This immediately allows us to cancel out the corresponding factors in Eq {4.7}. so we are left with is], + 1'1] = cg.- + 1J2 j. which can be rewritten as 151): "Ugf Z \"2.; 'L'13' {4.8) and this is equivalent to (4.4). So, in an elastic collision the speed at which the two objects move apart is the same as the speed at which they came together, whereas. in what is clearly the opposite extreme, in a totally inelastic collision the nal relative speed is zerothe objects do not move apart at all after they collide. This suggests that we can quantify how inelastic a collision is by the ratio of the nal to the initial magnitude of the relative velocity. This ratio is denoted by e and is called the coeicient of restitution. Formally, {'12:} _ _'EJ2'f \"1] 1-121 1'29; \"1i {4.9) e: For an elastic collision. e = 1. as required by Eq. (4.4]. For a totally inelastic collision. like the one depicted in Fig. 3, e = 0. For a collision that is inelastic, but not totally inelastic, e will have some value in between these two extremes. This knowledge can be used to "design" inelastic collisions (for homework problems, for instance!}: just pick a value for e, between '3 and 1, in Eq. (4.9], and combine this equation with the conservation of momentum requirement (4.5}. The two equations then allow you to calculate the nal velocities for any values of ml, mg. and the initial velocities. Figure 4.4 below, for example, shows what the collision in Figure 4.1 would have been like. if the coefcient of restitution had been 0.6 instead of 1. You can check. by solving (4.5) and (4.9) together, and using the initial velocities, that '1'\" = 1/' 15 n1 / s = D.0667 n1 / s. and 1J2 f = 8,! L":- m/ s = 0.533 m f s. 1.2 0.8 05 v (WE) 0:4 0.2 {1.2 {1.4 t{m5) Figure 4.4: 31.11 c = 0.6 collie-inn between objects with the same illcrtias and initial velocities as in Figure 1. Problem 3: For the collision depicted in Figure 4.4, what is the velocity of cart 1 relative to cart 2 before and after the collision? Select One of the Following: 0.0667m/s, Ugly = 0.533 m/s 1 m/s, 1321,; = 0 m/s Hgl' = ]. m/S,1J21.f = 0.6m/s fHw-HfHf-'x D" '-_-"-_-"-.-"-.-' 1'" :3 rA a || 1921;; = 1 m/5.1J21.f = 0.6m/s Problem 4: For the same collision, what are the kinetic energies of the two carts separately, and of the system as a whole, before the collision? Take ml = 1, m2 = 2 kg. Select One of the Following: a) K1,, = 1J, K2,, 2 OJ, K3,,\" = 1J b) K1,; = 0.1, K2,, = 0.5.1, K,.,,,_., = 0.5J c) K1,; = 0.5.1, K2,, = 0.1, K,.,,,_., = 0.5J d) K1,, = OJ, K2,, = 1J, K3,,\" = 1.! f ( ( ( Problem 5: For the same collision, what are the kinetic energies of the two carts separately, and of the system as a whole, after the collision? Select One of the Following: (a) K\" = 0.0022 J, K1Jr = 0.2844J. KWMr = 0.2er (h) K\" = 0.0044 J, K\" = 0.2844J. 51'er = 0.2ng (c) K\" = 0.0022 J, K\" = 0.1422J. myMr = 0.144.] (d) K\" = 0.0022 J, K1Jr = 0.5689J, Km\" 2 0.571 J