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PREACE Der Student, Hertiest congrtultions on mking up your mind nd deciding to be n engineer to serve the society. As you re plnning to tke vrious Engineering Entrnce Exmintions, we re sure tht this STUDY PACKAGE is going to be of immense help to you. At NARAYANA we hve tken specil cre to design this pckge ccording to the Ltest Pttern of AIEEE, which will not only help but lso guide you to compete for AIEEE & other Stte Level Engineering Entrnce Exmintions. The slient fetures of this pckge include :! Power pcked division of units nd chpters in scientific wy, with correltion being there.! Sufficient number of solved exmples in Physics, Chemistry & Mthemtics in ll the chpters to motivte the students ttempt ll the questions.! All the chpters re followed by vrious types of exercises (Level-I, Level-II, Level-III nd Questions sked in AIEEE nd other Engineering Exms). These exercises re followed by nswers in the lst section of the chpter. This pckge will help you to know wht to study, how to study, time mngement, your weknesses nd improve your performnce. We, t NARAYANA, strongly believe tht qulity of our pckge is such tht the students who re not fortunte enough to ttend to our Regulr Clssroom Progrms, cn still get the best of our qulity through these pckges. We feel tht there is lwys scope for improvement. We would welcome your suggestions & feedbck. Wish you success in your future endevours. THE NARAYANA TEAM ACKNOWLEDGEMENT While prepring the study pckge, it hs become wonderful feeling for the NARAYANA TEAM to get the wholeherted support of our Stff Members including our Designers. They hve mde our job relly esy through their untiring efforts nd constnt help t every stge. We re thnkful to ll of them. THE NARAYANA TEAM
CONTENTS C O N T E N T S LAWS O MOTION Theory Solved Exmples Exercises Level I Level II Level III Questions sked in AIEEE nd other Engineering Exms Answers
Physics : Lws of Motion NARAYANA LAWS O MOTION AIEEE Syllbus orce nd inerti Newton s Lws of Motion. Conservtion of liner momentum nd its pplictions, rocket propulsion, friction lws of friction CONTENTS " Inerti " orces " undmentl orces of Nture " Liner Momentum " Lw of conservtion of momentum " Impulse " ree body digrm INTRODUCTION orce is the mesure of the interction between different bodies which re either in contct or prt. When force is pplied on body, either it chnges or tries to chnge the bodies position. orce is vector quntity. Every force hs definite direction nd the result of its ction depends on the direction nd the mgnitude of the force. " rme of Reference " Pseudo orce NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion INERTIA It is tht property of the body by virtue of which it opposes ny chnge in stte of rest or uniform motion. Inerti of body is directly proportionl to its mss. Inerti is of three types (i) inerti of rest (ii) inerti of motion nd (iii) inerti of direction Exmples :! The pssenger stnding in sttionry motor cr or trins flls bck when it suddenly moves.! The pssenger stnding in moving motor cr or trin inclines forwrd when it suddenly stops.! If coin is plced on crd bord which is plced on tumbler, then on pulling the crd bord suddenly the coil flls into the tumbler. ORCE A force is push or pull. " Since push or pull hs both mgnitude nd direction, so tht force is vector quntity. " orces occur in pirs. If object A exerts force on object B, then B lso exerts force on A. or exmple, when bt strikes bll the bt exerts force on the bll, but the bll exerts force on the bt lso. " A force cn cuse n object to ccelerte. If you kick foot bll the bll s velocity chnges while your foot is in contct with it. " A force cn deform n object. As you cn see from fig. The bll when hits the floor, is deformed by the contct force exerted on it by the floor. The floor is deformed too, but since it is hrder thn the bll, its deformtion is not s noticeble. " Property 4, tht force cuses n object to be deformed is often used to mesure force this is the principle of spring scle. " A spring scle indictes the mount of the spring is stretched or compressed. The mgnitude of this force is proportionl to the mount of the spring stretched (or compressed), nd the direction of the force is long the spring. Newton 0.0 0.5.0.5.0.5 UNIT O ORCE Unit of force in SI system in Newton. One Newton force is tht force which when cted on body of mss one kilogrm produces on ccelertion of m/sec. Unit of force in C.G.S system is dyne. One dyne force is tht force which when cted on body of mss one grm produces n ccelertion of cm/sec newton = 0 5 dynes kg wt = g newton Dimensions of force is = [M L T ] NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA GRAVITATIONAL UNIT O ORCE In SI system, kilogrm weight (kg-wt) or kilogrm force (kgf) kg wt = kgf = 9.8 newton. In C.G.S system, grm-weight (gm - wt) or grm - force (gmf) UNDAMENTAL ORCES O NATURE Different forces or interctions cn be divided in the following four fundmentl forces on the bsis of their nture - () Grvittionl force or interction (b) Electromgnetic force or interction (c) Nucler force or interction (d) Wek force or interction (A) GRAVITATIONAL ORCE This force cn be obtined from Newtons grvittionl lw. According to Newton s grvittionl lw m m r or mm or r mm = G r In vector nottion, the force cting on m due to m mm = G (r) ˆ r G is constnt which is clled Grvittionl constnt. It is universl constnt. Its vlue is G = 6.67 0 N-m /kg nd dimensions re [M L 3 T ] The rnge of this force is very lrge. " This force cts in between the plnets nd its mgnitude is very lrge. This force lso cts in between tomic prticles (e e), (e p), (p p), (p n) but its mgnitude is very smll. This force is wekest mong ll forces in nture " Energy prticle of this force is Grviton. " This force is importnt in those circumstnces in which one body tking prt in the interction is hving stronomicl size. " Grvittionl force of 6.67 0 newton cts between two bodies of mss kg ech plced t distnce m prt. 3 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830 m r m
