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Homework Help: What is the velocity after the collision?

  1. Oct 4, 2004 #1
    help please :(

    A 15.0 kg object is moving east at a velocity of 7.0 m/s when it collides with a 10.0 kg statinary object. After the collision the 15.0 object is travelling at a velocity of 4.2 m/s 20.0° S of E.

    What is teh velocity of the 10.0 kg object after the collsion.

    so far this is what i did, can you please tell me what i am doing worng? Thank-you.

    15 kg object → a y→y component
    10 kg object → b x→x component

    Pa + Pb = P'a + P'b

    Before Collision
    Pa=mv
    Pa=(15.0 kg)(7.0 m/s E)
    Pa=105 Kgm/s

    After Collsion
    P'a=mv
    P'a=(15.0 Kg)(4.2 m/s 20° S of E)
    P'a=63 Kgm/s

    P'b=mv
    P'b=(10.0 Kg)(v)
    P'b=?

    Sin20 = P'ay/63 kgm/s
    P'ay=21.5 kgm/s

    Cos20°=P'ax / 63 kgm/s
    P'ax = 59.2 kgm/s

    Px=Squareroot[(105 kgm/s)^2 - (21.5 kgm/s)^2]
    Px=102.8 kgm/s

    P'bx=102.8 kg m/s - 59.2 kgm/s
    P'bx= 43.6 kgm/s

    P'b= squareroot[(43.6 kgm/s)^2 + (21.5 kgm/s)^2]
    P'b= 48.6 kgm/s

    P'b=mv
    v=P'b/m
    v=(48.6 kgm/s) / (10.0kg)
    v= 4.86 m/s

    CosX=(43.6 kgm/s)/(48.6 kgm/s)
    X= 26° N of E
     
  2. jcsd
  3. Oct 4, 2004 #2

    HallsofIvy

    User Avatar
    Science Advisor

    You are writing this in a very strange way: I would not, for example, say that "Pa=105 Kgm/s" since Pa is a vector quantity. Other than that, I get very close to what you do- perhaps a roundoff error problem.

    The way I would do this is write out the momentum vectors:
    before the collision they are
    <15*7, 0>= < 105, 0> and <0, 0> which adds, or course, to <105, 0>

    After the collision, the momentum vector of the 15 kg mass is
    <15*4.2*cos(20), 15*4.2*sin(-20)>= <59.2, -21.55>
    We must have <px, py>+ <59.20, -21.55>= <105, 0> so
    px= 105- 59.20= 45.80 and py= 0+ 12.55= 21.55
    Since the momentum vector of this 10 kg mass is <45.80, 21.55>, its velocity vector is <4.580, 2.155>. If you want that in terms of speed and angle, the speed is
    [itex]\sqrt{4.580^2+ 2.155^2}= 5.06[/itex] m/s and the angle is [itex]tan^{-1}\frac{2.155}{4.58}= 25.2[/itex] degrees north of east.
     
  4. Oct 4, 2004 #3
    thanks...i figured out my error. I should have used 105 Kgm/s for Px.
     
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