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Average force/impulse/collision problem

  1. May 13, 2008 #1
    [SOLVED] Average force/impulse/collision problem

    1. The problem statement, all variables and given/known data

    A 75-kg ice skater moving at 10 m/s crashes into a stationary skater of equal mass. After the collision, the two skaters move as a unit at 5 m/s. The average force that a skater can experience without breaking a bone is 4500 N. If the impact time is 0.1 s, does a bone break?

    2. Relevant equations

    [tex]\overrightarrow{I} = \Delta \overrightarrow{p} = \overline{F} \Delta t[/tex]
    [tex]\Rightarrow \overline{F} = \frac{\Delta \overrightarrow{p}}{\Delta t}[/tex]

    Also ...

    [tex]\Delta p = p_f - p_i = m_1v_{1f} + m_2v_{2f} - m_1v_{1i} - m_2v_{2i}[/tex]

    However, for a perfectly inelastic collision, [itex]v_f = v_{1f} = v_{2f}[/itex]. Therefore ...

    [tex]\Delta p = (m_1 + m_2)v_f - m_1v_{1i} - m_2v_{2i}[/tex]

    3. The attempt at a solution

    The answer given in the back of the book says that the average force is 3750 N so that no, bones do not break.

    [itex]m_1[/itex] = moving skater; [itex]m_2[/itex] = stationary skater

    [tex]\overline{F} = \frac{(75 kg + 75 kg)(5 m/s) - (75kg)(10m/s) - (75kg)(0 m/s)}{0.1s} = 0 N[/tex]

    What have I done wrong? Thank you for your help.
  2. jcsd
  3. May 13, 2008 #2


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    Staff Emeritus
    Science Advisor
    Gold Member

    Here you have calculated the change in momentum of the entire system (i.e. the change in momentum of both skaters), which is zero as it should be since momentum is conserved!

    Instead, what you need to calculate is the change in momentum of one of the skaters.
  4. May 13, 2008 #3

    Wow ... That was really foolish of me! Thank you!
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