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6.System of Particles and Rotational Motion
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A ball of mass $1 \,kg$ is projected with a velocity of $20 \sqrt{2}\,m / s$ from the origin of an $x y$ co-ordinate axis system at an angle $45^{\circ}$ with $x$-axis (horizontal). The angular momentum [In $SI$ units] of the ball about the point of projection after $2 \,s$ of projection is [take $g=10 \,m / s ^2$ ] ( $y$-axis is taken as vertical)
A
$-400 \hat{k}$
B
$200 \hat{i}$
C
$300 \hat{j}$
D
$-350 \hat{j}$
Solution
(a)
Time of flight $T=\frac{2 u \sin \theta}{g}=\frac{2(20 \sqrt{2}) \frac{1}{\sqrt{2}}}{10}=4$ second
$\Rightarrow$ After $2$ second particle will be at maximum height of the projectile $L=m v r_{\perp}$
$r_{\perp}=H_{\max }=\frac{u^2 \sin ^2 \theta}{2 g}=20 \,m$
So $L=(1)(20)(20)=400(-\hat{k})$
Standard 11
Physics
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