NARAYANA Physics : Lws of Motion " The interction time of this force is bout 0 sec. " A force of.3 0 0 newton cts in between the moon nd the erth wheres force of 4.054 0 47 newton cts in between n electron nd proton of hydrogen tom. (B) ELECTROMAGNETIC ORCE " This force cts between chrged or mgnetised objects. " The force cting between sttionry or moving chrges in electromgnetic force. " Source of electricl forces re sttionry nd moving chrges. " Mgnetic force cts only on moving chrges. " It cn be ttrctive or repulsive. There is repulsive force between the chrges of similr nture nd ttrctive force between the chrges of dissimilr nture. " Coloumb s lw q q r qq or r qq = K r where K is constnt which depends upon the nture of the medium. or vccum K = 4π 0 = 9 0 9 N m / C 0 is clled permittivity of vcuum nd its vlue is 8.85 0 C /N - m The rnge of this force is lso very lrge. Electromgnetic force is 0 36 times more stronger thn the grvittion force but times weker 37 thn the nucler force. (C) NUCLEAR ORCE " This force cts within the nucleus of n tom. " This force cts only mong protons nd neutrons within the nucleus of n tom. " This force is mximum inside the nucleus nd zero outside the nucleus. " The rnge of the force is very smll (~ 0 5 m) " This force is strongest force of nture 4 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA " Its energy prticle is pion. " The mgnitude of this force cn be obtined from Yukw potentil : U (r) = U = r /r e 0 U0 r where U 0 nd r 0 re constnts. Nucler force does not depend on chrge i.e., it is chrge independent. This force cts between proton-proton, neutron-neutron nd neutron-proton in equl strength. (D) WEAK ORCES " β decy in rdioctive disintegrtion is explined by this force. " This force is weker in comprison to nucler nd electromgnetic forces nd is 0 5 times the nucler force. The rnge of this force is of the order of 0 5 m. The energy prticle of this force is boson. Neutron nd proton in the nucleus decy due to wek forces s n p +β +ν (Antineutrino) 0 p n +β +ν (Neutrino) 0 + This force is relted with the trnsformtion of neutron to proton or proton to neutron in β decy. The nture of wek forces is still unknown. COMPARATIVE STUDY O ORCES S.No. orces Nture Rnge Reltive Energy prticle strength. Grvittionl ttrctive infinite grviton. Electromgnetic ttrctive infinite 0 36 photon or repulsive 3. Nucler ttrctive very short 0 39 pion 4. Wek unknown very short 0 4 boson 5 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion NEWTON S LAWS O MOTION " irst lw : Every body continues to remin in its stte of rest or of uniform motion unless n externl force is pplied on it i.e. If prticle is t rest, it will remin t rest nd if it is moving with constnt speed, it will continue to move in the sme direction with sme constnt speed unless n externl force is pplied on it. " irst lw is lso clled lw of inerti or Glilens lw " irst lw defines force. According to it the device used for chnging the stte of body or prticle is clled force. " Second lw : The rte of chnge of momentum of prticle is equl to the pplied force nd chnge in momentum is in the direction of pplied force, i.e. = dp dt () (b) d p d v = = m = m dt dt irst lw cn be obtined from second lw. rom this, lw of conservtion of liner momentum cn be obtined, i.e., If = 0, then dp = 0 dt p = mv = constnt or v = constnt (since m = constnt) Exmples:! While ctching bll the cricketer moves hnds bckwrds.! Uses of springs nd shockers in motor crs. " Third lw - To every ction there is n equl nd opposite rection, i.e., = () (b) (c) Action nd rection ct on different objects but ct simultneously. This lw cn be obtined from second lw. Rel force lwys exists in pir known s ction - rection pirs. Exmples:! While ctching bll the cricketer moves hnds bckwrds.! On firing bullet from the gun, jerk cts in the opposite direction.! Lunching of rocket or motion of jet eroplne.! Pulling of horse-crt.! Moving of bot in the opposite direction due to jumping of mn from the bot. 6 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA LINEAR MOMENTUM " The product of mss of body (m) nd velocity ( v ) is clled liner momentum, i.e., p = m v " It is vector quntity nd its direction is in the direction of velocity. " In SI system its unit is kg-m/s nd dimensions re [M L T ]. " If m nd v re the mss nd velocity of one body nd m nd v re the mss nd velocity of nother body, then p = p m v m v If v = v nd m > m then p > p Thus the momentum of hevier body will be greter thn the momentum of lighter body. " If p = p, then m v = m v or v = v m m If m > m then v > v i.e. momentum being sme, the velocity of hevier body will be lesser thn the velocity of lighter body LAW O CONSERVATION O MOMENTUM If no externl force is cting on body, then its liner momentum remins conserved or constnt i.e. if = 0, then d p dt = 0 or = p m v = constnt IMPULSE " The impulse of constnt force is equivlent to the product of the force nd the time during which the force cts on the body. " Impulse = (mgnitude of force) time = t " If the force chnge with time, then the impulse = dt " It is vector quntity its unit is newton-sec. " Reltion between impulse nd liner momentum t 0 y b d c t t t x 7 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion Impulse = t t t dp dt = dt = dp dt = Chnge in momentum = (m v m u ) 0 0 0 If grph is drwn between force nd time, then re between ( t) curve nd time xis is equl to the impulse. Impulse between t nd t = re bcd. REE BODY DIAGRAM " A digrm showing in ll externl forces cting on n object is clled ree body digrm.b.d. " In specific problem, first we re required to choose body nd then we find the number of forces cting on it, nd ll the forces re drwn on the body, considering it s point mss. The resulting digrm is known s free body digrm (BD) " or exmple, if two bodies of msses m nd M re in contct nd force on M is pplied from the left fig. (), the free body digrms of M nd m will be s shown in fig. (b) nd (c) R R M m M f f M () (b) Mg (c) mg IMPORTANT POINT " Two forces in Newton s third lw never occur in the sme free-body digrm. This is becuse free-body digrm shows forces cting on single object, nd the ctionrection pir in Newton s third lw lwys ct on different objects. MOTION O BODIES IN CONTACT orce of contct - When