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Momentum/kinetic energy w/ rifle

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data

    A rifle with a mass of 3.22 kg fires an 11.9 g bullet at a velocity of 590 m/s. What is the recoil velocity of the rifle? Compare the change in momentum of the bullet and the rifle. Compare the change in kinetic energy of the bullet and the rifle. Which is bigger, by what factor?


    2. Relevant equations

    m_1v_1 + m_2v_2 = m_1v_1 + m_2v_2
    (initial) (final)

    K = 1/2mv²

    3. The attempt at a solution

    (a)What is the recoil velocity of the rifle?

    We know

    mass of rifle = 3.22 kg
    mass of bullet = .0119 kg
    initial velocity of rifle = 0
    initial velocity of bullet = 0 (this one I thought would be the 590m/s but my prof said they both start from rest)
    final velocity of bullet = 590 m/s
    final velocity of rifle = V_1 (what we want)

    m_1v_1(initial) + m_2v_2(initial) = m_1v_1(final) + m_2v_2(final)


    So this would be solving for v_1 final so rearrange for that

    V_1 = [m_1v_1(initial) + m_2v_2(initial) - m_2v_2(final)]/ m_1

    V_1 = [3.22 kg(0) + .0119 kg(0) - .0119 kg(590m/s)] / 3.22 kg
    (the answer would be a negative number which makes sense because the direction is opposite)

    so V_1 = -2.18m/s


    (b) Compare the change in momentum of the bullet and the rifle.

    I'm not sure about this one


    (c) Compare the change in kinetic energy of the bullet and the rifle

    I was thinking

    1/2mV_f² - 1/2mV_i² for the bullet and for the rifle

    so

    mass of rifle = 3.22 kg
    mass of bullet = .0119 kg
    initial velocity of rifle = 0
    initial velocity of bullet = 0 (this one I thought would be the 590m/s but my prof said they both start from rest)
    final velocity of bullet = 590 m/s
    final velocity of rifle = V_1 (what we want)

    Rifle

    ΔK_r = 1/2mV_f² - 1/2mV_i²
    = 1/2(3.22kg)(-2.18m/s)² - 1/2(3.22kg)(0)²

    K_r = 7.65J


    Bullet

    ΔK_b = 1/2mV_f² - 1/2mV_i²
    = 1/2(.0119 kg)(590 m/s)² - 1/2(.0119 kg)(0)²

    K_b = 2071.2J



    Are these correct? and how would I go about (b)? I'm just not too sure what it is asking
     
  2. jcsd
  3. Apr 18, 2013 #2

    CWatters

    User Avatar
    Science Advisor
    Homework Helper

    The change in momentum of each is calculated much like you did for the change in KE eg...

    mV_f - mV_i
     
  4. Apr 18, 2013 #3
    Oh...I thought I had to go through all of the
    m_1v_1 + m_2v_2 = m_1v_1 + m_2v_2
    again...and I didn't get what I would be solving for


    Okay so

    change in momentum of Rifle

    3.22(-2.18m/s) - 3.22kg(0) = -6.5836
    (only negative to show direction)

    change in momentum of Bullet

    .0119(590) - .0119(0) = 7.021

    Is that correct?
     
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