Average Speed - Introductory Physics - Solved Paper, Exams of Physics

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FROM : FAX NO. = May. 23 2683 8:86AM PL SotuTt eons Physics 8A Section 2 Final Tame May 21, 2003 5 pm - 8 pm SID Number discussion session number / GSI name a) DON’T OPEN THIS EXAM UNTIL INSTRUCTED TO BEGIN b) Sit one seat away from anyone else. c) Do all your work on the page (front/back) indicated for each problem. d) Show all work; don’t just write an answer without showing your reasoning. e) This is a closed book exam but calculators and one sheet of notesare allowed. f) To simplify the math, take the acceleration due to gravity as g = 10 m/s’. g) There are NENE problems on the exam. h) Possibly useful equations are shown on the next page. 1 5 2 10 , _I5 sfafs la 7 8 9 (0 total (qo) FROM : FAM NO. : F =dp/dt=ma p=mv F.=mv/r X=EX,tVott+ 2a’ F, = pyN W=Fx U=mgh K=l2mv l=mr 1=1/2mr T= Loon + mh? t=la T=IXF v=dLidt L=Io v=or a=ar K=1210° v, Ay = V2 Ay p+l/2pv'+pgy =constant wo = (k/m)"? T=2x (Lig)'” AE=Q-W AE =nCy AT heat of fusion of H,O = 333 kJ/kg specific heat of ice = 2220 J/kg K p (H,0) = 1000 kg/m? P=kA(T,-T)/L PeoeA(T/-T o=5.7 x 10° Wim?/K* pV=nRT Cy =3/2R Cys=Cyt+R May. 23 26683 GB:86AM PS R=83]/mol y=Cp/Cy AS = AQ/T pV’=constant - adiabatic W=nRTIn(V;/V,)_ - isothermal £=W/Q AS = nR In (V;/ V)) + 0Cy In (T/T FROM FAX NO. + May. 23 2083 B8:a7AM PS 2) Asmall block with mass m, = 2 kg rest on the left edge of a second block of mass m, = 8 kg and Jength L = 3 m. A coefficient of kinetic friction a = 0.3 exists between the two blocks. The surface on which the large block sits is frictionless. A constant horizonta! force F = 10 N is applied to block m,, setting it and the larger block in motion to the right. (10 points) (a) How long does it take for block m, to reach the right edge of block m,? (b) How far does block m, move in the process? aa ‘ —_- (a) F B. 1) Le each block. . N= F, table, If I f. |__ gp Fepplied Ma, |_—_—_1> fi, Woe M2) Ture, , LWs "9 Newt's 22 \aus: Aewbin's 3 lacs Fret =O -P tg= Fir ° Fret, y= Pag = N ” Feet x= Aa = $4 “Thetix= 1,8, = Fapplied - fi, 2 F ~ BY applied Ve "Frage lied “A M9 So the — accelerahen ot the blocks is FR, _ MS G3) (2g YC to a] 54) a,= ae ~7G Q, = rid (a) = oh -(3 Siem |) = (OFS m/s? = 2 ae 4 FROM : FAX NO. 2 May. 23 2683 @8:@6AM P6 The position ok each bleck as a fiw fie ot pre K,t4\= 24, ‘Vs Y= bat whee %a ie He positien ef the TeFt side of black 2 Ubine the relat posihen petween ‘C4 ave et n ; ; a "le : eqvets L, the RP mass has reached Wig side of fle bottom mass - x, (8 )- &, CH) = b= Z (4,-42 ) al oe ae | 2 (2m) - t _ = = . See Auras mfr .0F5m|5* | a4 (b) Durigy +kis V1 geonds, the Second block Moves 2 ax x (48)2 Lar tt = £ ( 07smle)(1 FFs) FROM : FAX NO. f May. 23 28035 BB:B8AM Pr 3) Two masses are placed on a frictionless track. Mass m, = 1 ke is given an initial velocity v = 3 m/s to the right, It collides elastically with mass m, = 2 kg with was initially at rest. Mass m, runs into a massless spring that has a force constant k = 50 N/m and is attached to a wall at the far right. (15 points) (a) What is the velocity of each mass just after the collision? (b) What is the maximum compression of the spring? (c) How long is mass m, in contact with the spring? = vos fe] bre (=) before fle cllisten Ny Po = mv wie iv E> mvt after the ellisiog “h] fat Beth mowniem and every y ave elastiee tollision, Conceryetion of uv 44 if WI,V,- wv, 2 Zz Ep= tha Ve +S NHN, conserved ducing een Pp Dives MV = MV - mV) —- Va = ™, (vev,) Ma (onservatio. 