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--- 183_notes:examples:deer_slug_example [2014/09/26 05:06] – pwirving
+++ 183_notes:examples:deer_slug_example [2014/10/01 05:24] (current) – pwirving
@@ Line 27: / Line 27: @@
 Surroundings: Nothing
-{{183_notes:examples:momentum_example_2_upload.jpg?400|}}
+{{183_notes:examples:momentum_example_2_upload.jpg?600|}}
  $\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$
@@ Line 39: / 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.
  $P_{tot,x} = 0$
-Because
+This is because the initial momentum of the system is 0 and therefore the final momentum of the system is zero.
  $P_{tot,i,x} = 0$
- $0 = M_G * V_G + m_S * V_S \longrightarrow M_G * V_G$  is negative and  $m_S * V_S$  is positive
+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.
  $V_G = - {\dfrac{0.22kg}{3.5kg}}{500m/s} = -31.4m/s$
-Need to find what kind of force that is on your shoulder.
+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.
  $\vec{F}_{net} =\dfrac{(3.5kg)(-31.4m/s + 0m/s)}{(1/24s)}$