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Problem regarding conservation of momentum in different systems

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data

    A bullet of mass m_{1} with initial horizontal speed v_{0} hits a wooden block of mass m_{2} initially at rest on an incline of angle β and embeds itself into the wooden block, how far up the incline will the bullet and block travel?

    2. Relevant equations

    [itex]\bar{P}_{initial}[/itex] = [itex]\bar{P}_{final}[/itex]

    U = ΔT+Δ[itex]V_{g}[/itex]+Δ[itex]V_{e}[/itex]

    3. The attempt at a solution

    [itex]\bar{P}_{initial}[/itex] = m_{1}*v_{1}*[itex]\bar{e}_{i}[/itex]
    [itex]\bar{P}_{final}[/itex] = (m_{1}+m_{2})v_{2}*(cos(β)[itex]\bar{e}_{i}[/itex]+sin(β)*[itex]\bar{e}_{j}[/itex])
    v_{2}*sin(β) = 0 v_{2}*cos(β) = m_{1}*v_{1}/(m_{1}+m_{2}) ?
     
    Last edited: Mar 25, 2013
  2. jcsd
  3. Mar 25, 2013 #2
    What makes you think conservation of momentum is relevant in this problem?
     
  4. Mar 25, 2013 #3
    Because the bullet embeds itself in the block of wood.
     
  5. Mar 25, 2013 #4
    But then it travels along an incline, exchanging its momentum with the surroundings. It stops in the end, which means the momentum of the block and the bullet is zero. I cannot see how you could use conservation of momentum here.
     
  6. Mar 25, 2013 #5
    First the entire system has kinetic energy (m_1+m_2)*v_final^2/2 This is then converted to gravitational potential energy.
     
  7. Mar 25, 2013 #6
    @voko

    You can use conservation of momentum between the bullet (horizonal motion) and the block instananeously after the collision, then it switches to conservation of energy.
     
  8. Mar 25, 2013 #7
    Well, it could be easier to note that initially only the bullet has non-zero kinetic energy. Then the bullet and the block end up having some potential energy. This way you don't have to figure out the velocity of the system after the impact.

    EDIT: this is incorrect. See below.
     
    Last edited: Mar 25, 2013
  9. Mar 25, 2013 #8
    This is true. But that is not enough to solve the problem. And because it can be avoided, as I just remarked above, one should apply Occam's razor.

    EDIT: this is incorrect. See below.
     
    Last edited: Mar 25, 2013
  10. Mar 25, 2013 #9
    the collision isn't eleastic, can you use conservation of energy from start to finish?
     
  11. Mar 25, 2013 #10
    Yes, you are quite right. One has to apply conservation of momentum to find out the velocity of the system after the impact, because the impact is not elastic.
     
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