Chapter 7, Problem 72GP

To determine

The distance by which the boat moves and the direction of movement.

Answer to Problem 72GP

Boat moves in right direction by $0.224\text{m}$

Explanation of Solution

Given:

Mass of the each ball $=m=80\text{kg}$

Mass of the one people $=m_1=75\text{kg}$

Mass of the other people $=m_2=60\text{kg}$

Length of the boat $=L=3.2\text{m}$

Initial location of centre of mass of one people relative to left side of boat $=x_{1i}=0\text{m}$

Final location of centre of mass of one people relative to left side of boat $=x_{1f}=3.2\text{m}$

Initial location of centre of mass of other people relative to left side of boat $=x_{2i}=3.2\text{m}$

Final location of centre of mass of other people relative to left side of boat $=x_{2f}=0\text{m}$

Location of centre of mass of the boat relative to left side $=x=\frac{L}{2}=\frac{3.2}{2}=1.6\text{m}$

Formula Used:

Centre of mass of system of masses is given as

  $x_{cm}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3+\mathrm{...}}{m_1+m_2+m_3+\mathrm{...}}$

Calculation:

Before exchanging the seats, centre of mass of the system of two people and boat is given as

  $\begin{array}{l}{x}_{cmi}=\frac{m_1{x}_{1i}+m_2{x}_{2i}+mx}{m_1+m_2+m_3}\\ x_{cmi}=\frac{(75)(0)+(60)(3.2)+(80)(1.6)}{75+60+80}\\ x_{cmi}=1.488\text{m}\end{array}$

After exchanging the seats, centre of mass of the system of two people and boat is given as

  $\begin{array}{l}{x}_{cmf}=\frac{m_1{x}_{1f}+m_2{x}_{2f}+mx}{m_1+m_2+m_3}\\ x_{cmf}=\frac{(75)(3.2)+(60)(0)+(80)(1.6)}{75+60+80}\\ x_{cmf}=1.712\text{m}\end{array}$

Since there is no net force on the system, hence the centre of mass must remain at the same location. Hence the boat moves in left direction and distance by which it move is given as

  $\begin{array}{l}\Delta x=x_{cmf}-x_{cmi}\\ \Delta x=1.712-1.488\\ \Delta x=0.224\text{m}\end{array}$

Conclusion:

Therefore, boat moves in right direction by 0.224 m.