4 E gives ttl od zd 2 2mVO= SW. + BMY, 7 au =i m 4 -3m. (2 (vewy) + 3MNy With Some adgebin this beco ws 2 2 2 o-[2 Roum te avdy ef A pe! TS ky Vet (15 ky) Vy, + (205 Wy) FROM : : FAX NO. + May. 23 2083 G8:29AM Pla 4) A mass m= 0.5 kg hangs from a stiff, massless cable of length L = 1m. As it passes through the lowest part of its path it has a speed v=3 m/s. (15 points) (a) What is the tension in the cable at this lowest part of the masses motion? (b) When the mass reaches its highest point, what angle does the cable make with the vertical? (c) What is the tension in the cable when the mass reaches its highest point? — +t (2) FRD at lowest pet | Newton's BE lac in vertical divectio.: GW = img Femas T- ma -> T= m (avg) ZB Bece $ ‘ 3 t eee fhe mass ut mevity a circle, we must hare he verticad accelertio~ Bou ok te contend aAdcelertivn: av= Q.= Wr So 1 Te mC Bagd= (5) (EN. soahe) (b) low geint high gorst 10 FROM : FAX NO. = May. 23 2683 @8216AM P11 Ok +e |e pent E= T= $mv? Ok Pe Mnigh yond E= Us mgh ~ mg (L- Leos) Cons . of energy gives tmv = mg 2) (Ci~co sO) No ws =| -~ a0 OF BS = 108 4* -[50,6° {c) FRD ot Pa Wghest port Newba's law in the radial Rivecken : +r . Mag = T- Wees& LD Becese fle mass is moving ho ’ 4 Circle ths radical actcelemdder ’ Wem abuse s Coutved acceleretion Ac= acs vir Oe the Wigs got Ge. =O so Aro This gives ‘T= \N coc 8 = mg cos Sbib = (S¥a)CeWlsC88) = 295 pi] 11 FROM : FAM NO. : May. 23 2065 B8:1B8AM Pi2 5) A horizontal pipe has water flowing in it to the right at a rate of 4 x 107 m/s, Ina wide-diameter part of the pipe (with radius r, = 2.5 cm} a vertical tube supports a column of water of height h, = 10 cm above the pipe. In a narrow-diameter part of the pipe (with radius 1,} a vertical tube supports a column of water of height h, = 5 cm above the pipe. (5 points) (a) What is the radius 1,? AJ “TUVTY ay alge, Ca\\ Fi the pressor. ak the ele + the thie le ere , where the weaker is Md ving ok sped Vy, ot height C, ( mea sond relawe tw the pre axis ) Pte eg, th Cafe Pr th presse ok he ve ecly2 of 4), thin Fire , whe phe woke 14 moving ok spad Vy a height or Berwaoli's equation Gives Pra par, +E = Pit pgtat Leva oO The conbnvity equokion VS QavA= va ke | MAMEPEELE © iv FTV = ITN te 12 FROM : FAX NO. = May. 23 2683 @8211AM P1iS (e) P= 0. + piga = [0° Pa 4 (i000 kg fw? Yrs m}s*) (23 FS) = 234.59 Po 15 FROM : i FAX NO. : May. 23 2403 @8:12AM Pi6 7) A ladder of length L = 15m and mass m, = 50 kg rests against a wall at an angle of 45 degrees with respect to the horizontal, A man of mass m, = 80 kg stands on the ladder at a distance d, = 4 m from the bottom of the ladder. {10 points) (a) What are the horizontal and verlical forces on the bottom of the ladder? (b) If the ladder is on the verge of slipping when the person is at a distance d, = 9m from the bottom, what is the coefficient of friction between the ladder and the ground? i \¥ (6) Assome the wath Lrchon less. Then the Fp. fer th laden is Fw force on ladder from tooth ™ Wie wert of addr = mg N F4 = fore on ladda flow Mm, = Mag Ww N= nocmak Bree Rom ge vind f = SOTO (oe<€ Fa F fe ¥ — Ys «Nl Newtrs SA law iw th y Ri vecion: ety 02 N= We-Fis (m,+m. ) {B00 # ‘p N= Wet Fite Newton's aa lak th th BR diveten: Trey see Fu- £ —D Pa F cw ce Actewmine Fy, congich- th Ae rg wes bog + + bottom of th ladder . 16 FROM : FAx NOL: May. 23 2683 @8:12AM P17 ye baeo- L (Fi sed) - 5 (Wie 46) -4d, (Fie cos 45) be OF Le Fiw tm 46 - So m.q-dimig Fas = CE mit dima) 3 / Ltan 48 = (‘82 ZS0kS 4 Ym # BOK, ) 10 mmfs iSm * 7 = 463 N SPs Fas = Hew) tb) Using d= Am instead na d, =4m in the above. eq vekion . em Flaw f= (3 “80 kg + Loe Im * BO) 10 m| 5” iSm ¥l = 730 N The vertical fore on th bottom is shill N= [300 N Using phe equetion Ge moximem stetc Friction = ree Fenax = Ms No -? Ms = “N= \- 06 17