Question

Two particles of equal mass have velocities ${\overrightarrow{v}}_1=4\hat{i}{\text{ms}}^{-1}$ and ${\overrightarrow{v}}_2=4\hat{j}{\text{ ms}}^{-1}$ . First particle has an acceleration ${\left(\overrightarrow{a}\right)}_1=\left(5\hat{i}+5\hat{j}\right){\left(\text{ ms}\right)}^{-2}$ , while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

• A. Straight line
• B. Parabola
• C. Circle
• D. Ellipse

Answer 1

Answer: (A)

Since, $m_1=m_2=m$
${\left(\overrightarrow{v}\right)}_{\text{cm}}=\frac{m_1{\left(\overrightarrow{v}\right)}_1+m_2{\left(\overrightarrow{v}\right)}_2}{m_1+m_2}=\frac{m}{2m}\left({\left(\overrightarrow{v}\right)}_1+{\left(\overrightarrow{v}\right)}_2\right)$
$=\frac{1}{2}\times (4\hat{i}+4\hat{j})$
$=2(\hat{i}+\hat{j})$
Similarly ${\left(\overrightarrow{a}\right)}_{\text{cm}}=\frac{m_1\overrightarrow{a_1}+m_2\overrightarrow{a_2}}{m_1+m_2}=\frac{m}{2m}(\overrightarrow{a_1}+\overrightarrow{a_2})$
$=\frac{1}{2}\times (5\hat{i}+5\hat{j}+0)$
$=\frac{5}{2}(\hat{i}+\hat{j})$
Both ${\overrightarrow{v}}_{cm}$ and ${\overrightarrow{a}}_{cm}$ are parallel
Centre of mass will move in a straight line because velocity and acceleration of centre of mass are in same direction.