JURONG JUNIOR COLLEGE 2010 JC1 H1 8866 hysics utoril : Dynmics Lerning Outcomes Sub-topic utoril Questions Newton's lws of motion 1 1 st Lw, b, e f 2 nd Lw, including drwing FBDs nd solving problems by reclling & pplying F = m 2, c, d, f, g, 5c, 6, 7, 13(iii) 3 rd Lw 5, 5b, 8f b Mss 8, 8b c Concept of weight d Liner momentum nd Impulse 3, 8c, 8d, 9, 13(i) e Force s rte of chnge of momentum 8e, 9, 13(ii) g Stting rinciple of Conservtion Of Momentum 11, 13(iv) h Applying C.O.M to solve simple problems 10, 11b, 12 including elstic nd inelstic interctions i Reltive speed of pproch & seprtion for 13c elstic collisions j Momentum nd KE for collisions Not covered. 1 Stte Newton s lws of motion. 2 () A resultnt force of 1 N cts on sttionry mss of 1 kg for durtion of 1 s. How fr will the mss move? [0.5 m] A force of 30 N hlves body s velocity in 900 m. If the mss of the body is 5 kg, clculte the originl velocity nd the durtion for which the force cts on it. [120 m s -1, 10 s] 3 he vrition of the force F with time t cting on body of mss 200 g is s shown in the grph below. F / N 80 60 0 20 0 50 100 150 200 t / m s Given tht the velocity of the body t t = 75 ms is 15 m s -1, wht is the velocity of the body t t = 150 ms? A 15 m s -1 B 30 m s -1 C 35 m s -1 D 39 m s -1 [D] A lift hs mss of 1.2 10 3 kg. king g = 10 m s -2, clculte the tension in the supporting cble when the lift is ge 1 of 9
() t rest. [1.20 10 N] descending t uniform velocity. [1.20 10 N] (c) ccelerting downwrds t. [0.96 10 N] (d) decelerting downwrds t 1 m s -2. [1.32 10 N] (e) scending t uniform velocity. [1.20 10 N] (f) ccelerting upwrds t. [1. 10 N] (g) decelerting upwrds t 1 m s -2. [1.08 10 N] 5 Newton's third lw of motion cn be stted in the form when object X exerts force on object Y, then object Y exerts force of the sme type tht is equl in mgnitude nd opposite in direction on object X. () Explin wht is ment by of the sme type in this sttement of the lw. (c) Object X is book resting in equilibrium on tble (object Y). Drw lbelled force digrms to show the forces on X nd on Y. Mke it cler which forces re equl in mgnitude nd opposite in direction. Now suppose object X is book tht hs fllen onto the tble (object Y) nd hs just lnded on it. For the instnt of rrivl, drw lbelled force digrms to show the forces on X nd on Y. Mke it cler which forces re equl in mgnitude nd opposite in direction, nd which forces re different in mgnitude from those in. 6 he totl mss of lift with mn of mss 80 kg inside is 1000 kg. he lift moving upwrds t speed of 8 m s -1 is brought to rest over distnce of 20 m. Clculte () the tension in the cbles supporting the lift during the decelertion. [8210 N] the force exerted on the mn by the lift floor. [657 N] 7 wo blocks A nd B re connected using n inextensible string on horizontl frictionless floor. Another inextensible string is tied to block A nd pulled by force so tht both the blocks hve n ccelertion of 2.0 m s -2. Given tht the msses of blocks A nd B re 50 kg nd 10 kg respectively, wht re the vlues of the force nd the tension in the string connecting the two blocks? [120 N; 20 N] 8 A helicopter of mss 1000 kg hovers by imprting downwrd velocity v to the ir displced by its rotors. he re swept out by the rotor bldes is 80 m 2. he density of ir is 1.3 kg m -3. Considering the ir displced downwrds per second, () wht is the volume of ir displced? [80v m 3 ] wht is the mss of ir displced? [10v kg] (c) wht is momentum of the ir displced? [10v 2 kg m s -1 ] (d) wht is the chnge in momentum of the ir displced? [10 v 2 kg m s -1 ] If force is directly proportionl to the rte of chnge of momentum, (e) wht is the force exerted on the ir by the rotor bldes? [10 v 2 N] (f) wht is the direction nd the force exerted on the rotor bldes by the ir displced? [upwrds, 10 v 2 N] (g) wht is the vlue of v? [9.71 m s -1 ] 9 A stedy 50 m s -1 wind hits rigid wll which is in plne perpendiculr to the wind direction nd is stopped completely by the wll. Estimte the pressure exerted by the wind on the wll. he density of ir is 1.25 kg m -3. [3.13 10 3 ] ge 2 of 9
