Question

Two particles of equal mass have velocities $\overset{ \rightarrow }{v}_{1}=4\hat{i} \, \text{ms}^{- 1}$ and $\overset{ \rightarrow }{v}_{2}=4\hat{j}\text{ ms}^{- 1}$ . First particle has an acceleration $\left(\overset{ \rightarrow }{a}\right)_{1}=\left(5 \hat{i} + 5 \hat{j}\right)\left(\text{ ms}\right)^{- 2}$ , 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, $m_{1}=m_{2}=m$
$\left(\overset{ \rightarrow }{v}\right)_{\text{cm}} = \frac{m_{1} \left(\overset{ \rightarrow }{v}\right)_{1} + m_{2} \left(\overset{ \rightarrow }{v}\right)_{2}}{m_{1} + m_{2}} = \frac{m}{2 m} \left(\left(\overset{ \rightarrow }{v}\right)_{1} + \left(\overset{ \rightarrow }{v}\right)_{2}\right)$
$=\frac{1}{2}\times \left(\right.4\hat{i}+4\hat{j}\left.\right)$
$=2\left(\right.\hat{i}+\hat{j}\left.\right)$
Similarly $\left(\overset{ \rightarrow }{a}\right)_{\text{cm}}=\frac{m_{1} \overset{ \rightarrow }{a_{1}} + m_{2} \overset{ \rightarrow }{a_{2}}}{m_{1} + m_{2}}=\frac{m}{2 m}\left(\right.\overset{ \rightarrow }{a_{1}}+\overset{ \rightarrow }{a_{2}}\left.\right)$
$=\frac{1}{2}\times \left(\right.5\hat{i}+5\hat{j}+0\left.\right)$
$=\frac{5}{2}\left(\right.\hat{i}+\hat{j}\left.\right)$
Both $\overset{ \rightarrow }{v}_{c m}$ and $\overset{ \rightarrow }{a}_{c m}$ are parallel
Centre of mass will move in a straight line because velocity and acceleration of centre of mass are in same direction.