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Homework Help: How far apart were the centres of the two tunnels?

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A competitive mini-golf player notices an opponent's ball situated 50cm in front of a wall containing two tunnels. One tunnel leads to the hole, the other to a bunker. The player skillfully putts their ball due East at the opponent's ball with a velocity of 4m/s. The opponent's ball is knocked 20 degrees to the South of East with a velocity of 1.7m/s into the tunnel leading to the bunker. Meaning the player's ball travels into the second tunnel located North of East (to great applause from spectators). If the player's ball weighed 40g, and the opponent's ball weighed 45g, how far apart were the centres of the two tunnels?

    2. Relevant equations

    Velocity = displacement/time

    Force = G*(mass1*mass2)/(distance^2)

    3. The attempt at a solution

    Using the above formula, transpose so distance can be found.

    I'm not sure where to get the value for force though if this is the correct method.

    I have tried drawing a diagram myself but am stuck and don't know what to do.

    I hope someone can help me with this problem, work through it with me or point me in the right direction.

    Any help would be appreciated!
  2. jcsd
  3. May 2, 2013 #2

    Doc Al

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    Staff: Mentor

    I seriously doubt that you need to worry about the gravitational force between the two balls. :smile:

    Instead, consider the collision. What's conserved?
  4. May 2, 2013 #3
  5. May 2, 2013 #4

    Doc Al

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    Staff: Mentor

    Good. Use conservation of momentum to figure out the angle that the player's ball traveled after the collision. The rest is a bit of trig.

    I assume that the wall is oriented along a North-South line and is due east of the balls. (You should have a diagram!)
  6. May 2, 2013 #5
    Does the attached diagram represent all the above information correctly?

    If so, what do I need to do next?

    The question mentions that the ball hits the opponent's ball while travelling at 4m/s.
    Using momentum formula:
    =160g m/s

    Conservation of momentum:

    mv (opponent's ball)=160g m/s
    v=(160 g m/s)/(45 g)
    v(opponent's ball) = 3.6 m/s

    So, the opponent's ball should be travelling 3.6 m/s? Why does it say 1.7 m/s in the question?

    Sorry, I'm still lost on how to calculate the distance between the centres of the two tunnels.

    How does knowing the conservation of momentum help in answering the final question.


    Attached Files:

    Last edited: May 2, 2013
  7. May 2, 2013 #6

    Doc Al

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    Staff: Mentor

    Almost. The 20° angle is in the wrong place.

    That's the momentum of the player's ball and is thus the total momentum of the system. Note that momentum is a vector. The direction is east.

    You are told the speed and thus momentum of the opponents ball.

    You are misapplying conservation of momentum.

    Since momentum is a vector, break it into components. x (east) and y (north) components of total momentum must be separately conserved.

    Once you have the angle that the player's ball bounces off at, you can use trig to find where the straight line trajectories hit the tunnel.
  8. May 2, 2013 #7
    could you give me more clues/hints on what to do???


    I need more help. :(
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