JEE Physics 2025 Question with Solution

Given: An object of mass $m$ is projected from the origin with initial speed $v_0$ at angle $45^\circ$ in the vertical $xy$-plane.

Find: The magnitude and direction of angular momentum about the origin at maximum height.

At projection, the velocity components are

$$v_{0x} = v_0 \cos 45^\circ = \frac{v_0}{\sqrt{2}}, \qquad v_{0y} = v_0 \sin 45^\circ = \frac{v_0}{\sqrt{2}}$$

At maximum height, the vertical component becomes zero, so

$$v_y = 0, \qquad v_x = \frac{v_0}{\sqrt{2}}$$

Time to reach maximum height is

$$t = \frac{v_{0y}}{g} = \frac{v_0}{\sqrt{2}g}$$

Hence the x-coordinate at maximum height is

$$x = v_{0x} t = \frac{v_0}{\sqrt{2}} \cdot \frac{v_0}{\sqrt{2}g} = \frac{v_0^2}{2g}$$

The y-coordinate at maximum height is

$$y = \frac{v_{0y}^2}{2g} = \frac{v_0^2}{4g}$$

Angular momentum about the origin is

$$L_z = m(xv_y - yv_x)$$

Substituting $v_y = 0$,

$$L_z = -m \left(\frac{v_0^2}{4g}\right) \left(\frac{v_0}{\sqrt{2}}\right) = -\frac{mv_0^3}{4\sqrt{2}g}$$

The negative sign indicates direction along the negative z-axis.

Therefore, the angular momentum is $\frac{mv_0^3}{4\sqrt{2}g}$ along the negative z-axis. The correct option is D.

Given: The projectile is launched at $45^\circ$ with speed $v_0$ from the origin.

Find: Angular momentum at the highest point.

At the highest point, the momentum is purely horizontal because the vertical velocity becomes zero. So the momentum vector is along $+x$, while the position vector has both $x$ and $y$ components.

Write the position vector and momentum vector as

$$\vec{r} = x\hat{i} + y\hat{j}, \qquad \vec{p} = m v_x \hat{i}$$

Then

$$\vec{L} = \vec{r} \times \vec{p} = (x\hat{i} + y\hat{j}) \times (m v_x \hat{i})$$

Using cross product,

$$\vec{L} = x\hat{i} \times m v_x \hat{i} + y\hat{j} \times m v_x \hat{i}$$ $$\vec{L} = 0 + y m v_x (\hat{j} \times \hat{i})$$ $$\vec{L} = -y m v_x \hat{k}$$

Now,

$$y = \frac{v_0^2}{4g}, \qquad v_x = \frac{v_0}{\sqrt{2}}$$

So,

$$\vec{L} = - \left(\frac{v_0^2}{4g}\right) m \left(\frac{v_0}{\sqrt{2}}\right) \hat{k}$$ $$\vec{L} = -\frac{mv_0^3}{4\sqrt{2}g} \hat{k}$$

Thus the magnitude is $\frac{mv_0^3}{4\sqrt{2}g}$ and the direction is negative z-axis.