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### Ph1a-2007-HW5-soln

Course: PH 1a, Fall 2007
School: Caltech
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Word Count: 1152

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Solutions Ph1a 5 Chefung Chan (cchan@caltech.edu), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 5.1 5.1.a QP14 (5 points) (1 point) Use the definition of the center of mass: yCM = 5.1.b (1 point) i mi y i md + (2m)(0) m1 y 1 + m2 y 2 = = = m1 + m 2 m + 2m mi i 1 3 d. Use conservation of energy to find v1i by equating...

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Solutions Ph1a 5 Chefung Chan (cchan@caltech.edu), Fall 2007 Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself. Use these instead. 5.1 5.1.a QP14 (5 points) (1 point) Use the definition of the center of mass: yCM = 5.1.b (1 point) i mi y i md + (2m)(0) m1 y 1 + m2 y 2 = = = m1 + m 2 m + 2m mi i 1 3 d. Use conservation of energy to find v1i by equating the energies at the top and bottom of the arc: 1 2 Ei = mgd = 2 mv1i = Ef v1i = 2gd = v. The lower mass m2 is at rest, so v2i = 0. Momentum is conserved in the collision. Because the collision is elastic the kinetic energy is also conserved. The initial momenta and energies are p1i p2i p K1i K2i K = mv0 = 0 p1i + p2i = mv0 1 = mv 2 2 0 = 0 1 2 K1i + K2i = mv0 2 The final momentum of the two pendula are given by p1f = mv1f and p2f = 2mv2f and the final kinetic 1 2 2 energies by K1f = 2 mv1f and K2f = mv2f . Applying the conservations laws: p1f + p2f mv1f + 2mv2f K1f + K2f 1 2 mv 2 + mv2f 2 1f = p = mv0 = K 1 = mv 2 2 0 2 Solving these equations yields v1f = - 1 v0 and v2f = + 3 v0 . 3 5.1.c (2 points) The energies of the masses after the collision are E1f = 1 1 2 mv1f = mv 2 2 18 0 4 mv 2 9 0 2 E2f = mv2f = 1 2 Apply conservation of energy to each mass, noting that v0 = 2gd: E1f = E2f 1 mv 2 = mgh1 18 0 4 2 = mv0 = 2mgh2 9 1 -3 v 2g 2 3 2 2 h1 = = 1 9 d v h2 = = 4 d. 9 2g To find the height of the center of mass, apply the definition again: yCM = i mi y i m1 h 1 + m2 h 2 = = mi m1 + m2 i 1 9 md + 4 (2m)d 9 = 3m 1 3 d. 1 16 The corresponding answers for the values given in case you couldn't solve (b) are h1 = 19 and yCM = 48 d. Bonus (1 point) d, h2 = 9 16 d, The stated assumption that the masses reach their maximum height simultaneously is invalid except in the limit of small displacements from the bottom. Bonus points should be awarded to students who realize this. Another way around this faulty assumption was given by Christiaan Huygens in 1673: the solution to the tautochrone ("equal time") problem. See for example the website http://mathworld.wolfram.com/TautochroneProblem.html for a nice explanation and animation. 5.1.d (1 point) Energy is no longer conserved, but momentum is still conserved: pi = mv = (m + 2m)vf = pf h= 1 The final energy is (3m) g ( 9 d) = 1 3 2 vf = 2g 1 9 d. vf = 1 3 v0 . mgd, and initially it was mgd, so the fractional change is 1 mgd E Ef = -1= 3 - 1 = -2, 3 Ei Ei mgd so two-thirds of the initial energy is lost in the collision (dissipated as heat, sound, or physical deformation). 