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Homework Help: Linear Momentum - One-Dimensional Internal Explosion

  1. May 15, 2008 #1
    [SOLVED] Linear Momentum - One-Dimensional Internal Explosion

    1. A body is travelling at 2.0m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion separates the body into two parts, each of 4.0 kg, and increases the total kinetic energy by 16 J. The forward part continues to move in the original direction of motion. What are the speeds of (a) the rear part and (b) the forward part?

    2. I know that, since the system is isolated (no net external force acts on the system), the total linear momentum is conserved. What I'm not sure of, is if the system can also be treated as an inelastic (since kinetic energy isn't conserved) collision. Lastly, since the question asks for the speeds of the two pieces, we know that only magnitude need to be calculated and that direction isn't of consequence.

    3. Note: I tried the Latex references, but it's a bit time-consuming, so I'll just write my train of thought down as clearly as possible. Please bear with me.

    The initial mass of the body is 8.0 kg (two 4.0 kg pieces) and initial velocity is 2.0 m/s. This gives the initial linear momentum as 16 kg m/s. Since the body separates into two equal parts of 4.0 kg each, we can set this momentum equal to 4.0kg(v1 + v2) which finally gives the value of the combined velocities of the two pieces as v1 + v2 = 4.0 m/s.

    This is where I get stuck. With no additional information other than that the forward part continues in the original direction of motion, I am unsure as to what portion of the 4.0 m/s I should allocate to which piece.

    At first, I thought I could use the given increase in the total kinetic energy of the system (16 J) in another equation involving the relevant velocities and then to solve the two equations simultaneously, but can't seem to get that right either. This is what I got

    Tot KE before explosion + increase in KE = KE of piece 1 + KE of piece 2 (after explosion)
    which gives
    16 J + 16 J = 2(v1^2 + v2^2)
    v1^2 + v2^2 = 16 J

    and as you can see, this doesn't really bring us any closer to anything useful since it basically boils down to the values we got for the conservation of linear momentum.

    Any pointers or ideas will be greatly appreciated. So far I've only managed to discover examples of explosions in two dimensions or ones where a lot more information is provided.

    I have this feeling that I'm standing and staring at the obvious, but that it's very well camouflaged against the background of my inexperience :)
  2. jcsd
  3. May 15, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    You are doing fine.

    You have two equations (one from momentum and one from energy) and two variables (v1 and v2). Solve!

    Hint for solving: Express one speed in terms of the other (using the momentum equation) and then substitute that expression into the other equation. Solve the quadratic.
  4. May 16, 2008 #3
    Thanks Doc! I guess re-learning the basics is the price I'll have to pay for only starting a degree when I'm well on my way to 30... :smile:
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