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Momentum: A bullet passing through a wooden block on a frictionless surface

  1. Oct 29, 2009 #1
    1. The problem statement, all variables and given/known data

    A wooden block with a length of 10 cm and a mass of 1 kg lies on an icy plane. A projectile with mass of 2 g hits the wooden block with velocity of 300 m/s and breaks through its center of gravity. How much are the final velocities of the wooden block and the projectile? While moving through the wooden block, the projectile worked on it with a force of 500 N. This is a frictionless system.

    2. Relevant equations

    KE= ½ mv²
    p=mv

    3. The attempt at a solution

    KE= F*d=50 J
    KE= ½ m(1)v(final)² → v(1-final)²= 2KE/ m(1)
    v(1-final)= 223.6 m/s

    Conservation of the momentum:
    m(1)v(1-initial) + m(2) v(2-initial) = m(1)v(1-final) + m(2) v(2-final)
    v(2-final)= [m(1)v(1-initial) + m(2) v(2-initial) - m(1)v(1-final)] / m(2)
    v(2-final)= 0.15 m/s

    Is my approach to the problem and calculations correct?
    Thank you for helping!
     
  2. jcsd
  3. Oct 29, 2009 #2

    kuruman

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    Your energy balance equation is incorrect. The initial kinetic energy of the bullet KE0, which you can calculate is a certain number of Joules. That number of Joules is divided into three parts

    1. Work done by bullet on block (F*d)
    2. Final kinetic energy of block.
    3. Final kinetic energy of bullet.

    Your momentum conservation equation is also incorrect.

    Pbefore=Momentum of bullet only
    Pafter=Momentum of bullet + momentum of block

    "Before" means before the bullet hits the block; "after" means after the bullet has made it through.
     
  4. Oct 29, 2009 #3
    So, if I write it like this:

    ½ m(1)v(1-initial)²= ½ m(1)v(1-final)² + ½m(2)v(2-final)²+F*d

    And

    m(1)v(1-initial)= m(1)v(1-final) + m(2)v(2-final)

    I have the right equations to work with?

    Thank you for helping!
     
    Last edited: Oct 30, 2009
  5. Oct 29, 2009 #4

    kuruman

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    Yes, these are the correct equations to start from.
     
  6. Oct 29, 2009 #5
    OK, but how do I continue.:confused:
    I tried to express v(2-final) with the components of the second formula (conservation of the momentum) and insert it into the first formula (KE) but it just got very complicated and weird.:redface:

    Can you give me another hint, please?
     
  7. Oct 29, 2009 #6

    kuruman

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    Weird or no weird, that's exactly what you have to do. There are no other hints. You have to solve a system of two equations and two unknowns and you described the correct way to solve it.
     
  8. Oct 30, 2009 #7
    Thank you for helping! :smile:
    Back to solving my weirdness of equation!:biggrin:
     
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