Work done by man jumping off boat: is it just 1/2(M+m)v²?

[Delzac]

A man of mass "m" on an initially stationary boat gets off the boat by jumping to the left in an exactly horizontal direction. Immediately after he leaps, the boat of mass "M", is observed to be moving to the right at speed v. how much work did the man do during the leap ( both to his own body and on the boat)?

is the ans as simple as " 1/2(M+m)v^2 " , or is there on to it?

 

[nrqed]

A man of mass "m" on an initially stationary boat gets off the boat by jumping to the left in an exactly horizontal direction. Immediately after he leaps, the boat of mass "M", is observed to be moving to the right at speed v. how much work did the man do during the leap ( both to his own body and on the boat)?

is the ans as simple as " 1/2(M+m)v^2 " , or is there on to it?

Careful. It is true that the answer will be the sum of the kinetic energies of the boat and man but the man will not have the same speed as the boat. Use conservation of momentum to find the speed of the man then add the kinetic energies of the boat and man.
Patrick

 

[Delzac]

got it! the ans is 1/2(M+(M^2/m))v^2 right? thanks a bunch!