A 45.0-kg woman stands up in a 60.0-kg canoe 5.00 m long. Se walks from a point 1.00 m from one end to a point 1.00 m from the other end (Fig. P8.92). If you ignore resistance to motion of the canoe in water, how far does the canoe move during this process?

The given data is listed below as-

• The mass of woman is $m_A=45kg$
• The mass of the canoe is $m_B=60.0kg$
• Length of Canoe is $L=5.0m$

The x-coordinate of the center of mass of a system of particles is given by-

${\mathit{x}}_{\mathbf{c}\mathbf{m}}\mathbf{=}\frac{\mathbf{\sum }_{\mathbf{}}{\mathbf{m}}_{\mathbf{i}}{\mathbf{x}}_{\mathbf{i}}}{\mathbf{\sum }_{\mathbf{}}{\mathbf{m}}_{\mathbf{i}}}$

Where m is the mass of the particle and x is the distance of the particle.

When the centre of mass is shifted from one end to the other, the canoe also moves.

Now apply the centre of mass equation to the woman standing on the canoe to obtain the below equation as:

$$\begin{gathered} x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2} \\ {x}_{cm}=\frac{60kg\times \left({\displaystyle \frac{5m}{2}}\right)}{60kg+45kg} \\ =1.429m \end{gathered}$$

Now. Add the two centre of masses.

$1.429+1.429=2.857$

So, the distance covered by Canoe while woman is walking will be obtained by subtracting 2.857 from the length of the canoe which is 5 m,

Therefore, $5-2.875=2.143m$

Thus, the distance covered by Canoe while the woman is walking is 2.143 m.