Question

A particle P with a mass 2.0 kg has position vector r = 3.0 m and velocity v = 4.0 m/s as shown. It is accelerated by the force = 2.0 N. All these vectors lie in a common plane. The angular momentum vector is

• A. 12 kg $m^2/s$ out of the plane of the figure
• B. 12 kg $m^2/s$ into the plane of the figure
• C. Zero
• D. 24 kg $m^2/s$ into the plane of the figure

Answer 1

The correct option is A 12 kg $m^2/s$ out of the plane of the figure

Given,

$m=2kg$

$r=3m$

$v=4m/s$

$F=2N$

The angular momentum is,

$L=rmvsin\theta$

where, $\theta =$ angle between $v$ and $r$

$\theta ={150}^0$

$L=3\times 2\times 4sin{150}^0$

$L=12kg{m}^2/s$ out of the Plane.