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Ballistic Spring System Problem

  1. Jan 16, 2015 #1
    1. The problem statement, all variables and given/known data
    A 10.1 g bullet is fired horizontally into a
    44.1 g wooden block that is initially at rest on
    a rough horizontal surface and connected to a
    massless spring of constant 97.8 N/m.
    If the bullet-block system compresses the
    spring by 0.441 m, what was the speed of
    the bullet just as it enters the block? The
    acceleration of gravity is 9.8 m/s
    2
    . Assume
    that the coefficient of kinetic friction between
    the block surface is 0.554.
    Answer in units of m/s.


    2. Relevant equations
    The equation our teacher gave us to solve it is: 1/2 mv^2= 1/2kx^2 + Fd
    but I've used the equation and it hasn't worked for me.



    3. The attempt at a solution
    Using the equation above. Is the equation right?
     
  2. jcsd
  3. Jan 16, 2015 #2

    Orodruin

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    This depends on how you are using it. Can you show your work please?
     
  4. Jan 16, 2015 #3
    Well to start off I plugged in everything that I had from the information given:
    1/2(.0101)v^2= 1/2(97.8)(.441)^2 + F(.441)

    Then I found F by using F= u(coefficient of friction) x Fn(Normal Force)
    so F= (.554)Fn which I found normal force by using Fn= (m+M)g
    Fn= (.0101 + .0441)(9.8)= .53116
    So when you plug that back in you get
    F=(.554)(.53116)
    1/2(.0101)v^2= 1/2(97.8)(.441)^2 + F(.441)
    After i work this equation out solving for v I always get a consistent answer but it's not right.
     
  5. Jan 16, 2015 #4

    haruspex

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    There's a step before this.
    Consider the process in two stages:
    - merging of bullet and block
    - subsequent movement of combined bullet and block
    Both stages involve loss of mechanical energy, the first in impact, the second in friction.
    So the first thing to find out is the speed with which the combined bullet and block start moving.
     
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