What is the velocity of the boat after Batman lands in it?

Batman (mass = 91kg) jumps straight down from a bridge into a boat (mass=150kg) in which a criminal is fleeing. The velocity of the boat is initially +11 m/s. What is the velocity of the boat after Batman lands in it?What is the velocity of the boat after Batman lands in it?
conservation of momentum in the direction of the boat,



150 * 11 + 90 * 0 = (150 + 91) * v



v = 6.85 m/s

the velocity of the boat after Batman lands in it

answerWhat is the velocity of the boat after Batman lands in it?
The intent of the question is to demonstrated conservation of linear momentum, though the example is pretty bad.

Assuming everything is ideal, the boat pretty much instantly increases in mass. Before batman jumps on, it has a linear momentum in the forward direction of

p = mv = (150)(11) = 1650 kg m / s

After batman jumps on, because he exerts no force in the forward or backward direction, linear momentum in that direction must be conserved. That is, it must be the same as before

p = 1650 kg m / s = m' v'

Where m' and v' are the mass and speed, respectively, of the boat after batman jumps on. We know the mass of batman and the boat, just compute the speed v'.

p = 1650 kg m / s = (150+91) v'

v' = 6.85 m/sWhat is the velocity of the boat after Batman lands in it?
Conservation of momentum



initial momentum of boat is 150 x 11

momentum of bat is zero

final mass is 150+91 = 241



241v = 150*11

v = 6.85 m/s



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