Dale said:
So have two “pinewood derby” vehicles both on a straight slope, both with identical rocket engines. One fires the engine at the top of the track and the other fires the engine in the middle. Determine the final KE of each at the end.
Thank you for all of the above input on my formatting, I will try to present everything from now on using the appropriate equations, variables, formatting and # of significant digits. (I obtained
@Dale 's permission before posting this)
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Firstly a vertical drop depth of ##29## seconds at ##9.8 m/s^2## in a vacuum is chosen because it is roughly equivalent in depth to the world's current deepest goldmine ( ##4.1km## ) in hopes it can be achievable with present technology. There is a vertical drop, a flat section, followed by another vertical ramp, connected by smooth curved sections, in vacuum.
The drop depth, time and final velocity can be calculated with:
##a=(v_f−v_i)/Δt##
##a=2∗(Δd−v_i∗Δt)/Δt²##
##a=F/m##
where:
##a## = the acceleration,
##v_i## = initial velocity,
##v_f## = final velocity,
##Δd## = distance traveled during acceleration,
##Δt## = acceleration time,
##F## = is the net force acting on an object that accelerates,
##m## = is the mass of this object.##Δd=4.12km##
##v_f=284m/s##
##m=m_{tan}+m_{pas}=10.1*10^3kg##
##m_{tan}:1*10^4 kg##
##m_{pas}:100kg##
##E_{impulse}:202MJ##
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The ##E_{impulse}## is calculated from a desired ##2km/s## velocity boost for the passenger vehicle, above and beyond its velocity from free fall at the bottom of the ramp from momentum conservation and kinetic energy equations:##m_1v_1=m_2v_2##
##v_1=(m_2v_2)/m_1##
##m_{1KE}=(1/2)mv^2##
##m_{2KE}=(1/2)mv^2##
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Results
Impulse on Top ( ##202MJ## ):
Covering ##4.1km## with initial ##2km/s## initial velocity and ##9.8m/s## acceleration takes ##2.07## seconds.
##2.07s## with ##9.8m/s## acceleration is a velocity change of ##20.3m/s##
##m_1## velocity at ramp bottom with impulse on top: ##2.02km/s##
##m_{1KE}## (bottom of ramp, impulse on top) = ##204MJ##
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Impulse on Bottom ( ##202MJ## ):
Free-fall Velocity at Bottom of Ramp: ##284m/s##
##V_1##'s actual velocity is velocity obtained from the impulse in addition to the free-fall velocity at the bottom of the ramp:
##m_1## velocity at ramp bottom with impulse on bottom = ##2.28km/s##
##m_{1KE}## (bottom of ramp, impulse on bottom) = ##260MJ##
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Conclusions:
Max Velocity with no ramp: ##2km/s## @ ground
Energy with ##202 MJ## Impulse on top: ##200 MJ## <----
Max Velocity "Standard Car" pushing off ground: ##2.009km/s## @ ground
Energy with ##202MJ## Impulse on top: ##202MJ## <----
Max Velocity with impulse on top of ramp: ##2.02km/s## @ bottom
Energy with ##202MJ## Impulse on top: ##204MJ##
Ground Level Exit Velocity with impulse on top of ramp: ##2km/s## @ ground
Exit Energy with ##202MJ## impulse on top of ramp: ##200MJ## <----
Max Velocity with impulse at bottom of ramp: ##2.28km/s## @ bottom
Energy with ##202MJ## Impulse on bottom: ##260MJ##
Ground Level Exit Velocity with impulse at bottom of ramp: ##2.26km/s## @ ground
Exit Energy with ##202MJ## impulse on bottom of ramp: ##256MJ## <----------------
Conclusions
Using the same ##202MJ## amount of mechanical impulse energy at the top or bottom of the ramp, impulse at the top of the ramp resulted in ##204MJ_{KE}## at the bottom of the ramp while impulse at the bottom of the ramp resulted in ##260MJ_{KE}## at the bottom of the ramp.
After returning to the surface on a second ramp, the "impulse at bottom vehicle" had more kinetic energy than the total energy in the mechanical impulse, while the "impulse at top vehicle" had less.
The "impulse at top" vehicle did worse in terms of kinetic energy obtained per Joule of mechanical impulse energy than a "standard car" pushing off the ground after returning to the surface, while the "impulse on bottom" vehicle did better.
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Questions:
Are the conclusions correct? I'm having a hard time understanding what difference does it make if the passenger vehicles uses the mechanical impulse to push off the ##284m/s## tank at the bottom after free fall as opposed to pushing off the ground after free fall?