Answer:
The boat moves away from the dock at 0.5 m/s.
Explanation:
Hi there!
Since no external forces are acting on the system boy-boat at the moment at which the boy lands on the boat, the momentum of the system is conserved (i.e. it remains constant).
The momentum of the system is calculated as the sum of the momentum of the boy plus the momentum of the boat. Before the boy lands on the boat, the momentum of the system is given by the momentum of the boy.
momentum of the system before the boy lands on the boat:
momentum of the boy + momentum of the boat
m1 · v1 + m2 · v2 = momentum of the system
Where:
m1 and v1: mass and velocity of the boy.
m2 and v2: mass and velocity of the boat.
Then:
50 kg · 2.0 m/s + 150 kg · 0 m/s = momentum of the system
momentum of the system = 100 kg m/s
After the boy lands on the boat, the momentum of the system will be equal to the momentum of the boat moving with the boy on it:
momentum of the system = (m1 + m2) · v (where v is the velocity of the boat).
100 kg m/s = (50 kg + 150 kg) · v
100 kg m/s / 200 kg = v
v = 0.5 m/s
The boat moves away from the dock at 0.5 m/s.
