Question

13. A particle of mass 1 kg is moving along the line y=x+2
(here, x and y are in metres) with speed 2 m/s. The
magnitude of angular momentum of particle about origin
is :
(a) 4 kg - mº/s (b) 2/2 kg - m/s
(c) 4/2 kg - m/s (d) 2 kg - m/

..​

Answer 1

The linear momentum of the particle is,

$\longrightarrow\sf{p=mv}$

$\longrightarrow\sf{p=1\times2}$

$\longrightarrow\sf{p=2\ kg\ m\,s^{-1}}$

which acts along the motion of the particle, i.e., along the line $\sf{x-y+2=0.}$

We know that, if d is the perpendicular distance of a point $\sf{(h,\ k)}$ from the line $\sf{Ax+By+C=0,}$  then,

• $\sf{d=\dfrac{|Ah+Bk+C|}{\sqrt{A^2+B^2}}}$

Hence the line $\sf{x-y+2=0}$ is away from the origin $\sf{(0,\ 0)}$ at a distance,

$\longrightarrow\sf{r=\dfrac{|0-0+2|}{\sqrt{1^2+(-1)^2}}}$

$\longrightarrow\sf{r=\sqrt2\ m}$

This is the perpendicular distance of the particle from the origin.

Hence the magnitude of the angular momentum of the particle about the origin is,

$\longrightarrow\sf{L=pr}$

$\longrightarrow\sf{\underline{\underline{L=2\sqrt2\ kg\,m^2\,s^{-1}}}}$