Question

A system consists of two small spheres of mass $1\text{kg}$ and $2\text{kg}$ interconnected by a weightless spring. At $t=0$, the spheres are set in motion with initial velocities as shown in the figure. The position vector of centre of mass at $t=0$ is $\left(20\hat{j}\right)\text{m}$. Initially, spring is in its natural length. The system moves in earth's gravitational field. The position vector of centre of mass of the system in $\text{m}$ at $t=1\text{s}$ is

$[g=10{\text{m/s}}^2,\text{com is initial centre of mass}]$

• A. $4\hat{i}+\frac{24}{25}\hat{j}$
• B. $6\hat{i}+\frac{39}{2}\hat{j}$
• C. $4\hat{i}+\frac{23}{3}\hat{j}$
• D. $6\hat{i}+\frac{41}{3}\hat{j}$

Answer 1

The correct option is D $6\hat{i}+\frac{41}{3}\hat{j}$


Momentum of system at $t=0$ is,

$\overrightarrow{p_i}=1(8\hat{i}+6\hat{j})+2(5\hat{i}-5\hat{j})=18\hat{i}-4\hat{j}$

So, initial velocity of centre of mass,

$v_i=\frac{p_i}{m_1+m_2}=\frac{18\hat{i}-4\hat{j}}{3}=6\hat{i}-\frac{4}{3}\hat{j}$

Let $\overrightarrow{r}$ be the position of centre of mass at $t=0$ and $\overrightarrow{r^{\prime }}$ be the position of centre of mass at $t=1\text{s}$.

$\therefore \overrightarrow{r^{\prime }}=\overrightarrow{r}+{\overrightarrow{v}}_{i}t+\frac{1}{2}\overrightarrow{a}{t}^2$

$\Rightarrow \overrightarrow{r^{\prime }}=20\hat{j}+(6\hat{i}-\frac{4}{3}\hat{j})\times 1+\frac{1}{2}\times (-10\hat{j})\times (1{)}^2=6\hat{i}+\frac{41}{3}\hat{j}$

Hence, option $\left(D\right)$ is the correct answer.