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Collision/Momentum Question

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data
    A steel ball of mass 0.50 kg, moving with a velocity of 2.0 m/s [E], strikes a second ball of mass 0.30 kg, initially at rest. The collision is a glancing one, causing the moving ball to have a velocity of 1.5m/s [30 degrees N of E] after the collision. Determine the velocity of the second ball after the collision.


    2. Relevant equations
    m1v1 + m2v2 = m1v1' + m2v2'


    3. The attempt at a solution
    I've tried using the y-component to solve for the speed on v2y' but there's no direction given so there are two variables to solve for. The final speed of the second ball and the direction of the second ball after the collision. Using the y-component i had the equation:

    0 = (0.50)(1.5sin33) + (0.30)(v2')(siny)

    How can I solve for both the speed and the direction? I've tried subbing in v2 in terms of siny but I just can't seem to get the right answer. Can someone solve this problem and show me how you solved it? Thanks.
     
  2. jcsd
  3. Jul 13, 2008 #2

    Doc Al

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    Staff: Mentor

    Since momentum is a vector you can get two conservation equations by considering horizontal (E) and vertical (N) components separately.
     
  4. Jul 13, 2008 #3
    Yes, I've noticed that. I've tried separating the x and y components and subbing in for v2'. I've reached an answer however, the answer does not match the correct solution for the problem. It's either i'm doing something wrong or the book had a missprint. My answer was 18m/s [E 48 degrees S] while the answer in the book was 17m/s [E 47 degrees S] I've also carried at least four sig figs to ensure my accuracy.
     
  5. Jul 13, 2008 #4

    Doc Al

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    Staff: Mentor

    Please show how you got your answer. Those answers seem rather unlikely, as the final KE would be much greater than the initial KE. (Quite interesting that your answer is close to that of the book, though!)
     
  6. Jul 13, 2008 #5
    As the question does not mention the collision as an elastic one, it would be wrong to assume that it is. Therefore, I excluded the use of the kinetic energy theorem.
     
  7. Jul 13, 2008 #6

    alphysicist

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    Homework Helper

    Hi konceptz,

    Your result for the speed seems too large, which is what I believe Doc Al was referring to, and why he asked to see your work. Notice that if you plug your speed and angle into the y-equation you have in your original post, they don't work, so something seems to have gone wrong.

    (Also you have an angle of 30 degrees in the problem statement, but an angle of 33 degrees in that equation. Which one is right?)
     
  8. Jul 14, 2008 #7

    Doc Al

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    As alphysicist already explained, your answers are not consistent with conservation of momentum, which is all you need to solve this problem--so you must have made an error somewhere. (I did not mean to imply that you had to use conservation of KE to solve this problem--you don't.)
     
  9. Jul 14, 2008 #8
    Sorry, my speed was 1.7 not 17, forgot to type in the decimal. As for the degrees the correct one was 30. I've solved the problem now, thank you reminding me to include the x and y components.
     
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