Question

A particle of mass 6
kg is moving along y
= x + 4. Speed of
particle is
4 m/s.
Areal velocity (in
of position
vector of particle
with respect to the
origin is
m/s)
D 2.4
4.8
5.6
D 3.2​

Answer 1

Answer:

5.6

Explanation:

This is very simple mark me

Answer 2

Answer:

Explanation:

Given mass of the particle, $m=6kg$

direction of motion of particle is along, $y=x+4$

speed of the particle, $v=4m/s$

Drop a perpendicular to origin from the curve, $y=x+4$

Therefore , the distance $r=acos45^{o} =4\times\frac{1}{\sqrt{2}}$

Angular momentum is given by

                           $L=mvr$

                               $=6\times\ 4 \times\frac{4}{\sqrt{2} }=\frac{96}{\sqrt{2} }$

Areal velocity, $A=\frac{L}{2m}$

                            $=\frac{\frac{96}{\sqrt{2} }}{2\times 6}=4\sqrt{2}$

Hence the areal velocity of the particle is $4\sqrt{2} m^{2}/s$