A dog, weighing 10.0 lbs, is standing on a flatboat so that he is 20 ft from the shore. He walks 8.0 ft on the boat toward shore and then halts. The boat weighs 40 lb, and one can assume there is no friction between it and the water. How far is he from the shore at the end of this time?
...I proved that the center of mass does not change, which to me, seems like the most entirely unhelpful thing ever. All I know is that there is a difference of 2.0 ft between the initial boat position and the final boat position, but I have no idea whether or not that actually effects where the dog is or not.
I'm really frustrated, because this seems like a simple question, but I just can't conceptually grasp at it.
as you said, the location of the center of mass does not change. So, since the boat weighs 4 times as much as the dog, it moves backward 1 foot for every 4 feet the dog moves forward.
So, as you say, the boat has moved back 2 feet and the dog has moved forward 8 feet.
That means his net motion is 6 ft toward the shore.
To solve this problem, you need to consider the principle of conservation of momentum. Since there is no external force acting on the boat-dog system, the momentum of the system will remain constant.
Let's denote the initial position of the dog as x1 and the final position as x2. The initial position of the boat is x1 = 0 ft, and the final position is x2 = 2 ft (since the dog moves 8 ft toward the shore).
The momentum of the system before the dog starts moving is given by:
Initial momentum = (Mass of the boat + Mass of the dog) * Initial velocity
Let's assume the dog moves with a constant velocity v, and the direction of motion is toward the shore. The final momentum of the system is given by:
Final momentum = (Mass of the boat + Mass of the dog) * Final velocity
Since the momentum of the system is conserved:
Initial momentum = Final momentum
(Mass of the boat + Mass of the dog) * Initial velocity = (Mass of the boat + Mass of the dog) * Final velocity
Now, let's plug in the given values:
Mass of the boat = 40 lb
Mass of the dog = 10 lb
Initial velocity = 0 ft/s (since the boat is stationary)
Final velocity = v ft/s (since the dog is moving with velocity v)
Simplifying the equation, we get:
40 * 0 = 50 * v
0 = 50v
This means that the velocity of the boat-dog system after the dog starts moving is zero.
Therefore, the boat remains stationary, and the dog will be at the same distance from the shore as it was initially, which is 20 ft. The movement of the dog on the boat does not affect its distance from the shore.
I understand your frustration. Let me try to help you conceptually grasp the situation and solve the problem.
First, let's consider the initial state of the dog and the boat. The dog weighs 10.0 lbs and is standing 20 ft from the shore. The boat weighs 40 lbs. Since the boat is flat and there is no friction between it and the water, the dog is distributed on the boat in such a way that their center of mass is in the middle of the boat.
Now, as the dog walks 8.0 ft towards the shore, the boat will experience a shift in its center of mass as the dog's weight distribution changes. But you correctly pointed out that the overall center of mass of the system (dog + boat) will not change. This means that the boat will move in the opposite direction of the dog's movement, compensating for the shift in the center of mass.
To find the final position of the dog relative to the shore, we need to consider how the boat moves in response to the dog's movement. Since the dog moves 8.0 ft towards the shore, the boat will move 8.0 ft away from the shore. This means that the dog will end up 8.0 ft + 2.0 ft = 10.0 ft from the shore at the end of this time.
So, the answer is that the dog is 10.0 ft from the shore at the end of this time.
I hope this explanation helps you understand the concept behind solving this problem.