1. VELOCITY AND ACCELERATION

Size: px

Start display at page:

Download "1. VELOCITY AND ACCELERATION"

Maurice Chandler
5 years ago
Views:

1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.

1N. Block se in moion from X wih speed 3ms 1. I his verical surface a Y s laer. Block rebounds from wall direcly owards X and sops a Z. The insan ha block his wall i loses 0.07J of is kineic energy.

Then skech a velociy-ime graph of he moion of he small block. ii. Displacemen of block from X, in he XY direcion is s m a ime s. Skech a displacemen-ime graph.

8ms Calculae v a Y using relevan kinemaics equaion 0.8 = v 3 v = 1.4ms 1 Calculae kineic energy a Y E K = 1 (0.15)(1.4) = 0.147J Calculae energy los: Iniial Change = Final 0.147 0.07 = 0.

1 1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under graph = change in displacemen {S1-P4} Quesion 7: The small block has mass 0.15kg. The surface is horizonal. The fricional force acing on i is 0.1N. Block se in moion from X wih speed 3ms 1. I his verical surface a Y s laer. Block rebounds from wall direcly owards X and sops a Z. The insan ha block his wall i loses 0.07J of is kineic energy. The velociy of he block from X o Y direcion is v ms 1 a ime s afer i leaves X. i. Find values of v when he block arrives a Y and when i leaves Y. Also find when block comes o res a Z. Then skech a velociy-ime graph of he moion of he small block. ii. Displacemen of block from X, in he XY direcion is s m a ime s. Skech a displacemen-ime graph. On graph show values of s and when block a Y and when i comes o res a Z. Calculaing deceleraion using Newon s second law: 0.1 = 0.15a a = 0.1 = ms Calculae v a Y using relevan kinemaics equaion 0.8 = v 3 v = 1.4ms 1 Calculae kineic energy a Y E K = 1 (0.15)(1.4) = 0.147J Calculae energy los: Iniial Change = Final = 0.075J Calculae speed as leaving Y using k. E. formula: = 1 (0.15)v v = 1ms 1 Calculae when paricle comes o res: 0.8 = 0 1 = 1.5s Draw velociy-ime graph wih daa calculaed: Calculae displacemen from X o Y s = (3 ) + 1 ( 0.8)() Calculae displacemen from Y o Z s = 4.4m s = (1 1.5) + 1 ( 0.8)(1.5) s = 0.65m in he opposie direcion Draw displacemen-ime graph wih daa calculaed: 1.4 Average Velociy For an objec moving wih consan acceleraion over a period of ime, hese quaniies are equal: o The average velociy o The mean of iniial & final velociies o Velociy when half he ime has passed Page of 8 Visi: --> for more classified papers and answer keys

1.5 Relaive Velociies Le s A be he disance ravelled by A and s B for B s A = u + 1 a s B = u + 1 a If a collision occurs a poin C s A + s B = D This gives you he ime of when he collision occurred

2 1.5 Relaive Velociies Le s A be he disance ravelled by A and s B for B s A = u + 1 a s B = u + 1 a If a collision occurs a poin C s A + s B = D This gives you he ime of when he collision occurred Same analysis if moion is verical. FORCE AND MOTION Newon s 1 s Law of Moion: Objec remains a res or moves wih consan velociy unless an exernal force is applied Newon s nd Law of Moion: F = ma 3. VERTICAL MOTION Weigh: direcly downwards Normal conac force: perpendicular o place of conac 3.1 Common Resuls of Verical Moion Finding ime aken o reach maximum heigh by a projecile ravelling in verical moion: Le v = 0 and find The ime aken o go up and come back o original posiion would be double of his Finding maximum heigh above a launch poin use: as = v u Le v = 0 and find s Finding ime inerval for which a paricle is above a given heigh: Le he heigh be H and use s = u + 1 a Le s = H There will be a quadraic equaion in Solve and find he difference beween he s o find he ime inerval {S04-P04} Quesion 7: Paricle P 1 projeced verically upwards, from horizonal ground, wih speed 30ms 1. A same insan P projeced verically upwards from ower heigh 5m, wih speed 10ms 1 Page 3 of 8 i. Find he ime for which P 1 is higher han he op of he ower ii. Find velociies of he paricles a insan when hey are same heigh iii. Find he ime for which P 1 is higher han P and moving upwards Subsiue given values ino displacemen equaion: 5 = (30) + 1 (10) = 0 Solve quadraic for = 1s or 5s P 1 reaches ower a = 1 hen passes i again when coming down a = 5s Therefore ime above ower = 5 1 = 4 seconds Displacemen of P 1 is S 1, and of P is S & relaionship: S 1 = 5 + S Creae equaions for S 1 and S S 1 = ( 10) Subsiue back ino iniial equaion S = ( 10) ( 10) = ( 10) Simple cancelling = 1.5s Find velociies V 1 = 30 10(1.5) = 17.5ms 1 V = 10 10(1.5) =.5ms 1 Par (iii) We know when P 1 and P a same heigh = 1.5s. Find ime aken o reach max heigh for P 1 V is 0 a max heigh 0 = = 3s Time for P 1 above P = = 1.75 seconds 4. RESOLVING FORCES If force F makes an angle θ wih a given direcion, he effec of he force in ha direcion is F cos θ F cos(90 θ) = F sin θ F sin(90 θ) = F cos θ Forces in equilibrium: resulan = 0 If drawn, forces will form a closed polygon Visi: --> for more classified papers and answer keys

