Final answer:
The initial recoil speed of the boat is found using conservation of momentum. By considering the momentum before and after the rock is thrown, we calculate the boat's recoil velocity, which is 0.2685 m/s in the opposite direction of the thrown rock.
Step-by-step explanation:
To answer this question, we need to apply the principle of conservation of momentum, since the man throwing the rock and the boat are an isolated system with no external horizontal forces. According to this principle:
Initial Total Momentum = Final Total Momentum
Before the rock is thrown, both the man and the boat are at rest, so the initial momentum is zero. When the man throws the rock, the boat moves in the opposite direction to conserve momentum. The man's mass is 86.1 kg, the boat's mass is 156 kg, and the rock has a mass of 5 kg.
Let v be the recoil velocity of the boat and the man together. Then the momentum of the boat and the man is (156 kg + 86.1 kg) × v, and the momentum of the rock is 5 kg × 13 m/s. Conservating momentum we have:
0 = (156 kg + 86.1 kg) × v + 5 kg × 13 m/s
Solve for v:
v = - (5 kg × 13 m/s) / (156 kg + 86.1 kg)
v = - (65 kg·m/s) / (242.1 kg)
v = -0.2685 m/s
The initial recoil speed of the boat is therefore 0.2685 m/s in the direction opposite to the rock's motion. The negative sign indicates the direction is opposite to that of the thrown rock, but when asked for speed, we typically report the magnitude, which would simply be 0.2685 m/s.