A 100kg man stands initially on the left end of a 60 kg, flat-bottomed fishing boat that is 10 meters long. This end of the fishing boat is 60 meters from shore. No external forces act on the boat, such as air or water friction. If the man walks to the opposite end (right end) of the boat:
a. Does the center of mass of the man-boat system move or remain stationary? If the boat does move, which way does it move?
b. After casually strolling to the opposite (right) end of the boat, how far is he from shore?
Help me. I am lost.
There are no external forces on the boat/man system. Therefore if it was at rest there is no acceleration and the center of mass does not move. If the man moves away from shore in the boat, the boat moves toward the shore to keep the center of mass stationary.
To solve this question, we need to consider the conservation of momentum and the concept of the center of mass.
a. According to the law of conservation of momentum, the total momentum of an isolated system remains constant unless acted upon by external forces. In this case, since no external forces act on the boat, the total momentum of the system remains constant.
The center of mass will move if there is a net external force acting on the system. In this scenario, when the man walks from one end to the other end of the boat, he exerts a force on the boat in the opposite direction, causing the boat to move in the opposite direction. Therefore, the center of mass of the man-boat system will move towards the side where the man walks.
b. To determine how far the man is from the shore after he walks to the opposite end of the boat, we can apply the concept of the center of mass.
The location of the center of mass can be calculated using the formula:
Center of mass = (m1 * x1 + m2 * x2) / (m1 + m2)
where m1 and m2 are the masses of the objects (man and boat) and x1 and x2 are the positions (initial and final) of the respective objects.
Man's mass (m1) = 100 kg
Boat's mass (m2) = 60 kg
The initial position of the man (x1) is 0 meters from the shore, and the final position (x2) is 10 meters from the shore (opposite end of the boat).
Plugging these values into the formula:
Center of mass = (100 kg * 0 m + 60 kg * 10 m) / (100 kg + 60 kg)
= (600 kg*m) / 160 kg
= 3.75 m
Therefore, when the man walks to the opposite end of the boat, he is 3.75 meters from the shore.
Remember, in order to get accurate results, it is important to consider the assumption that no external forces such as air or water friction act on the boat.