Results 1 to 4 of 4

Math Help - one or two dimensions of collisions

  1. #1
    Senior Member
    Joined
    Jan 2007
    Posts
    477

    one or two dimensions of collisions

    Kevin has a mass of 78.5 kg and is skating with in-line skates.

    He sees his 22.50 kg younger brother up ahead standing on the sidewalk,
    with his back turned.

    Coming up from behind, he grabs his brother and rolls off at a speed of 1.95 m/s.

    Ignoring friction, find Kevin's speed just before he grabbed his brother.
    in m/s

    i'm not sure if this is for collisions in one or two dimensions, or both?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Right cant remember my physics very well but pretty sure its just...

    m_1(v_1)^2 = m_2(v_2)^2

    which in his case is 100(1.95^2)= 78.5{(v_2)^2}

    which gives v_2 = 2.2m/s
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2007
    Posts
    477
    i got it.

    i used (m1 + m2) / m1 * vf

    and got 2.51 m/s

    thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by rcmango View Post
    Kevin has a mass of 78.5 kg and is skating with in-line skates.

    He sees his 22.50 kg younger brother up ahead standing on the sidewalk,
    with his back turned.

    Coming up from behind, he grabs his brother and rolls off at a speed of 1.95 m/s.

    Ignoring friction, find Kevin's speed just before he grabbed his brother.
    in m/s

    i'm not sure if this is for collisions in one or two dimensions, or both?
    Quote Originally Posted by Deadstar View Post
    Right cant remember my physics very well but pretty sure its just...

    m_1(v_1)^2 = m_2(v_2)^2

    which in his case is 100(1.95^2)= 78.5{(v_2)^2}

    which gives v_2 = 2.2m/s
    We aren't told that energy is conserved, so we cannot assume it. Also, since this is an archetypal inelastic collision problem, I can guarantee that energy is not conserved.

    So consider momentum conservation. There are two objects before the collision and (effectively) only one afterward. Set a coordinate system with +x in the direction that Kevin is initially moving in. At the collision point there are no net external forces acting on the two brothers, so momentum is conserved during the collision. So:
    P_{0, tot} = P_{tot}

    m_1v_{01} + m2v_{02} = (m_1 + m_2)v_{12}

    Kevin's brother is not moving initially so we have that v_{02} = 0~m/s.

    m_1v_{01} = (m_1 + m_2)v_{12}

    v_{01} = \frac{m_1 + m_2}{m_1} \cdot v_{12}

    Thus
    v_{01} = \frac{78.5 + 22.50}{78.5} \cdot 1.95 = 2.50892~m/s

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Probability-collisions
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: May 4th 2010, 03:56 PM
  2. Expected number of collisions
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: December 1st 2009, 07:42 AM
  3. Mechanics, collisions help
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 26th 2009, 02:57 PM
  4. Successive Collisions
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: February 11th 2008, 04:39 PM
  5. Collisions
    Posted in the Advanced Applied Math Forum
    Replies: 0
    Last Post: February 10th 2008, 02:25 AM

Search Tags


/mathhelpforum @mathhelpforum