two bodies whose msses re m nd m, rest on frictionless surfce in contct with ech other, then the force with which one body presses the other body, is clled force of contct. orce of contct nd ccelertion produced in the bodies - (i) When two bodies re in contct with ech other () The ccelertion produced in both bodies will be sme. If be the ccelertion, then force ccelertion = or = mss (m + m ) Due to contct of mss m nd m let the pplied force be then for mss m ' = m ' = m = m m + m m = (m + m ) or ' = m m m m = force cting on mss m 8 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA (b) If force is pplied on the body of mss m, then in this cse the ccelertion will be = m + m If due to contct of msses m nd m the pplied force is " then for mss m ' = m or ' = m m = (m + m ) = m (m + m ) m m m or ' = m = force of contct cting on mss m (ii) If three bodies of msses m, m nd m 3 re kept in contct with ech other - If force cts on mss m : The ccelertion produced in the system = (m + m + m3 ) m m 3 m " If the force of contct between m nd m msses is ', then ' = m ' = m = = m (m + m + m3 ) (m + m 3) = (m + m + m ) (m + m 3 ) 3 If the force of contct between the bodies of msses m nd m 3 is ", then " = m 3 = m3 (m + m + m3 ) or the body of mss m = m m = (m + m + m3 ) 9 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion INERTIAL AND ACCELERATED RAME O REERENCE INERTIAL RAME O REERENCE " A non-ccelerting frme of reference is clled n inertil frme of reference. A frme of reference moving with constnt velocity is n inertil frme of reference. " All the fundmentl lws of physics hve been formulted in respect of inertil frme of reference. " All the fundmentl lws of physics cn be expressed s to hve the sme mthemticl form in ll the inertil frmes of reference. " The mechnicl nd opticl experiments performed in n inertil frme in ny direction will lwys yield the sme results. It is clled isotropic property of the inertil frme of reference. Exmples of inertil frmes of reference! A frme of reference remining fixed w.r.t. distnce strs is n inertil frme of reference.! A spce-ship moving in outer spce without spinning nd with its engine cut-off is lso inertil frme of reference.! or prcticl purposes, frme of reference fixed to the erth cn be considered s n inertil frme. Strictly speking, such frme of reference is not n inertil frme of reference, becuse the motion of erth round the sun is ccelerted motion due to its orbitl nd rottionl motion. However, due to negligibly smll effects of rottion nd orbitl motion, the motion of erth my be ssumed to be uniform nd hence frme of reference fixed to it my be regrded s inertil frme of reference. NON-INERTIAL RAME O REERENCE " An ccelerting frme of reference is clled non-inertil frme of reference. " Newton s lws of motion re not directly pplicble in non inertil frmes. Note : A rotting frme of reference is non-inertil frme of reference, becuse it is lso n ccelerting one due to its centripetl ccelertion. PSEUDO ORCE " Those forces which do not ctully ct on the prticles but pper to be cting on the prticles due to ccelerted motion of frme of reference, re clled pseudo forces. " ictitious force = (mss of prticle ccelertion of non-inertil frme of reference with respect to n inertil frme of reference) Note : Pseudo force should be pplied in non-inertil frmes only. Exmples : " The dditionl force cting in rockets or lifts moving with ccelerted velocity is pseudo force. " If body is plced on rotting frme of reference (the frme of reference on erth), the coriolis nd centrifugl forces pper to be cting due to rottion of frme of reference. They re not rel forces but pper to be cting due to rottion of frme of reference. Therefore they re pseudo forces. 0 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA MOTION IN A LIT The weight of body is simply the force exerted by erth on the body. If body is on n ccelerted pltform, the body experiences fictitious force, so the weight of the body ppers chnged nd this new weights is clled pprent weight. Let mn of weight W = Mg be stnding in lift. We consider the following cses : Cse () If the lift moving with constnt velocity v upwrds or downwrds. In this cse there is no ccelerted motion hence no pseudo force experienced by observer O in side the lift. So pprent weight W = Actul weight W. Cse (b) If the lift ccelerted (i.e. = constnt upwrds ccelertion) Then net force cting on the mn re (i) weight W = Mg downwrd (ii) fictitious force 0 = M downwrd So pprent weight W = W + 0 or W = Mg + M = M (g + ) Effective grvittionl ccelertion g = g + Cse (c) Cse (d) If the lift ccelerted downwrd with ccelertion < g : Then ficitious force 0 = M cts upwrd while weight of mss MW = Mg lwys cts downwrd, therefore So pprent weight W = W + 0 or W = Mg M = M (g ) Effective grvitionl ccelertion g = g Specil cse : If = g then W = 0 condition. of weightlessness Thus, in freely flling lift the mn will experience weightlessness. If lift ccelertes downwrd with ccelertion > g Then s in Cse (c) Apprent weight W = M(g ) = M ( g) is negtive, i.e, the mn will be ccelerted upwrd nd will sty t the ceiling of the lift. MOTION O BODIES CONNECTED TOGETHER BY STRINGS (A) OR TWO BODIES Suppose two bodies of msses m nd m re tied together nd bodies re pulled by pplying force on the body m. If T is the tension produced in the string nd is the ccelertion produced in the system, then m m ccelertion = force totl mss = (m m ) + Since ccelertion is sme for both bodies. NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion or the body m, T = m or the body m T = m or T = m = m Tension in the string (m m ) + m T = (m + m ) (B) OR THREE BODIES If three bodies of msses m, m nd m 3 re tied by strings nd pulled by force, the ccelertion produced in the system m T T m m = (m + m + m 3 ) or the body of mss m, T = m or the body of mss m, T T = m or the body of mss m 3 T = m 3 Solving these equtions T = m (m + m + m ) 3 nd (m m ) T + = (m + m + m ) 3 = (m + m ) ROPE LYING ON A HORIZONTAL SURACE A uniform rope of length L which is lying on frictionless tble is pulled by pplying constnt force Let the mss of the rope be m nd its length be L. So the mss per unit length of the rope is (M/L) A L T P T B (L!)