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6.System of Particles and Rotational Motion
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A particle of mass $100\,g$ is projected at time $t =0$ with a speed $20\,ms ^{-1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2\,s$ is found to be $\sqrt{ K }\,kg\,m ^2 / s$. The value of $K$ is $............$ $\left(\right.$ Take $\left.g =10\,ms ^{-2}\right)$
A$80$
B$800$
C$8$
D$0.8$
(JEE MAIN-2023)
Solution
$\text { Use } \Delta L =\int \limits_0^{ t } \tau dt$
$L _0=\int \limits_0^2 mg \left( v _{ x } t \right) dt$
$= mg _{ x } \frac{ t ^2}{2}=(0.1)(10)(10 \sqrt{2}) \frac{2^2}{2}$
$=20 \sqrt{2}$
$=\sqrt{800}\,kgm ^2 / s$
$L _0=\int \limits_0^2 mg \left( v _{ x } t \right) dt$
$= mg _{ x } \frac{ t ^2}{2}=(0.1)(10)(10 \sqrt{2}) \frac{2^2}{2}$
$=20 \sqrt{2}$
$=\sqrt{800}\,kgm ^2 / s$
Standard 11
Physics
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