a 70 kg man and 50 kg woman are in a boat when its motor fails. the man dives into the water with a horizontal speed of 3 m/s in order to swim ashore. if he changes his mind, can he swim back to the boat if his swimming speed is 1 m/s? If not, can the woman change the boats motion enough by diving off at 3 m/s in the opposite direction? Could she return to the boat herself if her swimming speed is also 1 m/s?
To determine whether the man can swim back to the boat and whether the woman can change the boat's motion enough to return to it, we need to analyze the conservation of momentum.
First, let's calculate the initial momentum of the system, considering only the boat and the people in it. Momentum (p) is given by the product of mass (m) and velocity (v).
The initial momentum of the system is:
Initial momentum = (mass of the man * velocity of the man) + (mass of the woman * velocity of the woman) + (mass of the boat * velocity of the boat)
Given:
Mass of the man (m1) = 70 kg
Velocity of the man (v1) = -3 m/s (negative because the man is diving in the opposite direction of motion)
Mass of the woman (m2) = 50 kg
Velocity of the woman (v2) = 0 m/s (since she hasn't moved yet)
Mass of the boat (m3) = unknown
Velocity of the boat (v3) = 0 m/s (since the motor has failed)
Initial momentum = (70 kg * -3 m/s) + (50 kg * 0 m/s) + (m3 * 0 m/s)
= -210 kg·m/s + 0 kg·m/s + 0 kg·m/s
= -210 kg·m/s
Now, let's consider the scenario where the man changes his mind and tries to swim back to the boat at a speed of 1 m/s. In this case, his velocity (v1) is now positive 1 m/s. We can recalculate the momentum of the system.
New momentum = (70 kg * 1 m/s) + (50 kg * 0 m/s) + (m3 * 0 m/s)
= 70 kg·m/s + 0 kg·m/s + 0 kg·m/s
= 70 kg·m/s
Comparing the new momentum to the initial momentum, we see that the new momentum is higher in magnitude than the initial momentum. Momentum must be conserved in an isolated system, so it implies that the boat and woman cannot return to the boat solely by the man swimming.
Now, let's analyze whether the woman can change the boat's motion enough to return to it. If the woman dives off the boat with a velocity of 3 m/s in the opposite direction, her velocity (v2) would be -3 m/s. Let's recalculate the momentum of the system with this new value.
New momentum = (70 kg * 1 m/s) + (50 kg * -3 m/s) + (m3 * 0 m/s)
= 70 kg·m/s - 150 kg·m/s + 0 kg·m/s
= -80 kg·m/s
Comparing the new momentum to the initial momentum, we see that the new momentum is lower in magnitude than the initial momentum. It implies that the boat and woman can potentially return to the boat if her swimming speed is also 1 m/s.
Therefore, based on the calculations, the woman could potentially change the boat's motion enough by diving off at 3 m/s in the opposite direction and could return to the boat herself if her swimming speed is also 1 m/s.