Solving Problems Involving Dog Walking on a Boat

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The problem involves a dog running on a boat that is initially adjacent to a dock, with the boat being H times heavier than the dog. When the dog moves towards the dock, the boat shifts in the opposite direction to maintain the center of mass. The key to solving this problem is understanding that for every unit the dog moves, the boat moves a fraction of that distance based on the weight ratio. The final distance the dog gets to the dock is calculated as L/(1+H), where L is the length of the boat. This illustrates the principles of conservation of momentum and center of mass in a frictionless environment.
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hi all, here's the problem
"A dog sits on the left end of a boat of length L that is initially adjacent to a dock to its right. The dog then runs toward the dock, but stops at the end of the boat. If the boat is H times heavier than the dog, how close does the dog get to the dock?

Ignore any drag force from the water. "

So, on the internet I found the answer that it's L/(1+H) but it didn't provide any explanation

could someone help me understand how to approach problems like this. Basically I'm really stuck on problems like " walking on a block on a friction less surface " . So I know it's got something to do with the fact that the Center of Mass doesn't change position, only the block ( boat in that case ) and the object moving( dog). Could someone help me out with this ? thank you so much if you do :)
 
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KKazaniecki said:
hi all, here's the problem
"A dog sits on the left end of a boat of length L that is initially adjacent to a dock to its right. The dog then runs toward the dock, but stops at the end of the boat. If the boat is H times heavier than the dog, how close does the dog get to the dock?

Ignore any drag force from the water. "

So, on the internet I found the answer that it's L/(1+H) but it didn't provide any explanation

could someone help me understand how to approach problems like this. Basically I'm really stuck on problems like " walking on a block on a friction less surface " . So I know it's got something to do with the fact that the Center of Mass doesn't change position, only the block ( boat in that case ) and the object moving( dog). Could someone help me out with this ? thank you so much if you do :)

If the dog moves 1 unit distance to the right, the boat beneath his feet must move 1/H unit distance to the left to keep the center of gravity at the same place. The net movement of the dog relative to the boat is 1+1/H. The net movement of the boat relative to the water is 1/H.

Express this ratio as boat movement (relative to dock) divided by dog movement (relative to boat)

\frac {\frac {1} {H}}{1 + \frac {1}{H}} = \frac {\frac {1} {H} }{\frac{H}{H} + \frac {1}{H}} = \frac {\frac {1} {H} }{\frac{H + 1}{H}} = \frac{1}{H+1}

If the dog moves distance L from the left end of the boat to the right, how far does the right end of the boat move away from the dock?
 
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