1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

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

    P'b=(10.0 Kg)(v)

    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

    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


    User Avatar
    Staff Emeritus
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?