An object initially at rest explodes into three fragments `A, B` and `C`.The momentum of `A` is `phati` and that of `B` is `sqrt(3)phatj`
where `p` is `a +ve` number. The momentum of `C` will be

Options

A. `(1+sqrt3)P` in a direction making `120^@` with A
B. 2P in a direction making `150^@` with A
C. 2P in a direction making `150^@` with B
D. `(1+sqrt3)P` in a direction making `150^@` with B

Answer

C

Explanation

Here `P_x=P and P_y=sqrt3 P`
`therefore` Resultant momentum of A and B
`P=sqrt(P_x^2+P_y^2)=sqrt(P^2+(sqrt 3P)^2)=2P`
It is along OC.
As is clear from figure

`tan theta= (BC.)/(OB)=(OA)/(OB)=P/(sqrt3 P)=1/sqrt3` or `theta=30^@`
As the object was initially at rest, the vector sum of linear momenta of A , B and C must be zero. Therefore, momentum of `C=2P` along OC opposite to OC.. It takes an angle with `B=angle YOC=angle YOX.+angle X.OC=90^@+60^@=150^@`