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6.System of Particles and Rotational Motion
hard
A ball of mass $160\, g$ is thrown up at an angle of $60^o$ to the horizontal at a speed of $10\, m\,s^{-1}$ . The angular momentum of the ball at the highest point of the trajectcry with respect to the point from which the ball is thrown is nearly ........ $kg\, m^2/s$ $(g\, = 10\, m\,s^{-2})$
A$1.73$
B$3.0$
C$3.46$
D$6.0$
(JEE MAIN-2014)
Solution
$\begin{array}{l}
Given\,:\,m = 0.160kg\\
\theta = {60^ \circ }\\
V = 10\,m/s\\
Angular\,momentu\,m\,L = \vec r \times m\,\vec V\\
= H\,mv\,\cos \theta \\
= \,mv\frac{{{V^2}{{\sin }^2}\theta }}{{2g}}\cos \theta \,\,\,\left[ {H = \frac{{{V^2}{{\sin }^2}\theta }}{{2g}}} \right]\\
= \frac{{{{10}^2} \times {{\sin }^2}{{60}^ \circ } \times \cos {{60}^ \circ }}}{{2 \times 10}} \times \left( {0.16 \times 10} \right)\\
= 3\,kg\,{m^2}/s
\end{array}$
Given\,:\,m = 0.160kg\\
\theta = {60^ \circ }\\
V = 10\,m/s\\
Angular\,momentu\,m\,L = \vec r \times m\,\vec V\\
= H\,mv\,\cos \theta \\
= \,mv\frac{{{V^2}{{\sin }^2}\theta }}{{2g}}\cos \theta \,\,\,\left[ {H = \frac{{{V^2}{{\sin }^2}\theta }}{{2g}}} \right]\\
= \frac{{{{10}^2} \times {{\sin }^2}{{60}^ \circ } \times \cos {{60}^ \circ }}}{{2 \times 10}} \times \left( {0.16 \times 10} \right)\\
= 3\,kg\,{m^2}/s
\end{array}$
Standard 11
Physics
