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
$40x_0+120\left(x_0+4\right)+60\left(x_0+8\right)=60\left(x_0+x\right)$
$+120\left(x_0+x+4\right)+40\left(x_0+x+8\right)$
$\Rightarrow (120)(4)+(60)(8)=60x+120x+(120)(4)+40x+40(8)$
$(60-40)(8)=220x$
$\Rightarrow x=\frac{20\times 8}{220}=\frac{8}{11}\approx 0.73m$