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183_notes:examples:deer_slug_example [2014/09/26 04:49] – pwirving | 183_notes:examples:deer_slug_example [2014/10/01 05:24] (current) – pwirving | ||
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Surroundings: | Surroundings: | ||
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+ | {{183_notes: | ||
$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$ | $\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$ | ||
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=== 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. | ||
${\Delta p_x} = 0$ | ${\Delta p_x} = 0$ | ||
+ | |||
+ | The total momentum of the system in x direction is also 0. | ||
$P_{tot,x} = 0$ | $P_{tot,x} = 0$ | ||
- | Because | + | This is because the initial momentum of the system is 0 and therefore the final momentum of the system is zero. |
$P_{tot, | $P_{tot, | ||
- | $0 = M_G * V_G + m_S * V_S \longrightarrow M_G * V_G$ 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 | ||
+ | |||
+ | 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. | ||
$V_G = {\dfrac{-m_s}{M_G}} V_S$ | $V_G = {\dfrac{-m_s}{M_G}} V_S$ | ||
+ | |||
+ | Fill in the values for the corresponding variables. | ||
$V_G = - {\dfrac{0.22kg}{3.5kg}}{500m/ | $V_G = - {\dfrac{0.22kg}{3.5kg}}{500m/ | ||
- | 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. |
$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$ | $\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$ | ||
+ | |||
+ | Fill in values for known variables. | ||
$\vec{F}_{net} =\dfrac{(3.5kg)(-31.4m/ | $\vec{F}_{net} =\dfrac{(3.5kg)(-31.4m/ |