Answer:
Step-by-step explanation:
Given that
Log mass Ml=29kg
Burt mass Mb=25.5kg
Ernest mass Me=26kg
Log length L=4m.
X coordinate of the center mass is given as
Xcm=ΣMi•xi / ΣMi
Let take the origin of Burts initial position and +x toward Ernie position, so the x coordinate of the Burts initial position is Xb1=0 and the x coordinate of the Ernie initial position is Xe1=L=4m.
The center of mass of the log is at L/2=4/2=2m
XL1=2m.
Applying the formula
Xcm, 1=(Mb•Xb1+Me•Xe1+ML1•XL1) / ( Mb+Me+ML)
Xcm, 1= (25.5×0+26×4+29×2)/(25.5+26+29)
Xcm, 1=(0+104+58)/(80.5)
Xcm, 1=162/80.5
Xcm, 1=2.01m
When Ernie moves to Burt, the log moves in opposite direction, so if the log moves a distance d, the final position of the system is
Xb2=d
Xe2=d
XL2=2+d.
Applying the equation again
Xcm, 2=(Mb•Xb2+Me•Xe2+ML1•XL2) / ( Mb+Me+ML)
Xcm, 2=(25.5×d+26×d+29×(2+d))/(25.5+26+29)
Xcm, 2=(25.5d+26d+58+29d)/(80.5)
Xcm, 2=(80.5d+58) / 80.5
Since there is no external force acting on the body, then, the total momentum is conserved.
Initially the log was at rest, then it will have 0 momentum at the beginning.
So,
Xcm, 1=Xcm, 2
2.01=(80.5d+58) / 80.5
Cross multiply
2.01×80.5=80.5d+58
162=80.5d+58
80.5d=162-58
80.5d=104
d=104/80.5
d=1.292m
Relative to the shore, the distance the log moved by the time Ernie reaches Burt is 1.292m