If the particle of mass `m` is moving with constant velocity `v` parallel to x-axis in x-y plane as shown in figure. Find its angular momentum with respect to origin at any time `t`.

Options

A. `mvbhat(k)`
B. `-mvbhat(k)`
C. `mvbhat(i)`
D. `mvhat(i)`

Answer

B

Explanation

We know that , Angular momentum
`vec(L)= vec(r )xxvec(p)` in terms of component becomes

`vec(L)= |(hat(i),hat(j),hat(k)),(x,y,x),(p_(x),p_(y),p_(z))|`
As motion is in x-y plane `(z=0 and p_(z)=0)`, so `vec(L)= vec(k)(xp_(y)-yp_(x))`
Here, `x=vt,y=b,p_(x)= mv and p_(y)=0`
`:. vec(L)= vec(k)[vtxx0-bmv]= -mvbhat(k)`