A 10m boat of mass 290 kg (of uniformly distributed mass), 100m away from shore has two people standing on its ends. The one standing on the end nearer to the shore is of 50kg and the other is of 60kg. If both of them start to walk and meet at the centre of the boat; then how far are they from the shore? The conditions are ideal and there is no resistance or breeze.
2007-02-24 02:12:20 · 4 answers · asked by Cephalic 3 in Science & Mathematics ➔ Physics
Hmmm, I'm not sure if you worded your question correctly.
I don't see the purpose of the given masses in the solution to the problem. You would need to give the initial velocity, or a constant acceleration to solve this.
Assuming they have the same constant velocity/acceleration, they would meet at the center of the boat, 150m from shore.
2007-02-24 02:24:47 · answer #1 · answered by ? 3 · 0⤊ 0⤋
â since total momentum of the system is 0 it will stay 0 no matter what men are doing in the boat; and position of the mass center of the system will also be unchangeable; thus the equation for mass center is:
⣠100*290 +(100 –10/2)*50 +(100 +10/2)*60 = x*290 +x*50 +x*60, hence
x=100.125m; and what says the book back?
2007-02-24 14:27:42 · answer #2 · answered by Anonymous · 0⤊ 0⤋
They will be a little nearer than 95m, due to more force exerted by the heavier guy pushing the boat slightly forward.
2007-02-24 10:33:36 · answer #3 · answered by Norrie 7 · 0⤊ 0⤋
There is no chance for zero resistance because there would be no life for those guys if there is no air for breathing and also pressure of the air for maintenance of content of mass in those bodies of the guys. there would be no condition as ideal in any experiment of any field of science !!!
2007-02-25 03:34:04 · answer #4 · answered by Anonymous · 0⤊ 2⤋