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Crash test dummies don't always wear seat belts.

  1. Mar 20, 2008 #1
    Crash test dummies don't always wear seat belts. If a 78 Kg dummy, in a car traveling at 53 Km/h, hits a concrete barrier, what is the force applied to the dummy?
     
  2. jcsd
  3. Mar 20, 2008 #2

    Doc Al

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    Staff: Mentor

    You don't have enough information. How quickly does the dummy come to rest?
     
  4. Mar 20, 2008 #3
    what other info should I look for?

    I don't have any additional info; just the 3 other questions from the page...do they all use the same type of formula?? what other info should I look for?

    1. What acceleration does Chris give his 0.48 kg Physics book, when he accidentally knocks it off a table with an impulse of 2.3 Ns?

    2. A force of 3420 N is applied to a 145 g baseball. If the baseball approaches the batter at 41.3 m/s and leaves at 44.2 m/s, how long was the bat in contact with the ball?

    #3: crash test dummies ?. {If a 78 kg dummy, in a car traveling at 53 Km/h, hits a concrete barrier, what is the magnitude of the force applied to the dummy?}

    4. How much force is applied to a 76 kg man’s legs when he jumps off a table 1.0 m above the floor, if it takes 0.11 s for him to stop? (HINT: Remember your kinematic equations.)
     
    Last edited: Mar 20, 2008
  5. Mar 20, 2008 #4
    Would it be this calc??
    Units of Momentum and Impulse
    Impulse, being the product of force and time, has units Ns. Momentum, we discovered this morning, has units of kgm/s. Since our equation states that changes in momentum are equal to the applied impulse, it follows that Ns must be equivalent to kgm/s. However, by convention, when talking about momentum we will always express our answers in units of kgm/s and when talking about impulse we will always use Ns.
    Calculations using the Momentum-Impulse Theorem
    Example:
    What is the force applied to a 6.56 kg bowling ball that is taken from rest to a speed of 1.16 m/s in 1.31 s?
    In this problem, we are given the following information:
    m = 6.56 kg
    vi = 0 m/s
    vf = 1.16 m/s
    t = 1.31 s
    and need to find the force. We can rearrange F t = mv, to get
    F = mv/t
    Remembering that v in this case represents the change in velocity, we can substitute to get:
    F = 6.56 kg • (1.16 – 0 m/s) / 1.31 s
    F = 5.81 kg • m/s/s (three significant figures)
    But, 1 kg ? m/s/s equals a Newton (N), our standard unit of force, so we would express our final answer as:
    F = 5.81 N
     
  6. Mar 20, 2008 #5

    Doc Al

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    Staff: Mentor

    Looks like these are impulse problems: [itex]I \equiv F\Delta t = \Delta (mv)[/itex]

    Insufficient information provided. You're given the impulse but not the time. The same impulse can be given quickly (high acceleration) or slowly (low acceleration).

    You have all the needed information.

    Insufficient information provided. Again, no time is given.

    You have all the needed information.
     
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