Two particles of equal mass have velocities V¯1=4i¯ and V¯2=4j¯ ms-1. First particle has an acceleration a¯1=(5i¯+5j¯) ms–2 while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

# Two particles of equal mass have velocities . First particle has an acceleration ${\overline{\mathrm{a}}}_{1}=\left(5\overline{\mathrm{i}}+5\overline{\mathrm{j}}\right)$ ms–2 while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

1. A

Straight line

2. B

Parabola

3. C

Circle

4. D

Ellipse

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### Solution:

At any time 't',
co-ordinates of CM = (XCM , YCM) 

We know

and

${\mathrm{X}}_{\mathrm{CM}}=\frac{\mathrm{m}\left[4\mathrm{t}+\frac{1}{2}\left(5\right){\mathrm{t}}^{2}\right]+\mathrm{m}\left(0\right)}{\mathrm{m}+\mathrm{m}}=2\mathrm{t}+\frac{5{\mathrm{t}}^{2}}{4}$

Similarly,
${\mathrm{Y}}_{\mathrm{CM}}=\frac{\mathrm{m}\left[0+\frac{1}{2}\left(5\right){\mathrm{t}}^{2}\right]+\mathrm{m}\left(4\mathrm{t}\right)}{\mathrm{m}+\mathrm{m}}=\frac{5{\mathrm{t}}^{2}}{4}+2\mathrm{t}$

Here, XCM = YCM for all 't'

The path of CM is straight line passing through origin making angle 45o

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