Romeo (82.0 kg) entertains Juliet (58.0 kg) by playing his guitar from the rear of their boat at rest in still water, 270 m away from Juliet, who is in the front of the boat. After the serenade, Juliet carefully moves to the rear of the boat (away from shore) to plant a kiss on Romeo's cheek, How far does the 74.0-kg boat move toward the shore it is facing? Part1 of 4 - Conceptualize As the girl makes her way toward the rear of the boat, the boat and the boy glide forward toward shore. The glide distance is shorter than the distance she moves in the boat since their mass is greater than her mass. L2.70 m>| Part 2 of 4 - Categorize No outside forces act on the boatplu5elovers system, so its momentum is conserved at zero and the center of mass of the boatpassengers system stays fixed. Part 3 of 4 -Analy1e Since the center of mass of the boatpassengers system stays fixed, we have XCM,i = XCM,r- Define K to be the point where they kiss, and Ax; and Axb as shown in the figure. Since Romeo moves with the boat, and thus Axkomeo = Axb. Let m, be the combined mass of Romeo and the boat. The front of the boat and the shore are to the right in this picture, and we take the positive x direction to the right. Then, we have the following. mJAxJ + 177be,; = 0 Choosing the x axis to point toward the shore, and substituting for the masses, we have (58 y /58 kg)AXJ+(82 w /82 kg+74.0kg)Axb=0, and solving for Ax] gives Ax] : 2.69 y / 2.69 Axb. Part 2 of 4 - Categorize No outside forces act on the boatplusIovers system, so its momentum is conserved at zero and the center of mass of the boatpassengers system stays fixed. Par! 3 of 4 -Analyze Since the center of mass of the boatpassengers system stays fixed, we have XCMJ = XCM,f- Define K to be the point where they kiss, and Ax; and Axb as shown in the figure. Since Romeo moves with the boat, and thus AxRomeO = Axb. Let my, be the combined mass of Romeo and the boat. The front of the boat and the shore are to the right in this picture, and we take the positive X direction to the right. Then, we have the following. mJAxJ + m|:,A>qJ = 0 Choosing the x axis to point toward the shore, and substituting for the masses, we have (58 v /}58 kg)AXJ+(82 v /}82 kg+74.0kg)Axb=0, and solving for Ax) gives Ax,=2.69 v /2. 9 Axb. As Juliet moves away from shore, the boat and Romeo glide toward the shore until the original 2.70 m gap between them is closed. We describe the relative motion with the equation. AxJ + Axb = 2,70 m. Substituting our result AX] = Axb, we find l Axb+Axb=l m. Solving for the distance the boat moves, we have that the boat moves 2,70 m toward the shore. Submit g (you cannot come back) Viewing Saved Work Reven (a Last Response Submit