Question

In Fig. a man stands on a boat floating in still water. The mass of the man and the boat is `60 kg` and `120 kg`, respectively.

a. If the man walks to the front of the boat and stops. what is the separation between the boat and the pier now?
b. If the man moves at a constant speed of `3 m//s` relative to the boat, what is the total kinetic energy of the system (boat `+` man)? Compare this energy with the kinetic energy of the system if the boat was tied to the pier.

Answer 1

a. displacement of man w.r.t boat `x_(m,b)=6m`
`x_(b)=x_(m,b)(m_(1)/(m_(1)+m_(2)))=-6xx30/(60+20)=-2m`
Therefore, boat displaces by `2 m` away from the pier.
`:.` Seperation `=2.5m`
b. `vecv_(mb)=vecvm-vecvb`
`m_(1)vecv_(m)+m_(2)vecv_(b)=0`
`m_(1)(vecv_(mb)+vevcb)+m_(2)vecv_(b)=0`
`m_(1)v_(mb)=(m_(1)+m_(2))v_(b)`
`v_(b)=(m_(1)/(m_(1)+m_(2)))v_(mb)`
`v_(b)=60/180xx3=1m//s`
`KE` of the man
`=1/2m_(m)v_(m)^(2)=1/2xx60xx(3-1)^(2)=120J`
`KE` of the boat `=1/2xx120xx(1)^(2)=60J`
Total `KE` energy `=120+60=180J`
`KE` of the sytem when the boat is tied
`=1/2x60xx3^(2)=270J`