Note: Take East as the positive direction.

Question

A(n) 86 kg fisherman jumps from a dock
into a 131 kg rowboat at rest on the West side
of the dock.
If the velocity of the fisherman is 5.2 m/s
to the West as he leaves the dock, what is the
final velocity of the fisherman and the boat?

Answer 1

We can solve this problem using the conservation of momentum principle. The momentum before the jump is equal to the momentum after the jump.

Initial momentum (before the jump):
Fisherman: mass (m1) = 86 kg, velocity (v1) = -5.2 m/s (west direction is negative)
Rowboat: mass (m2) = 131 kg, velocity (v2) = 0 m/s

Initial momentum (P_initial) = m1 * v1 + m2 * v2

After the jump, the fisherman and rowboat move together with a final velocity (V_final).

Final momentum (P_final) = (m1 + m2) * V_final

According to the conservation of momentum, P_initial = P_final, so:

m1 * v1 + m2 * v2 = (m1 + m2) * V_final

Now, solve for V_final:

V_final = (m1 * v1 + m2 * v2) / (m1 + m2)
V_final = (86 * -5.2 + 131 * 0) / (86 + 131)
V_final = (-446.72 + 0) / 217
V_final = -446.72 / 217
V_final ≈ -2.06 m/s

So, the final velocity of the fisherman and the boat is -2.06 m/s, which is approximately 2.06 m/s to the West.