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Batman (mass = 90 kg) jumps straight down from a bridge into a boat (mass = 530 kg) in which a criminal is fleeing. The velocit...Asked by Candice
Batman (mass = 75.6 kg) jumps straight down from a bridge into a boat (mass = 631 kg) in which a criminal is fleeing. The velocity of the boat is initially +10.4 m/s. What is the velocity of the boat after Batman lands in it?
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Answered by
Victoria
To get started you need to use the principle conservation of linear momentum equation:
(mass1)*(finalvelocity1)+(mass2)*(finalvelocity2)= (mass1)*(intialvelocity1)+ (mass2)*(initialvelocity2)
This problem only cares about the horizontal motion, so don't think about how Batman's vertical fall effects the boat. Since he is falling straight down he has no vertical velocity. The final velocity for Batman will be the same as for the boat.
mass1 = batman = 75.6 kg
mass2 = boat = 631 kg
initial velocity 1 = batman's initial velocity = 0 m/s
initial velocity 2 = boat's initial velocity = 10.4 m/s
finalvelocity1 = finalvelocity2
so you simply plug in the numbers and solve for their final velocity
(mass1)*(finalvelocity1)+(mass2)*(finalvelocity2)= (mass1)*(intialvelocity1)+ (mass2)*(initialvelocity2)
This problem only cares about the horizontal motion, so don't think about how Batman's vertical fall effects the boat. Since he is falling straight down he has no vertical velocity. The final velocity for Batman will be the same as for the boat.
mass1 = batman = 75.6 kg
mass2 = boat = 631 kg
initial velocity 1 = batman's initial velocity = 0 m/s
initial velocity 2 = boat's initial velocity = 10.4 m/s
finalvelocity1 = finalvelocity2
so you simply plug in the numbers and solve for their final velocity
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