Question

A particle of mass $1\text{kg}$ has velocity ${\overrightarrow{v}}_1=\left(2t\right)\hat{i}$ and another particle of mass $2\text{kg}$ has velocity ${\overrightarrow{v}}_2=\left(t^2\right)\hat{j}$.

$\begin{array}{cc}\text{Columm I}& \text{Column II}\\ \text{(a) Net force on centre of mass at}2\text{s}& \text{(p)}\frac{20}{9}\text{unit}\\ \text{(b) Velocity of centre of mass at}2\text{s}& \text{(q)}\sqrt{68}\text{unit}\\ \text{(c) Displacement of centre of mass at}2\text{s}& \text{(r)}\frac{\sqrt{80}}{3}\text{unit}\\ & \text{(s) None}\end{array}$

• A. a-q, b-r, c-p
• B. a-r, b-q, c-p
• C. a-r, b-p, c-r
• D. a-r, b-p, c-r

Answer 1

The correct option is A a-q, b-r, c-p

${\overrightarrow{v}}_{c.m}=\frac{2t\hat{i}+2t^2\hat{j}}{3}\therefore {\overrightarrow{a}}_{c.m}=\frac{2}{3}\hat{i}+\frac{4}{3}t\hat{j}$
$\overrightarrow{F}=m\overrightarrow{a}=2\widehat{i+4t\hat{j}}$
$\therefore \left|\overrightarrow{F}\right|=\sqrt{4+64}=\sqrt{68}$
Now at $t=2\text{s},\overrightarrow{F}=2\hat{i}+8\hat{j}$
again at $t=2\text{s},{\overrightarrow{V}}_{c.m}=\frac{4\hat{i}+8\hat{j}}{3}$
$\therefore \left|{\overrightarrow{V}}_{c.c}\right|=\frac{\sqrt{80}}{3}$
also $x_{cm}=\frac{t^2}{3}\hat{i}+\frac{2}{9}{t}^3\hat{j}$
$\therefore \text{at}t=2\text{s},$
${\overrightarrow{x}}_{cm}=\frac{4}{3}\hat{i}+\frac{16}{9}\hat{j}=\frac{4}{3}(\hat{i}+\frac{4}{3}\hat{j})$
$\left|{\overrightarrow{d}}_{cm}\right|=\frac{4}{3}\left(\sqrt{1+\frac{16}{9}}\right)=\frac{4}{3}\times \frac{5}{3}=\frac{20}{9}$