DEPARTMENT OF PHYSICS AND ASTRONOMY Dt Provided: A formul sheet nd tble of physicl constnts is ttched to this pper. Liner grph pper is vilble. Autumn Semester 2015-2016 PHYSICS : 3 HOURS Answer questions ONE nd FIVE plus TWO others from section A, nd TWO others from section B, SIX questions in ll. Answers to different sections must be written in seprte books, the books tied together nd hnded in s one. All questions re mrked out of twenty. The brekdown on the right-hnd side of the pper is ment s guide to the mrks tht cn be obtined from ech prt. Plese clerly indicte the question numbers on which you would like to be exmined on the front cover of your nswer book. Cross through ny work tht you do not wish to be exmined. 1 TURN OVER
SECTION A 1. COMPULSORY () Define in words Newton s three lws of motion. [5] (b) Explin the difference between elstic nd inelstic collisions. [2] (c) Stte the principle of conservtion of momentum. [3] (d) A prticle moves in circulr pth, rdius r, with constnt ngulr velocity. Write down expressions, in terms of r nd, for: (i) The liner velocity, v, of the prticle. [1] (ii) The centripetl ccelertion,, of the prticle. [1] (e) A cr, trvelling long stright rod, increses its velocity from 16 ms 1 to 36 ms 1 in 6.0 s. Clculte: (i) The displcement of the cr during the 6 second period. [2] (ii) The ccelertion of the cr during this period. [2] (f) A crne lifts pllet of bricks, mss 250 kg, through verticl height of 30 m in 40 s. (i) Clculte the work done by the crne. [1] (ii) One of the bricks ccidently drops off from the top of the pllet shortly fter the pllet lift hs been completed by the crne. Clculte the velocity of the brick just before it hits the ground. You my presume tht the pllet hs height of 1 m nd tht ir resistnce is negligible. [3] 2 CONTINUED
2. () A cnnonbll of mss 5 kg is fired from cnnon tht hs brrel length of 3 m. The cnnon is locted on the top of tll tower nd the cnnon bll leves the cnnon horizontlly with n initil velocity of 250 ms -1. The projectile remins irborne for 5 seconds. You my ssume tht there is no ir resistnce nd tht the ground below the tower is level. Clculte: (i) The rnge of the projectile. [1] (ii) The height of the tower. (Assume tht the cnnon is level with the top of the tower.) [3] (iii) The mgnitude nd direction of the velocity of the projectile when it contcts the ground. [4] (iv) The verge force pplied to the cnnonbll during firing. [3] (v) The kinetic energy of the cnnonbll just before it contcts the ground. [1] (b) The vectors s 1 nd s 2 represent two displcements, in m: Clculte: s 1 = 4x + 4y nd s 2 = 5x + 2y (i) The mgnitudes of both vectors. [2] (ii) the ngle between s 1 nd s 2. [2] (iii) the length of the vector (s 1 + s 2 ). [2] (c) Define mechnicl power nd stte its SI unit. [2] 3 TURN OVER
3. () Define ccelertion. [2] (b) A firground ride releses from rest crt with pssengers t height H bove the ground. The crt runs downhill on frictionless trck to pick up speed. As it reches ground level it then enters circulr loop of rdius 9 m. The pssengers feel momentrily weightless t the top of the loop. Drg nd friction cn be neglected. Clculte: (i) The speed of the crt t the bottom of the loop. [7] (ii) The mximum ccelertion tht the crt will experience. [4] (iii) The height H t which the crt ws relesed. [2] (c) Two gliders re plced on frictionless, horizontl ir trck. Glider A hs mss of 0.50 kg nd initilly is sttionry ner the middle of the trck. Glider B hs mss of 0.30 kg nd speed 0.8 ms -1 when it collides with A. The gliders stick together fter the collision. Clculte: (i) The speed of the gliders fter the collision. [2] (ii) The chnge in kinetic energy s result of the collision. [3] 4 CONTINUED
