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Speed and Direction after Collision

  1. Nov 27, 2007 #1
    A 20 g ball of clay traveling east at 2.0 m/s collides with a 30 g ball of clay traveling 300 south of west at 1.0 m/s . What are the speed and direction of the resulting 50g blob of clay?



    Relevant equations
    Pi = Pf



    The attempt at a solution
    (Px)i = mF(-Vfcostheta) + mP (Vp)
    = (0.03 kg)(-1.0)(cos30) + (0.02kg)(2.0)
    = 0.014019

    (Py)i = mf(Vfy)i + mP(Vpy)i = mf(-vfsintheta) + 0
    = (0.3)(1.0)(sin30) = 0.15 kg.m/s

    (Px)f = (Px)i
    (Vx)f = (Px)i / (mF + mP) = 0.014019 / (0.03 +0.02)
    = 0.28038 m/s
    (Py)f = (Py)i
    (Vy)f = (Py)i / (mF + mP) = 0.15 / (0.02 +0.03) = 3.0 m/s

    Direction = tan-1 [(Vy)f / (Vx)f] = tan-1 [ 3 / 0.28038] = 84.66 Degrees

    V = SquareRoot[ (Vx)f^2 + (Vy)f^2) = Squareroot[ 3^2 + 0.28038^2] = 3.013 m/s



    IS this Right??
     
  2. jcsd
  3. Nov 28, 2007 #2
    What does that mean "300 south of west"? 300 degrees?

    Start with simple conservation of momentum:

    [tex]m_1 v_1 + m_2 v_2 = (m_1 + m_2)v[/tex]

    Now in vector notation:

    [tex]m_1 \vec v_1 + m_2 \vec v_2 = (m_1 + m_2) \vec v [/tex]

    Given the specifics of the problem, we can conclude that the motion is limited to two dimensions: east-west and north-south. We'll let the east-west be the x component and the north-south be the y component.

    So:

    [tex]\vec v_1 =v_1\cos{\theta_1}\hat x+v_1\sin{\theta_1}\hat y[/tex]
    [tex]\vec v_2 =v_2\cos{\theta_2}\hat x+v_2\sin{\theta_2}\hat y[/tex]
    [tex]\vec v =v\cos{\theta_3}\hat x+v\sin{\theta_3}\hat y[/tex]

    Now we can write the whole mess out...

    [tex]m_1 (v_1\cos{\theta_1}\hat x+v_1\sin{\theta_1}\hat y) + m_2 (v_2\cos{\theta_2}\hat x+v_2\sin{\theta_2}\hat y) = (m_1 + m_2) (v\cos{\theta_3}\hat x+v\sin{\theta_3}\hat y) [/tex]

    Now we can group the individual vector components:

    [tex]m_1 v_1\cos(\theta_1)\hat x + m_2 v_2\cos(\theta_2)\hat x = (m_1 + m_2) v\cos(\theta_3)\hat x[/tex]
    [tex]m_1 v_1\sin(\theta_1)\hat y + m_2 v_2\sin(\theta_2)\hat y = (m_1 + m_2) v\sin(\theta_3)\hat y [/tex]

    Now you have two equations and two unknowns.

    Reduce:

    [tex]\frac{m_1 v_1\cos(\theta_1) + m_2 v_2\cos(\theta_2)}{m_1 + m_2} = v\cos(\theta_3)[/tex]

    [tex]\frac{m_1 v_1\sin(\theta_1) + m_2 v_2\sin(\theta_2)}{m_1 + m_2} = v\sin(\theta_3) [/tex]

    Now it's easy to find the angle. Just divide the equations, resulting in this:

    [tex]\frac{m_1 v_1\sin(\theta_1) + m_2 v_2\sin(\theta_2)}{m_1 v_1\cos(\theta_1) + m_2 v_2\cos(\theta_2)}=\tan{\theta_3}[/tex]

    To find [tex]v[/tex], square both equations and add them together...

    [tex]v^2\cos^2{\theta_3}+v^2\sin^2{\theta_3}=...[/tex]

    Use the trig identity:

    [tex]\cos^2{\theta}+\sin^2{\theta}=1[/tex]

    [tex]v^2(\cos^2{\theta_3}+\sin^2{\theta_3})=...[/tex]

    [tex]v^2=...[/tex]
     
  4. Nov 29, 2007 #3
    sorry it's 30 degrees
     
  5. Nov 29, 2007 #4
    i got an angle of 52 degrees and a velocity of 0.94 .. is that right:S
     
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