10 A sphere of mss m trvelling in stright line with speed u collides hed-on with sttionry sphere, lso of mss m. he collision is perfectly elstic. he finl speeds re v 1 nd v 2 respectively s shown below. Before collision u m m After collision v 1 rite down expressions in terms of the quntities shown bove to illustrte: () the principle of conservtion of momentum. the principle of conservtion of kinetic energy. Use these expressions to find v 2 in terms of u. ht hppens to the incoming sphere fter the collision? 11 () Stte the rinciple of Conservtion of Momentum. A 2 kg rifle fires bullet of mss 20 g with velocity of 350 m s -1. Clculte the recoil velocity of the rifle. If the mss of the mn holding the rifle is 70 kg, wht will be the combined recoil velocity of the mn nd the rifle? [3.5 m s -1 ; 0.0972 m s -1 ] 12 () Distinguish between n elstic collision, n inelstic collision nd completely inelstic collision. A prticle A of mss m nd initil velocity u mkes hed-on collision with nother sttionry prticle B of mss 2m. In terms of u, clculte the finl velocity of prticle A if the collision is (i) elstic (ii) completely inelstic. [-u/3, u/3] 13 () (i) Define liner momentum. (ii) Use your definition of momentum to define force. (iii) Show tht this definition leds to the eqution F = m x. In gs hydrogen molecule, mss 2.00 u nd velocity 1.88 x 10 3 m s -1, collides elsticlly nd hed-on with n oxygen molecule, mss 32.0 u nd velocity 05 m s -1, s illustrted in the figure below. v 2 Hydrogen molecule Oxygen molecule 1.88 x 10 3 m s -1 05 m s -1 In qulittive terms, wht cn be stted bout the subsequent motion s result of knowing tht (i) collision is elstic, (ii) the collision is hed on? (c) Using your nswers to, (i) determine the velocity of seprtion of the two molecules fter the collision, (ii) pply the lw of conservtion of momentum to the collision, (iii) determine the velocity of both molecules fter the collision. End ge 3 of 9
2010 JC1 H1 8866 hysics utoril : Dynmics (Solutions) 1 Newton s First Lw of Motion sttes tht every object continues in its stte of rest or uniform motion in stright line unless resultnt force cts on it. Newton s Second Lw of Motion sttes tht for system the rte of chnge of momentum is directly proportionl to the resultnt force cting on it nd the chnge tkes plce in the direction of tht resultnt force. Newton s hird Lw of Motion sttes tht when object A exerts force on object B, then object B exerts force of the sme type tht is equl in mgnitude nd opposite in direction on object B. 2 () F = m 1 = (1) = 1 m s -2 F = m 30 = (5) = 6 m s -2 s u = 0 m s -1 v t = 1 s +ve?? 1 m s -2 [s = ut + ½t 2 ] s = (0)(1) + ½(1)(1) 2 = 0.5 m s u v t =? +ve 900 m? ½u 6 m s -2 [v 2 = u 2 + 2s] [v = u + t] ¼u 2 = u 2 + 2(-6)(900) u = 120 m s -1 60 = 120 + (-6)(t) t = 10 s 3 impulse = chnge in momentum = re under the grph 25 50 = (80 + 60) + (80 + 0) 2 2 = (1750 + 3000) = 750 N ms =.75 N s Initil momentum of body = (200 x 10-3 )(15) = 3 N s Finl momentum of body = 3 +.75 = 7.75 N s 7.75-1 Finl velocity of body = = 38.75 m s 3 200 10 ge of 9
() (e) = = 1.2 10 N = 0 m s -2 1.2 10 N =(c) 3 = (1.2 10 )(2) motion 3 1.2 10 = (1.2 10 )(2) 0.96 10 N = 1.2 10 N (d) 3 = (1.2 10 )(1) motion 3 1.2 10 = (1.2 10 )(1) = 1.32 10 N 1 m s -2 1.2 10 N (f) 3 = (1.2 10 )(2) motion 3 1.2 10 = (1.2 10 )(2) = 1. 10 N 1.2 10 N (g) 3 = (1.2 10 )(1) motion 3 1.2 10 = (1.2 10 )(1) = 1.08 10 N 1 m s -2 1.2 10 N ge 5 of 9
5 () he sme type of force could be explined using n exmple. If the rod exerts forwrd frictionl force on n ccelerting cr's tires, then it is lso frictionl force tht Newton's third lw predicts for the tires pushing bckwrd on the rod. Force on book by tble book eight of book tble Force on tble by book eight of tble Force on book by tble is equl in mgnitude nd opposite in direction s the force on tble by book. Force on book by tble is equl in mgnitude nd opposite in direction s weight of book. (c) Force on book by tble book eight of book tble Force on tble by book eight of tble Force on book by tble on impct is equl in mgnitude nd opposite in direction s the force on tble by book on impct. Force on book by tble is different in mgnitude (greter) from the weight of the book compred to (). ge 6 of 9