! If T is the tension in the rope t distnce! from end B nd ccelertion of the rope is, then for prt PB T = mss of prt PB = M! L NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA or the prt AP T = mss of prt AP (L!) T = ( T)! M = (L!) L or T =! L PULLEY A single fixed pulley chnges the direction of force only nd in generl ssumed to be mssless nd frictionless. PULLEY-BLOCK PROBLEM Cse I m = m = m Tension in the string T = mg Accelertion = zero Rection t the pulley R = T = mg [Strength of support or suspension] Cse II m > m for mss m, m g T = m...() for mss m, T m g = m...() Solving, we get = m m g m m + or (mm )g T = (m + m ) rection t the pulley R = T = 4mm g (m + m ) Cse III The forces cting on the system re shown in the fig. or mss m : T m g = m...() or mss m : m g + T T = m...() or mss m 3 : m 3 g T = m 3...(3) Solving these equtions, we get m + m3 m = g m m m + + 3 Putting the vlue of in eqution (i) & (iii) tension, T & T cn be clculted. 3 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion Cse IV () Without friction : or mss m : T = m or mss m : m g T = m Solving, we get Accelertion mg = (m + m ) mm g T = (m + m ) (b) With friction : (friction in between. surfce nd block) or mss m : T µm g = m or mss m : m g T = m Solving we get (m Accelertion µ m) g = or T = (m + m ) Cse V (m > m ) m g T = m...() T m g = m...() T T = M...(3) Solving (i), (ii) nd (iii), we get = T ( m m) g ( m + m + M) = mg m+ m + M mm (+ µ ) g (m + m ) m + M, m + M T = m g m + m + M Cse VI Mss suspended over pulley from nother on n inclined plne. () Without friction : or m : m g T = m or m : T m g sinθ = m Solving, we get M Accelertion mm (+ sinθ)g T = (m + m ) (m m sinθ) g = (m + m ) 4 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA (b) With friction : or m : m g T = m N = m g cos θ µn = µm g cosθ or m : T µm g cos θ m g sin θ = m m m(sinθ + µ cosθ) Accelertion = g m + m mm [+ sinθ + µ cosθ]g Tension T = (m + m ) Cse VII Msses m & m re connected by string pssing over pulley (m > m ) or mss m : m g sin α T = m or mss m : T m g sin β = m g (m sinα m sinβ) Accelertion = (m + m ) Tension Cse VIII mm (sinα + sinβ) g T = (m + m ) rom cse (IV-) we know tht the tension If x is the extension in the spring, then T = kx x = T k mm g = k(m + m CONDITION OR BALANCE ) mm T = g ( m + m ) Let S nd S be the weights of scle pns nd nd b be the rms of the blnce. As the bem remins horizontl when pns re empty, moments bout fulcrum must be zero. S = S' b...() 5 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830 µn b S S Suppose now tht equl weights, W ech, re put on the pns of the blnce, the bem hs to remin horizontl for blnce to be true tking moments bout, we hve (S + W) = (S + W) b...() Using () or W = Wb or = b i.e., rms of blnce must be equl Putting = b, in eqution () We hve S = S, weights of scle pns be equl The method of Double weighing : The true weight of body cn be determined with the help of flse blnce s follows : (i) When rms re equl but weight of pns re unequl then the weight of body cn be determined with the help of flse blnce s follows :
NARAYANA Physics : Lws of Motion Plce the body hving the true weight W in the lift pn & counter poise it with stndrd weights = W, (S + W) = (S + W ) or S + W = S + W S S = W W...() Now plce the body in the right pn nd counter poise it with stndrd weights = W (S + W ) = (S + W) or S + W = S + W S S = W W...() equting R.H.S. of eq n () nd eq. () W = W + W S S' W W W W true weight of body is equl to the rithmetic men of the two pprent weights. (ii) When the bem remins horizontl but neither rms re equl nor weights of scle pns re equl S = S b...() Plce the body with true weight W in the left rm nd counter poise it with stndrd weights W then (S + W) = (S + W )b Using () we hve W = W b...() Now put the body in right pn nd stndrd weights W in the left rm to counter poise it. (S + W ) = (S + W) b b using () we hve W = Wb or Wb = W...(3) Multiply eq n. () nd eq n. (3), we hve W b = W W b or W = W W S W S' W W W Thus the true weight of the body in this cse is the geometric men of the two pprent weights. RICTION When the surfce of body slides over surfce of nother body, ech body exerts force of friction on the other. Such friction is prllel to the surfce. The force of friction on ech body is in direction opposite to its motion reltive to the other body. It is self djusting force, it cn djust its mgnitude to ny vlue between zero nd the limiting (mximum) vlue. i.e. 0 f f mx The friction force cting between ny two surfces t rest with respect to ech other is clled the force of sttic friction. The frictionl force cting between surfces when there is reltive motion with respect to ech other, is clled the force of kinetic friction or sliding friction. LAWS O RICTION The limiting force of friction is proportionl to the norml force tht keeps two surfce in contct with ech other, nd is independent of the re of contct nd the reltive velocities between the two surfces in contct. i.e. f = µ N mx f ( f s ) mx f k 6 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA PROPERTIES O RICTION. If the body is t rest, the sttic frictionl force f s nd the externl force prllel to the surfce re equl in mgnitude; nd f s is directed opposite to. If the externl force increses f s increses.. The mximum vlue of sttic friction is given by ( f ) =µ s mx s N Where, µ s = coefficient of sttic friction. N= norml rection. (f s ) mx (f s ) mx= At this condition, the body is just bout to move reltive to the surfce. 