Answer:
a) 0.618 ft/s
b) 3.04 ft/s
Explanation:
Givens:
Weight of swimmer A [tex]W_{A}[/tex] = 190 Ib.
Weight of swimmer B [tex]W_{B}[/tex]= 125 Ib.
Weight of the raft [tex]W_{R}[/tex] = 300 Ib.
Swimmer A walks toward swimmer B relative to the raft with a speed
[tex]V_{A/R}[/tex]= 2 ft/s
a) Conservation of linear momentum
[tex]m_{A} v_{A} +m_{B} v_{B} +m_{R} v_{R} =0..........(1)\\v_{A/R}=v_{A} -v_{R}\\\v_{A}=v_{A/R}+v_{R}.................(2)[/tex]
Since swimmer B does not move
[tex]v_{B} =v_{R}...............(3)[/tex]
Substitute from (2) and (3) into (1)
[tex]m_{A} (v_{A/R} +v_{R} )+m_{B} v_{R} +m_{R} v_{R}=0\\(m_{A}+m_{B}+m_{R})v_{R} =-m_{A}v_{A/R}\\v_{R}=\frac{-m_{A}v_{A/R}}{m_{A}+m_{B}+m_{R}} \\v_{R}=0.618ft/s[/tex]
b) if the raft not to move [tex]v_{R}=0[/tex]
from (2)
[tex]v_{A} =v_{A/R}[/tex]
substitute in (1)
[tex]m_{A} v_{A/R} +m_{B} v_{B}+m_{R} (0)=0\\v_{B}=\frac{W_{A}v_{A/R}}{W_{B}} \\v_{B}=3.04ft/s[/tex]
