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 ${\overrightarrow{v}}_2=\left(t^2\right)\hat{j}$

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

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

Answer 1

The correct option is C a-q; b-r; c-p

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