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I am lost, dont know how to solve part 2

  1. Oct 28, 2008 #1
    1. The problem statement, all variables and given/known data

    part -1) A 30-06 caliber hunting rifle fires a bullet of mass 0.0113 kg with a velocity of 765 m/s to the right. The rifle has mass of 4.1 kg.
    what is the recoil speed of the rifle as the bullet leaves the rifle? answer in units of m/s. answer is 2.1084 /s

    part -2) If the rifle is stopped by the hunter's shoulder in a diatance of 2.52 cm, waht is the magnitude of the average force exerted on the shoulder by the rifle? answer in units of N.

    2. Relevant equations

    not sure, new user on this website

    3. The attempt at a solution

    not sure, new user on this website
  2. jcsd
  3. Oct 28, 2008 #2


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    You need to show us some effort.

    What do you know about conservation of momentum?
  4. Oct 29, 2008 #3
    part1) V(bullet) = (m(bullet)v(bullet))/(M(rifle)) = 2.1084 m/s

    Part2) J = (F) (deltaT)
    J = P(final) - P(inital) = MV(final) - MV(initial) = M(V(final) - V(inital))
  5. Oct 29, 2008 #4


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    For the second part you might be better using conservation of energy.
    Energy = force * distance.
    The hunters shoulder is essentailly a spring receiving the kinetic energy of the moving rifle. You know the mass and initial velocity of the rifle and the final velocity=0.
  6. Oct 29, 2008 #5
    do u mean,
    k = (.5)(m)(v^2) or ENERGY = (.5)(m)(v^2)+(.5)(k)(x^2)
  7. Oct 29, 2008 #6


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    You don't need to calculate the spring - I was jsut trying to describe what happened.
    You know the velocity of the rifle when it first touches the shoulder, so you can find the KE.
    Since work (or energy) is force * distance and you know the distance you can work out how much force it takes to stop this energy in this distance.
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