Mehods of working ou forces in equilibrium: o Consruc a riangle and work ou forces o Resolve forces in x and y direcions; sum of each = 0 Lami s Theorem: For any se of hree forces P,Q and R in

FRICTION Fricion = Coefficien of Fricion Normal Conac Force F = μr Fricion always acs in he opposie direcion of moion Limiing equilibrium: on he poin of moving, fricion a max (limiing fricion) Smooh

3 Mehods of working ou forces in equilibrium: o Consruc a riangle and work ou forces o Resolve forces in x and y direcions; sum of each = 0 Lami s Theorem: For any se of hree forces P,Q and R in equilibrium P sin θ = Q sin β = R sin γ 5. FRICTION Fricion = Coefficien of Fricion Normal Conac Force F = μr Fricion always acs in he opposie direcion of moion Limiing equilibrium: on he poin of moving, fricion a max (limiing fricion) Smooh conac: fricion negligible Conac force: o Refers o boh F and N o Horizonal componen of Conac force = F o Verical componen of Conac force = N o Magniude of Conac force given by he formula: C = F + N {W11-P43} Quesion 6: The ring has a mass of kg. The horizonal rod is rough and he coefficien of fricion beween ring and rod is 0.4. Find he wo values of T for which he ring is in limiing equilibrium The ring is in limiing equilibrium in wo differen scenarios; we have o find T in boh: Scenario 1: ring is abou o move upwards Resulan = T sin 30 fricion Weigh of Ring Since he sysem is in equilibrium, resulan = 0: Conac Force = T cos 30 Fricion = 0.4 T cos 30 Subsiue relevan informaion in o iniial equaion 0 = T sin T cos 30 0 T = 68.5N Scenario : ring is abou o move downwards This ime fricion acs in he opposie direcion since fricion opposes he direcion of moion, hus: Resulan = T sin 30 + Fricion Weigh of Ring Using informaion from before: 0 = T sin T cos 30 0 T = 8.3N 5.1 Equilibrium Force required o keep a paricle in equilibrium on a rough plane Max Value Min Value The paricle is abou o move up Thus fricion force acs down he slope P = F + mg sin θ The paricle is abou slip down Thus fricional force acs up he slope F + P = mg sin θ {W1-P43} Quesion 6: Coefficien of fricion is 0.36 and he paricle is in equilibrium. Find he possible values of P The magniude of fricion on paricle in boh scenarios is he same bu acing in opposie direcions Calculae he magniude of fricion firs: Conac Force = 6 cos 5 Fricion = cos 5 Scenario 1: paricle is abou o move upwards P = 6 sin 5 + Fricion P = 4.49N Scenario : paricle is abou o move downwards P = 6 sin 5 fricion P = 0.578N 6. CONNECTED PARTICLES Newon s 3 rd Law of Moion: For every acion, here is an equal and opposie reacion Visi: --> for more classified papers and answer keys Page 4 of 8

{Exemplar Quesion} A rain pulls wo carriages: Diagram showing how o resolve forces: The forward force of he engine is F = 500N. Find he acceleraion and ension in each coupling.

08ms To find T 1, look a C F T 1 00 = 1000a 500 T 1 00 = 1000 1.08 T 1 = 10N To find T, look a A T 90 = 400a T 90 = 400 1.08 T 1 = 5N 6.

equaion ino firs: ( W sin 60 sin 40 ) cos 40 + W cos 60 = 5 Solve o find W : W = 3.6N Pu his value back ino firs equaion o find W 1 W 1 = 4.40N {S1-P41} Quesion 6: P has a mass of 0.

