question_answer

A particle of mass \[m=5\] is moving with a uniform speed \[\text{v=3}\sqrt{2}\] 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 units

B)  \[40\sqrt{2}\] units

C) zero

D)        \[7.5\]units

Correct Answer: A

Solution :

\[y=x+4\]line has been shown in the figure. When \[x=0,y=4\]So, \[OP=4\]. The slope of the line can be obtained by comparing with the equation of line \[y=mx+c\] \[m=\tan \theta =1\Rightarrow \theta =45{}^\circ \] \[\angle OQP=\angle OPQ=45{}^\circ \] Length of the perpendicular drawn = OR In \[\Delta OPR,\,\,\frac{OR}{OP}=\sin 45{}^\circ \] \[\Rightarrow \,\,OR=OP\,\,\sin 45{}^\circ =4\times \frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\] Angular momentum of particle going along this line \[=r\times m\text{v=2}\sqrt{2}\times 5\times 3\sqrt{2}=60\]units