A 0.027 kg bullet is firedstraight up at a falling...

A 0.027 kg bullet is firedstraight up at a falling wooden block that has a mass of4.0 kg. The bullet has a speed of710 m/s when it strikes the block. Theblock originally was dropped from rest from the top of a buildingand had been falling for a time t when the collision withthe bullet occured. As a result of the collision, the block (withthe bullet in it) reverses direction, rises, and comes to amomentary halt at the top of the building. Find the timet.
s

Answers ( 7 )

  1. lea jørgensen
    lea jørgensen
    Speed of block after falling, from rest, a distances    V^2 = 2 * g * s
    Momentum of block P = M * ?(2 * g * s)
    Bullet must have momentum of 2 * P (ignore mass of bulletin block since it is small)
    So m * v = 2 * M * ?(2 * g * s)
    s = m^2 * v^2 / (M^2 * 8 * g)     Also s =1/2 * g * t^2
    Then t = 1/2 * ?(m^2 * v^2 / (M^2 * g^2)) = m * v/ (2 * M * g) = .24 sec
  2. esther byrd
    esther byrd

    The velocity equation for the block dropped from rest for a timet is v1B = -gt. The mass of the block is m1=4 kg, that of the bullet is m2 = 0.027 Kg, and thevelocity of the bullet before the collision is v2B = 710m/s. Note that after the collision, the velocity of the block (nowwith the bullet in it) is exactly reversed to vf =gt

    This is a completely inelastic collision. Applyingconservation of momentum, we have

    m1v1B + m2v2B =(m1 + m2)vf

    Substitute the time dependency, we have
    m2v2B - m1gt = (m1+ m2)gt. Solving this for the time yeilds:
    t = (m2v2B)/(2m1 +m2)g = 0.24369 s
  3. jen weaver
    jen weaver
  4. joe vasquez
    joe vasquez
  5. ???? ???????
    ???? ???????
    here bullet and block are moving in opposite directions
    take upward direction as positive
    initial momentum of bullet = mu =   0.0027*710kgm/s
    initial momentum of block = -Mv
    the block has fallen for a time t then
    initial momentum of block = -Mgt = -4*9.8*t kg m/s
    then total initial momentum = mu -Mv =  0.0027*710kgm/s   - 4*9.8*t kg m/s
    final momentum of bullet and block =( M+m)v'
    let total block and bullet has travelled a distance s to thetop of the building
    then v2-v'2= -2gs---1
    here v = 0 at the top
    but we know that s = (1/2)gt2
    because block also travelleled a distance s in downwarddirection initially in time t
    therefore from 1 v' = gt = 9.8*t
    then initial momentum = final momentum
    0.0027*710 kgm/s   - 4*9.8*t kg m/s = ( M+m)v'
                                                                 =(4+0.027)*9.8*t
    solve for t     
  6. kayla patel
    kayla patel
    here bullet and block are moving in opposite directions
    take upward direction as positive
    initial momentum of bullet = mu =   0.0027*710kgm/s
    initial momentum of block = -Mv
    the block has fallen for a time t then
    initial momentum of block = -Mgt = -4*9.8*t kg m/s
    then total initial momentum = mu -Mv =  0.0027*710kgm/s   - 4*9.8*t kg m/s
    final momentum of bullet and block =( M+m)v'
    let total block and bullet has travelled a distance s to thetop of the building
    then v2-v'2= -2gs---1
    here v = 0 at the top
    but we know that s = (1/2)gt2
    because block also travelleled a distance s in downwarddirection initially in time t
    therefore from 1 v' = gt = 9.8*t
    then initial momentum = final momentum
    0.0027*710 kgm/s   - 4*9.8*t kg m/s = ( M+m)v'
                                                                 =(4+0.027)*9.8*t
    solve for t     
  7. julia roger
    julia roger
    here bullet and block are moving in opposite directions
    take upward direction as positive
    initial momentum of bullet = mu =   0.0027*710kgm/s
    initial momentum of block = -Mv
    the block has fallen for a time t then
    initial momentum of block = -Mgt = -4*9.8*t kg m/s
    then total initial momentum = mu -Mv =  0.0027*710kgm/s   - 4*9.8*t kg m/s
    final momentum of bullet and block =( M+m)v'
    let total block and bullet has travelled a distance s to thetop of the building
    then v2-v'2= -2gs---1
    here v = 0 at the top
    but we know that s = (1/2)gt2
    because block also travelleled a distance s in downwarddirection initially in time t
    therefore from 1 v' = gt = 9.8*t
    then initial momentum = final momentum
    0.0027*710 kgm/s   - 4*9.8*t kg m/s = ( M+m)v'
                                                                 =(4+0.027)*9.8*t
    solve for t     

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