Spring 2001 Physics 2048 Test 3 solutions

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1 Sping 001 Physics 048 Test 3 solutions Poblem 1. (Shot Answe: 15 points) a. 1 b. 3 c. 4* d. 9 e. 8 f. 9 *emembe that since KE = ½ mv, KE must be positive

2 Poblem (Estimation Poblem: 15 points) Use momentum-impulse theoem p ca = I ca v 0 Initial velocity I ca t net 1 = Fca dt = Fma t1 p = mv f mv 0 t p = mv f mv, note that the final velocity is in the opposite diection of the initial velocity so v f = - α v,, whee α = 90% p = m ( v f v ) = m[( α v ) v ] = (1 + α) mv Taking the magnitude of p and setting it equal to the magnitude of the impulse 1 ( 1 + α ) mv = Fma t F = (1 + ) mv / t ma α 0 Given: v 0 = 5 mi/h * 0.6 km/h * 1 h/3600 sec * 1000 m/ km = m/s α = 0.90 Estimate time inteval and mass of the ca Reasonable estimates 500 kg < m < 3000 kg fo mass 0. s < t < 1 s fo time inteval Let m = 1000 kg and t = 0.5 s Fma = (1 + α ) mv / t = ( ) (1000 kg) (4.306 m/s) / (0.5 s) = 33,000 = 33 k Given quantities ae good to significant digits

3 Poblem 3 (Essay 10 points) You may use diagams and equations but no calculations in you esponse fo this poblem. Student Response 1: hen the bakes of a ca lock up, then the ca slides. hen the ca is sliding, then kinetic fiction is involved. hen the wheels ae tuning, then static fiction is involved. In the case of ubbe on wet concete, the coefficient of kinetic fiction (sliding) is less than that of static fiction (olling). And when tying to stop it is best to have the most fiction possible fo faste deceleation. Theefoe, it would be best to have static fiction in this case (in the ain) which means the wheels need to be tuning. Anti-lock bakes keep the wheels tuning which means thee is moe fiction between the ties and the oad. This is good essay that hits most of the key points. The only points that ae missed ae that the fiction foce govens the baking of the ca (bing the ca to a stop) and why you want maimum deceleation in the ain. The following esponse addesses these points. Student Response Static fiction is lage than kinetic fiction. hen stopping, if the wheels lock up kinetic fiction is used to slow the ca down. Howeve, since static fiction is lage, what anti-lock bakes do pump the beaks so that the ties do not epeience kinetic fiction but static, which in tun slows the ca faste and in a shote distance. Since the coefficient of static and kinetic fiction dop damatically on a wet oad, antilock bakes ae paticulaly safe. The small fiction foce due to skidding in the ain also educes you contol ove the ca, so the anti-lock bakes educe you stopping distance and help you maintain contol of you ca in the ain. Key points: hen the wheels ae locked and skidding, baking is by kinetic fiction hen the wheels ae otating, baking is by static fiction The coefficient of static fiction is geate than that of kinetic fiction The lage the coefficient of fiction, the geate the fictional foce fom beaking. A lage fictional foce means the ca stops in less time and less distance. Both coefficients of fiction fo ubbe and concete ae educed when it is aining and the ca is wet. This means that in the ain, the diving will be moe slippey and it will take longe to stop. Since diving in the ain is moe slippey, you would like the lagest fictional foce possible to minimize you beaking distance and maimize you ability to keep the ca unde contol. Anti-lock bakes pevent the wheels fom locking up and skidding so that the baking foce is fom static fiction instead of kinetic fiction.

4 Poblem 4 (15 points) A. [6 pts] Daw sepaate fee-body diagams fo each of the thee blocks. Label you foces to make clea (1) the object on which the foce acts, () the object eeting the foce, and (3) the type of foce (nomal, fictional, gavitational, etc.) Fee-body diagam fo block A B A table A eath A Fee-body diagam fo block B table B C B A B eath B Fee-body diagam fo block C table C hand C eath C B C B. [4pts] In the spaces at ight, daw a vecto that epesents the net foce on each block. Make sue you vectos ae dawn with coect elative magnitudes. Eplain how you knew to daw the net foce vectos as you did. et foce on Block A et foce on Block B et foce on Block C Each block is acceleating to the left at the same ate so by ewton s nd law, the net foce fo each block is invesely popotional to the mass so that the atio of F net /m is constant fo each block. C. [5 pts] Suppose the mass of block B wee doubled (the othe blocks ae left unchanged) and the hand pushes with the same foce as in pat A. i. Has the magnitude of the acceleation of block A inceased, deceased, o emained the same? Eplain. net F Deceased, the thee blocks acceleate as one system and since a =, if you incease the m mass of the system and keep the net foce on the system the same (i.e. the foce of the hand), the acceleation of the system of thee blocks must decease ii. Has the magnitude of the net foce on block A inceased, deceased, o emained the same? Eplain. Deceased, if the acceleation of block A deceases and the mass of A doesn t change, then the net foce needed to cause that acceleation is less.

5 Poblem 5 (0 points) Pat A Looking at the foces acting on the motocycle moving though a cicula loop at constant speed: mv Cicula motion at constant speed => = R net At the top of the loop, F = ote that both foces point downwad. mv R F net tact motocycly + eveywhee on the loop. eath motocycle, F net = + mg = + mg, Since m, R, and g ae constant, v will be a minimum when = 0. y motocylce y tack motocycle eath motocycle v = Rg = (5.0m)(9.8m / s 3600s 1km = 7.0m / s = 5km / h h 1000m ote the bigge the adius R is, the highe the minimum speed v needed to go though the loop Pat B Only gavity acts on the motocycle + ide when they ae in the ai a y = -9.8 m/s and a = 0 m/s, since a = constant => use kinematics equations = v 0 t, y = v 0y t - 1/ g t If we knew t, we could find fo the jump => let y = 0 0 = v 0y t - 1/ g t, t = 0 o v 0y - 1/ g t = 0 Since t = 0 is the stat of the jump, t = v 0y /g = v 0 sin θ / g is the time it take the motocycle to make the jump. = v 0 t = (v 0 cos θ)( v 0 sin θ /g) = v 0 sin θ / g) = (50 km/h 1000m/km 1 h / 3600 s) sin (*30 ) / 9.8 m/s = 17 m The landing should be placed 17 m to the ight of the jump amp

PS113 Chapter 5 Dynamics of Uniform Circular Motion

PS113 Chapter 5 Dynamics of Uniform Circular Motion PS113 Chapte 5 Dynamics of Unifom Cicula Motion 1 Unifom cicula motion Unifom cicula motion is the motion of an object taveling at a constant (unifom) speed on a cicula path. The peiod T is the time equied