A 70-kg man and a 50-kg woman are in a 60-kg 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 boat’s motion enough by diving off it at 3 m/s in the opposite direction? Could she then 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, we can analyze the situation using the principles of conservation of momentum.

1. Man's Initial Momentum:
Mass of the man (m1) = 70 kg
Initial velocity of the man (v1) = 3 m/s
Momentum of the man before diving off = m1 * v1

2. Woman's Initial Momentum:
Mass of the woman (m2) = 50 kg
Initial velocity of the boat (v2) = 0 m/s (since the motor failed)
Momentum of the woman/boat system before diving off = (m1 + m2 + m3) * v2, where m3 represents the boat's mass (60 kg).

3. Man's Final Momentum:
The man changes his mind and swims back to the boat at a speed of 1 m/s.
Velocity of the boat after the man dives off = v2 - v1 (due to conservation of momentum)
Final velocity of the boat (v2f) = -2 m/s (opposite direction to the man)
Final momentum of the boat (p2f) = (m2 + m3) * v2f

To determine whether the man can swim back to the boat, we'll compare the man's final momentum with the boat's final momentum. If the magnitudes are the same, he can reach the boat.

So, let's compare the magnitudes of the two momenta:

Magnitude of the man's final momentum (|p1f|) = m1 * (v2 - v1) = 70 kg * (0 m/s - 1 m/s) = -70 kg*m/s
Magnitude of the boat's final momentum (|p2f|) = (m2 + m3) * |v2f| = (50 kg + 60 kg) * 2 m/s = 220 kg*m/s

Since |p1f| < |p2f|, the man cannot swim back to the boat due to the imbalance of momentum.

Now, let's analyze whether the woman can change the boat's motion enough by diving off at 3 m/s in the opposite direction:

4. Woman's Final Momentum:
The woman dives off the boat in the opposite direction at 3 m/s.
Final velocity of the boat (v2f) = v2 - v3 = -3 m/s (opposite direction to the woman)
Final momentum of the boat (p2f) = (m2 + m3) * v2f = (50 kg + 60 kg) * (-3 m/s)

To determine if the woman can return to the boat, we'll compare the magnitudes of the final momenta:

Magnitude of the woman's final momentum (|p2f|) = (50 kg + 60 kg) * 3 m/s = 330 kg*m/s
Magnitude of the boat's final momentum (|p2f|) = (50 kg + 60 kg) * 3 m/s = 330 kg*m/s

Since |p2f| = |p2f|, the woman can change the boat's motion enough to counteract the man's momentum. However, since both magnitudes are the same, the woman's swimming speed of 1 m/s is not enough to bring her back to the boat.

In conclusion, the man cannot swim back to the boat with a swimming speed of 1 m/s, but the woman can change the boat's motion with her dive at 3 m/s. However, she cannot return to the boat herself with a swimming speed of 1 m/s.

To answer these questions, we need to understand the concept of momentum and the principle of conservation of momentum.

Momentum is the product of an object's mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction. The law of conservation of momentum states that the total momentum of a system remains constant if no external forces act on it.

Let's consider the scenario described:

1. The man dives into the water with a horizontal speed of 3 m/s. Since the boat's motor has failed, the total momentum of the system (man + boat) is initially zero. As the man jumps, he exerts a forward force on the boat due to Newton's third law of motion (action and reaction forces are equal and opposite). This will cause the boat to move backward in the water.

However, the man's individual momentum changes when he jumps. The momentum before he jumps is given by the product of his mass (70 kg) and initial velocity (3 m/s), which is 210 kg*m/s. When he jumps, his momentum changes to zero as he is now in the water while the boat starts moving backward.

2. If the man changes his mind and wants to swim back to the boat with a speed of 1 m/s, he will exert a force on the water in the opposite direction, pushing himself forward. However, in order to reach the boat, he needs to overcome his initial momentum (210 kg*m/s in the opposite direction) while swimming against the boat's backward momentum. Since his swimming speed is only 1 m/s, he will not be able to swim back to the boat.

3. Now, let's consider the woman who also dives off the boat at 3 m/s in the opposite direction. As she jumps, she exerts a backward force on the boat, causing it to move forward.

The momentum before the woman jumps is given by the product of her mass (50 kg) and initial velocity (3 m/s), which is 150 kg*m/s. After jumping, her momentum changes to zero as she enters the water.

4. Can the woman return to the boat if her swimming speed is 1 m/s? Similar to the man's scenario, she needs to overcome her initial momentum (150 kg*m/s in the opposite direction) while swimming against the boat's forward momentum. Like the man, with a swimming speed of 1 m/s, she will not be able to reach the boat.

In conclusion, neither the man nor the woman can swim back to the boat if their swimming speed is 1 m/s, given their initial momentum and the boat's momentum.