Answer to Question #134005 in Mechanics | Relativity for Sridhar

Question

A ball of mass 100g is dropped at time t=0. A second ball of mass 200g is dropped from the same point at t=0.2s .The distance between the center of mass of two balls and the release point at t=0.4s is:.......(Assume g=10m/s^(2) )
Ans:0.4m

Answer 1

The distance between the center of mass of 2 balls:

$r_C=\frac{m_1r_1+m_2r_2}{m_1+m_2}$

$r_1=\frac{gt_1^2}{2}$ , where $t_1=0.4$ s - time of falling the first ball

$r_2=\frac{gt_2^2}{2}$ , where $t_2=0.2$ s - time of falling the second ball identically.

Substitute numbers.

$r_C=\frac{100*0.4^2*\frac{10}{5}+200*0.2^2*\frac{10}{5}}{100+300}=0.4$ m