Question

A particle of mass $200\text{g}$ is projected from origin $O$ with speed $10\text{m/s}$ at an angle ${60}^{\circ }$ with positive $x-$ axis. Find the magnitude of angular momentum of particle after $1\text{sec}$ about the origin. (Take $g=10{\text{m/s}}^2$)

• A. $25{\text{kg-m}}^2/\text{sec}$
• B. $0{\text{kg-m}}^2/\text{sec}$
• C. $250{\text{kg-m}}^2/\text{sec}$
• D. $125{\text{kg-m}}^2/\text{sec}$

Answer 1

The correct option is D $125{\text{kg-m}}^2/\text{sec}$

Given, mass of the particle $m=200\text{g}$
Speed of the particle $u=10\text{m/s}$
$\theta ={60}^{\circ },t=1\text{sec}$
Let the angular momentum of the particle after $t=1\text{sec}$ is $\overrightarrow{L}$
Angular momentum is given by
$\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$

Now, $\overrightarrow{r}\left(t\right)=x\hat{i}+y\hat{j}=\left(u\cos \theta \right)t\hat{i}+(u\sin \theta \times t-\frac{1}{2}g{t}^2)\hat{j}\dots \dots \left(1\right)$
where, $x$ and $y$ is the distance travelled by the particle after time $\left(t\right)$ in $x$ and $y$ direction respectively.
$\overrightarrow{r}\left(t\right)=\left(10\cos {60}^{\circ }\right)\times 1\hat{i}+(10\sin {60}^{\circ }\times 1-\frac{1}{2}\times 10\times 1^2)\hat{j}$
$r\left(t\right)=5\hat{i}+(5\sqrt{3}-5)\hat{j}\dots \dots \left(2\right)$

Now, velocity of particle
$\overrightarrow{v}\left(t\right)=v_x\hat{i}+v_y\hat{j}=\left(u\cos \theta \right)\hat{i}+(u\sin \theta -gt)\hat{j}$
$v\left(t\right)=5\hat{i}+(5\sqrt{3}-10)\hat{j}\dots \dots \left(3\right)$

$\therefore \overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})=0.2\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 5& (5\sqrt{3}-5)& 0\\ 5& (5\sqrt{3}-10)& 0\end{array}\right|$

$\Rightarrow \overrightarrow{L}=0.2[0\times \hat{i}-\hat{j}\times 0+(-25\left)\hat{k}\right]$
$\overrightarrow{L}=-25\hat{k}{\text{kg-m}}^2/\text{sec}$
So, magnitude of angular momentum
$|\overrightarrow{L}|=25{\text{kg-m}}^2/\text{sec}$