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1) Consider a central in-line head-on collision of two vehicles with rebound which is described show that the following relationships are true. WR1 _ kuz WR2 kui Wrkeql WR1 Wrkegl WR2 = ku2 kegl WR = WT kegu kegl e2 = kequ kui 1 where kegu kui kuz

1) Consider a central in-line head-on collision of two vehicles with rebound which is described show that the following relationships are true. WR1 _ kuz WR2 kui Wrkeql WR1 Wrkegl WR2 = ku2 kegl WR = WT kegu kegl e2 = kequ kui 1 where kegu kui kuz

College Physics
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Recall that Fmax is the magnitude of the crush force acting on both vehicles at maximum deformation (see Figure 3). Then, the total crush energy absorbed by the structures of the two vehidles is given by Wr in Equation (17). Fmax = k1X71 = kax7a (16) W, = kx,/2 + kı2X,/2 = FmaxXT1/2 + Fmaz*r2/2 (17) Wr1 = FmaxXr1/2 . Wrz = Fmas*rz/2 (18) Wri and Wrz are the total crush energies absorbed by vehicles 1 and 2 respectively. Equation (17) can be written as, W, = Fmax /(2ka) + Fmar /(2k12) = Fimar /(2keqı) (18a) where kegl = kukı2/(kLı + k12) is the equivalent loading stiffness of the system. The common velocity V. = Vi can be determined by applying conservation of momentum principle to the two-vehicle system since the net force acting on this system is zero. Then. m,V, + m2V½ = (m, + m2)V, (19) Since the structures of the two vehicles do not undergo perfectly plastic deformation, part of the energy absorbed is transformed into kinetic energy due to spring-back effect in the unloading phase. A similar work-energy analysis can be made for the unloading phase and it can be shown that the crush energy retumed as kinetic energy in the unloading phase is given by Wa in Equations (20) and (21). Wa1 = ku, (Xr1 - Xp1)*/2 = Fmaz (Xr1 – Xp1)/2 WR2 = kuz (Xr2 -Xp2)/2 = Fmar(Xrz - Xp2)/2 WR = WR1 + WR2 We = kui (X71 – Xp1)*/2 + kuz(X72 – Xpa)* /2 (20) - m, (V;)*/2 + m,(V£D* /2 - (m, + m2)V? /2 WR = (21) Wan is the area under the unloading line with the slope ku1- Waz is the area under the unloading line with the slope kyz. The energy lost in the collision is equal to the difference between the initial and final kinetic energies of the system and it is given by W, in Equation (22). Wi = m,vi/2 + m2vš/2 - m: (vi)* /2 – m2 (v^)° /2 (22) Using Equations (15). (17). (21) and (22), it can easily be shown that W, - WL = WR (23)
Transcribed Image Text:Recall that Fmax is the magnitude of the crush force acting on both vehicles at maximum deformation (see Figure 3). Then, the total crush energy absorbed by the structures of the two vehidles is given by Wr in Equation (17). Fmax = k1X71 = kax7a (16) W, = kx,/2 + kı2X,/2 = FmaxXT1/2 + Fmaz*r2/2 (17) Wr1 = FmaxXr1/2 . Wrz = Fmas*rz/2 (18) Wri and Wrz are the total crush energies absorbed by vehicles 1 and 2 respectively. Equation (17) can be written as, W, = Fmax /(2ka) + Fmar /(2k12) = Fimar /(2keqı) (18a) where kegl = kukı2/(kLı + k12) is the equivalent loading stiffness of the system. The common velocity V. = Vi can be determined by applying conservation of momentum principle to the two-vehicle system since the net force acting on this system is zero. Then. m,V, + m2V½ = (m, + m2)V, (19) Since the structures of the two vehicles do not undergo perfectly plastic deformation, part of the energy absorbed is transformed into kinetic energy due to spring-back effect in the unloading phase. A similar work-energy analysis can be made for the unloading phase and it can be shown that the crush energy retumed as kinetic energy in the unloading phase is given by Wa in Equations (20) and (21). Wa1 = ku, (Xr1 - Xp1)*/2 = Fmaz (Xr1 – Xp1)/2 WR2 = kuz (Xr2 -Xp2)/2 = Fmar(Xrz - Xp2)/2 WR = WR1 + WR2 We = kui (X71 – Xp1)*/2 + kuz(X72 – Xpa)* /2 (20) - m, (V;)*/2 + m,(V£D* /2 - (m, + m2)V? /2 WR = (21) Wan is the area under the unloading line with the slope ku1- Waz is the area under the unloading line with the slope kyz. The energy lost in the collision is equal to the difference between the initial and final kinetic energies of the system and it is given by W, in Equation (22). Wi = m,vi/2 + m2vš/2 - m: (vi)* /2 – m2 (v^)° /2 (22) Using Equations (15). (17). (21) and (22), it can easily be shown that W, - WL = WR (23)
1) Consider a central in-line head-on collision of two vehicles with rebound which is described show that the following relationships are true. keql kequ kegl WR1 _ ku2 WR2 ku1 Wrkeql Wr1 Wrkegl Wr2 WR = Wr- kequ kui kuz 1 1 1 where kegu kui ' ku2
Transcribed Image Text:1) Consider a central in-line head-on collision of two vehicles with rebound which is described show that the following relationships are true. keql kequ kegl WR1 _ ku2 WR2 ku1 Wrkeql Wr1 Wrkegl Wr2 WR = Wr- kequ kui kuz 1 1 1 where kegu kui ' ku2
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