4. () A stellite of mss 900 kg is in geo-synchronous orbit round the erth. (i) Define wht is ment by the term geo-synchronous orbit. [1] (ii) By considering the forces cting upon the stellite nd its period of rottion round the Erth show tht the distnce of the orbit from the centre of the erth is pproximtely 42,000 km. [6] (iii) Clculte both the velocity nd the ngulr velocity of the stellite. [3] (b) The Aswn Dm in Egypt hs height of 111 m nd contins twelve 175 MW genertors for totl power genertion of 2.1 GW. (i) Assuming tht the wter flls the full height of the dm, clculte the rte t which wter must flow through one of the genertors to generte 175 MW of power if the efficiency of the conversion is 80%. [3] (ii) Assuming tht on verge seven of the twelve genertors re running t full cpcity ll yer round clculte the totl volume of wter tht psses through the dm s genertors in one yer. Express your nswer in km 3. [3] (c) A bll is thrown verticlly upwrds (from ground level) nd flls bck to the ground. Air resistnce is negligible. Represent the motion of the bll by sketching grphs of: (i) Accelertion ginst time. [2] (ii) Velocity ginst time. [2] 5 TURN OVER
5. COMPULSORY SECTION B ()(i) Define electric field strength nd stte whether it is vector or sclr quntity. [2] (b) (ii) (i) Stte the formul for electric field strength t distnce r from point chrge Q nd give its SI unit. [2] A hollow metllic sphere of rdius 5 cm crries positive chrge tht is distributed uniformly over its surfce. Sketch grph showing the vrition of voltge with distnce from the centre of the sphere. Ensure you include both positions inside nd outside the sphere. [3] The positive chrge on the sphere is mesured nd found to be 5 nc. (ii) Clculte the voltge t distnce of 1 cm from the surfce of the sphere. [3] (c) A constntn wire is 5.0 m long nd hs circulr cross section. The wire is 0.25 mm in dimeter. Clculte the resistnce of the conductor. (The resistivity of constntn is 4.5 10-7 Ω m.) [3] (d) The formul tht gives the mgnetic flux density in the centre of flt coil is: B = μ 0NI 2r A flt coil is composed of 200 turns nd hs rdius of 5 cm. It is found to hve resistnce of 5 Ω. Clculte the mgnetic flux density t the centre of the coil when voltge of 6 V is pplied. [2] (e) Stte Frdy s lw of electromgnetic induction. [2] (f) Two identicl copper tubes, pproximtely 1 m long, re mounted verticlly side by side. Two cylinders (A nd B), hving the sme mss nd size, re dropped into the tubes t the top. Cylinder A is mde from non-mgnetic lloy nd B is strong neodymium mgnet. A is observed to fll through its tube much more quickly thn B. Explin why this is so. [3] 6 CONTINUED
