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Inelastic collision

  1. Jan 20, 2010 #1
    A 1900 kg car moving at 20.0 m/s crashes into a brick wall and is stopped. The driver's mass is 70.0 kg, and although the driver is not wearing his seatbelt, the air bag thankfully slows him to a stop in 0.100 s. What is the magnitude of the average force that the air bag exerts on the driver?
    ______________

    (Side Note: Air bags and seat belts act to slow you to a rest in a longer time. By increasing the time, the force on your body decreases. An average force exceeding roughly 2.00 x 105 N will probably kill you.)

    How would you do this problem?
     
  2. jcsd
  3. Jan 20, 2010 #2
    Have you done impulse and change in momentum? The driver has an impulse supplied by the air bag that causes him to change momentum.
     
  4. Jan 20, 2010 #3
    so what would the equation look like? how would you solve it?
     
  5. Jan 21, 2010 #4
    Have you studied F*t...?
     
  6. Jan 24, 2010 #5
    So i read up on this and did this. It was marked wrong can someone help me figure out why?

    m*v = f*t
    (1900 + 70)(20) = F * (0.1)
    39400= 0.1F
    F=394000
     
  7. Jan 24, 2010 #6
    I believe they were asking about the force on the driver (70 kg) and he was moving at 20 m/s and then 0 m/s (stopped). So the driver's change in momentum is due to a force acting over a 0.1 second time period.
     
  8. Jan 24, 2010 #7
    so would it be like this then?

    m*v = f*t
    (70)(20) = F * (0.1)
    1400= 0.1F
    F=14000
     
  9. Jan 24, 2010 #8
    yeps...

    I am assuming on the left hand side of your equation you had final momentum being zero. So really you end up with a force that is (-). Which basically means the driver was accelerated in the opposite direction as his initial motion.
     
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