4 {Exemplar Quesion} A rain pulls wo carriages: Diagram showing how o resolve forces: The forward force of he engine is F = 500N. Find he acceleraion and ension in each coupling. The resisance o moion of A, B and C are 00, 150 and 90N respecively. To find acceleraion, regard he sysem as a single objec. The inernal Ts cancel ou and give: 500 ( ) = 1900a a = 1.08ms To find T 1, look a C F T 1 00 = 1000a 500 T 1 00 = T 1 = 10N To find T, look a A T 90 = 400a T 90 = T 1 = 5N 6.1 Pulleys Equaion 1: No backward force T = a Equaion : 3g T = 3a Resolving forces a A verically: W 1 cos 40 + W cos 60 = 5 Resolving forces a A horizonally: W 1 sin 40 = W sin 60 Subsiue second equaion ino firs: ( W sin 60 sin 40 ) cos 40 + W cos 60 = 5 Solve o find W : W = 3.6N Pu his value back ino firs equaion o find W 1 W 1 = 4.40N {S1-P41} Quesion 6: P has a mass of 0.6kg and Q has a mass of 0.4kg. The pulley and surface of boh sides are smooh. The base {W05-P04} Quesion 3: of riangle is horizonal. I is given ha sin θ = 0.8. Iniially paricles are held a res on slopes wih sring au. Paricles are released and move along he slope i. Find ension in sring. Find acceleraion of paricles while boh are moving. ii. Speed of P when i reaches he ground is ms 1. When P reaches he ground i sops moving. Q coninues moving up slope bu does no reach he pulley. Given his, find he ime when Q reaches is maximum heigh above ground since he insan i was released The srings are in equilibrium. The pegs are smooh. All he weighs are verical. Find W 1 and W Page 5 of 8 Visi: --> for more classified papers and answer keys

Effec of weigh caused by P in direcion of slope: Effec of weigh = mg sin θ where sin θ = 0.8 Effec of weigh = 4.8N Effec of weigh caused by Q in direcion of slope: Effec of weigh = 0.4 10 0.8 = 3.

4 Subsiue back ino iniial equaions o find a: 4.8 3.84 = 0.6a a = 1.6ms Use kinemaics equaions o find he ime which i ake P o reach he ground: a = v u and = 0 1.6 1 = 1.

5 Effec of weigh caused by P in direcion of slope: Effec of weigh = mg sin θ where sin θ = 0.8 Effec of weigh = 4.8N Effec of weigh caused by Q in direcion of slope: Effec of weigh = = 3.N Body P has greaer mass han body Q so when released P moves down Q moves up on heir slopes 4.8 T = 0.6a T 3. = 0.4a Solve simulaneous equaions: 4.8 T = T 3. T = 3.84N Subsiue back ino iniial equaions o find a: = 0.6a a = 1.6ms Use kinemaics equaions o find he ime which i ake P o reach he ground: a = v u and = = 1.5s When P reaches he ground, only force acing on Q is is own weigh in he direcion of slope = 3.N F = ma 3. = 0.4a a = 8ms Now calculae he ime aken for Q o reach max heigh This occurs when is final velociy is 0. 8 = 0 = 0.5s Now do simple addiion o find oal ime: Toal Time = = 1.5s 6. Force Exered by Sring on Pulley Pulley Case 1 Pulley Case Force on pulley = T Acs: downwards Force on pulley = T Acs: along doed line Page 6 of 8 Pulley Case 3 Force on pulley = T cos ( 1 θ) Acs: inwards along doed line which bisecs θ 6.3 Two Paricles {S10-P43} Quesion 7: A and B are recangular boxes of idenical sizes and are a res on rough horizonal plane. A mass = 00kg and B mass = 50kg. If P 3150 boxes remains a res. If P > 3150 boxes move i. Find coefficien of fricion beween B and floor ii. Coefficien of fricion beween boxes is 0.. Given ha P > 3150 and no sliding occurs beween boxes. Show ha he acceleraion of boxes is no greaer han ms iii. Find he maximum possible value of P in he above scenario F = μn F = o max P ha does no move he boxes N = o conac force of boh boxes acing on floor 3150 = μ ( ) μ = 0.7 Find fricional force beween A and B: F = F = 400N Use Newon s Second Law of Moion o find max acceleraion for which boxes do no slide (below F) 400 = 00a a = ms Par (iii) P has o cause an acceleraion of ms on B which will pass on o A as hey are conneced bodies Simply implemen Newon s Second Law of Moion P = ( )() The 3150 comes from he force required o overcome he fricion P = P = 4050N Visi: --> for more classified papers and answer keys

