System Of Particles And Rotational Motion Ques 4
4. A particle of mass $m=5$ is moving with a uniform speed $v=3 \sqrt{2}$ in the XOY plane along the line $y=x+4$. The magnitude of the angular momentum of the particle about the origin is
[1991]
(a) $60$ units
(b) $40 \sqrt{2}$ units
(c) zero
(d) $7.5$ units
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Answer:
Correct Answer: 4.(a)
Solution: (a)
$y=x+4$ line has been shown in the figure when
$x=0, y=4$. So, $O P=4$.
The slope of the line can be obtained by comparing with the equation of the straight line
$ y=m x+c $
$ m=\tan \theta=1 $
$ \Rightarrow \theta=45^{\circ} $
$ \angle O Q P=\angle O P Q=45^{\circ}$
If we draw a line perpendicular to this line, length of the perpendicular $=O R$
$\text { In } \triangle O P R, \frac{O R}{O P}=\sin 45^{\circ} $
$\Rightarrow O R =O P \sin 45^{\circ} $
$= 4 \times \frac{1}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2 \sqrt{2}$
Angular momentum of particle going along this
$ \text { line }=r \times m v=2 \sqrt{2} \times 5 \times 3 \sqrt{2}=60 \text { units } $
BETA