3. If externl force exceeds ( f s ) mx, then the body slides on the surfce. Now frictionl force rpidly decreses to constnt vlue given by fk =µ kn Where µ k is the co-efficient of kinetic friction. ANGLE O RICTION At point of rough contct, where slipping is bout to occur, the two forces cting on ech object re the norml rection N nd frictionl force µ N. The resultnt of these two forces is nd it mkes n ngle λ with the norml where, f k N v µ N tn λ= =µ N λ= tn ( µ ) λ N This ngle λ is clled the ngle of friction. ANGLE O REPOSE µn Suppose block of mss m is plced on n inclined plne whose inclintion θ cn be incresed or decresed. Let µ be the co-efficient of friction between the block nd the plne. At generl ngle θ, Norml Rection is N = mgcos θ Limiting friction ( fs )=µ N =µ mgcos θ m nd the pulling force, = mgsin θ There is criticl vlue of ngle θ, clled ngle of repose ( α ) t which pulling force nd limiting friction re equl. Now if θ is further incresed, the pulling force becomes more thn the limiting friction f s nd the block strts sliding. Thus, ( ) = f t θ=α s mx µ mg cos α= mg sin α µ = tn α θ 7 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion α= tn ( µ ) Here, we see tht ngle of friction ( λ ) is numericlly equl to the ngle of repose. i.e. λ=α POINTS TO REMEMBER! A system is sid to be isolted, when no externl forces re cting on the system. or such system, liner momentum remins constnt.! No externl force is required to move body uniformly, provided there is no friction. Externl force is required to stop moving body; to chnge the speed of the body nd lso to chnge the direction of motion of the body.! or body of mss m, moving with velocity v # # #, its momentum p = mv. As m is constnt during # # dv # motion, we cn write, force = m = m, where dt # is the instntneous ccelertion.! When body moves with high velocity, very ner to the velocity of light, m vries considerbly with velocity. In such cse, we cn not write # = m # # dp #, but the reltion = is vlid. dt! To find the ccelertion of the body, note ll the forces cting on body under considertion nd # # # then find the vector sum of ll the forces nd finlly use = m.! If body is in equilibrium, then it does not men tht no force cts on the body, but it simply mens tht the net force (resultnt of ny number of forces) cting on the body is zero. urther, for equilibrium, body need not necesrily be t rest. A body in uniform motion long stright line is lso in equilibrium.! An ccelerted motion of body is lwys due to force. It cn occur in two wys. (i) Due to chnge in its speed only : In this sitution, the force must ct on the body either in the direction of its motion or opposite to its direction of motion. (ii) Due to chnge in direction only : In this sitution, the force must ct perpendiculr to the direction of motion of the body. Such force mkes the body move long circulr pth.! Action nd Rection re lwys equl nd opposite nd ct on two different bodies.! The forces of interction between bodies forming system re clled internl forces. The resultnt of internl forces is zero due to ction nd rection. These forces do not cuse ny motion in the system. The forces exerted on bodies of given system by bodies situted outside re clled externl forces. The externl forces cuse the motion or chnge in the system.! The coefficient of limiting friction is slightly greter thn the coefficient of kinetic friction.!! 8 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA SOLVED EXAMPLES Exmple - Solution : The position x of body of mss kg vries with time t s x = (t + 3t 4) m orce cting on the body is - () 4N () 8N (3) N (4) 6 N x = (t + 3t 4) m dx Velocity v = = (4t + 3) m/s dt d x Accelertion f = = 4 m/s dt orce = mss ccelertion = 4 = 8N Answer is () Exmple - A body of mss kg moving with speed of 00 m/s hits wll nd rebounds with the sme speed. If the contct time is (/50)s, the force pplied on the wll is () 0 4 N () 0 4 N (3) 4N (4) 8N Solution : Initil momentum of the body = mv = 00 = 00 N-s inl momentum = 00 N-s Chnge of momentum of the body = 00 (00) = 400 m/s Momentum given to the wll = +400 N-s Time = s 50 Momentum given per second 400 = 0 4 N = orce pplied on the wll / 50 Answer is () 9 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion Exmple - 3 or motion in stright line the force-time grph is shown in figure. The prts of the curve re circulr. The momentum gined in the time intervl t = 0 to t = 8s, in N-s will be (N) () zero () 4π (3) π (4) 6π Solution : The totl momentum gined = Exmple - 4 Solution : t t dt = Are between -t curve nd time xis. = zero, becuse the re in negtive direction is equl to the re in positive direction. Answer is () A cricket bll of mss 50 gm moves t speed of m/s nd fter hitting by the bt it is deflected bck t the speed of 0 m/sec. If the bt nd the bll remined in contct for 0.0 sec then clculte the impulse nd verge force exerted on the bll by the bt. [Assume the bll lwys moves norml to the bt]. According to the problem the chnge in momentum of the bll p = p f p i = m (v u) = 50 0 3 [0 ( )] = 4.8 impulse = chnge in momentum = 4.8 N-s + 0-4 6 8 t(s) nd p 4.80 v = = = 40 N t 0.0 Exmple - 5 A cricket bll is deflected by btsmn t 45º without chnging its speed which is 54 km/hr. If the mss of the bll is 0.5 kg, then wht is the impulse given to bll? Solution : Initil momentum p i in component form is p i = î (mu sin.5) ĵ (mu cos.5º) Similrly the finl momentum is p f = î (mu sin.5) + ĵ (mu cos.5º) f p i p = ( mu cos.5º) î f i So the mgnitude of impulse = p p = 0.5 3 54 0 3600 cos.5º = 4.5 N Exmple - 6 Solution : Two blocks of mss m = kg nd M = 5kg re in contct on frictionless tble. A horizontl force = (35 N) is pplied to m. ind the force of contct between the block, will the force of contct remin sme if is pplied to m? As the blocks re rigid under the ction of force, both will move with sme ccelertion = m + M = 35 + 5 = 5 m/sec 0 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA Now s the mss of lrger block is M nd its ccelertion, m M so force of contct = ction on it or f m = M = 5 5 = 5 N If the force is pplied to M then its ction on m will be f m = m = 5 = 0 N rom this problem, it is cler tht ccelertion does not depends on the fct tht whether the force is pplied to m or M, but force of contct does. Exmple - 7 Three blocks of msses m = kg, m =.5 kg nd m 3 = kg re in contct with ech other on friction less surfce s shown in fig. find () horizontl force needed to push the block s on unit with n ccelertion of 4 m/sec (b) The resultnt force on ech block nd (c) The mgnitude of contct force between blocks. Solution : () = (m + m + m 3 ) or = ( +.5 + ) 4 = 4.5 4 = 8 N (b) for m resultnt force f = m = 4 or f = 4N... () for m resultnt force f f = m or =.5 4 = 6 or f f = 6N...() nd resultnt force on m 3, f = m 3 = 4 or f = 8 N...