A 4.5 kg dog stands on an 18 kg flatboat, and is 6.1 m from the shore. He walks 2.4 m along the boat towards the shore, and then stops. Assuming there is no friction between the boat and water, find how far the dog is from the shore.
Imagine the dog moves leftward and the boat moves rightward, but does the center of mass of the boat + dog system move?
2006-10-19 16:41:47 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The c.g. of the system remains unchanged, so
4.5*2.4 = 18x
x = 0.6
6.1-2.4+0.6 = 4.3
The boat moves 0.6 m farther from shore, moving the dog's original position to 6.7 m from the shore. The dog walked 2.4 m towards the shore, so he ends up 4.3 m from the shore.
2006-10-19 17:12:33 · answer #1 · answered by Helmut 7 · 1⤊ 0⤋
As the dog moves toward the shore the boat will move away. When the dog stops- its momentum will cancel the momentum of the boat- the dog will still be 6.1 meters from the shore!
2006-10-19 23:49:10 · answer #2 · answered by Anonymous · 0⤊ 1⤋
6.1m, the boat moved under the dogs feet because there was no friction.
2006-10-19 23:44:19 · answer #3 · answered by nkarasch 2 · 0⤊ 1⤋
8.5 m at zero friction
2006-10-19 23:43:49 · answer #4 · answered by Anonymous · 0⤊ 0⤋