Question

A particle of mass $1\text{kg}$ is projected with a velocity $20\sqrt{2}\text{m/s}$ from the origin of an $x-y$ co- ordinate axis system at an angle ${45}^{\circ }$ with $x$ axis (horizontal). The angular momentum [in ${\text{kg m}}^2\text{/s}$] of the ball about the point of projection after $1\text{second}$ of projection is [Take $g=10{\text{m/s}}^2$ and $y$ - axis is taken as vertical]

• A. $200\hat{k}$
• B. $-100\hat{k}$
• C. $300\hat{j}$
• D. $-350\hat{j}$

Answer 1

The correct option is B $-100\hat{k}$

Given, Mass of particle, $m=1\text{kg}$
Projected velocity $u=20\sqrt{2}\text{m/s}$
Projected angle with horizontal $\theta ={45}^{\circ }$
As we know, Time of flight $T=\frac{2u\sin \theta }{g}=\frac{2\times 20\sqrt{2}\times \sin {45}^{\circ }}{10}$
$=\frac{2\times 20\sqrt{2}\times \frac{1}{\sqrt{2}}}{10}=4\text{s}$


From time of flight, we can say particle will reach maximum height at time $t=2\text{s}$, and at time $t=1\text{s}$ particle will be going upward.
At time $t=1\text{s}$, co-ordinates of particle:
$x=u\cos \theta \times t=20\sqrt{2}\times \cos {45}^{\circ }\times 1=20\text{m}$
$y=u\sin \theta t-\frac{1}{2}g{t}^2=20\sqrt{2}\times \sin {45}^{\circ }\times 1-\frac{1}{2}\times 10\times 1^2$
$=15\text{m}$
Hence, $\overrightarrow{r}=x\hat{i}+y\hat{j}=20\hat{i}+15\hat{j}$

Velocity components at $t=1\text{s}$.
${\overrightarrow{v}}_y=u_y+a_{y}t=u\sin \theta -g\times t$
$=20\sqrt{2}\times \frac{1}{\sqrt{2}}-10\times 1=10\text{m/s}$
and ${\overrightarrow{v}}_x=u\cos \theta =20\sqrt{2}\times \cos {45}^{\circ }=20\text{m/s}$
$\overrightarrow{v}=(20\hat{i}+10\hat{j})\text{m/s}$

Angular momentum about point $\text{O}$, $\overrightarrow{L}=m(\overrightarrow{r}\times \overrightarrow{v})$
$\overrightarrow{L}=1\times \left[\right(20\hat{i}+15\hat{j})\times (20\hat{i}+10\hat{j}\left)\right]$
$\overrightarrow{L}=200\hat{k}+300(-\hat{k})=100(-\hat{k})$