question_answer

A particle of mass\[m=5\]units is moving with a uniform speed\[v=3\sqrt{2}\]units in the XOY plane along the line\[y=x+4\]. The magnitude of the angular momentum of the particle about the origin is:

A)  60 unit         

B) \[40\sqrt{2}\]unit

C)  zero            

D)  7.5 unit

Correct Answer: A

Solution :

 Momentum of the particle = mass\[\times \]velocity \[=(5)\times (3\sqrt{2})=15\sqrt{2}\] The direction of momentum in the XOY plane is given by \[y=x+4\] Slope of the line\[=1=tan\text{ }\theta \] i.e.,         \[\theta =45{}^\circ \] Intercept of its straight line = 4 Length of the perpendicular z from the origin of the straight line \[=4\sin 45{}^\circ =\frac{4}{\sqrt{2}}=2\sqrt{2}\] Angular momentum = momentum\[\times \]perpendicular length \[=15\sqrt{2}\times 2\sqrt{2}=60\,unit\]