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A Linear Momentum Problem

  1. Jan 5, 2008 #1
    1. The problem statement, all variables and given/known data

    A 10.0g bullet is fired horizontally into a 106 g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring of constant 155 N/m. If the bullet-block system compresses the spring by a maximum of 78.0 cm, what was the velocity of the bullet at impact with the block?

    2. Relevant equations

    where p is momentum, m is mass (kg), and v is velocity (m/s)

    (1/2 m1 v1^2 + 1/2 m2 v2^2)initial = (1/2 m1 v1^2 + 1/2 m2 v2^2)final + 1/2kx^2
    Where m is mass (kg), v is velocity (m/s), k is the spring constant (N/m), and x is the distance of compression (meters).

    (1/2 m1 v1^2 + 1/2 m2 v2^2)initial = (1/2m3 v3^2)final + 1/2kx^2
    Where m is mass (kg), v is velocity (m/s), k is the spring constant (N/m), x is the distance of compression (meters), m3 is the sum of the masses, and v3 is the combined, new velocity. Not even sure if this one makes sense, I kind of combined some equations.

    3. The attempt at a solution

    To sum up my attempt, I plugged all of my variables into my "combined" equation. Thus I had two variables, v1 and v3^2. I decided to make everything equal to v1 and plug that equation into the original "combined" equation. (I can't really copy that onto here because it had a gigantic square root symbol.) So then I solved for v3 and got a ridiculously low answer, 1.0217m/s. According to my text book, which has the same problem with different variables, the answer is 237m/s.

    Can someone help me out? Also, is this considered a perfectly inelastic collision?
  2. jcsd
  3. Jan 5, 2008 #2

    Doc Al

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    Staff: Mentor

    Yes, so energy is not conserved during the collision. But what else is?

    After the collision, then you can use energy conservation.
  4. Jan 5, 2008 #3
    Momentum is, also. My main problem is that I can't set up a good equation (or two, if need be) or get rid of the double unknowns.
  5. Jan 5, 2008 #4

    Doc Al

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    Staff: Mentor

    Think of the situation as having two phases:
    (1) The collision of bullet and block
    (2) The compressing of the spring after the collision

    Set up equations for each phase. Each phase uses a different conservation law. Hint: You might want to analyze phase 2 first and work backwards.
  6. Jan 6, 2008 #5
    OK I tried this.

    (1/2*m1*v1^2 + 1/2*m2*v2^2)final = ½*k*x^2
    And solved for v2
    I got 97.11 m/s, understanding that the final velocity of the bullet is 0 because it’s lodged into the block.
    Then I plugged it into (1/2*m1*v1^2 + 1/2*m2*v2^2)initial = (1/2*m1*v1^2 + 1/2*m2*v2^2)final + ½*k*x^2
    And solved for v1 initial.
    I got 137.317 m/s. This seems much more reasonable. Am I close?
  7. Jan 7, 2008 #6
    bro..u r missing the equilibrium condition of the spring....i.e. when the spring is in its natural state...think about it and then proceede..:p

    also..never go on answers..they may be wrong ..

    have self belief and proceed...if ur concepts are right..u may even deny the text book's answer..
  8. Jan 7, 2008 #7

    Doc Al

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    Staff: Mentor

    The left side of this equation is incorrect. After the collision, the bullet and block move together with a single speed. Their KE = 1/2 (m1+m2)V^2. Find that speed.
    The final speed of the bullet (immediately after the collision) is not zero--it's the same speed as the block.
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