A neutron having a mass of `1.67 xx 10^(-27) kg` and moving at `10^(8) m//s`collides with a deuteron at rest and sticks to it. If the mass of the deuteron is `3.33 xx 10^(-27) kg` then the speed of the combination is

Options

A. `0.25xx10^(8)m//s`
B. `0.5xx10^(8)m//s`
C. `0.33xx10^(8)m//s`
D. `0.65xx10^(8)m//s`

Answer

C

Explanation

We have to apply the principle of conservation of linear momentum . The resultant momentum of A and B is
`P' = sqrt(((M)/(4 ) xx v_(1))^(2)+ ((M)/(4) xxv_(2))^(2))`
`= [ ( M^(2))/( 16) xx 25 + ( M^(2))/( 16) xx 144]^(1//2)`
`= [ ( M^(2))/(16) xx 169]^(1//2) = ( 13M)/( 4)`
Let `v_(3)` be the velocity of C.
The momentum of `C = (M)/(2)V_(3)`
As the body was at rest its initial momentum was 0.
`:.` Momentum of ( A & B ) `+` Momentum of C = 0
`:. ( 13M)/( 4) + ( M)/( 2) v_(3) = 0 :. (M)/(2)v_(3) = - ( 13M)/( 4)`
`:. v_(3) =- ( 13)/( 2)`
or `v_(3) =6.5 m//s `is magnitude.