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 13:48:45 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
If it's a frictionless system with no initial velocity, the center of mass doesn't move. (Conservation of momentum, which = 0 in this case.) As the dog applies to his body the force F needed to reach some speed, the reaction force -F is applied to the boat. With 4 times the dog's mass, the boat moves with 1/4 the acceleration, speed and distance the dog does. So he moves 0.8*2.4 m toward the shore while the boat moves 0.2*2.4 m away from it. His distance is 6.1 - 0.8*2.4 m.
2006-10-20 12:14:05 · answer #1 · answered by kirchwey 7 · 1⤊ 0⤋