Question

1. Things One and Two are sitting in a symmetric boat near the shore. Originally, each Thing is a third of the way from the end of the boat he is closest to, and the right edge of the boat is exactly aligned with the shore, as shown here. Then, eager to get out, both Things move all the way to the right end of the boat (i.e., the end nearest to the shore). Assume that the water is frictionless. If each thing has a mass that is 3.9 times the mass of the boat, and the length of the boat is L = 470 cm, how far are the Things from the shore when they both reach the end of the boat?
   

Answer 1

VVU Given the boat is symmetric. The center of mass of the boat lies at the center of the boat. Center of mass of the system of boat & Tuis Things when the Things were sitting &mini - Xon = {mi Xem = m 1/3 + M. 1/4 + m2 1 x * 24 - mt Motm Xem = mL + More - 2m7 Mb Using m= 3.9 Mb Xem = 3.9 M₂ L + Mo, 12 - 2x3.9 M₂ + My Xem = 4.4 M, L 8.8 M₂ The center of mass of the system when two things were sitting is at the center the boat the of

Since the water is frictionless, there is no external horizontal force on the system which con displace the center of mass of the system. So, the center of mass of the system remains at x=24 even which the two things move around in the boat. Let the two things are at distance & from the shore, a t the center of mass of the bout will be at distance x + ½ <x+ → from the shore. The centers of mass of the system is Emiri = h Emi - mx + mx + Mo (x + 1 ) m + m + M 2m2 + M, 2 + M, L (2m+. My), x (2m + Mb) = M, L + m2 + M, L x= mL 2m+ Mi

Substituting m= 3.9 Mb, L=470 cm x - 3.44 x 470 см 233-49, + Н. - 2,0.5 см