6 7. WORK, ENERGY AND POWER Principle of Conservaion of Energy: Energy canno be creaed or desroyed, i can only be changed ino oher forms Work Done: W = Fs Kineic Energy: E k = 1 mv Graviaional Poenial Energy: E P = mgh Power: P = W.d T and P = Fv 7.1 Changes in Energy ε f ε i = (Work) engine (Work) fricion ε f is he final energy of he objec ε i is he iniial energy of he objec (Work) engine is he energy caused by driving force acing on he objec (Work) fricion is he energy used up by fricional force or any resisive force {S05-P04} Quesion 7: Car ravelling on horizonal sraigh road, mass 100kg. Power of car engine is 0kW and consan. Resisance o moion of car is 500N and consan. Car passes poin A wih speed 10ms 1. Car passes poin B wih speed 5ms 1. Car akes 30.5s o move from A o B. i. Find acceleraion of he car a A ii. Find disance AB by considering work & energy Use formula for power o find he force a A P = Fv 0000 = 10F Driving force = 000N We mus ake ino accoun he resisance o moion F = Driving Force Resisance = F = 1500 Use Newon s Second Law o find acceleraion: 1500 = 100a a = 1500 = ms Use power formula o find work done by engine: w. d. P = 0000 = w.d. w. d. = J 30.5 There is change in kineic energy of he car so ha means some work done by he engine was due o his: k. E. a A = 1 100(10) k. E. a B = 1 100(5) Change in k. E. = k. E. a B k. E. a A Change in k. E. = = There is also some work done agains resisive force of 500N; due o law of conservaion of energy, his leads us o he main equaion: w. d. by engine = change in k. E + w. d. agains resisance = s s = = = 590m 8. GENERAL MOTION IN A STRAIGHT LINE s displacemen v velociy Paricle a insananeous res, v = 0 Maximum displacemen from origin, v = 0 Maximum velociy, a = 0 a acceleraion {W10-P4} Quesion 7: Paricle P ravels in sraigh line. I passes poin O wih velociy 5ms 1 a ime = 0s. P s velociy afer leaving O given by: v = v of P is increasing when: 0 < < T 1 and > T v of P is decreasing when: T 1 < < T i. Find he values of T 1 and T and disance OP when = T ii. Find v of P when = T and skech velociyime graph for he moion of P Find saionary poins of v; maximum is where = T 1 and minimum is where = T dv d = Saionary poins occur where dv = 0 d = 0 Solve for in simple quadraic fashion: = 30 and 10 Naurally T 1 comes before T T 1 = 10s and T = 30 Finding disance OP by inegraing 30 s = 0 30 v d s = ( ) d s = 0 30 DIFFERENTIATE INTEGRATE 0[ ] s = 85m Visi: --> for more classified papers and answer keys Page 7 of 8

7 Do basic subsiuion o find v v = = 30 v = 5 To draw graph find v of P a T 1 using subsiuion and plo roughly v a T 1 = 13 Graph: {S13-P4} Quesion 6: Paricle P moves in a sraigh line. Sars a res a poin O and moves owards a poin A on he line. During firs 8 seconds, P s speed increases o 8ms 1 wih consan acceleraion. During nex 1 seconds P s speed decreases o ms 1 wih consan deceleraion. P hen moves wih consan acceleraion for 6 seconds reaching poin A wih speed 6.5ms 1 i. Skech velociy-ime graph for P s moion ii. The displacemen of P from O, a ime seconds afer P leaves O, is s meres. Shade region of he velociy-ime graph represening s for a value of where 0 6 iii. Show ha for 0 6, s = and (ii) Since he disance before 0 seconds has already been aken ino consideraion: = 0 a = a = 0.75 s = ( 0) + 1 (0.75)( 0) s = s = Finally add boh o give you s s = s 1 + s s = s = Visi: --> for more classified papers and answer keys Firs find s when = 0, his will produce a consan since 0 6 s 1 = 1 (8)(8) + 1 (8 + )(1) = 9m Finding s when 0 6: s = u + 1 a Page 8 of 8

IB Physics Kinematics Worksheet

IB Physics Kinematics Worksheet IB Physics Kinemaics Workshee Wrie full soluions and noes for muliple choice answers. Do no use a calculaor for muliple choice answers. 1. Which of he following is a correc definiion of average acceleraion?