JEE Main 2025 Jan 24 Shift 1, Physics Q8: Angular Momentum, Its Conservation and Angular Impulse

JEE Main2025Jan 24, Shift 1Physics

An object of mass ' $m$ ' is projected from origin in a vertical $x y$ plane at an angle $4 5^{\circ}$ with the $x$ -axis with an initial velocity $v_{0}$ . 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
$\frac{m v _{0}^{3}}{2 2 g}$ along negative $z$ -axis
B
$\frac{m v _{0}^{3}}{2 2 g}$ along positive $z$ -axis
C
$\frac{m v _{0}^{3}}{4 2 g}$ along positive $z$ -axis
D
$\frac{m v _{0}^{3}}{4 2 g}$ along negative $z$ -axis

Solution

At the highest point of the trajectory, the vertical velocity is zero, so the velocity is purely horizontal: $v_{x} = v_{0} cos 4 5^{\circ} = \frac{v _{0}}{2}$ .

Maximum height (using $v_{y}^{2} = v_{y 0}^{2} - 2 g h$ ):

$H = \frac{( v _{0} sin 4 5 ^{\circ} ) ^{2}}{2 g} = \frac{v _{0}^{2}}{4 g}$

Angular momentum about origin: $L = r \times m v$ . At the apex, $r = (x_{H}, H, 0)$ and $v = (v_{0} / 2, 0, 0)$ . The cross product has only a $z$ -component:

$L_{z} = m [x_{H} \cdot 0 - H \cdot (v_{0} / 2)] = - \frac{m H v _{0}}{2} = - \frac{m v _{0}}{2} \cdot \frac{v _{0}^{2}}{4 g} = - \frac{m v _{0}^{3}}{4 2 g}$

The negative sign indicates the direction is along the negative $z$ -axis (consistent with motion in the $+ x$ direction at height $+ y$ , using the right-hand rule the angular momentum about the origin points into the page).

$∴$ Magnitude $\frac{m v _{0}^{3}}{4 2 g}$ , direction $- \overset{z}{^}$ . Option (4).

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JEE Main 2025 Jan 24 Shift 1, Physics Q8: Angular Momentum, Its Conservation and Angular Impulse

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