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Bullet and Block

  1. Nov 13, 2011 #1
    1. The problem statement, all variables and given/known data
    40-g Bullet is fired with a velocity of magnitude v0=700 m/s into a 5kg block of wood. Coefficient of kinetic friction = 0.6.
    a) how far will the block move

    b) how long after impact does it come to rest

    c) percentage of energy lost on impact

    (block has no velocity until bullet is completely embedded in block)


    2. Relevant equations



    3. The attempt at a solution

    mass bullet: m = 40g = .04kg
    mass block: M = 5kg
    velocity bullet = 700 m/s
    uk = 0.6


    KE of bullet
    1/2*0.04*700^2 = 9800J

    Conservation of Momentum Work Energy

    mv+0 = (M+m)v -uk(m+M)gd=-1/2(m+M)v^2
    0.04(700) = 5.04v ukgd=1/2v^2

    v = 5.56m/s 0.6(9.81)d=0.5(5.56)^2
    d = 2.63

    a) 2.63m

    b) D=v*t t=d/v
    t = 2.63/5.56
    t= 0.473s

    c) kinetic energy bullet = 9800J
    Kinetic energy of block and bullet = 77.9J

    %= 77.9/9800 == 0.07%

    I am most confused on the percentage of energy lost?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 14, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Take care. The momentum is conserved during the impact, as we can assume it so fast that the force of friction does practically no work during that time. After the impact, there is force on the block, so its momentum will change.

    Momentum is not equal to energy.

    Energy is not conserved.


    ehild
     
  4. Nov 14, 2011 #3
    Also, take care with the units. Bullet are generally measured in grains (abbreviated "gr"), not grams (abbreviated "g").

    40 g (grams) would be a rather large bullet, but not out of the realm of possibility. A .50 BMG shell will have a bullet weighing anywhere from about 42 g to 52 g which equates to around 647 gr to 800 gr.

    By contrast, a .44 magnum bullet can weigh around 340 gr (22 g)
     
  5. Nov 15, 2011 #4
    The weight of the bullet is 40 grams, sounds large to me as well.

    Will work energy equations work better here? Or am I on the right track?
     
  6. Nov 15, 2011 #5

    gneill

    User Avatar

    Staff: Mentor

    If you determine the acceleration of the block+bullet (due to friction), then for part b you have initial velocity, final velocity (zero) and acceleration. A simple kinematic equation relates them with time.
     
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