Differences
This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
183_notes:examples:deer_slug_example [2014/09/26 04:30] – pwirving | 183_notes:examples:deer_slug_example [2014/10/01 05:24] (current) – pwirving | ||
---|---|---|---|
Line 11: | Line 11: | ||
=== Lacking === | === Lacking === | ||
+ | |||
+ | $\vec{F}_{net}$ on shoulder | ||
=== Approximations & Assumptions === | === Approximations & Assumptions === | ||
+ | |||
+ | ${\Delta t} \longrightarrow | ||
+ | |||
+ | $\vec{V}_{Slug} \longrightarrow 500m/s$ This is a conservative estimate based on an internet search. | ||
+ | |||
=== Representations === | === Representations === | ||
Line 19: | Line 26: | ||
Surroundings: | Surroundings: | ||
+ | |||
+ | {{183_notes: | ||
$\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$ | $\vec{F}_{net} = \dfrac{\Delta\vec{p}}{\Delta t}$ | ||
Line 30: | Line 39: | ||
=== 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, | ||
+ | |||
+ | 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/ | ||
+ | |||
+ | 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/ | ||
+ | |||
+ | $\vec{F}_{net} = 2637.6N$ (at least) | ||
+ | |||
+ | |||
+ | |||
+ |