Question

An 80.0 kg fisherman jumps into a 37.5kg boat that is at rest. If the velocity of the fisherman is 5.60m/s when he jumps from the dock, what will be the velocity of the boat and fisherman after he lands in the boat?

Answer 1

The velocity of the boat and fisherman after he lands in the boat is 3.812m/s

Step-by-step explanation:

We will solve this question on the general Derivation for general case of a man jumping n the boat

m = Mass of the man = 80kg

M= mass of the boat 37.5kg

V(m, g) = Velocity of man with respect to ground = 5.6m/s

V(b , g) = Velocity of boat with respect to ground = Vo = 0m/s ( rest)

velocity of the boat and fisherman after he lands in the boat = V1

As there is no external force acting on the man then we can apply conservation of the Linear Momentum.

m*V(m, g) + M*Vo = (M+m)V1

`V1 = ( m*V(m, g) + M*Vo)/(M+m)`

On substituting the values we get

V1 =
`(80*5.6 + 37.5*0)/(80+ 37.5)`

V1 =
`(448)/(117.5)` = 3.812 m/s

Therefore the velocity of the boat and fisherman after he lands in the boat is 3.812m/s.