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--- 183_notes:examples:deer_slug_example [2014/09/26 04:30] – pwirving
+++ 183_notes:examples:deer_slug_example [2014/10/01 05:20] – pwirving
@@ Line 11: / Line 11: @@
 === Lacking ===
+$\vec{F}_{net}$ on shoulder
 === 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 ===
@@ Line 19: / Line 26: @@
 Surroundings: Nothing
+{{183_notes:examples:momentum_example_2_upload.jpg?600|}}
 $\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$
@@ Line 30: / Line 39: @@
 === 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).
+$V_G = {\dfrac{-m_s}{M_G}} V_S$
+$V_G = - {\dfrac{0.22kg}{3.5kg}}{500m/s} = -31.4m/s$
+Need to find what kind of force that is on your shoulder.
+$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$
+$\vec{F}_{net} =\dfrac{(3.5kg)(-31.4m/s + 0m/s)}{(1/24s)}$
+$\vec{F}_{net} = 2637.6N$ (at least)