5.2 5.2.a QP22 (5 points) (1 point) Conserving energy from the top to the bottom of the incline gives (where W is the work done by friction): Ui = Kf + W 1 2 M vM + FN d 2 The work done by friction depends on the length d of the incline, where d sin = h, and the normal force FN = M g cos . h 1 2 M gh = M vM + M g cos 2 sin M gh = vM = 2gh(1 - cot ) Assume for the rest of the problem that vM = gh. 2 5.2.b (2 points) The velocity of the mass m is found from energy. conserving This gives: Ui = Kf mgL = vm = 1 mv 2 2 m 2gL For an inelastic collision, we conserve x momentum, with right positive. M vM - mvm = (M + m)vC gh - m 2gL = (M + m)vC 1 M gh - m 2gL vC = M +m The combined object will come to rest when vC = 0, which gives: M M gh = m m M 2gL 2 h = 2L 5.2.c (2 points) Let MT = M + m be the total mass. For the combined object to remain in circular motion at the top of the loop, equating the forces gives: T + Fg = Fc The critical case is T = 0 since T is not allowed to become negative. MT g = 2 MT vtop L vtop = gL Now we conserve energy between the initial point (right after the collision) and the top of the loop. Ki = Kf + Uf 1 1 2 2 MT vC = MT vtop + MT g(2L) 2 2 2 vC = gL + 4gL vC = 1 M M +m 5gL On the other hand, given M = 2m, we calculate vC from part (b). vC = gh - m 2gL 1 2m gh - m 2gL 3m 2 1 vC = gh - 2gL 3 3 Setting these equal, and solving for h gives: vC = 1 2gL = 5gL 3 2 h = 2L + 3 5L L 47 + 6 10 16.5L h= 4 gh - 3 2 3 5.3 5.3.a QP23 (5 points) (1 point) The amount of sand in midair is not weighing on the scale. mfalling = t W = g(M + m - mfalling ) W = g(M + m - t) 5.3.b (1 point) We calculate the impact force F that the scale feels by stopping the falling sand. Fimpact = d dm dp = (mv) = vf = vf dt dt dt where vf = gt1 is the speed at which the sand hits the scale. Also note that during this time, there will still be sand remaining at the top, and also a small pile accumulating at the bottom. mbottom = (t - t1 ) W = g(M + m - mfalling + mbottom ) + Fimpact W = g(M + m - t + (t - t1 )) + vf W = g(M + m - t1 ) + gt1 W = g(M + m) 5.3.c (1 point) Here there is no more sand at the top, so we have: W = g(M + mbottom ) + Fimpact W = g(M + (t - t1 )) + vf W = g(M + t) 5.3.d (1 point) For all times t > t3 , the sand is all sitting at the bottom. W = g(M + m) 5.3.e (1 point) We can simplify the previous results by noting the sand takes a time t2 to completely leave the top reservoir: = m t2 Given that m < M , we have the following qualitative graph of W (t) in Fig. ??. 4 This simplifies the above 4 equations to the following: t g(M + m - m t2 ) g(M + m) W = t g(M + m t2 ) g(M + m) for for for for t [0, t1 ) t [t1 , t2 ] t [t2 , t3 ) t t3 W(t) (M+m)g Mg t t1 t2 t3 Figure 1: QP23e 5.4 5.4.a FP2 (5 points) (2 points) 1 1 2 m1 v1 = mv 2 2 2 1 1 2 m2 v2 = (ab2 )mv 2 2 2 1 2 mv 1 + ab2 2 Prior to the collision the total energy and linear momentum of the two-mass system is E1i E2i Ei p1i p2i pi = = = = m1 v1 = m(v^) x = m2 v2 = (am) [bv(cos ^ + sin ^)] x y = mv [(1 + ab cos )^ + (ab sin )^] x y 5.4.b (3 points) In an elastic collision momentum is conserved but kinetic energy is not. pi pf vf vf Ef = mv [(1 + ab cos )^ + (ab sin )^] x y = (m1 + m2 )vf pf pi = = m(1 + a) m(1 + a) v = [(1 + ab cos )^ + (ab sin )^] x y 1+a 1 2 (m1 + m2 )vf = 2 1 2 = m(1 + a)vf 2 mv 2 = (1 + ab cos )2 + (ab sin )2 2(1 + a) mv 2 = 1 + a2 b2 + 2ab cos 2(1 + a) 1 + a2 b2 + 2ab cos 1 mv 2 1 + ab2 - = 2 1+a 1 a = (1 + b2 - 2b cos ) mv 2 2 1+a Ef E 5
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