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Collisions in 2 dimensions (please help)

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data

    A 0.321 kg puck, initially at rest on a horizon-
    tal, frictionless surface, is struck by a 0.227 kg
    puck moving initially along the x axis with
    a speed of 2.51 m/s. After the collision, the
    0.227 kg puck has a speed of 1.51 m/s at an
    angle of 31 degrees to the positive x axis.
    Determine the velocity of the 0.321 kg puck
    after the collision. Answer in units of m/s.

    2. Relevant equations

    Final Momentum in the x direction=Original Momentum in the x direction

    3. The attempt at a solution

    I do not understand collisions at all. My teacher didn't go over this and reading about this online is confusing the hell out of me, especially since I can't find an example of hitting something at rest and going off at an angle online so I have no idea how I'm supposed to solve this.
     
  2. jcsd
  3. Dec 5, 2007 #2
    No octopus?
     
  4. Dec 8, 2007 #3
    Why would there be an octopus?
     
  5. Dec 9, 2007 #4
    A sketch showing the overall geometry would be a good start. Momentum has both magnitude and direction, thus momentum is a vector quantity. When we deal with momentum equations we need to keep in mind that we are working with vector equations.

    For this problem and assuming momentum is conserved, the before colision momentum and the after colision momentums will be equal. This is indicated as follows:

    [tex]\underbrace{m_1 \cdot v^b_1 + m_{2}\cdot v^{b}_{2}}_{Before\;Colision} = \underbrace{m_{1}\cdot v^{a}_{1} + m_{2}\cdot v^{a}_{2}}_{After \; Colision} [/tex]

    The superscripts indicate efore and [a]fter the colision values.

    Consider the "before colision" side of the equation. The problem itself gives "A 0.321 kg puck, initially at rest on a horizontal, frictionless surface, is struck by a 0.227 kg puck moving initially along the x axis with a speed of 2.51 m/s." Therefore:

    [tex] m_1 \cdot v^b_1 = 0 \;[/tex] and the before colision momentum is:
    [tex] m_2\cdot v^b_2 \;[/tex] Which is a vector directed along the x-axis.

    Consider the "after colision" side of the equation. The problem indicates "After the collision, the 0.227 kg puck has a speed of 1.51 m/s at an angle of 31 degrees to the positive x axis." This puck has a momentum vector given as:

    [tex] m_2\cdot v^a_2 \;[/tex]

    However, this momentum vector is oriented at 31 degrees which means it will have x and y components (that you will have to figure out). Hint: The [itex]\sin [/itex] and [itex]\cos [/itex] functions may be of some use here.

    The other mass, [itex]m1 [/itex] will also have a momentum vector as a result of the colision. Which is [itex]m_1 \cdot v^a_1\;\; [/itex] directed at some unknown angle. Let's call that angle [itex] \vartheta [/itex]. Again, x and y componets of this vector are required.

    If we project all the "before colision" and all "after colision" vectors to the coordinate axes x and y and apply the conservation of momentum equation along each axis we'll end up with 2 equations "x-axis" momentum and "y-axis" momentum which will look like:

    x-axis [itex] \{ m_2 \cdot v^b_2 = m_2 \cdot v^a_2\cos{31} + m_1 \cdot v^a_1\cos{\vartheta} \}[/itex]

    y-axis [itex]\{ 0 = m_2 \cdot v^a_2\sin{31} + m_1 \cdot v^a_1\sin{\vartheta} \}[/itex]

    I've probably said to much already but assign the variables as follows:
    [tex] m_1 =0.321 kg[/tex]
    [tex] v^{b}_{1} = 0.0 m/s [/tex]
    [tex] m_2 = 0.227 kg [/tex]
    [tex] v^{b}_{2} = 2.51 m/s [/tex]

    [tex] v^{a}_{2} = 1.51 m/s [/tex]

    Solve the equations however you care to (simultaneously) and there ya go...

    I'm not sure why latex is generating a white background? perhaps one of the moderators could help ?
     
    Last edited: Dec 9, 2007
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