1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Using Newton's Second Law for a System of Particles

  1. Mar 2, 2008 #1
    The Problem:
    A 4 kg dog stands on a 16 kg flatboat at distance 12 m from the shore. The dog walks 5 m along the boat toward the shore and then stops. Assuming no friction between the boat and the water, find how far the dog is then from the shore.
    The dog's displacement is towards the shore. The boat's displacement is away from the shore.

    My work:
    I set the shore as the origin.
    Then, ((4 kg x 12 m) + (16 kg x 12 m))/(4 kg + 16 kg) = 12 m for original center of mass. ((4 kg x 7 m) + (16 kg x 12 m))/(4 kg + 16 kg) = 11 m.
    So I get a 1 m difference.

    My question: does it mean that the dog is currently 7 m + 1 m away from the shore, or does it mean that the dog is 12 m - 1 m away from the shore?
     
  2. jcsd
  3. Mar 2, 2008 #2
    i think the answer will be 8.25 m since when the dog walks towards the shore. the boat will move away from the shore according to the principle of momentum conservation.
     
  4. Mar 2, 2008 #3

    kdv

    User Avatar

    Is the dog initially in the middle of the boat? You are using this here.
    The last calculation is wrong. How di dyou get 7 meters?? And you are assuming that the boat has not moved! No, you must call the new position of the dog "x". It's an unknown. The center of the boat has moved but since the dog walked 5 meters relative to th eboat, the center of the boat is now at x+5 meters from the shore.
    Use that the center of mass has not moved and solve for x.
     
  5. Mar 2, 2008 #4
    If I set the new position of the dog as:

    ((4 kg x (X+5) m) + (16 kg x 12 m))/(4 kg + 16 kg) = ? m

    Then solve for X?
     
  6. Mar 2, 2008 #5

    kdv

    User Avatar

    No. You have again treated the boat as if it had not moved!

    As for "x", use it to represent the position of the dog. Then the position of the center of the boat is x+5 .
     
  7. Mar 2, 2008 #6
    Alright, I think I have my answer.

    Total distance of the dog moving toward the shore is: 5 m - [tex]\Delta[/tex]X_boat;
    The final distance of the dog is: X_dog_initial - 5 m + [tex]\Delta[/tex]X_boat;
    X_dog_intial = 12 m
    After some algebra: [tex]\Delta[/tex]X_boat = [M_dog/(M_dog + M_boat)] x (5 m);

    [tex]\Delta[/tex]X_boat = [(4 kg)/(4 kg + 16 kg)] x (5 m) = 1 m;

    X_dog_initial - 5 m + [tex]\Delta[/tex]X_boat;
    12 m - 5 m + 1 m = 8m
     
  8. Mar 2, 2008 #7

    kdv

    User Avatar

    And you can double check your answer by calculating the final position of the center of mass ans see that it is the same as before.

    (and note that the dog has moved with respect to the shore by 4 times more distance than the boat has moved which makes sense since the boat is 4 times more massive)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Using Newton's Second Law for a System of Particles
Loading...