MT1L2 - Physics 7A Midterm #1, Lecture Section 2, Fall...

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Unformatted text preview: Physics 7A Midterm #1, Lecture Section 2, Fall 2006. ©A. Zettl All Problems Worth 20 points. One side of one sheet notes (8-1/2" x 11") allowed. Good luck. 1. A small crazy squirrel runs along the nearly flat ground at speed 2.0m/s onto a large, flat, frozen mud puddle whose surface is absolutely frictionless. Somewhere near the middle of the puddle the squirrel jumps up and reaches a maximum height 1.23m above the ice surface. a) Determine the total time the squirrel is in the air (i.e. time between jumping and landing). b) Determine the horizontal distance between the jumping and landing spots on the ice. c) Determine the speed of the squirrel just before it lands. 2. A car of mass M will be towed along a roadway uniformly inclined at angle (F)1 by means of a (massless) cable attached to a powerful tow truck. The cable makes an angle 92 with respect to the roadway. The tow truck starts from rest. What is the greatest distance Lmax along the roadway that the car can be towed in the firstbt seconds if the cable has a breaking strength Tm? Remember, M, 91, 62,me, and 5t are all known quantities. Also, neglect frictional forces on the car (you may treat the car as if it slides without friction). 3. A crate slides down a right-angled trough inclined at angle 9 as shown in the figure. The coefficient of kinetic friction between the crate and the material composing the trough is yk. Find the magnitude of the acceleration of the crate down the trough. /.90°\ 4. Five blocks all of different mass are connected by massless (non-stretching) string segments as shown: The pulley is massless and frictionless, and the horizontal table on which M1 and M2 slide is frictionless. Grav1ty g, vertically directed, acts on the blocks. A 8) Detcrfnine the horizontal acceleration vector of M1. W M 1 AN b) Find the tension T in the string segment connecting M3 and M4. 1 /\ ————-————-— > ’X M3 . t’\ q 1 Mb“ 5. The plot below ShOWS the velocity versus time of a point particle moving in one dimension. Since v is always positive, you may assume the particle always moves in the +x direction. Assume that at t=0 the particle is at x=0. Make corresponding plots of a) position x (in meters) and b) acceleration a (in m/secz) versus time for the partide, for the same time range (0 to 10 seconds). In both cases you should mark the axes explicitly with numerical values. Y0ur plots must also clearly show what happens at t = 2, 4, and 6 seconds. 60 A £40 i v > 20 IO 8 1(5) [ AX ) w “B $3 ‘4‘“ VXB _ Vila—o =/ 23m V c "’ VXA=2.o"/5 - ~ ' S W! as W» “WWW Wag? W@ h K15 5%gLfl—Wfl-jfile. les "("the, 3* CL °Yfi A—(:’/:»_' 30 A'tAQB = A—tvb: Air“ 34¢ Artmc ; 25%: at. "g; C331 MM‘VZL IXQVL '12; CoV\B\C1eY\l$¢\—HTQ tile-{10147 we» flittwvzl "(122M "Each. A34 “* Voikb/z + 2'4 55v: l’leva AA: -A C€)_76—¥v1c1'7%l& aged Cffi‘e Qium’E/l/ 7g); 17”?” 1:709 w 0-?‘14/13 squm'e/L 64K?!” J:an FF V :/\Z\ {gal-6v 5° Vc‘qc. :' _ fl = “1/5 314%“ We 81:52:} vcz. M: +v,5 ; 5.3 “1/5? Flayst 3% , M.‘Aterm#l., Scol ~ A. Zet-t{. Choose. coorolmwtes So Hmt Hm X-axfs 1.5 along “flu: accelem‘h‘on Vector 1 11¢. abng, 'Hne. MC "he, , F :: Tmax C0592 " Scne’ : M axlmm‘. 7<,Ma\>< Sch/e, for 1%; woodman acceLerwtCon. CLxlmng: TmM C03 92 _. g 3(119‘ Physms :m Fan 200Lo Sachem Z Muier i 2e'VH A,” ' 39W .HAC oym‘ wou\+o (>th 44mg ‘3 ‘ wobka ‘ _,'_._L8_________. ._ . Y _ I: V6 . 2 ER, '— N.LosHs + Nzcogbpg - mficose ; MON :0 iF23’Ném‘15 +N25mH$:mai—.O —47 N\;NL2N Fmd Nf-Nf-N wH’h iFVmc-k 2N cos HS : MO) L08 6 N 2 m3 C039 ZCOSWS E‘: El: EC; MLN : Mngcose: 2. (95 RS VIM mto ifizmm " I —— 1: .. 1 _Mkrm$(r0$6 max , W32.an B” F“ W5st 23: mane LOSL‘SG 3 m O A Okf— casme ' “£03 “’36 .1 COSHS gsme - TzMLcfico5® =0\ Fkyss’cs TrA . Msokterm 4*! , Sec 2—. A. Zettt (0.), Because +l'te, string 1's non—sweating, “LL masscs hon/e, {-Ltc same; acwtemflon} 0., Here. are Newton’s Equarh’ons for +116 blocks: m3 +11 ~T3 = W 6L M5 3" "‘" Adel up all, Hue ec‘uOd't'ons G +@+@+@+@ awoL cancel. 6L(L +ke Tensn’ons , we. 39,1; M33 + Mug—l" M53 = M.Q+Mza+M3a+Mtp4+/V15a, & (M3+M4+Ms>3 Z. M— M|+M1+M3+M¢+M$ CD @ M33 +TS“T7_ 1" Msa @ ® @ (b). The fenSt'OrL in. “W Sffl'ma Segment ConfleC’Hné M3 and Mq, 1': just T: T3 , Fron’L Eq'n ) We gag sol/‘18. for TL‘L : Mr‘I'Mz. 58' . fl - Z: TIL ” Mgg Msa Mu+M1+M3+Mu+Ms J From Eq'rL (LB , We COU’L solve. for T3 .~ (M.+M7_)Ms& _(M3+M4+M:)M43 Ml+ML1Msfln¢Mk T3 Zeltl.nb Fa 5 400 100 ~ -10 fi ...
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MT1L2 - Physics 7A Midterm #1, Lecture Section 2, Fall...

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