Question

Two boys of masses 40 kg and 60 kg are standing at the opposite ends of a boat of length 8 m and mass 120 kg in a river. What would be the displacement of the boat relative to water if the boys interchange their positions?

• A. 0.12 m
• B. 0.36 m
• C. 0.73 m
• D. 0.96 m

Answer 1

Answer: (C)

Let the boat be at a distance $x_{0} m$ from the shore
$x_{ cm }=\frac{40\left(x_{0}\right)+120\left(x_{0}+4\right)+60\left(x_{0}+8\right)}{(40+60+120)}$ _____(1)
When the boys interchange their positions, let the boat move by $x$, then

$x_{ cm }=\frac{60\left(x_{0}+x\right)+120\left(x_{0}+x+4\right)+40\left(x_{0}+x+8\right)}{(60+120+40)}$ ____(2)
Equating (1) and (2), we get
$40 x_{0}+120\left(x_{0}+4\right)+60\left(x_{0}+8\right)=60\left(x_{0}+x\right) $
$+120\left(x_{0}+x+4\right)+40\left(x_{0}+x+8\right)$
$\Rightarrow (120)(4)+(60)(8)=60 x+120 x+(120)(4)+40 x +40(8) $
$(60-40)(8)=220 \,x$
$\Rightarrow x=\frac{20 \times 8}{220}=\frac{8}{11} \approx 0.73\, m$