Mr. Verma (50Kg) and Mr. Mathur $(60 \mathrm{Kg})$ are sitting at the two extremes of a $4 \mathrm{~m}$ long boat (40Kg) standing still in water. To discuss a mechanic problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move on the water during the process?
V-Verma
M-Mathur
B-Boat
$x_{C O M}=\frac{0(50)+2 \times 40+60 \times 4}{150}$
$=\frac{320}{150} \mathrm{~m}$
When they come to center
$x_{\text {COM }}=\frac{2(50)+2(40)+60(2)}{150}$
$=2\left(\frac{150}{150}\right)=2$
Shift in $x_{\mathrm{COM}} \Rightarrow\left|x_{\mathrm{COM}}-x_{\mathrm{COM}}\right|$
$=\frac{32}{15}-2$
$=\frac{2}{15}=0.13$
So, the boat would shift $0.13 \mathrm{~m}$ so that COM of system remains same.
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