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Rise in temperature after a collision

  1. Mar 24, 2013 #1
    1. The problem statement, all variables and given/known data
    "A body of mass m, moving with velocity v, collides with a body of mass 2m at rest, in a head-on collision. The coefficient of restitution is 1/3. If the 2m body has a specific heat c, and if it is assumed that the two bodies share the heat generated in the collision equally (not a very reasonable assumption), and that no heat is lost (a ridiculous assumption), how much does the temperature of the 2m body rise? (Keep fractions throughout in solution.)"
    2. Relevant equations
    $$\text{Coefficient of restitution}\ =~e~= \frac{v_{2F} - v_{1F}}{v_{1I} - v_{2I}}$$
    $$p_I = p_F$$
    $$\Delta K_2 = Q_2$$
    3. The attempt at a solution
    Since it is head on, this is a one dimensional problem, with ##v_{2I} = 0## and ##v_{1I} = v##. I started off by resolving my momentum equation into
    $$mv = -mv_{1F} + 2mv_{2F}$$
    (1) ##v_{1F} = 2v_{2f} - v##
    I then set my restitution equation equal to 1/3 and substituted (1) in.
    [tex] \frac{1}{3}\ = \frac{v_{2f} - 2v_{2f} + v}{v} [/tex]

    Which simplifies to

    (2) $$ v_{2F} = \frac{2v}{3} $$

    I then expand my Q equation to

    (3) $$ \frac{1}{2}\ 2 m (v_{2f})^2 = 2 m c \Delta T $$

    I then plug (2) into (3) and solve for ## \Delta T ## :

    $$ \Delta T\ = \frac{2v^2}{9c} $$

    However, the answer has a fraction of 2/27 rather than 2/9. I'm off by a factor of a third, and I don't know why.
     
    Last edited: Mar 24, 2013
  2. jcsd
  3. Mar 25, 2013 #2
    Why the minus sign?
     
  4. Mar 25, 2013 #3
    Because it is a one dimensional collision. The first object has to bounce backwards because there is no other place for it to go. Even if I omit the minus sign, I get 8/81 for the fraction.
     
  5. Mar 25, 2013 #4
    It does not have to go backward. It can continue forward at a reduced speed. You do not have to guess what is going to happen. Use the equations.

    Then you must have made another error somewhere. Show your new derivation.
     
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