6 s u 20 m +ve 8 m s -1 R 1.6 m s -2 v = 0 m s -1? t =? [v 2 = u 2 + 2s] 0 2 = 8 2 + 2()(20) = -1.6 m s -2 L (920)(9.81) N R 1.6 m s -2 M (80)(9.81) N (920)(9.81) + R = (920)(1.6) --(1) (80)(9.81) R = (80)(1.6) --(2) Solving simultneously, = 8210 N R = 657 N () 8210 N 657 N OR: () = (1000)(1.6) = (1000)(1.6) = 9810 1600 = 8210 N m R = (80)(1.6) R = m (80)(1.6) = (80)(9.81) (80)(1.6) = 657 N 7 B A 10 kg 50 kg = (10)(2) = 20 N B 10 kg A 50 kg = (50)(2) 20 = 100 = 120 N 8 Consider the ir displced downwrds per second. Volume = (80)(v) m 3 Mss = (80)( v)(1.3) kg = 10v kg Momentum = (80)( v)(1.3)( v) kg m s -1 = 10v 2 kg m s -1 Force = rte of chnge in momentum = (80)( v)(1.3)( v) = (1000)(9.81) v = 9.71 m s -1 9 Consider the wind hitting the wll per second. Let the re of the wll be A. Volume = (50)(A) m 3 Mss = (50)(A)(1.25) kg Momentum = (50)(A)(1.25)(50) kg m s -1 Force = rte of chnge in momentum = (50)(A)(1.25)(50) N ressure = force per unit re = (50)(1.25)(50) = 3.13 10 3 ge 7 of 9
10 () Initil momentum = Finl momentum mu = mv 1 + mv 2.. (1) Initil k.e. = Finl k.e. mu 2 /2 = mv 1 2 /2 + mv 2 2 /2.. (2) Subst (1) into (2) u 2 = (u - v 2 ) 2 + v 2 2 2v 2 (u - v 2 ) = 0 v 2 = u or 0 m s -1 Since the 2nd sphere ws initilly t rest nd it will move fter the collision, v 2 = u. From (1), v 1 = u v v 1 = 0 m s -1 hus, the incoming sphere will come to rest fter the collision. 11 () he totl momentum of closed system is constnt. Consider rifle nd bullet only. rifle (m r = 2 kg) bullet (m b = 20 g) Before firing: u r = 0 m s -1 u b = 0 m s -1 After firing: v r v b otl finl momentum = otl initil momentum m r v r + m b v b = m r u r + m b u b (2)(v r ) + (0.020)(350) = 0 v r = -3.5 m s -1 herefore the recoil velocity of the rifle is 3.5 m s -1. 350 m s -1 Consider mn nd rifle only. mn (m m = 70 kg) rifle (m r = 2 kg) otl finl momentum = otl initil momentum (m m + m r ) v m+r = m m u m + m r u r (70 + 2)(v m+r ) = 0 + (2)(-3.5) v m+r = -0.0972 m s -1 herefore the combined recoil velocity of the mn nd the rifle is 0.0972 m s -1. u m = 0 m s -1 u r v m+r 3.5 m s -1 ge 8 of 9
12 () (i) he totl kinetic energy of the intercting objects is conserved in elstic collision but not in inelstic collision nd completely inelstic collision. In completely inelstic collision the objects move off s single unit. otl finl momentum = otl initil momentum m A v A + m B v B = m A u A + m B u B mv A + 2mv B = mu + 0 v A + 2v B = u --- (1) otl finl kinetic energy = otl initil kinetic energy ½m A v A 2 + ½m B v B 2 = ½m A u A 2 + ½m B u B 2 m A v A 2 + m B v B 2 = m A u A 2 + m B u B 2 mv A 2 + 2mv B 2 = mu 2 + 0 v A 2 + 2v B 2 = u 2 --- (2) (ii) Solving simultneously for v A, v A = -u/3 otl finl momentum = otl initil momentum m A+B v A+B = m A u A + m B u B 3mv A+B = mu + 0 3v A+B = u v A+B = u/3 13 () (i) Liner momentum of n object is defined s the product of the object s mss nd its liner velocity. () (ii) Force is defined s the rte of chnge of momentum nd the chnge occurs in the direction of the force. () (iii) dp d( mv) dv dm F = = = m + v dt dt dt dt dm dv For constnt mss ( is zero), F = m dt dt dv Since ccelertion =, F = m dt () (iv) he totl momentum of system of intercting bodies is constnt provided no externl resultnt force cts on it. (i) otl momentum nd totl kinetic energy re conserved. (ii) he motion fter the collision will be colliner with the motion before the collision. (c) (i) Reltive velocity of seprtion fter collision = Reltive velocity of pproch before collision = 1.88 10 3 (-05) = 2285 m s -1 = 2290 m s -1 (c) (ii) otl finl momentum = otl initil momentum m O v O + m H v H = m O u O + m H u H king the positive direction s towrds the right, (32.0u)(v O ) + (2.00u)(v H ) = (32.0u)(-05) + (2.00u)(1880) 32v O + 2v H = -9200 16v O + v H = -600 --- (1) (c) (iii) v O - v H = 2285 --- (2) Solving (1) nd (2) simultneously, v O = -136 m s -1 nd v H = -221 m s -1 = -220 m s -1 - he END - ge 9 of 9