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Collision/Impulse question

  1. Mar 16, 2008 #1
    I'm unsure if I set this up correctly. It's a mulitple choice homework question and it doesn't match any possible answers.

    Question:
    A 0.605 kg ball drops vertically onto a floor, hitting with a speed of 34 m/s. It rebounds with an initial speed of 11 m/s. (a) What impulse acts on the ball during the contact? (b) If the ball is in contact with the floor for 0.0171 s, what is the magnitude of the average force on the floor from the ball?

    Attempt:
    Part A:
    I = mvf - mvi
    I = m(vf - vi)
    I = 0.605(11-34)
    I = -13.915 kg-m/sec

    Attempt:
    Part B:

    If the contact time = 0.0171 s, then:
    F = Δp/Δt = I/Δt = -13.915 kg-m/sec/0.0171 s
    F = -813.743 Nt
     
  2. jcsd
  3. Mar 16, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Incorrect: Realize that momentum is a vector. Direction (sign) counts! (Call up + and down -.)

    Right idea, but you need to fix part A first. Also: magnitudes are always positive.
     
  4. Mar 16, 2008 #3

    You're right about the magnitude! Thanks, I forgot that.

    So I have...

    Re-Attempt:
    Part A:

    I = mvf - mvi
    I = m(vf - vi)
    I = 0.605(11-(-34))
    I = 27.225 kg-m/sec

    Re-Attempt:
    Part B:

    If the contact time = 0.0171 s, then:
    F = Δp/Δt = I/Δt = 27.225 kg-m/sec/0.0171 s
    F = 1592.11 Nt
     
  5. Mar 16, 2008 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Looks good! (The abbreviation for Newtons is just N, not Nt.)
     
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