Law of Conservation of Momentum and Jumping Off a Boat with Constant Velocity

[pierce15]

If a person jumps off of a boat with constant velocity, should the equation resulting from the law of conservation of momentum be (where 1 denotes the mass/velocity of the person, 2 denotes those of the boat, and f denotes a final velocity):

m1 * 0 + m2 * v2 = m2 * v2f + m1 * v1f

or

m1* 0 + m2 * v2 = (m2-m1) * v2f + m1* v1f

 

[Nugatory]

If a person jumps off of a boat with constant velocity, should the equation resulting from the law of conservation of momentum be (where 1 denotes the mass/velocity of the person, 2 denotes those of the boat, and f denotes a final velocity):

m1 * 0 + m2 * v2 = m2 * v2f + m1 * v1f

or

m1* 0 + m2 * v2 = (m2-m1) * v2f + m1* v1f

Neither. The left hand side is wrong because the initial momentum is $(m_1+m_2)v_2$ and that's what has to be equal to the final momentum $m_1v_{1f}+m_2v_{2f}$.

These problems are easiest to work if you start with the total momentum equal to zero, which in this case means that the initial velocity of the boat and person is zero. You can make the problem turn out that way just by using the speeds measured by a second boat that is pacing the first boat (relative speed zero) until the person jumps. Working with that definition, it's easy: $m_{1}v_{1f} + m_{2}v_{2f} = 0$. Then you just have to remember to add the speed of the observer back in if you want to get the speed of the boat and/or the person relative to the shore.