(2))/169)/(v_(0)^(2))=(169-121)/169 =48 /169 = 0.28`
` :. P_(d) = 0.89 ,P_(c ) = 0.28` .