Baptiste the boatman with mass m1=86.7 kg is standing at the back of his boat. The boat has a total mass m2=111.8 kg which is uniformly distributed along its length L=2.9 m. The boat is initially touching the dock. As he walks towards the dock, the boat moves away from the dock in reaction. Find the distance between the edge of the dock and the center of mass of Baptiste and his boat (x-com), and also find how far Baptist is from the dock when he reaches the other end of his boat (d). Select one answer for the position of the center of mass (x-com) and one for the distance from the dock (d).

d=0.68 mx-com=1.95 mx-com=0.57 mx-com=2.70 md=2.13 md=1.95 mx-com=1.35 md=1.19 m

a) Xcom = 2.083 m

b) 1.266 m

Explanation:

Given

m1 = 86.7 kg

m2 = 111.8 kg

L = 2.9 m

a) The expression for the center of mass is given as

Xcm = (m1*x1+m2*x2)/(m1+m2)

where

x1 = L =2.9 m

x2 = (2.9/2) m = 1.45 m (the edge of the dock)

⇒ Xcm = (86.7 kg*2.9 m+111.8 kg*1.45 m)/(86.7 kg+111.8 kg)

⇒ Xcom = 2.083 m

b) We use the formula

2*Xoc - L = 2*2.083 m - 2.9 m = 1.266 m

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.