1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Oct 30, 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


    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)

    Correct formula thanks to help from "kuruman"- PF Contributor and Homework Helper Homework Helper:smile: Thank you again!

    Then 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.

    Then I did some more research and find an easier approach of solving my weirdness:

    First, I applied:
    x= v(1-final)/ / v(1-initial)
    y= v(2-final) / v(1-initial)
    and
    m(2)/m(1)= 1kg/ 0.002kg= 500
    F*d= 50J

    After I plugged in the above, I obtained:
    1. For the conservation of the moment: 1= x + 500y → x= 1- 500y
    2. For the KE: 1= x² + 500y² + 50

    Now, I plugged x value from the first equation into the second equation and after arranging I got:
    0= 20y (250.5y – 1)

    We now that velocity cannot be 0, so the correct solution is y= 1/ 250.5

    Last step:

    y= v(2-final) / v(1-initial) → v(2-final)= y* v(1-initial)= 1.19 m/s

    v(2-final)= 1.19 m/s → velocity of the wooden block after the bullet passes through

    x= 1- 500y= -0.996
    x= v(1-final) / v(1-initial) → v(1-final)= x* v(1-initial)= -298.8 m/s

    v(1-final)= 298.8 m/s → velocity of the bullet after it passes through the block

    Please tell me that this is correct?!
    Thank you!
     
  2. jcsd
  3. Oct 30, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

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

    Try putting your final values of 1.19 and 298.8 back into the original equation for momentum:
    m(1)v(1-initial)= m(1)v(1-final) + m(2)v(2-final)
    .002*300 = .002*298.8 + 1*1.19
    0.6 = 1.7876
    According to that, your solution does not conserve momentum!
    Better check my work - I often make calculating mistakes.

    It seems to me the x and y substitution just adds complexity. Could you show us what you got when you solved the momentum equation and substituted into the energy equation? It should be just a quadratic equation which can easily be solved on a calculator or spreadsheet.
     
  4. Oct 30, 2009 #3
    Re: Momentum: A bullet passing through a wooden block on a frictionless surface- AGAI

    So my steps were correct?


    :redface: Ooops! I exactly know where I went wrong- 50/50 is not 0 but 1. So what i should have gotten is:

    0= 5010y² + 20y + 1

    I think I got it right this time:smile:

    And now it's just a quadratic equation which is a little complicate to solve, so back to work!
     
  5. Oct 30, 2009 #4
    Re: Momentum: A bullet passing through a wooden block on a frictionless surface- AGAI

    Not only patience but also persistence is a virtue. In the end it all pays off:biggrin:

    I took your advice and plugged the numbers into the equations and here is the result of my hard labor:wink::

    Conservation of the moment:
    m(1)v(1-initial)= m(1)v(1-final) + m(2)v(2-final)
    (0.002)(300)= (0.002)v(1-final) + (1)v(2-final)
    0.6= 0.002v(1-final) + v(2-final) → v(2-final)= 0.6 - 0.002v(1-final)

    Energy:
    ½ m(1)v(1-initial)²= ½ m(1)v(1-final)² + ½m(2)v(2-final)²+F*d
    ½ (0.002)(300)²= ½ (0.002) v(1-final)² + ½ (1) v(2-final)² + 50
    90= 0.001 v(1-final)² + 0.5 v(2-final)² + 50
    0= 0.001 v(1-final)² + 0.5 (0.6 - 0.002v(1-final))² - 40
    0= 0.001002 v(1-final)² - 0.0012v(1-final) – 39.82

    v(1-final)= [-(-0.0012) ± sqrt[(0.0012)² - 4(0.001002)(-39.82)]] / 2(0.001002)
    v(1-final)= (0.0012 ± 0.3995) / 0.002004

    First solution: v(1-final)= -198.75 m/s → v(2-final)= 0.9975 m/s
    Second solution: v(1-final)= 199.95 m/s → v(2-final)= 0.2001 m/s

    As the bullet continues to fly forward, I chose the second solution to be the solution of the problem.

    Quick recheck for the conservation of the moment:
    m(1)v(1-initial)= m(1)v(1-final) + m(2)v(2-final)
    (0.002)(300)= (0.002)(199.95) + (1)(0.2001)
    0.6 = 0.6

    Are my calculations correct?
    Thank you for helping!:smile:
     
  6. Oct 30, 2009 #5

    Delphi51

    User Avatar
    Homework Helper

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

    Good work, mmoadi! I had a different answer and your solution showed me where my mistake was. Thank you for showing the work.
     
  7. Oct 31, 2009 #6
    Re: Momentum: A bullet passing through a wooden block on a frictionless surface- AGAI

    Thank you for helping and have a spooky Halloween weekend!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Momentum: A bullet passing through a wooden block on a frictionless surface- AGAIN
Loading...