sysprog
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It does help -- you've given me something to think about -- thanks for disambiguating ##\dots##
Let me be certain that my understanding is clear. We are talking about a trailer full of water behind a passenger vehicle. The water is not ejected in this scenario. It remains with the trailer throughout. There are no frictional losses. The whole assembly drops and gains kinetic energy. The passenger vehicle is then ejected at a velocity such that conservation of momentum brings the trailer to a stop.metastable said:##W_i=M_tV_iT_dG_e+(M_tT_d^2G_e^2+M_tV_i^2)/2+(M_t^2(V_i+T_dG_e)^2)/(2M_v)##
^I haven't gone through your whole analysis yet but I had a question about this line... shouldn't it be:jbriggs444 said:##M_{tot}V_b=M_vV_v##
No. It was correct as written. Momentum before = momentum after. The momentum before is the total mass times the velocity of the complete assembly.metastable said:^I haven't gone through your whole analysis yet but I had a question about this line... shouldn't it be:
##M_{trailer}V_b=M_vV_v##
As I understand this, you want ##V_b## to be the forward velocity of the trailer+vehicle assembly.metastable said:Effectively what I used was this:
##m_1v_1=m_2v_2##
^If the mass of the trailer acquires the ramp-bottom velocity in an energetic interaction / push-off with the passenger section (bringing the trailer to a halt), then the passenger vehicle velocity is ##v_2## + ##V_{bottom}##
The energy of the interaction between trailer and tank would be:
##W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2+V_b)^2##
Was it wrong?
metastable said:##W_i## = ##12.8∗10^6## = Mechanical Impulse Energy (##Joules##) That Brings Trailer to 0 Velocity At Ramp Bottom
##M_t## = ##240## = Trailer Tank + Water Mass (##Kilograms##)
##M_v## = ##10^3## = Passenger Vehicle Mass (##Kilograms##)
##T_d## = ##30## = Free Fall Duration (##Seconds##)
##V_i## = ##0## = Initial Velocity Before Free Fall (##m/s##)
##G_e## = ##9.80## = Gravity Acceleration at Earth’s Surface (##m/s^2##)
I am not willing to run the numbers if you are not willing to do the algebra.metastable said:^Using your separate method do you calculate the same value for ##W_i##? The reason I ask is I don't understand part of your method at the beginning.
metastable said:Effectively what I used was this:
##m_1v_1=m_2v_2##
^If the mass of the trailer acquires the ramp-bottom velocity in an energetic interaction / push-off with the passenger section (bringing the trailer to a halt), then the passenger vehicle velocity is v2v2v_2 + VbottomVbottomV_{bottom}
The energy of the interaction between trailer and passenger section would be:
##W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2+V_b)^2##
Was it wrong?
Yes. As I explained, you need to pick a frame of reference for your energy balance.metastable said:In my above calculation, ##v_2## was intended to be the velocity of the passenger section, relative to the ##V_{bottom}## rest frame of the trailer+vehicle at the bottom of the tunnel before the push off, so ##v_2 + V_{bottom}## is the final velocity of the passenger vehicle while still on the flat section at the bottom of the tunnel.
metastable said:The energy of the interaction between trailer and passenger section would be:
##W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2+V_b)^2##
Was it wrong?
Yes, that would be a correct calculation for incremental work done.metastable said:I misspoke here, I meant:
##W_i=(1/2)m_1V_b^2+(1/2)m_2(v_2)^2##
Thanks for that.metastable said:bringing all the equations together, we have:
##m_1v_1=m_2v_2##
##v_2=(m_{trailer}V_{bottom})/m_{passenger}##
##W_i=(1/2)m_{trailer}V_{bottom}^2+(1/2)m_{passenger}(v_2)^2##
##V_{passenger}=v_2+V_{bottom}##
##W_{KEpassenger}=(1/2)m_{passenger}(v_2+V_{bottom})^2##
metastable said:bringing all the equations together, we have:
##m_1v_1=m_2v_2##
##v_2=(m_{trailer}V_{bottom})/m_{passenger}##
##W_i=(1/2)m_{trailer}V_{bottom}^2+(1/2)m_{passenger}(v_2)^2##
I don’t believe so because when I input it in a calculator it looks like this:sysprog said:?