6. () Three cpcitors re chosen with vlues of 100 µf, 220 µf, nd 470 µf. These cpcitors re wired together in series nd then connected to 9V DC power supply. Clculte: (i) The combined series cpcitnce of the three cpcitors. [2] (ii) The voltge drop (potentil difference) cross ech cpcitor. [6] The sme cpcitors re now disconnected, dischrged nd then wired together in prllel. This prllel circuit is connected bck up to the 9V DC power supply. Clculte: (iii) The combined prllel cpcitnce of the three cpcitors. [2] (iv) The totl energy stored in the circuit when the cpcitors re fully chrged. [4] (b) A step down trnsformer hs n RMS input voltge of 240 V AC t 60 Hz nd produces n RMS output voltge of 150 V AC. The primry winding hs 600 turns. Clculte: (i) The number of turns on the secondry winding. [2] (ii) The frequency of the output wveform. [1] (iii) The period of the input wveform. [1] (iii) The pek output voltge. [2] 7 TURN OVER
7. () As prt of lbortory demonstrtion you hve been given fixed 6 V DC power supply with negligible internl resistnce. (i) Clculte the minimum resistnce of resistor tht cn be connected cross the power supply if the totl power dissipted in the resistor is not to exceed 0.25 W. [2] You re now given 20 Ω resistor nd second resistor of unknown vlue. You build potentil divider using these two resistors. After connecting the potentil divider to the power supply you mesure 4 V to ground cross the second resistor. (ii) Wht is the resistnce of the second resistor? [3] The circuit is disconnected from the fixed 6 V DC power supply nd connected to 6 V bttery. The voltge cross the second resistor drops to 3.95 V. (iii) Clculte the internl resistnce of the bttery. [4] (b) Two strips of luminium of width 5 cm nd length 100 cm re used to construct prllel plte cpcitor. The gp between the pltes is filled with n insulting dielectric mteril with reltive permittivity of 4. The cpcitnce of this cpcitor is found to be 50 nf. (i) Write down the formul tht gives the cpcitnce of prllel plte cpcitor, describing the mening of ech of the symbols used. [4] (ii) Clculte the thickness of the dielectric mteril. [3] The cpcitor is dischrged nd then the dielectric mteril is removed from the cpcitor. The cpcitor is then connected to 12 V DC power supply nd chrged. (iii) Sketch the Electric Field between the pltes of the cpcitor. [2] (iv) Clculte the Electric Field strength in the gp between the pltes. [2] 8 CONTINUED
8. () (b) Clculte the force cting on conductor 12 cm long crrying current of 500 ma when plced perpendiculr to uniform mgnetic field of flux density 0.75 T. [2] A bem of electrons moving t 2.0 10 6 ms 1 moves through region of uniform mgnetic field of flux density 0.1 T t n ngle of 40 to the mgnetic field lines. (i) Wht is the shpe of the pth followed by the electrons nd explin why the electrons follow this pth. [2] (i) Clculte the rdius of the motion of ech electron. [4] (iii) Describe how the pth would chnge if the electrons were replced by bem of protons moving t the sme speed s the electrons. [2] (c) A cell of emf E nd internl resistnce r is set up in the circuit s shown. E r A I V V Vlues of potentil difference (V) nd corresponding vlues of current (I) re mesured nd tbulted. V / V 8.5 7 6 5.5 4 3.0 I / A 2.5 10 15 17.5 25 30 Use the dt nd grphicl method to clculte the emf (E) nd internl resistnce (r) of the cell. [10] END OF EXAMINATION PAPER 9 TURN OVER
