Question

Two particle of equal masses have velocities $\overrightarrow{V_1}=2im/s$ and the second particle has velocity $2\hat{j}m/s$. The first particle has an acceleration $\overrightarrow{a_1}=(3\hat{i}+3\hat{j})\frac{m}{s^2}$ while the acceleration of the other particle is zero. The center of mass of the particle moves in a :

• A. circle
• B. parabola
• C. straight line
• D. ellipse

Answer 1

The correct option is C straight line

Given that,

The velocities of both particles

${\overrightarrow{v}}_1=2\hat{i}m/s$

${\overrightarrow{v}}_2=2\hat{j}m/s$

The acceleration of both particles

${\overrightarrow{a}}_1=(3\hat{i}+3\hat{j})m/s^2$

${\overrightarrow{a}}_2=0$

Now, the center of mass

$v_{com}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

$v_{com}=\frac{2m(\hat{i}+\hat{j})}{2m}$

$v_{com}=(\hat{i}+\hat{j})m/s$

Now,

$a_{com}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$

$a_{com}=\frac{m(3\hat{i}+3\hat{j})+0}{2m}$

$a_{com}=\frac{3}{2}(\hat{i}+\hat{j})$

so, $v_{com}$is parallel to $a_{com}$

The position of CM of the particle will be given by:

$s=ut+\frac{1}{2}a{t}^2$

$s\left(t\right)=(\hat{i}+\hat{j})t+\frac{1}{2}\times \frac{3}{2}(\hat{i}+\hat{j})t^2$

$s\left(t\right)=(t+\frac{3}{4}{t}^2)\hat{i}+(t+\frac{3}{2}{t}^2)\hat{j}$

Therefore,

$\overrightarrow{x}\left(t\right)=\overrightarrow{y}\left(t\right)$

Hence, the path is straight line