An object of mass ' m ' is projected from origin in a vertical xy plane at an angle 45^\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]

\fracm v_o^34 \sqrt2 g along negative z-axis

\fracm v_o ^32 \sqrt2 g along negative z -axis

\fracm v_o^32 \sqrt2 g along positive z -axis

\fracm v_o^34 \sqrt2 g along positive z -axis

Motion in A Plane PYQ — JEE Main 2025 · 24 Jan · Shift 1 · Q51 with Solution

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Correct answer: D

Initial velocity components

The particle is projected at angle $45^\circ$ with speed $v_0$ .

So,

v_{x}=v_0\cos45^\circ=\frac{v_0}{\sqrt2}, \qquad v_{y}=v_0\sin45^\circ=\frac{v_0}{\sqrt2}

Coordinates at maximum height

At maximum height, vertical velocity becomes zero.

Time to reach maximum height:

t=\frac{v_y}{g}=\frac{v_0/\sqrt2}{g}=\frac{v_0}{\sqrt2 g}

Hence the horizontal coordinate at that instant is

x=v_x t=\frac{v_0}{\sqrt2}\cdot \frac{v_0}{\sqrt2 g}=\frac{v_0^2}{2g}

Vertical coordinate at maximum height:

y=\frac{v_y^2}{2g}=\frac{\left(\frac{v_0}{\sqrt2}\right)^2}{2g} =\frac{v_0^2/2}{2g}=\frac{v_0^2}{4g}

So,

\vec r = x\hat i + y\hat j = \frac{v_0^2}{2g}\hat i + \frac{v_0^2}{4g}\hat j

Velocity at maximum height

At maximum height, vertical component is zero, while horizontal component remains unchanged:

\vec v = \frac{v_0}{\sqrt2}\hat i

Therefore linear momentum is

\vec p = m\vec v = m\frac{v_0}{\sqrt2}\hat i

Angular momentum about the origin

Angular momentum is

\vec L = \vec r \times \vec p

Substitute:

\vec L = \left(\frac{v_0^2}{2g}\hat i + \frac{v_0^2}{4g}\hat j\right) \times \left(m\frac{v_0}{\sqrt2}\hat i\right)

Now,

$\hat i \times \hat i = 0$
$\hat j \times \hat i = -\hat k$

So only the second term contributes:

\vec L = \frac{v_0^2}{4g}\hat j \times m\frac{v_0}{\sqrt2}\hat i

\vec L = \frac{m v_0^3}{4\sqrt2 g}(\hat j \times \hat i)

\vec L = -\frac{m v_0^3}{4\sqrt2 g}\hat k

Magnitude and direction

Thus,

|\vec L|=\frac{m v_0^3}{4\sqrt2 g}

and its direction is along the negative $z$ -axis.

Option check

A: wrong magnitude
B: wrong magnitude and wrong direction
C: correct magnitude, wrong direction
D: correct magnitude and correct direction

Therefore, the correct option is D.

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View written solutionFree

Correct answer: D

Initial velocity components

The particle is projected at angle $45^\circ$ with speed $v_0$ .

So,

v_{x}=v_0\cos45^\circ=\frac{v_0}{\sqrt2}, \qquad v_{y}=v_0\sin45^\circ=\frac{v_0}{\sqrt2}

Coordinates at maximum height

At maximum height, vertical velocity becomes zero.

Time to reach maximum height:

t=\frac{v_y}{g}=\frac{v_0/\sqrt2}{g}=\frac{v_0}{\sqrt2 g}

Hence the horizontal coordinate at that instant is

x=v_x t=\frac{v_0}{\sqrt2}\cdot \frac{v_0}{\sqrt2 g}=\frac{v_0^2}{2g}

Vertical coordinate at maximum height:

y=\frac{v_y^2}{2g}=\frac{\left(\frac{v_0}{\sqrt2}\right)^2}{2g} =\frac{v_0^2/2}{2g}=\frac{v_0^2}{4g}

So,

\vec r = x\hat i + y\hat j = \frac{v_0^2}{2g}\hat i + \frac{v_0^2}{4g}\hat j

Velocity at maximum height

At maximum height, vertical component is zero, while horizontal component remains unchanged:

\vec v = \frac{v_0}{\sqrt2}\hat i

Therefore linear momentum is

\vec p = m\vec v = m\frac{v_0}{\sqrt2}\hat i

Angular momentum about the origin

Angular momentum is

\vec L = \vec r \times \vec p

Substitute:

\vec L = \left(\frac{v_0^2}{2g}\hat i + \frac{v_0^2}{4g}\hat j\right) \times \left(m\frac{v_0}{\sqrt2}\hat i\right)

Now,

$\hat i \times \hat i = 0$
$\hat j \times \hat i = -\hat k$

So only the second term contributes:

\vec L = \frac{v_0^2}{4g}\hat j \times m\frac{v_0}{\sqrt2}\hat i

\vec L = \frac{m v_0^3}{4\sqrt2 g}(\hat j \times \hat i)

\vec L = -\frac{m v_0^3}{4\sqrt2 g}\hat k

Magnitude and direction

Thus,

|\vec L|=\frac{m v_0^3}{4\sqrt2 g}

and its direction is along the negative $z$ -axis.

Option check

A: wrong magnitude
B: wrong magnitude and wrong direction
C: correct magnitude, wrong direction
D: correct magnitude and correct direction

Therefore, the correct option is D.

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