PHYSICAL CONSTANTS & MATHEMATICAL FORMULAE Physicl Constnts electron chrge e = 1.60 10 19 C electron mss m e = 9.11 10 31 kg = 0.511 MeV c 2 proton mss m p = 1.673 10 27 kg = 938.3 MeV c 2 neutron mss m n = 1.675 10 27 kg = 939.6 MeV c 2 Plnck s constnt h = 6.63 10 34 J s Dirc s constnt ( = h/2π) = 1.05 10 34 J s Boltzmnn s constnt k B = 1.38 10 23 J K 1 = 8.62 10 5 ev K 1 speed of light in free spce c = 299 792 458 m s 1 3.00 10 8 m s 1 permittivity of free spce ε 0 = 8.85 10 12 F m 1 permebility of free spce µ 0 = 4π 10 7 H m 1 Avogdro s constnt N A = 6.02 10 23 mol 1 gs constnt R = 8.314 J mol 1 K 1 idel gs volume (STP) V 0 = 22.4 l mol 1 grvittionl constnt G = 6.67 10 11 N m 2 kg 2 Rydberg constnt R = 1.10 10 7 m 1 Rydberg energy of hydrogen R H = 13.6 ev Bohr rdius 0 = 0.529 10 10 m Bohr mgneton µ B = 9.27 10 24 J T 1 fine structure constnt α 1/137 Wien displcement lw constnt b = 2.898 10 3 m K Stefn s constnt σ = 5.67 10 8 W m 2 K 4 rdition density constnt = 7.55 10 16 J m 3 K 4 mss of the Sun M = 1.99 10 30 kg rdius of the Sun R = 6.96 10 8 m luminosity of the Sun L = 3.85 10 26 W mss of the Erth M = 6.0 10 24 kg rdius of the Erth R = 6.4 10 6 m Conversion Fctors 1 u (tomic mss unit) = 1.66 10 27 kg = 931.5 MeV c 2 1 Å (ngstrom) = 10 10 m 1 stronomicl unit = 1.50 10 11 m 1 g (grvity) = 9.81 m s 2 1 ev = 1.60 10 19 J 1 prsec = 3.08 10 16 m 1 tmosphere = 1.01 10 5 P 1 yer = 3.16 10 7 s
Polr Coordintes x = r cos θ y = r sin θ da = r dr dθ 2 = 1 ( r ) + 1r 2 r r r 2 θ 2 Sphericl Coordintes Clculus x = r sin θ cos φ y = r sin θ sin φ z = r cos θ dv = r 2 sin θ dr dθ dφ 2 = 1 ( r 2 ) + 1 r 2 r r r 2 sin θ ( sin θ ) + θ θ 1 r 2 sin 2 θ 2 φ 2 f(x) f (x) f(x) f (x) x n nx n 1 tn x sec 2 x e x e x sin ( ) 1 x ln x = log e x 1 x cos 1 ( x sin x cos x tn ( 1 x cos x sin x sinh ( ) 1 x cosh x sinh x cosh ( ) 1 x sinh x cosh x tnh ( ) 1 x ) ) 1 2 x 2 1 2 x 2 2 +x 2 1 x 2 + 2 1 x 2 2 2 x 2 cosec x cosec x cot x uv u v + uv sec x sec x tn x u/v u v uv v 2 Definite Integrls 0 + + x n e x dx = n! (n 0 nd > 0) n+1 π e x2 dx = π x 2 e x2 dx = 1 2 Integrtion by Prts: 3 b u(x) dv(x) dx dx = u(x)v(x) b b du(x) v(x) dx dx
Series Expnsions (x ) Tylor series: f(x) = f() + f () + 1! n Binomil expnsion: (x + y) n = (1 + x) n = 1 + nx + k=0 ( ) n x n k y k k n(n 1) x 2 + ( x < 1) 2! (x )2 f () + 2! nd (x )3 f () + 3! ( ) n n! = k (n k)!k! e x = 1+x+ x2 2! + x3 x3 +, sin x = x 3! 3! + x5 x2 nd cos x = 1 5! 2! + x4 4! ln(1 + x) = log e (1 + x) = x x2 2 + x3 3 n Geometric series: r k = 1 rn+1 1 r k=0 ( x < 1) Stirling s formul: log e N! = N log e N N or ln N! = N ln N N Trigonometry sin( ± b) = sin cos b ± cos sin b cos( ± b) = cos cos b sin sin b tn ± tn b tn( ± b) = 1 tn tn b sin 2 = 2 sin cos cos 2 = cos 2 sin 2 = 2 cos 2 1 = 1 2 sin 2 sin + sin b = 2 sin 1( + b) cos 1 ( b) 2 2 sin sin b = 2 cos 1( + b) sin 1 ( b) 2 2 cos + cos b = 2 cos 1( + b) cos 1 ( b) 2 2 cos cos b = 2 sin 1( + b) sin 1 ( b) 2 2 e iθ = cos θ + i sin θ cos θ = 1 ( e iθ + e iθ) 2 nd sin θ = 1 ( e iθ e iθ) 2i cosh θ = 1 ( e θ + e θ) 2 nd sinh θ = 1 ( e θ e θ) 2 Sphericl geometry: sin sin A = sin b sin B = sin c sin C nd cos = cos b cos c+sin b sin c cos A
Vector Clculus A B = A x B x + A y B y + A z B z = A j B j A B = (A y B z A z B y ) î + (A zb x A x B z ) ĵ + (A xb y A y B x ) ˆk = ɛ ijk A j B k A (B C) = (A C)B (A B)C A (B C) = B (C A) = C (A B) grd φ = φ = j φ = φ x î + φ y ĵ + φ z ˆk div A = A = j A j = A x x + A y y + A z z ) curl A = A = ɛ ijk j A k = ( Az y A y z φ = 2 φ = 2 φ x + 2 φ 2 y + 2 φ 2 z 2 ( φ) = 0 nd ( A) = 0 ( A) = ( A) 2 A ( Ax î + z A ) ( z Ay ĵ + x x A ) x y ˆk