JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
An object of mass ' m ' is projected from origin in a vertical \(x y\) plane at an angle \(45^{\circ}\) with the x axis with an initial velocity \(\mathrm{v}_0\). The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]
- A \(\frac{m v_o^3}{2 \sqrt{2} g}\) along negative \(z\)-axis
- B \(\frac{m v_o^3}{4 \sqrt{2} g}\) along positive \(z\)-axis
- C \(\frac{m v_o^3}{4 \sqrt{2} g}\) along negative \(z\)-axis
- D \(\frac{m v_o^3}{2 \sqrt{2} g}\) along positive \(z\)-axis
Answer & Solution
Correct Answer
(C) \(\frac{m v_o^3}{4 \sqrt{2} g}\) along negative \(z\)-axis
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{H}=\frac{\left(\frac{\mathrm{v}_0}{\sqrt{2}}\right)^2}{2 \mathrm{~g}}=\frac{\mathrm{v}_0^2}{4 \mathrm{~g}} \\ & \mathrm{~L}=\mathrm{mvh} \\ & \mathrm{~L}=\mathrm{m} \frac{\mathrm{v}_0}{\sqrt{2}} \frac{\mathrm{v}_0^2}{4 \mathrm{~g}} \end{aligned}\)
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