(3) (c) Contct force between, m nd m 3 = f = 8N from eqution () nd eqution (3) f = f + 6 = 8 + 6 = 4N Exmple - 8 Solution : The mss of n elevtor (lift) is 500 kg. Clculte the tension in the cble of the elevtor when the elevtor is (i) sttionry, (ii) scending with n ccelertion of.0 ms, (iii) descending with the sme ccelertion. (g = 9.8 N kg ) The grvity - force on the elevtor is mg = 500 9.8 = 4900 N When the elevtor is sttionry (ccelertion is zero), the net force on it will be zero (fig.) tht is, T mg = 0 T or T = mg = 4900 N T T When the elevtor is ccelerted upwrd, the net force on it will Elevtor be in the upwrd direction. Therefore, the tension T will be greter thn the grvity force mg (fig.b). According to Newton s second lw, the net force (T mg) will be equl to mss m ccelertion. Tht is, mg mg mg T mg = m () (b) (c) or T = mg + m = 4900 + 500.0 = 5900 N When the elevtor is ccelerted downwrd, net force will be in the downwrd direction m f m f M M NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion Exmple - 9 Solution : nd the tension T will be less thn mg (fig.c) Agin by Newton s lw, we hve mg T = m or T = mg m = 4900 500.0 = 3900 N An elevtor nd its lod weight totl of 600!b. ind the tension T in the supporting cble when the elevtor, originlly moving downwrd t 0 fts is brought to rest with constnt ccelertion in distnce of 50 ft. The elevtor moving downwrd stops fter distnce of 50 ft. Therefore, by the formul v = u + s, we hve 0 = 0 + 50 = 4ft s Let T be the tension in the cble. Then, we hve T = m (g ) Here m = 600!b. In PS system, g = 3 ft s 57600 T = 600 {3 ( 4)} = 57600 poundls = = 800!b. wt. 3 Exmple - 0 A lift of mss 000 kg is supported by thick steel ropes. If the mximum upwrd ccelertion of the lift be. ms nd the breking stress for the ropes be.8 0 8 N m, wht should be the minimum dimeter of the ropes? (g = 9.8 ms ) Solution : When the lift is ccelerted upwrd, the tension T in the rope is greter thn the grvityforce mg. If the ccelertion of the lift be, then by Newton s lw, we hve T mg = m or T = mg + m = m (g + ) Mss of the lift m = 000 kg nd mximum ccelertion =. ms. Hence mximum tension in the rope is T = 000 (9.8 +.) =, 000 N. If r be the rdius of the rope, then the stress will be T/πr T πr =.8 0 8 Nm (given) or r = T π (.8 0 8 ) = 000 ( / 7) (.8 0 8 ) = 4 4 0 r = 00 = 0.005 m. Exmple - A lift is going up, the totl mss of the lift nd the pssenger is 500 kg. The vrition in speed of lift is s shown in fig. Wht will be the tension in the rope pulling the lift t (i) sec, (ii) 6 sec, (iii) sec. Solution : A slope of v t curve gives ccelertion At, t = sec = 3.6 0 =.8 m/s v 3.6 m/s 0 4 6 8 0 t NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA lift is moving up T = m (g + ) = 500 (9.8 +.8) = 7400 N t t = 6 sec = 0 T = mg = 500 9.8 = 4700 N t t = sec = 0 3.6 =.8 m/s lift is moving down. T = mg (g ) = 500 (9.8.8) = 000 N Exmple - A lift is moving upwrds with ccelertion 0. An inclined plne is plced in this lift. Wht is the time tken by body of mss m in sliding down from the top of this plne to the bottom (If length of the bse of the plne is! nd ngle is θ) Solution : As the lift is moving up so the pprent weight of the body is m (g + 0 ) nd its component long the plne is m (g + 0 ) sin θ due to which the ccelertion down the plne is (g + 0 ) sin θ length of inclined plne AB =!/cosθ using 0 s = ut + t nd u = 0, = (g + 0 ) sin θ m A! cosθ = (g + 0 ) sinθ t B θ! C t =! (g + 0 )sinθcosθ Exmple - 3 Two bodies whose msses re m = 50 kg nd m = 50 kg re tied by light string nd re plced on friction less horizontl surfce. When m is pulled by force, n ccelertion of 5 ms is produced in both the bodies. Clculte the vlue of. Wht is the tension in the string? m m Solution : The force is pulling both the bodies together. Hence if the ccelertion produced in the direction of force be, then by newton s lw of motion, we hve (net) force = mss ccelertion T T m m = (m + m ) = (50 + 50) 5 ms - = 000 N To determine the tension in the string, we hve to consider the force cting on the bodies seprtely. When m is pulled by the force, then m pulls m through the string by force T. This force is the tension in the string which cts on m in the forwrd direction (see fig) m lso pulls m by the sme (rectionry) force T. Hence the tension T of the string cts on m in the bckwrd direction. Thus, net force T cts on m in the forwrd direction. Hence,by Newton s lw pplied for m lone, we hve T = m or T = m = 000 (50 5) = 750 N 3 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion We cn determine T lso by pplying Newton s lw for the motion of m lone. On m there is net force T in the forwrd direction T = m = 50 5 = 750 N Exmple - 4 The msses m, m nd m 3 of the three bodies shown in fig. re 5,, 3 kg respectively. Clculte the vlues of the tension T, T nd T 3 when () the whole system is going upwrd with n ccelerting of ms, (b) the whole system is sttionry (g = 9.8 ms ) T Solution : () All the three bodies m, m nd m 3 re moving upwrd together. The force pulling the system upwrd is T nd the downwrd grvity-force is (m + m + m 3 ) g. Hence the net force on the system is T (m m + m + m 3 ) g. According to Newton s second lw, this force is equl to totl mss ccelertion. If be the ccelertion of the system in the upwrd direction, then T T = (m + m + m 3 ) g = (m + m + m 3 ) or T (5 + + 3) 9.8 = (5 + + 3) m T = 0 + 98 = 8 N The force pulling m nd m 3 in the upwrd direction is T nd the grvityforce T 3 on them is (m + m 3 ) g. Hence the net force in the upwrd direction is T (m + m 3 ) g. Agin, by Netwon s lw, we hve T (m + m 3 ) g = (m + m 3 )g m T ( + 3) 9.8 = ( + 3) T = 0 + 49 = 59 N The net force on m 3 in the upwrd direction is T 3 m 3 g. Hence by Newton s lw, we hve T 3 m 3 g = m 3 or T 3 3 9.8 = 3 T 3 = 6.0 + 9.4 = 35.4 N (b) If the whole system is sttionry (or moving with uniform velocity), then = 0. Hence from eqution. (i), (ii) nd (iii), we hve T = (m + m + m 3 ) g = 0 9.8 = 98 N T = (m + m 3 ) g = 5 9.8 = 49 N T 3 = m 3 g = 3 9.8 = 9.4 N Exmple - 5 Two blocks of msses.9 kg nd.9 kg re suspended from rigid support S by two inextensible wires ech of length m. The upper wire hs negligible mss nd the lower wire hs uniform mss of 0. kg m. The whole system of blocks, wires nd support hve n upwrd ccelertion of 0. ms. Accelertion due to grvity is 9.8 ms. (i) ind the tension t the mid-point of the lower wire. (ii) ind the tension t the mid-point of the upper wire. Solution : Suppose, the tension t the point A is T A. Then T A mg = m S or T A = m ( + g) B where m =.9 kg + mss of the wire of length AD =.9 + 0. =.0 kg 4 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830 C D A.9 kg.9 kg
