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Two particles A and B of masses 2 kg and 3 kg are separated by 10 m. If they move towards each other under a mutual force of attraction, the distance of the point where they meet from the initial position of A is

\[\begin {array} {1 1} (a)\;5\;m & \quad (b)\;4\;m \\ (c)\;6\;m & \quad  (d)\;8\;m \end {array}\]

1 Answer

They will meet at their center of mass.
distance of centre of mass from A
$r_A= \large\frac{m_B r}{m_{A}+m_{B}}$
$\qquad=\large\frac{3 \times 10}{5}$
$\qquad=6m$
answered Nov 21, 2013 by meena.p
 

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