183_notes:examples:deer_slug_example

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183_notes:examples:deer_slug_example [2014/09/26 04:14] – created pwirving183_notes:examples:deer_slug_example [2014/10/01 05:24] (current) – pwirving
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 ===== Example: Firing a deer slug. ===== ===== Example: Firing a deer slug. =====
 +
 +How much force does a 12 gauge exert on your shoulder when firing a deer slug?
  
 === Facts === === Facts ===
 +
 +Mass of gun = 3.5kg
 +
 +Mass of slug = 0.22kg
 +
  
 === Lacking === === Lacking ===
 +
 +$\vec{F}_{net}$ on shoulder
  
 === Approximations & Assumptions === === Approximations & Assumptions ===
 +
 +${\Delta t} \longrightarrow  1/24s$ - Based on when a gun is fired in a movie, it usually occurs at about one movie frame, therefore, the collision time is less than 1/24s.
 +
 +$\vec{V}_{Slug} \longrightarrow 500m/s$ This is a conservative estimate based on an internet search.
 +
  
 === Representations === === Representations ===
 +
 +System: Gun + Slug
 +
 +Surroundings: Nothing
 +
 +{{183_notes:examples:momentum_example_2_upload.jpg?600|}}
 +
 +$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$
 +
 +$\vec{p}_{sys,f} = \vec{p}_{sys,i}$
 +
 +$\vec{p}_{1,f} + \vec{p}_{2,f} = \vec{p}_{1,i} + \vec{p}_{2,i}$
 +
 +$m_1\vec{v}_{1,f} + m_2\vec{v}_{2,f} = m_1\vec{v}_{1,i} + m_2\vec{v}_{2,i}$
 +
  
 === Solution === === Solution ===
 +
 +We know that the momentum of the system (gun + slug) does not change due to their being no external forces acting on the system, therefore, the change in momentum in the x-direction is 0.
 +
 +${\Delta p_x} = 0$
 +
 +The total momentum of the system in x direction is also 0.
 +
 +$P_{tot,x} = 0$
 +
 +This is because the initial momentum of the system is 0 and therefore the final momentum of the system is zero.
 +
 +$P_{tot,i,x} = 0$
 +
 +We can relate the momentum before to the momentum after then giving us the following equation.
 +
 +$0 = M_G * V_G + m_S * V_S \longrightarrow M_G * V_G$ is negative and $m_S * V_S$ is positive (see diagram).
 +
 +To find the force acting on the shoulder of the shooter me need to know $V_G$ in order to find change in momentum for the gun and relate this to the force using $\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$. Rearrange the previous equation.
 +
 +$V_G = {\dfrac{-m_s}{M_G}} V_S$
 +
 +Fill in the values for the corresponding variables.
 +
 +$V_G = - {\dfrac{0.22kg}{3.5kg}}{500m/s} = -31.4m/s$
 +
 +Use the value found for $V_G$ to find the change in momentum and hence find what kind of force that is on your shoulder.
 +
 +$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$
 +
 +Fill in values for known variables.
 +
 +$\vec{F}_{net} =\dfrac{(3.5kg)(-31.4m/s + 0m/s)}{(1/24s)}$
 +
 +$\vec{F}_{net} = 2637.6N$ (at least)
 +
 +
 +
 +
  • 183_notes/examples/deer_slug_example.1411704889.txt.gz
  • Last modified: 2014/09/26 04:14
  • by pwirving