Physics 6 - Moth #4.. axe, ction. . Assume the particle...

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Unformatted text preview: Moth #4.. axe, ction. . Assume the particle model for the two blocks, and the model of kinetic and static Model: m m m P n l a k w. My 3 m x .n m 0 0 C IV. Y s m m a Dr. . m cm 3 m y r. w r. l a .n m C H 9 = 20° = 0.50 533%.? 4 N ‘ Hagan mnxfiwht, , .C t 8-28 Chapter 8 Solve: (a) If the mass m is too small, the hanging 2 kg mass will pull it up the slope. We want to find the small mass that will stick because of friction. The smallest mass will be the one for which the force of static friction is its maximum possible value: fx = (f8)max = [,tsn. As long as the mass m is stuck, both blocks are at rest w F m = O N. Newton’s second law for the hanging mass M is (Fact),=TM-Mg=0N=>TM=Mg=19.6N For the smaller mass m, (FML = 7:" —fs — mgsin9 = 0 N (an)y = n —mgcos0 => n == mgcosB —~ For a massless string and frictionless pulley, forces T and T‘M act as if they are an action/reaction pair. Th m TIn = TM. Mass m is a minimum when (f5 )m = flsn = psmgcose. Substituting these expressions into the x-equation TM T -— cosB—m sin9=0N=>m=——————— M flsmg g (#5 cos9+sin9)g = 1.83 kg (b) Because pk < #3, the 1.83 kg block will begin to slide up the ramp, and the 2 kg mass will begin to fall, if“ block is nudged ever so slightly. Now the net force and the acceleration are not zero. Notice how, in the pict representation, we chose different coordinate systems for the two masses. This gives block M an acceleration only a y»component and block m an acceleration with only an x—component. The magnitudes of the acceleratx are the same because the blocks are tied together. But block M has a negative acceleration component ay (vecto ‘ points down) whereas block m has a positive ax. Thus the acceleration constraint is (am); = — (aM)y = a, where a w have a positive value. Newton’s second law for block M is (F )y = T—Mg = M(aM)v = -Ma net For block m we have (Fm)x = T—fk — mgsinO = T—ykmgcose—mgsine = m(am) = ma I In writing these equations, we used Newton’s third law to write Tm = TM = T. Also, we noticed that the y—equa and the friction model for block m don’t change, except for [.L5 becoming pk, so we already know fl from part Notice that the tension in the string is not the weight Mg. We have two equations in the two unknowns T and a: T—Mg=—Ma T—(,uk cosO+sin6)mg=ma Subtracting the second equation from the first to eliminate T, —Mg +01k 0089 + sin6)mg = —Ma -— ma = —(M + m)a M — (pk c059 + sin 9)m M+ m =94: g=1.32 m/s2 . out; fixaa‘nlumw‘n‘mmfiqlfi if, A NC «a “N 5 x pulleys frictionless Physical representation Pictorial representation tramt Frictiouless fodel: Use the panicle model for the two blocks. Assume a massless rope and massless, Acceleration cons Qfifi , $me l ‘5 Mfiuqé \wn 1 8—32 Chapter 8 For every one meter that the l-kg block goes down, each rope on the 2—kg block will be shortened by one-half meter. Thus the acceleration constraint is all: —2a2. Solve: Newton’s second law for the two blocks is 2T: mza2 T— wl = m,al Since a1 = -2a2, the above equations become 2T: mza2 T— mlg = m,(— 2a,) 2mg _ 2(1kg)(9.8 m/sz) a =>m2—22—+m,(2a2)=m1g =>a2 =——————-— = 3.27 m/s2 m2 + 4m1 (2 kg + 4 kg) Assess: If m: 0 kg, then a2 = O m/sz, which is expected. 8-36 Chapter 8 8.48. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic equations Visualize: Pictorial representation Physical representation Player off floor Player while jumping ‘~-----.... -.__......-..-.. m.