A man of mass 50 kg is standing at one end of a boat of mass 300 kg floating on still water. He walks towards the other end of the boat with a constant velocity of `2 m s^(-1)` with respect to a stationary observer on land. What will be the velocity of the boat, (a) with respect to the stationary observer on land? (b) with respect to the man walking in the boat?
[Given: There is friction between the man and the boat and no friction between the boat and water.]

Explanation

Mass of the man `(m_(1)) is m_(1)=50 kg`
Mass of the boat `(m_(2))` is, `m_(2)- 300 kg `
With respect to a stationary observer The man moves with a velocity. `v_(1)-2ms^(-1)` and the boat moves with a velocity `v_(2)` (which is to be found) To determine the velocity of the boat with respect to a stationary h er own land: As there is no external force acting on the system the man and beat move due to the friction, which is an internal force in the boat man system. Hence, the velocity of the center of mass is zero `(v_(CM)=0)`
Using equation
`0=( sum m_(i)v_(i))/(sum m_(i))=(m_(1)v_(1)+m_(2)m_(2))/(m_(1)+m_(2))`
`0=m_(1)v_(1)+m_(2)v_(2)`
`-m_(2)v_(2)=m_(1)v_(1)`
`v_(2)=-(m_(1))/(m_(2))v_(1)`
`v_(2)=-(50)/(300)xx2=-(100)/(300)`
`v_(2)=-0.33 ms^(-1)`
The negative sign in the answer implies that the boat moves in a direction opposite to that of the walking man on the boat to a stationary observer on land.
(i) To determine the velocity of the boat with respect to the walking man We can find the relative velocity as
`v_(21)=v_(2)-v_(1)`
where, `V_(21)`, is the relative velocity of the boat with respect to the walking man.
`v_(21)=(-0.33)-(2)`
`v_(21)=2.33 ms^(-1)`
The negative sign in the answer implies that the boat appears to move in the opposite direction to the man walking in the boat.