JEE PYQ: Motion in a Plane - Question ID c233fa211a86 (JEE Main 2025)

To determine the angular momentum of the projectile at its maximum height, we rely on the following key concepts and formulas:

• Angular Momentum Definition: The angular momentum $\vec{L}$ of a particle of mass $m$ moving with velocity $\vec{v}$ at a position $\vec{r}$ relative to the origin is given by: $\vec{L} = \vec{r} \times \vec{p} = m (\vec{r} \times \vec{v})$ where $\vec{p} = m\vec{v}$ is the linear momentum.
• Projectile Motion in a Plane: When an object is projected at an angle $\theta$ with initial velocity $v_0$, its motion can be resolved into horizontal ($x$) and vertical ($y$) components: $v_{0x} = v_0 \cos \theta, \quad v_{0y} = v_0 \sin \theta$ The horizontal motion is uniform (no acceleration), while the vertical motion is under gravity ($-g$).
• Maximum Height: At maximum height, the vertical component of velocity becomes zero. The time to reach maximum height is: $t = \frac{v_{0y}}{g} = \frac{v_0 \sin \theta}{g}$ The horizontal distance covered in this time is: $x = v_{0x} \cdot t = v_0 \cos \theta \cdot \frac{v_0 \sin \theta}{g} = \frac{v_0^2 \sin \theta \cos \theta}{g}$ The vertical position at maximum height is: $y = v_{0y} \cdot t - \frac{1}{2} g t^2 = \frac{v_0^2 \sin^2 \theta}{2g}$
• Cross Product for Angular Momentum: The angular momentum $\vec{L}$ is perpendicular to both $\vec{r}$ and $\vec{v}$. For motion in the $xy$-plane, $\vec{L}$ will be along the $\pm z$-axis, depending on the direction of $\vec{r} \times \vec{v}$.

Step 1: Resolve Initial Velocity
The object is projected at $\theta = 45^\circ$ with initial velocity $v_0$. The components are: $v_{0x} = v_0 \cos 45^\circ = \frac{v_0}{\sqrt{2}}, \quad v_{0y} = v_0 \sin 45^\circ = \frac{v_0}{\sqrt{2}}$

Step 2: Determine Time to Reach Maximum Height
At maximum height, $v_y = 0$. The time taken is: $t = \frac{v_{0y}}{g} = \frac{v_0 / \sqrt{2}}{g} = \frac{v_0}{\sqrt{2} g}$

Step 3: Calculate Position at Maximum Height
The horizontal distance ($x$) and vertical height ($y$) at time $t$ are: $x = v_{0x} \cdot t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{\sqrt{2} g} = \frac{v_0^2}{2g}$ $$y = v_{0y} \cdot t - \frac{1}{2} g t^2 = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{\sqrt{2} g} - \frac{1}{2} g \left( \frac{v_0}{\sqrt{2} g} \right)^2 = \frac{v_0^2}{2g} - \frac{v_0^2}{4g} = \frac{v_0^2}{4g}$$ Thus, the position vector at maximum height is: $\vec{r} = x \hat{i} + y \hat{j} = \frac{v_0^2}{2g} \hat{i} + \frac{v_0^2}{4g} \hat{j}$

Step 4: Determine Velocity at Maximum Height
At maximum height, the vertical velocity is zero, and the horizontal velocity remains unchanged (no air resistance): $\vec{v} = v_{0x} \hat{i} = \frac{v_0}{\sqrt{2}} \hat{i}$

Step 5: Compute Angular Momentum
The angular momentum is: $\vec{L} = m (\vec{r} \times \vec{v}) = m \left( \left( \frac{v_0^2}{2g} \hat{i} + \frac{v_0^2}{4g} \hat{j} \right) \times \left( \frac{v_0}{\sqrt{2}} \hat{i} \right) \right)$ Using the cross product $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$: $\vec{L} = m \left( \frac{v_0^2}{4g} \cdot \frac{v_0}{\sqrt{2}} (-\hat{k}) \right) = -\frac{m v_0^3}{4 \sqrt{2} g} \hat{k}$ The negative sign indicates the direction is along the negative $z$-axis.

Step 6: Match with Given Options
The magnitude of $\vec{L}$ is $\frac{m v_0^3}{4 \sqrt{2} g}$, and its direction is along the negative $z$-axis. This matches Option D. --- Common Traps & Exam Tip:

1. Incorrect Position at Maximum Height:
Students often miscalculate $y$ by forgetting the $\frac{1}{2} g t^2$ term in the vertical displacement. Always use: $y = v_{0y} t - \frac{1}{2} g t^2$ not just $y = v_{0y} t$.

2. Direction of Angular Momentum:
The cross product $\vec{r} \times \vec{v}$ yields a vector perpendicular to the plane of motion. For a projectile launched in the $xy$-plane, $\vec{L}$ is along $\pm z$. The sign depends on the order of $\vec{r}$ and $\vec{v}$ in the cross product. Here, $\vec{r} \times \vec{v}$ gives a negative $z$-component.

3. Forgetting Horizontal Velocity at Maximum Height:
At maximum height, $v_y = 0$, but $v_x$ remains $v_{0x}$. Students sometimes assume $\vec{v} = 0$, leading to $\vec{L} = 0$, which is incorrect.

4. Angle-Specific Simplifications:
For $\theta = 45^\circ$, $\sin \theta = \cos \theta = \frac{1}{\sqrt{2}}$. Students may incorrectly use $\sin 45^\circ = \cos 45^\circ = \frac{1}{2}$ or forget to simplify $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$.

Exam Tip:
Always draw a diagram to visualize $\vec{r}$ and $\vec{v}$ at the point of interest. For angular momentum, focus on the perpendicular distance from the origin to the line of motion (lever arm) and the component of velocity perpendicular to $\vec{r}$. Here, the lever arm is the $y$-coordinate, and the relevant velocity is $v_x$.