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Question: Two particles of equal mass have velocities \({\overrightarrow{v}}_{1} = 2\widehat{i}ms^{- 1}\) and ...

Two particles of equal mass have velocities v→1=2i^ms−1{\overrightarrow{v}}_{1} = 2\widehat{i}ms^{- 1} and v→2=2j^ms−1{\overrightarrow{v}}_{2} = 2\widehat{j}ms^{- 1} First particle has an acceleration a→1=(3i^+3j^)ms−2{\overrightarrow{a}}_{1} = \left( 3\widehat{i} + 3\widehat{j} \right)ms^{- 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

Straight line

Explanation

Solution

Here, v→1=2i^ms−1,v→2=2j^ms−1.{\overset{\rightarrow}{v}}_{1} = 2\widehat{i}ms^{- 1},{\overset{\rightarrow}{v}}_{2} = 2\widehat{j}ms^{- 1}.

a→1=(3i^+3j^)ms−2,a→2=0ms−2{\overset{\rightarrow}{a}}_{1} = (3\widehat{i} + 3\widehat{j})ms^{- 2},{\overset{\rightarrow}{a}}_{2} = 0ms^{- 2}

}{(\because m_{1} = m_{2}) = (\widehat{i} + \widehat{j})ms^{- 1}}$$ Similarly, $${\overset{\rightarrow}{a}}_{CM} = \frac{{\overset{\rightarrow}{a}}_{1} + {\overset{\rightarrow}{a}}_{2}}{2} = \frac{3\widehat{i} + 3\widehat{j} + 0}{2} = \frac{3}{2}(\widehat{i} + \widehat{j})ms^{- 2}$$ Since,${\overset{\rightarrow}{v}}_{CM}$is parallel to ${\overset{\rightarrow}{a}}_{CM}$The path will be a straight line