Question

A particle is projected at time $t = 0$ from a point $P$ on the ground with a speed $v_0$, at an angle of $45^{\circ}$ to the horizontal. The angular momentum of the particle about $P$ at time $t = v_0/g$ is

• A. $\frac{mv_{0}^{3}}{2\sqrt{2}g}$
• B. $\frac{mv_{0}^{3}}{\sqrt{2}g}$
• C. $3\frac{mv_{0}^{3}}{\sqrt{2}g}$
• D. $\frac{\sqrt{2}mv_{0}^{3}}{g}$

Answer 1

Answer: (A)

Let $L =$ angular momentum of the particle about $P$ at time $t = v_0/g$
At time $t = v_0/g$ :
Let $p_x = x$-component of momentum of particle
$v_x = x - $ component of velocity of particle
$x =$ displacement along $x$-axis.
Let $p_y , v_y$ and $y$ denote the quantities along $y$-axis To find $v_x$ and $v_y$:
$v_x =$ Horizontal velocity of the projectile
$or v_{x} = v_{0} cos \,45^{\circ} = \frac{v_{0}}{\sqrt{2}}, \quad.....\left(ii\right) $
It remains constant all along.
$v_{y} =$ vertical velocity of particle
or $v_{y} = \left(v_{0} sin\, 45^{\circ}\right) - g \times \left(\frac{v_{0}}{g}\right) \quad\left[\because v= u+at\right] $
or $v_{y} = \left(\frac{v_{0}}{\sqrt{2}} - v_{0}\right) \quad......\left(iii\right) $
To find $x$ and $y$:
$ x= v_{x}\times$ time
or $x= \frac{v_{0}}{\sqrt{2}} \times\frac{v_{0}}{g} = \frac{v_{0}^{2}}{\sqrt{2}g} $
$ \therefore x= \frac{v_{0}^{2}}{\sqrt{2}g} \quad.....\left(iv\right) $
$y =v_{0} sin \,\theta t - \frac{gt^{2}}{2} $
$ y = \frac{v_{0}}{\sqrt{2}} \times\frac{v_{0}}{g} -\frac{g}{2}\left(\frac{v_{0}}{g}\right)^{2} $
or $y = \left(\frac{v_{0}^{2}}{\sqrt{2}g} - \frac{v_{0}^{2}}{2g}\right) \quad.....\left(v\right) $
$\therefore $ Substitute $\left(ii\right), \left(iii\right), \left(iv\right)$ and $\left(v\right)$ in $\left(i\right) $
$ L= m\left[\frac{v_{0}^{2}}{\sqrt{2} g}\times\left(\frac{v_{0}}{\sqrt{2}} -v_{0}\right) - \left(\frac{v_{0}^{2}}{\sqrt{2}g} - \frac{v_{0}^{2}}{2g}\right)\frac{v_{0}}{\sqrt{2}}\right]$
or $L =\frac{mv_{0}^{3}}{g}\left[\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)-\left(\frac{1}{2}-\frac{1}{2\sqrt{2}}\right)\right]$
or $L =\frac{mv_{0}^{3}}{g}\times\left(\frac{-1}{2\sqrt{2}}\right) = -\frac{mv_{0}^{3}}{2\sqrt{2}g}$
$L$ is $\bot$ to plane of motion and is directed away from the reader i.e., along -ve $z$ direction.