-- - _ --.. -ae (10 Floor y Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straights his legs. He exerts a force F1, on P on the floor. The player experiences a weight force $171, as well as a normal 1’ by the floor 13m P. The only force that the floor experiences is the one exerted by the player in“? (b) The player standing at rest exerts a force F P m P on the floor. The normal force ii”. P is the reaction force to Fm But 11”“, = P0”, so F "a = 0 N. When the basketball player accelerates upwards by straightening his legs, , speed has to increase from zero to vly with which he leaves the floor. Thus, according to Newton’s second la there must be a net upward force on him during this time. This can be true only if rim”, > wp. In other words,‘ ‘ player presses on the floor with a force FP on F larger than his weight. The reaction force fig 0,, 1, then exceeds weight and accelerates him upward until his feet leave the floor. (c) The height of 80 cm = 0.80 m is sufficient to determine the speed v‘y with which he leaves the floor. Once feet are off the floor, he is simply in free fall, with a1 = —— g. From kinematics, v22y = vlzy + 2a,(y2 —- y,) => 0 mzls2 = v12y + 2(—g)(0.80 m) => vly = 1/2 g(0.80 m) = 3.96 m/s (d) The basketball player reached v1), = 3.96 m/s by accelerating from rest through a distance of 0.60 m. Assumin- an to be constant during the jump, we find ' vfy _ (3.96 m/s)2 2yl 2(0.60 m) 1)?y = vgy +2ao(yl —yo) = 0 mZ/s2 +2a(,(yl —0 m) ==> a0 = = 13.1 m/s2 (e) The scale reads the value of nFmP, the force exerted by the scale on the player. Before jumping, um? — WP = o N => nm = w? = (100 kg)(9.8 m/sz) = 980 N While accelerating up, nmn, -—wP = ma0 => nFonP = mao +mg = m l+—§- = (980 N)(l+—l—9§£1)=1710 N ‘10 After leaving the scale, 11m P = 0 N because there is no contact with the scale. 8-38 Chapter 8 8.50. Model: The hanging masses m1, m1, and m3 are modeled as particles. Pulleys A and B are massless frictionless. The strings are massless. Visualize: Pictorial representation ,-----...._ Solve: (a) The length of the string over pulley B is constant. Therefore, (YB “Y3)+(YB ‘1“): LB :9 VA 2 23’]; ‘Y3 “LB The length of the string over pulley A is constant. Thus, ‘ (yA -yz)+(yA ~y1)=LA =2yA -y1-yz => 2(2ya —y3 - 110%. -y2 = LA =9 2% +y2 +y1= constant This constraint implies that gain in dt dt dt = 0 m/s = 2v3y +122}, + v”, Also by differentiation, 2a3y + a2y + a” = 0 ml s2. (b) Newton’s second law for the masses m3, mz, m,, and pulley A is Tn— msg = m3a3y TA “ "'93 = "My TA “ mtg = mlaly TB—ZTA=ON The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint are five equations for the five unknowns (two tensions and three accelerations). To solve for TA, multiply the m3 equation by 2, substitute 2TB = 4TA, then divide each of the mass equations by the mass. This gives the three equations 4TA/m3 -2g = 2a3y TA/mz —g=a2y TA/ml —g=a,y 35:", m a: x q Newton’s Third Law 8-39 If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus 4:; (4lm3 +1/m2 +1/m2) (4/m3 +1/m2 + 1/m2 )TA —4g= 0 => TA = (c) Using numerical values, we find T A = 18.97 N. Then a1y = TA/ml —g = —2.21 m/s2 a2 = TAlm2 ——g = 2.85 m/s2 a3 = ZTAIm3 —-g =—0.315 m/s2 (d) m3 = ml + m2, so it looks at first like m3 should hang in equilibrium. For it to do so, tension TB would need to equal m3g. However, TB is not (m1 + nag because masses m1 and m2 are accelerating rather than hanging at rest. Consequently, tension TB is not able to balance the weight of m. .i, ,-.:g£,~.—.