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Plz could you checkif i got the 2d collisions right?

  1. Feb 6, 2013 #1
    1. The problem statement, all variables and given/known data
    A 0.30 kg puck A, initially at rest on a frictionless horizontal surface, is struck by a 0.20 kg puck B
    that is initially moving along the x-axis with a velocity of 2.0 m/s. After the collision, the 0.20
    kg puck has a speed of 1.0 m/s at an angle of θ = 530 to the positive x-axis.
    a) Determine the velocity of the 0.30 kg puck after the collision.


    2. Relevant equations

    total mommentum before collision + total momentum after collision



    3. The attempt at a solution

    total momentum before collision in x

    puck A

    0.2 x 2 = 0.4

    puck B

    0.3 x 0

    total = 0.7kgm/s


    total momentum before collision in y = 0 for all

    Total momentum after collision x

    puck A

    0.2 x 1 x cos 53=0.12

    puck B
    0.3 x V x cos 53

    Total =0.7kgm/s

    Total momentum after collision y

    puck A

    0.2 x 1 sin 53 = 0.1597

    puck B

    0.3 x V x sin53

    Total =0

    Resolve in x for puck B

    0.3 x V x cos 53
    Let Vcos 53 = Vx

    0.3 x Vx + 0.12 = 0.7
    Vx = 1.93333m/s

    Resolve in y for puck B

    0.3 x V x sin 53
    Let Vsin 53 = Vy

    0.3 x Vy + 0.1597 = 0
    Vy = -0.53

    V2=Vx2 + Vy2

    V2 = 1.933332+(-0.53)2

    V=√4.02

    v=2.005m/s


    Is this right???
     
  2. jcsd
  3. Feb 6, 2013 #2

    PeterO

    User Avatar
    Homework Helper

    Probably not correct - see red above.
     
  4. Feb 6, 2013 #3
    actually i knew there was something wrong
    I did all of this all again since the pc froze.
    previously i got it right
    i knew there was a glitch since the answers looked different.
     
  5. Feb 6, 2013 #4
    total momentum before collision in x

    puck A

    0.2 x 2 = 0.4

    puck B

    0.3 x 0 = 0

    total = 0.4kgm/s


    total momentum before collision in y = 0 for all

    Total momentum after collision x

    puck A

    0.2 x 1 x cos 53=0.12

    puck B
    0.3 x V x cos θ

    Total =0.4kgm/s

    Total momentum after collision y

    puck A

    0.2 x 1 sin 53 = 0.1597

    puck B

    0.3 x V x sinθ

    Total =0

    Resolve in x for puck B

    0.3 x V x cos θ
    Let Vcos θ = Vx

    0.3 x Vx + 0.12 = 0.4
    Vx = 0.93333m/s

    Resolve in y for puck B

    0.3 x V x sin θ
    Let Vsin θ = Vy


    0.3 x Vy + 0.1597 = 0
    Vy = -0.53

    V2=Vx2 + Vy2

    V2 = 0.933332+(-0.53)2

    V=√0.5902

    v=0.768m/s


    is this better?
     
  6. Feb 6, 2013 #5

    PeterO

    User Avatar
    Homework Helper

    I don't expect it to be, from the way you calculated.

    Step 1: calculate initial momentum. check you have done that.

    Step 2: calculate components of final momentum of the block you have been given. Check you did that.

    Step 3: deduce the components of the momentum of the other block. - Not the way you did.

    Step 4: Add those components (as vectors) to find the actual momentum of that other block.

    Step 5: Calculate the velocity of that block.



    eg: I am about to randomly select some values for example - so as not to solve your actual question for you.

    Let's pretend the the momentum of the incoming block was 5 units in the x direction.

    Lets also assume that the components after collision were 3.5 in the x direction, and 1.5 in the y direction.

    The components of MOMENTUM of the other block would be
    1.5 in the x direction [since 1.5 + 3.5 = 5; the original value was 5]
    -1.5 in the y direction [since 1.5 + -1.5 = 0; the original value was 0]

    So the net momentum = 1.5√2 at 45 degrees

    Given the second mass was 0.3 kg, that meant the velocity must have been 5 m/s at an angle of 45 degrees with the positive x-axis.

    Note: I chose values that were very easy to work with - even to havng equal components so I could make use of the 1,1,√2 Pythagorean triangle.

    I expect your values to be a bit more messy

    Note: the use of an angle of 53 degrees usually implies to use sin and cosine values of 0.8 and 0.6 rather that the complicated/inconvenient, accurate figures, as the angles in a 3,4,5 Pythagorean are just over 53 degrees and just under 37 degrees.


    NOTE ALSO: there is a possibility that when you have found the final answer using momentum conservation, it may be inconsistent with kinetic energy calculations [there may appear to be an increase in total kinetic energy] The problem was that the question setter may have just chosen the speed and angle of the 0.2 kg mass unwisely.
     
  7. Feb 6, 2013 #6
    thank you very much that was very detail
     
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