Question

a particle of mass 80 units is moving with a uniform speed v=4√2 units in xy plane ,along a line Y=X+5. the magnitude of angular momentum of the particle about the origin is

Answer 1

angular momentum of an object moving with some linear speed is given by

$L = mvr_p$

here

m = mass = 80 units

v = speed = $4\sqrt2$

$r_p$ = perpendicular distance from reference to the momentum

perpendicular distance of origin from given line Y = X + 5 is given by

$r_p = 5cos45 = \frac{5}{\sqrt2}$

now the angular momentum is

$L = 80 * 4\sqrt2 * \frac{5}{\sqrt2}$

$L = 1600$

So the angular momentum about origin will be 1600 units