P A Q

• Q : A 10.0-lb dog is standing on a flatboat so that he is 20.0 ft from the shore. He walks 8.00 ft on the boat toward the shore and then halts. The boat weights 40.0 lb and one can assume there is no friction between it and the water. How far is the dog then from the shore?
  
  Note1: The answer is not 20-8 = 12 ft from the shore because as the dog moves forward on the boat he pushes the boat a certain distance back.
  Note2: There is probably something to do with momentum going on here.
• A : The length covered by the dog on the boat is its velocity vd (wrt the boat) times the time T during which is walks. If vb is the boat velocity relative to the water, the boat recedes from the shore by vbT The total momentum is zero so Mb*vb + md(vd+vb) = 0 (the dog's velocity relative to the shore is vb+vd, vb and vd being obviously anti parallel)
  So (Mb+md)vb = -mdvd = -md 8/T
  The boat has backed by vbT = -8md/(Mb+md) so that the dog position wrt the shore is closer by
  8(1-md/(Mb+md)) = 8Mb/(M+md) and its new distance is 20 -8Mb/(M+md) = 20 -8*40/50= 13.6 ft
  
  Note: exactly the same calculation can be made using distances instead of velocities and bypassing T. It gives the same result and this is the correct version of the other responder's idea which is wrongly implemented in his answer

Labels: Momentum

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