Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 4.0 m long and has mass 290 kg. Burt has mass 25.5 kg and Ernie has mass 26.0 kg. Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it.

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

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