This is a classic physics problem that involves the conservation of the center of mass. Since there is no external force acting on the system (no friction between the boat and water), the center of mass of the system must remain in the same position before and after the child walks.

Step 1: Define the system

• Mass of the boat: $M_b = 96 \, \text{kg}$
• Mass of the child: $M_c = 35 \, \text{kg}$
• Length of the boat: $L = 6.6 \, \text{m}$
• Initial distance from the pier to the boat: $d_{\text{initial}} = 7.9 \, \text{m}$

Step 2: Determine the initial center of mass

Initially, the child stands at the end of the boat closest to the pier. We assume that the boat's center of mass is at its center (i.e., at $\frac{L}{2}$) and that the child is at the edge closest to the pier. We will place the pier at $x = 0$ for convenience.

• Position of the child initially: $x_c = 7.9 \, \text{m}$ (since the child is at the closest end)
• Position of the boat's center of mass: $x_b = 7.9 + \frac{6.6}{2} = 7.9 + 3.3 = 11.2 \, \text{m}$

Now, the initial center of mass of the system is given by: $x_{\text{cm, initial}} = \frac{M_c \cdot x_c + M_b \cdot x_b}{M_c + M_b}$ Substitute the values: $x_{\text{cm, initial}} = \frac{35 \cdot 7.9 + 96 \cdot 11.2}{35 + 96}$ $x_{\text{cm, initial}} = \frac{276.5 + 1075.2}{131} = \frac{1351.7}{131} \approx 10.32 \, \text{m}$

Step 3: Final position of the center of mass

Since there is no external force, the center of mass does not change, so $x_{\text{cm, final}} = 10.32 \, \text{m}$.

Step 4: Analyze the child’s final position

When the child moves to the far end of the boat (the end farthest from the pier), the boat will move to maintain the center of mass. Let $x_b'$ be the final position of the center of mass of the boat. Since the boat moves, its center will be at $x_b' + \frac{L}{2}$, and the child will now be at the far end of the boat (6.6 m away from $x_b'$).

Let the new position of the child be $x_c' = x_b' + 6.6$. The new center of mass equation is: $x_{\text{cm, final}} = \frac{M_c \cdot x_c' + M_b \cdot x_b'}{M_c + M_b}$ Substitute the values, with $x_{\text{cm, final}} = 10.32 \, \text{m}$: $10.32 = \frac{35 \cdot (x_b' + 6.6) + 96 \cdot x_b'}{131}$ Multiply both sides by 131 to eliminate the denominator: $1351.7 = 35(x_b' + 6.6) + 96x_b'$ $1351.7 = 35x_b' + 231 + 96x_b'$ $1351.7 = 131x_b' + 231$ Solve for $x_b'$: $131x_b' = 1351.7 - 231 = 1120.7$ $x_b' = \frac{1120.7}{131} \approx 8.55 \, \text{m}$

Step 5: Calculate the final position of the child

Now that we know the final position of the boat’s center, the child is at the far end of the boat, which is 6.6 m away from $x_b'$: $x_c' = x_b' + 6.6 = 8.55 + 6.6 = 15.15 \, \text{m}$

Thus, the child is 15.15 m from the pier when she reaches the far end of the boat.

Would you like further details or have any other questions?

Relative questions:

1. How does the conservation of the center of mass affect movement in different systems?
2. What would happen if there was friction between the boat and water?
3. How can we calculate the center of mass for systems with more than two objects?
4. How does the mass of the boat affect the final position of the child?
5. What would change if the child had a different mass?

Tip: Conservation of the center of mass is key in systems where no external forces act, as it ensures that the system's overall balance remains unchanged.