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6. system of particles and rotational motion — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics6. system of particles and rotational motion1 Mark
MCQ
A particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momnetum of particle about $P$ at $t = \frac{V_0}{g}$ is:-
  • ✓
    $\frac{1}{2 \sqrt 2} \frac{mV_0^3}{g}$
  • B
    $\frac{1}{2 \sqrt 2} \frac{mV_0^2}{g}$
  • C
    $\frac{1}{2} \frac{mV_0^3}{g}$
  • D
    $\frac{1}{2} \frac{mV_0^2}{g}$
✓

Answer

Correct option: A.
$\frac{1}{2 \sqrt 2} \frac{mV_0^3}{g}$
a
$\tau=\operatorname{mg} \mathrm{x}=$ torque

$\tau=\operatorname{mg}\left(V_{0} \cos 45\right) t$

$\frac{d J}{d t}=\frac{m g V_{0}}{\sqrt{2}} t$

$\int_{0}^{\mathrm{J}} \mathrm{d} \mathrm{J}=\int_{0}^{\mathrm{t}} \frac{\mathrm{mg} \mathrm{V}_{0}}{\sqrt{2}} \mathrm{t} \mathrm{dt}$

$J=\frac{m g V_{0}}{\sqrt{2}} \frac{t^{2}}{2}$

$J=\frac{m g V_{0}^{3}}{2 \sqrt{2} g^{2}}$

$J=\frac{1}{2 \sqrt{2}} \frac{m V_{0}^{3}}{g}$

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