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 its origin is

Question

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 its origin is (A) 4 kg m 2 s -1 (B) 2 √2 kg m 2 s -1 (C) 4 √2 kg m 2 s -1 (D) 2 kg m 2 s -1

Tardigrade Admin · Accepted Answer

The given eqn. of the line is y=x+2. At x=-2, y=0 and at x=0, y=2 The line is drawn as shown in the figure. Length of perpendicular drawn from origin to this line is O N=r perp=(0+0+2/√12+12)=√2 Angular momentum of particle about L=m v r1=(1)(2) √2=2 √2 kg m 2 s -1

Q. A particle of mass $1 k g$ 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 its origin is

$4 k g m^{2} s^{- 1}$

$22 k g m^{2} s^{- 1}$

$42 k g m^{2} s^{- 1}$

$2 k g m^{2} s^{- 1}$

Q. A particle of mass 1kg is moving along the line y=x+2 (here x and y are in metres) with speed 2m/s The magnitude of angular momentum of particle about its origin is

4kgm2s−1

22​kgm2s−1

42​kgm2s−1

2kgm2s−1

The given eqn. of the line is y=x+2. At x=−2,y=0 and at x=0,y=2 The line is drawn as shown in the figure. Length of perpendicular drawn from origin to this line is ON=r⊥​=12+12​0+0+2​=2​ Angular momentum of particle about L=mvr1​=(1)(2)2​=22​kgm2s−1

Q. A particle of mass $1 k g$ 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 its origin is

$4 k g m^{2} s^{- 1}$

$22 k g m^{2} s^{- 1}$

$42 k g m^{2} s^{- 1}$

$2 k g m^{2} s^{- 1}$