£ , a: y $525; 9.32. Model: Model the train cars as particles. Since the train cars stick together, we are dealing with perfect] inelastic collisions. Momentum is conserved in the collisions of this problem. Pictorial representation V23 Solve: In the collision between the three—car train and the single car: mle + (3m)v2x = 4mv3x => v” + 3v“ = 4v3x = (4.0 m/s) + 3(2.0 m/s) = 4va => v3x = 2.5 m/s \ In the collision between the four~car train and the stationary car: ’ " 4 (4m)v3x + mv4x = (5m)v5x => 41131 + 0 m/s = 5va => v5,‘ = v“ = (0.8)(25 m/s) = 2.0 m/s 9-18 Chapter9 ’ 9.41. Model: This is a two—part problem. First, we have an explosion that creates two particles. The momentum of the system, comprised of two fragments, is conserved in the explosion. Second, we will use kinematic equations and the model of kinetic friction to find the displacement of the lighter fragment. Visualize: Pictorial representation Physical representation Y "11., = 7 m mL = m (V093 = 0 (V091. = 0 Before ' y _, Free-body diagram " of mg a x Just after WE fl; collision y _. Free-body diagram f» " of mL k After 1: some time H _, xzn’ (V2x)n = 0 1521., (V2191. = 0 W Solve: The initial momentum is zero. Using momentum conservation pg =11“ during the explosion, mH(le)H + mL(le)L = mH(V0x)H + mL(v0x)L => 7m("1x)u + m(le)L = 0 kg m l 5 => (V143,; = f(‘;f)(v1x)L Because mH slides to x2H = —8.2 m before stopping, we have fk = luknn = llka = #kag = mnau => an = “kg Using kinematics, (vhf = (11”): + 2a,,(x2H — xm) => 0 1112/52 = (%)2(v1x): + 2pkg(——8.2 m—O m) =:>(v1x)L = «88.74 pk m/s H How far does mL slide? Using the information obtained above in the following kinematic equation, (v2x)2 (v13): + 2aL(xZL " le) =3 0 m2/52 = #k(88'74)2 “’2”:ng => sz = 402 m ‘ J L— Assess: Note that aH is positive, but aL is negative, and both are equal in magnitude to M. Also, xm is negati «i but sz is positive. . and particle 3 mg west), lastic. The momentum of I rfectly ine IS pe rth), particle 2 (mov After which ing no llismn, 11' CO Pictorial representation . The three stick together during the theast) :3 system is conserved. mung SOL! .55. Model: Model the three balls of clay as particle 1 (mov » l “Zr. ( 5 c3: .33. e « ex: W llll : ha WWW. w...me aafuflfimfisfifis t (s 38.x , 4 . t. m l , . u» M. m x% mm W: L we” mamas, . H. t. «with finfimt «@mmmuwnt 72.1.?» : . A mkfikiwvfififih u mwmwmflwfi a . c $Wwéwmwxax 9-26 Chapter 9 Solve: The three initial momenta are ,3“ = min = (0.02 kg)(2 m / 3)} = 0.04} kg m / 5 [3‘2 = min = (0.03 kg)(—3 m / s i) = —0.092 kg m / s 1313 = mfifl = (0.04 kg)[(4 m / s)cos45°€ — (4 m / s)sin45°j'] = (0.113? —0.113j) kg ml s Since if = 13, = 1—7“ + 1312 + [313, we have (ml + m2 + m3)17f = (0.023? — 0.073;) kg m / 8 => at = (0.256? —0.811}) m / s => v, = (0.256 m/s)2 +(—0.811m/s)2 = 0.850 m/s _. lvfyl = my, 0811 v, 0.256 X 6 = tan = 72.5° below +x 1. Model: Let the system be rocket + bullet. This is an isolated system, so momentum is conserved. 'alize: The fact that the bullet’s velocity relative to the rocket is 139,000 can be written (vf B = (129R + ______ M vi AZ W vIf + 139,000 m/s ._ _ W —‘ "" 5 kg Before After Consider the firing of one bullet when the rocket has mass M and velocity v,. The conservation of momentum =Pt is (M — 5kg)vf + (5 kg)(vf + 139,000 m/s) = MvI => Av = vf —vi = —%g—(139,000 m/s) rocket starts with mass M = 2000 kg, which is much larger than 5 kg. If only a few bullets are needed, M will change significantly as the rocket slows. If we assume that M remains constant at 2000 kg, the loss of speed per at is Av = —347.5 m/s = —1250 km/h. Thus exactly 8 bullets will reduce the speed by 10,000 km/h, from 25,000 /h to 15,000 km/h. If you’re not sure that treating M as a constant is valid, you can calculate Av for each bullet reduce M by 5 kg after each shot. The loss of mass causes Av to increase slightly for each bullet. An eight-step nlation’then finds that 8 .bullets will slow the rocket to 14,900 km/h. Seven bullets wouldn’t be enough, and mild slow the rocket far too much. ...
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Physics 6 - Moth #4.. axe, ction. . Assume the particle...

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