A 85 kg fisherman jumps from a dock
into a 129 kg rowboat at rest on the West side
of the dock. If the velocity of the fisherman is 3.4 m/s
to the West as he leaves the dock, what is the
final velocity of the fisherman and the boat?
Answer in units of m/s
thx
into a 129 kg rowboat at rest on the West side
of the dock. If the velocity of the fisherman is 3.4 m/s
to the West as he leaves the dock, what is the
final velocity of the fisherman and the boat?
Answer in units of m/s
thx
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For this problem you can use the conservation of momentum. The initial momentum is the mass of the fisherman multiplied by his initial velocity. The boat is at rest and has no initial momentum
p(initial) = (85kg)*(3.4 m/s) + 0 = 289 kg*m/s
the final momentum is the total mass (sum of the fisherman and boat) times the final velocity, which we do not know yet.
p(final) = (85kg+129kg)*v = 214kg*v
The initial momentum of the system is equal to the final momentum of the system (conservation of momentum) so:
p(initial) = p(final)
289 kg*m/s = 214kg*v
solve for v:
v = [289kg*m/s] / [214kg] = 1.35 m/s
p(initial) = (85kg)*(3.4 m/s) + 0 = 289 kg*m/s
the final momentum is the total mass (sum of the fisherman and boat) times the final velocity, which we do not know yet.
p(final) = (85kg+129kg)*v = 214kg*v
The initial momentum of the system is equal to the final momentum of the system (conservation of momentum) so:
p(initial) = p(final)
289 kg*m/s = 214kg*v
solve for v:
v = [289kg*m/s] / [214kg] = 1.35 m/s