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Center of Mass Movement in a perfectly inelastic collision

  1. May 2, 2009 #1
    1. There are two 50 kg skaters in a friction-less horizontal surface. Both have velocities of |V|=10 m/s, in opossite directions (that is, skater 1's velocity is 10 m/s i, while skater 2's velocity is -10 m/s i). They both skate parallel to each other with a distance of 1,5m. Skater 1 has a stick, whose mass is [tex]\approx[/tex]0kg and 1,5m in lenght. In one point, Skater 2 will grab the stick.

    The question is to describe the position and velocity of the center of mass just before and just after the collision.




    2. Because the only external forces of the skater 1-skater 2 system are each skater's weight P and normal N, I found that [tex]\sum[/tex]F[tex]_{external}[/tex]=0. Therefore, [tex]\vec{P}[/tex] (momentum) is constant and therefore conserves. Because after one of the skaters holds the stick, both skaters will have the same speed, I assumed this is a perfectly inelastic collision. So, if skater 1's mass is m1, and skater 2's mass is m2[tex]\Rightarrow[/tex] m1[tex]\vec{V1i}[/tex] + m2[tex]\vec{V2i}[/tex] = (m1+m2)[tex]\vec{Vf}[/tex].

    I can find the position of the center of mass using [tex]\vec{r}[/tex][tex]_{CM}[/tex]= (m1[tex]\vec{r1}[/tex] + m2[tex]\vec{r2}[/tex])/(m1+m2).

    And its velocity with [tex]\vec{V}[/tex][tex]_{CM}[/tex]= (m1[tex]\vec{V1}[/tex] + m2[tex]\vec{V2}[/tex])/(m1+m2)



    3. The attempt at a solution

    Here's the thing. The problem asks to give the position of the center of mass and its velocity before and after the collision. In any case, using the conservation of momentum and the equations to find [tex]\vec{r}[/tex][tex]_{CM}[/tex] and [tex]\vec{V}[/tex][tex]_{CM}[/tex], I find that the position and velocity of the center of mass are the same just before and just after the collision, and particularly, the velocity is 0 m/s.

    Am I doing something wrong? Because, intuitively, in a situation like this I expect the skaters to begin a circular motion with center in the center of mass. What am I missing?
     
  2. jcsd
  3. May 3, 2009 #2

    Redbelly98

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    You're correct that the CM velocity is 0 m/s. The two skaters have opposite momenta of ±500 kg*m/s, making vCM zero. And yes, the skaters will move in a circle after the collision.
     
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