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Mass question

  1. Oct 18, 2005 #1
    Okay in an inelastic collision the formula the book taught me to use is this one:
    m1v1 + m2v2 = (m1 + m2)V'

    Now i've done every single problem in the book including the extra exercises, but of course when it comes to doing the homework they have to ask you questions which they haven't taught you how to work out. It feels to me like a 12 years old, if given the formulas could do any of these problems so the only thing they can do to make it harder is to ask you questions which you've never worked on before. Every question in the book relating to collisions involved 2 objects.
    One of them with a velocity, coming from an angle hitting a stationary one, or both of them coming from different sides of the X-axis. But how do you work out a question where they are both coming from the same side of the axis?

    Here's the question:
    A football player on Team A moving at a velocity of 8m/s and an angle of 49 degree below the horizontal (below the x-axis) collides with a player on Team B. The player on Team B is moving at 7.5 m/s at an angle of 41 degree below the horizontal and has a mass of 110 kg. After the collision, the two players remain in contact and move along the horizontal. what is the mass of the player on Team A?

    Now you can't work this problem out along the X-axis as you'd have 2 unknown variables: m1 & V'.
    So then you try to do it along the Y-axis:
    m1*8*sin-49 + 110*7.5*sin-41 = 0
    -6.04*m1 - 541.25 = 0
    -6.04*m1 = 541.25
    m1 = -89.61

    Obvioustly this isn't right unless this guy has figured out how to float around.

    I'm sure it's something simple like you have to add a negative somewhere in the formula, but that's not written in my book.

    Anyone? Thanks in advance
  2. jcsd
  3. Oct 18, 2005 #2


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    If both the players come from below the x-axis and collide at a point on the x-axis, then they both have positive momentum in the y-direction (upwards). So, using conservation of momentum, you know you can never end up with zero momentum in the y-direction. No wonder you ended up with negative mass.
    Do you have a drawing of the problem? The vertical momenta should cancel out.
  4. Oct 18, 2005 #3
    Unfortunately there is no drawing for the problem.

    So since they are coming from the same quadrant, how does one solve this problem without getting a negative mass?
  5. Oct 19, 2005 #4


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    You don't. Your approach was correct, but this problem does not have a solution the way interpret it. As I said: A system with positive momentum in the y-direction will have the same positive momentum in the y-direction after collision. So I think the problem is a bit different, more likely we have for object (player) A:
    [tex]p^A_x=m_A v^A_x = 8 m_A \cos (49^\circ), \qquad p^A_y=8 m_A \sin(49^{\circ})[/tex]
    and for B:
    [tex]p^B_x=m_B v^A_B = (7.5)(110) \cos (41^\circ), \qquad p^B_y=-(7.5)(110)\sin(41^{\circ})[/tex]

    So A is moving in the positive x and y directions, but B is moving in the positive x- and negative y-direction. Do you have a way to check our answer?
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