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Impulse and Momentum of a falling student

  1. Nov 13, 2004 #1
    Can anyone assist, I have tried various formulas on both problems with no luck. Thanks for any assistance or feedback.

    1) A student (m = 59 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.020 s. The average force exerted on him by the ground is +21000 N. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground. The answer should be 2.59m

    2) A 0.295-kg projectile is fired with a velocity of +585 m/s at a 1.00-kg wooden block that rests on a frictionless table. The velocity of the block, immediately after the projectile passes through it, is +40.4 m/s. Find the velocity with which the projectile exits from the block. The answer should be 352.19 m/s
  2. jcsd
  3. Nov 13, 2004 #2


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    Please show what work you have done. Here's how you should be approaching these:

    1: -Find the student's velocity just before he hits the ground (use the idea that the impulse is equal to the change in momentum and realize that the student's velocity is 0 immediately after the collision (we'll assume he doesn't bounce)).
    -Next, use this velocity to find the initial height. You can use the conservation of energy or the kinematic equations you learned for contant acceleration motion.

    2. - This is a straight application of the conservation of momentum. Initially, only the bullet has the momentum. After the bullet passes through the block, both the block and the bullet have momentum:

    [tex] m_bv_{bi} = m_bv_{bf} + m_Bv_{Bf} [/tex]

    (b = bullet; B = Block; i = initial; f = final)

    I believe the number you gave for what "should" be the answer is incorrect.
  4. Nov 13, 2004 #3
    I also think that your official answer for 2) is wrong.
  5. Nov 14, 2004 #4

    Andrew Mason

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    Use: [itex]mgh = 1/2mv^2[/itex] so [itex] v = \sqrt{2gh}[/itex]
    Since [itex]F\triangle t = mv[/itex] we have:
    [tex]h = F^2t^2/2gm^2[/tex]
    [tex]h = 2.59 m.[/tex]

    The answer should be 448.05 m/s. The real question is why we should assume that the wood block still has a mass of 1 kg after the projectile passes through it. A .295 kg projectile is a small canonball that would make a huge hole in the block if not destroy it. :smile:

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