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Bullet block problem again

  1. Oct 14, 2005 #1
    An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height (Fig. P6.30). The bullet remains in the block, and after the impact the block lands d = 1.5 m from the bottom of the table. Determine the initial speed of the bullet.


    this is what i did:

    time for block to hit floor:
    x = ut + 0.5at^2
    1.0 = 0.5*9.8t^2
    t = 0.4517 seconds

    acceleration needed to move block 1.5 meters in t seconds

    1.5 = 0.5a*0.4517^2
    a = 14.7


    velocity at time of hitting the ground:
    v=u+at
    v=6.63999 m/s

    mv = (m+M)V
    .008v = (.258)6.63999
    v = 214.139m/s


    this is the wrong answer; could someone give me a hand?

    thanks

    bill
     
  2. jcsd
  3. Oct 14, 2005 #2
    While the block falls, it is not accelerating in the x direction, but merely has a velocity. You know it travels 1.5m in 0.45 seconds, so find the velocity of the block+bullet system after the collision. From there I would try energy methods, since its a perfectly inelastic collision.
     
  4. Oct 14, 2005 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This is your error. This formula assumes a constant acceleration throughout the motion. That is not true here. The bullet gives an "instantaneous" acceleration to the block. After that, there is no horizontal acceleration.

    Assume the block(and bullet) has some initial horizontal velocity v0 m/s and acceleration 0. Then the distance the block moves in t seconds is given by v0t. Assuming your value for t is correct (I didn't check that) you can solve v0(0.4517)= 1.5 to find v0.

    Now, use conservation of momentum with the velocity you just found to find the velocity of the bullet just before hitting the block.
     
  5. Oct 14, 2005 #4
    Thank you guys so much. I am a moron; wasted an hour on this question.
     
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