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 - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

At any time 't',
co-ordinates of CM = (X_CM , Y_CM)

We know $x = ut + \frac{1}{2} {at}^{2}$

and $X_{cm} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}$

$X_{CM} = \frac{m [4 t + \frac{1}{2} (5) t^{2}] + m (0)}{m + m} = 2 t + \frac{5 t^{2}}{4}$

Similarly,
$Y_{CM} = \frac{m [0 + \frac{1}{2} (5) t^{2}] + m (4 t)}{m + m} = \frac{5 t^{2}}{4} + 2 t$

Here, X_CM = Y_CM for all 't'

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

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