Physics : Lws of Motion NARAYANA (ii) T A =.0 (0. + 9.8 ) = 0 N Suppose, the tension t the point B is T B, then T B = M ( + g) where M =.9 kg +.9 kg + mss of the wire CD =.9 +.9 + 0. = 5.0 kg T B = 5.0 (0. + 9.8) = 50 N Exmple - Solution : A bucket of mss 5kg is rised by 50 kg mn in two different wys s shown in fig. Wht is the ction on the floor by the mn in the two cses? If the floor yields to norml force of 700 N, which mode should the mn dopt to lift the bucket without the floor yielding? Here, mss of the bucket, m = 5 kg mss of the mn, M = 50 kg orce pplied to lift the bucket, = mg = 5 9.8 = 45 N Weight of the mn, Mg = 50 9.8 = 490 N fig () When the bucket is rised by the mn by pplying force in upwrd direction, rection equl nd opposite to will ct on the floor in ddition to the weight of the mn. Therefore, ction on the floor = Mg + = 490 + 45 = 735 N Since the floor yields to norml force of 700 N, So mn should not dopt this mode to lift the bucket ig. (b) When the bucket is rised by the mn by ppling force over the rope (pssed over the pulley) in downwrd direction, rection equl nd opposite to will ct on the floor. Therefore, ction on the floor = Mg = 490 45 = 45 N The mode (b) should be dopted by the mn to lift the bucket. Exmple - 7 The pulley rrngement of figures () nd (b) re indenticl. The mss of the rope is negligible. In figure () the mss m is lifted by ttching mss m to the other end of the rope. In figure (b) mss m is lifted up by pulling the other end of the rope with constnt downwrd force = mg. The rtio of ccelertion in the two cses will be m m m =mg () : () : 3 () (b) (3) 3 : (4) : Solution : Let in the cse shown in figure (), then tension in the string be T nd ccelertion produced is, then T mg = m...() nd mg T = m...() rom eqution () nd () 5 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion g = 3 Let in the cse shown in figure (b), the tension in the string be T nd ccelertion be, then T mg = m...(3) T mg = 0...(4) from eqution (3) nd (4) = g ' = g/ 3 g = 3 Answer is () Exmple - 8 A body of mss 0 kg is plced on horizontl smooth tble. A string is tied with it which psses over frictionless pulley. The other end of the string is tied with body of mss 5 kg. When the bodies move, the ccelertion produced in them, is m = 0 kg Solution : () 9.8 m/s () 4.8 m/s (3) 4.5 m/s (4) 3.7 m/s m = 5 kg Due to grvittionl force 5g cting on the body of mss 5kg. both bodies move. Let the ccelertion produced in the system be, then m g = (m + m ) = mg 5g = (m + m ) (5 + 0) = g 9.8 = 3 3 = 3.7 m/s Answer is (4) Exmple - 9 The following figure shows pinter in pltform suspended long the building. When the pinter pulls the rope the force exerted on the floor is 450 N while the weight of the pinter is 000N. If the weight of the pltform is 50 N, the ccelertion produced in the pltform will be (g = 0 m/s ) () 4 m/s () m/s (3) 5 m/s (4) 6 m/s Solution : 000 Let the ccelertion be nd mss of the pinter = 00 kg. If the pull pplied to 0 the rope by the pinter is, then the rope will lso pply sme mount of force. rom Newton s lw + 450 000 = 00 or 550 = 00...() 50 Mss of pltform = = 5 kg 0 or 450 50 = 5 700 = 5 6 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA from eqution () nd () = m/s Answer is () Exmple - 0 In the following figure, if the tble nd pulley re frictionless nd strings re weightless, then the ccelertion of the system will be - () 3. m/s ().5 m/s (3).8 m/s (4).4 m/s Solution : rom the figure, m g T = m...() T m g = m...() or T T = M...(3) solving (), () nd (3) =.4 m/s Exmple - Two kg weights re suspended from spring blnce s shown in figure. The reding in the scle of the spring blnce will be - () zero () 4 kg-wt (3) kg-wt (4) kg-wt Solution : Spring blnce is rest therefore tensions will be sme in both strings. Let it be T. Since the weights re lso t rest. So T = g = kg-wt. Here one weight, cts s support of nother weight. Answer is (3) Exmple - A tringulr block of mss M with ngles 30º, 60º nd 90º rests with its 30º 90º side on horizontl smooth tble. A cubicl block of mss m is plced on 30º 60º side of the block. With wht ccelertion should M be moved reltive to sttionry tble so tht 60º m the mss m remins sttionry reltive to the tringulr block? () 5.66 m/s () 4.3 m/s (3) 9.8 m/s (4) 4.9 m/s M 90º 30º Solution : Accelertion of the block of mss m long the inclined plne in downwrd direction = g sin θ = g sin 30º If the block m is to remin sttionry, then g sin 30 = cos 30º 60º m 30º or sin30º = g = gtn 30º cos 30º M cos 30 o 90º 9.8 30º = 9.8 = 3.73 = 5.66 m/s Answer is () Exmple - 3 A cube hving mss m nd side is plced on horizontl surfce s shown in the figure. A horizontl force is cting perpendiculr to one of the surfce, the force is t point t height of 3/4 from the bse. The minimum force required tht the cube turns bout one of its edges is () (3/) mg () mg (3) (/3) mg (4) 3/4 mg g sin 30 o 7 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006 Ph.: (0) 3003/3 x : (0) 48830 m mg / O 3/4
