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Collisions homework problem

  1. Apr 30, 2007 #1
    1. The problem statement, all variables and given/known data
    A bullet of mass m moving with velocity V_o crashes into a block of mass M at rest on a frictionless horizontal surface the block is sitting at the edge of a table that is H high. When the bullet hits the block it flies off the table and it lands a distance x from the bottom of the table. In each case below, determine the initial velocity of the bullet in the terms of the give quantities.

    A) in the first case assume that the bullet sticks to the block
    B) in the second case assume that the bullet bounces back directly with half its original speed
    C)In the Third case the bullet crashes through the block emerging on the other side with half its original speed but still moving in the same direction
    D) which makes the block move farther, the sticky bullet or the bouncy bullet?
    [​IMG]
    2. Relevant equations
    Was thinking i can use the conservation of momentum for part A

    M_total*V=M_1*V_1+M_2*V_2


    3. The attempt at a solution

    A) So first i solve for the Final velocity
    V=(m*v_o)/(M+m)

    Now i know X = t*V (so i must solve for the time it takes for the box to hit the ground to get t)

    h=(.5)at^2
    a=g
    So
    t= (h/2g)^(1/2) (don't know how to make the square root symbol)

    Now i just plug this in and get

    X= ((m*v_o)/(M+m))*(h/2g)^(1/2)

    Then solve for V_o

    So V_0= (X(M+m))/(m*(h/2g)^(1/2))


    B) so i did the same thing i did for part A and used the same t

    m(V_0)=m(.5V_0)+M(V)
    V=(3m(v_0))/(2M)

    X=t*(3m(v_0))/(2M)
    t=(h/2g)^(1/2)

    V_0= (2M*X)/(3m*(h/2g)^(1/2))



    C)And i still haven't had time to start on Part C



    D)For this i just got to compare the x= answers i got from part A and b and see what one is bigger (not sure how to do this either)
     
    Last edited: May 1, 2007
  2. jcsd
  3. May 1, 2007 #2
    Just curious if anyone could tell me if i have the right idea -Thanks Tom
     
  4. May 1, 2007 #3

    mukundpa

    User Avatar
    Homework Helper

    A

    h=(.5)at^2
    a=g
    So
    t= (2h/g)^(1/2)
     
  5. May 1, 2007 #4
    so are u saying that is right?
     
  6. May 1, 2007 #5
    One thing I noticed for part b, is that you left a sign off, the bullet recoils and so its contribution is -1/2mv_0. But keep those results I think they are useful for c.
     
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