NARAYANA Physics : Lws of Motion Solution : The cube will turn bout n edge O if the torque t O stisfies the following condition (3/4) > mg / or > 3 mg minimum = (/3) mg Hence the nswer is (3) Exmple - 4 Wht should be the vlue of m (mss of suspended block) so s to prevent the smller block m from sliding over the tringulr block of mss M. All surfces re friction less nd the string nd pulley re light. M m θ T T m' () m+ M m' = cot θ () cot θ m' = m+ M (3) m+ M m' = tnθ (4) M m m' = cot θ Solution : Writing force equtions m = m g T...() T = (m + M) - cosθ...() m = m g (m + M) (m + m + M) = m g θ or m' g g sinθ = g m' + m + M θ or the block hving mss m, not to slide it is necessry tht cos θ = g sin θ m' g cosθ = gsinθ m' + m + M or m = (m + m + M) tnθ m ( tnθ) = (m + M) tnθ (m + M) tnθ m ' = = tnθ m + M cot θ Hence the nswer is () Exmple - 5 A mss of 400 kg is suspended by two ropes from points A nd B on the roof nd the wll. The tension in rope OA is. () 00 kg weight () 300 kg weight (3) 400 kg weight (4) 600 kg weight 8 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006 Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA Solution : On resolving tension T, the horizontl nd verticl components for T re T sin 45º = Mg = 400 g T 45º B nd T cos45º = T T tn45º = 400 g T A T O T = 400 g = 400 kg wt. 400kg Hence the nswer is (3) Exmple - 6 A rod AB whose length is 3m rests on two perpendiculr surfces s shown in the figure nd strts sliding. At prticulr time end B is t distnce of m from verticl surfce nd its velocity is 0 m/s. Wht is the velocity of A t this time : () 0 m/s (upwrds) () 4 m/s (upwrds) (3) 4 m/s (downwrds) (4) 0 m/s (horizontl) Solution : If length of rod is! nd t ny time the coordintes of A nd B re (0, y) nd (x, 0) then! = x + y now when Differentiting with respect to time dx dy 0 = x + y dt dt dy x dx = dt y dt dy dx = V, = V dt dt or A B V A = x VB y 0 m/s 3 m B m A O or V B = 0 m/s, x = nd y = 5 V A = /5 0 = 4 m/s negtive ( ) sign indictes tht velocity V A is downwrds, Hence the nswer is (3) Exmple - 7 A block of weight 00 N lying on horizontl surfce just begins to move when horizontl force of 5 N cts on it. N Determine the coefficient of sttic friction. Solution : = 5 N µ N s As the 5 N force brings the block the point of sliding, the frictionl force = µ N. s w = 00 N from free body digrm : N = 00 µ N = 5 s µ s = 0.5 Exmple - 8 A 5 kg block slides down plne inclined t 30 o to the horizontl. ind () the ccelertion of the block if the plne is frictionless (b) the ccelertion if the coefficient of kinetic friction is 0.. Solution : 9 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion () The only force cusing ccelertion is mg sin30 mg sin 30 o = m = g sin 30 o, = g/ 4.9 m/s (b) Here friction force ( µ N) cts opposite to mg sin30 N = mg cos 30 o mg sin 30 o µ N k = m = g sin 30 o - µ g k cos 30 = 3.0 m/s Exmple - 9 Solution : ind the time required by block to come to rest from speed of 0 m/s moving on horizontl surfce where µ k = 0.. Wht is the distnce covered before stopping? If the block is moving towrds right nd slowing down, the force of friction nd hence ccelertion is directed towrds left. force µ mg = = = µ g mss m Now u = 0 m/ s, v = 0 v = u + t 0 = 0 µ gt k Let displcement = s v u 0 0 s = = = 5.5m ( µ g) Exmple - 30 Solution : A block of mss m is plced on nother block of mss M lying on smooth horizontl surfce. The coefficient of sttic friction between m nd M is µ s. Wht is the mximum force tht cn be pplied to M so tht the blocks remin t rest reltive to ech other? Drw the free body digrms of blocks t the moment when is t its mximum vlue nd m is just bout to slide reltive to M. N rictionl force between m nd M = µ N s (N = norml rection between the blocks) Due to the friction force, M will try to drg m towrds right nd hence frictionl force will ct on m towrds right. Let = ccelertion of ech block (s they move together) R = norml rection between M nd surfce. rom free body digrm of m : N = mg m M m mg µ N s 30 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
Physics : Lws of Motion NARAYANA µ N s = m...(i) rom free body digrm of M : N + Mg = R µ N = M...(ii) mx s Combining these two equtions(i) nd (ii), we get ( M) =µ m+ g mx s µ N s R M N Mg mx Hence, ( M) µ s m+ g is the criticl vlue of force. If is greter thn this criticl vlue, m begins to slip reltive to M nd their ccelertions will be different. If is smller thn this criticl vlue, m nd M move together, with sme ccelertion, i.e. without ny reltive motion. 3 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830
NARAYANA Physics : Lws of Motion EXERCISES LEVEL - I Which of the following reltions is wrong - dr d r () = m () = m dt dt dv (3) m(v u) = m (4) = dt t A pssenger in uniformly moving trin throws bll upwrds. The bll will return - () in the pssenger s hnd () by the side of the pssenger (3) in front of the pssenger in direction of motion (4) Behind the pssenger in the direction opposite to the motion of trin 3 A force of 0 N is pplied to mss of 0gm for 0 seconds. The chnge of momentum in kgm/sec units will be- () 0 () 00 (3) 000 (4) 0.0 4 A bll moving with velocity 0 m/sec hs mss of 50 gm. If collides ginst wll normlly nd is rebounded normlly with the sme speed. If the time of impct of the bll nd the wll is 40 milli seconds, the verge force exerted on the wll in dynes is - () zero ().5 0 7 (3).5 0 7 (4) 5 0 6 5 A boy holds hydrogen filled blloon with string. He is sitting in trin moving with uniform velocity on stright trck. The string is verticl. On pplying brkes the blloon will - () be thrown forwrd () be thrown bckwrd (3) remin verticl (4) fll downwrds 6 Which of the following sttements is flse - () the mss of two bodies will be sme if they exhibit sme inerti () the weight of body t the centre of the erth is zero (3) if bll is thrown verticlly upwrds by mn sitting in uniformly moving cr, then the bll will fll behind the mn (4) the mss of body remins constnt t ll plces. 7 The correct sttement in the following is - () body cn hve ccelertion even if its velocity is zero () if the velocity of the body be zero, it will necessrily hve ccelertion (3) if the speed of body is constnt its ccelertion lso will be constnt (4) body moving with vrible speed hs zero ccelertion 8 Wht will be the ngle between two forces of equl mgnitude, if the mgnitude of resultnt force is equl to ny one of the two - () 0º () 60º (3) 90º (4) 0º 9 A beker filled with liquid is plced in the pn of spring blnce. If you dip your finger in it, the reding of the blnce will be - () incresed () unchnged (3) decresed (4) my increse or decrese ccording to the nture of liquid 0 A body of mss m is thrown verticlly upwrds with velocity v,. It reches to height h nd returns fter t seconds. Then, the totl chnge in its momentum is - () zero () mv (3) mgt (4) mght 3 NS House, 63, Klu Sri Mrket, Srvpriy Vihr, New Delhi-006! Ph.: (0) 3003/3 x : (0) 48830