Laws of Motion TOPIC 1 5 1000550 69 ten Newton's Laws of Motion and ot 800. Conservation of Momentum os nisi eet ae a OF The initial mass of a rocket is = 1000 kg. Calculate at what rate the 02 Apparticle of mass M originally at uel should be burnt, so that the rest is subjected to a force whose “ey rocket is given an acceleration of direction is constant but aoe 20s". The gases come out at a ‘magnitude varies with time ae relative speed of 500ms" with according to the relation 03 AforceF =(40i+10j)N acts ona 2 aT a F=Fo -() body of mass 5 kg. If the body 9: fee ieer starts from rest its position vector Noigs*—(eDkas" here, Fo anaT are constants, rae eo 25 uy Si) Ans. (3) The force acts only for the time (2)(1001 + 400j)m (by{1004 + 10oj)m eee interval2T. The velocity vof the vias v kg.0=20m/e particle after time 2T is (it400% + Toor (aN400i + 4005) ten = SOD Wg = HO 12021, 27 Jay Shi). () he given situation is shown below DET Fl (esMFat. ta Eh opal yh (eySEAL cay Given, fore Frew Ce nr seen “ Ans. (c) initial speed, u =O ly] te Given, force on partite, Tinietuont=108 J a rali-(| n travelled bythe body | Since, F=mo so=F - Acceleration of balwit ganas : on ty whereas ace Ltt) oi +10 (2) From Egs.(ijand{il we get c ( Now y Newton's second law of motion = sa0xios hea +2)no = Fae MaMa ae (si lesa +2) = ao (SE) nto Thiet a = =(400i + 100)Im - 190010 = 1000 «20 aa at Integrate above equation, we get 04 Two billiard balls of equal mass 30.9 = 100010 +1) evans fifi Yl strike a rigid wall with same speed * for of 108 km/h (as shown) but at a dierent angles. the bal get [L-The force acts only for time intervaZT] reflected with the same speed,Laws of Motion then the ratio of the magnitude of impulses imparted to ball a and ball by the wall along, x. direction is (202%, 25 July Shit Bal) fst (any2 010" kg Speed of approach, u=180 kmph =180x5=50me" 8 {and speed of approachiy,) =speed of separation(y,] Baia) As we know that, impulse(!)=change in momentum(n) 32 for balla J, emty, —uy) i, =mu+ u)=2mu = 2x30%10"°x50= 3kg-ms" ‘and impulse for bali = rlucos®+vc0s8) Bal (o) 05 A bullet of 4 g massis fired from a un of mass 4 kg. If the bullet moves with the muzzle speed of '50 ms”, the impulse imparted to the gun and velocity of recoll of gun are (2021, 22 July Shit) (2)0.4K¢-me"1,0.1ma") {0)0.2kg-me" 0.05 me’ (0.2 hq-ms", 0.4m" (400.4 kg-ms™", 0.05 ms“! Ans. (b) siven. rags of bull, y= 49=4x 107g Mass of gun, me = 4g bullet vj = 80ms 1 velocity of quo bey By using law of conservation of ‘omentum for recoiling of un vp wattle = 41060 im, 4 =-80x10' = -005ms Impulse = Change in momentum of gun =m, (0-V)=4x005=0.2kgms"* 06 A bullet of mass 0.1 kg is fired on a ‘wooden block to pierce through it, but it stops after moving a distance ‘of 50 cm into it. Ifthe velocity of bullet before hitting the wood is 10 m/s and it slows down with Uniform deceleration, then the magnitude of effective retarding force on the bullet is x\N. The value of xto the nearest integer is. (202%, 18 Merch Shite Ans. (10) Given, Themass ofthe bulet, m=d.1kg The initia velocity ofthe bullet before hiting the wooden w= 10mis The final velocity ofthe bullet atter hitting the wooden. ‘The distance travel by the bullet before coming to the rests = 80cm Using the equation the motion, Fou =20s, = -(0F =201050) = o=-100m/¢ The uniform retardation of the bullets 100 m/s" . The magnitude of the effective retarding force on the bullet, Hence, the value of xto the nearest integeris 0. 7 A body of mass 2 kg moves under a force of (21 +3]+5K)N. It starts from rest and was at the origin initially. After 4s, its new coordinates are (6,b, 20). The value of bis (Round off to the nearest integer) (2021, 16 March Shit) ‘Ans. (12) Given, Force.F=(2i+3]+ 510N Mass, m=2kg, Time, t 43 We know that Force «mass xacceleration = Famxasce . 2i+3)+ 5 o 2 From second equatian af mation, 1 aro tat? 7) 2 From €asandi we get 1 di +5 yo 20+ lai +s + si? rs sure [eueQand 24s) +81 1+ 20% mois ijaz0 Let sexistyjtzk axis yjechesi+ 2j+20k (i) ‘According te question,new coordinates 2re(8.b,20) it means wityl+ chesi+b]+20k Liv) Comparing Eqs. itandivl we get b= 08 A boy pushes @ box of mass 2 kg with a force F =(20i +10j) Nona frictionless surface. If the box was initially at rest, then ...... mis displacement along the X-axis after 10s. (2021, 26 Feb shite ‘Ans. (500) Given, mass of box, m=2 kg Force, F =20% + 10)N Inti speed ot box, u=Oms"* Time.t=10s Let accoleation of boxisaand sisplacement along Xaxisatter Osis, As, Fema g=Fim 201+ 10} =(10i+ Sjims 2 By second equation of mation slong X- axis, 8, 2ujt+to,%=0+) x10x110=500m ae Hence, displacement along X-axis after 10 sis 500m 89 A small ball of mass mis thrown upward with velocity from the ‘groun(d) The ball experiences a resistive force mkv’, where visits44 JEE Main Chapterwise Topicwise Physics speed) The maximum height Lf a? Ans. (6) attained by the ball is xo Given, rag force. F =e? o (2020,4 867 9H acy, option is comrect. |Aswe know, generalequationof force ae 1, Me : =o (iy a) tan to) ure } ==; ; stationary interplanetary dust.AS@ per etardation ofthe ball when ‘aut its: nee Increases at a rate thrown vertically upwards: abv Ue where wists oue-lgevia® instantaneous velocity. The el dt cae instantaneous acceleration of the grr =n satellite is (2020, § sep shit-M ‘By integrating both sides of Eq. i) in ta) -bv'tt () knowa limits. HCO) When the bal thrown upward with gt yb! velocity vg and then reaches tots ta a) Zen zformasimanbeit atime ‘Ans. () ae Given that, mass inereases atthe rate, Mt pvt) at ‘Now. trom Newton's second aw, ab Fore t=g+hi Fy resistance of 2.5x10-*N, the T7 Aba is thrown upward with an : acamen 9+ HOF =t >t Illcyeg ome sutoce ng ther ah wal® rai eal i veg of the earth. The motion of the ball fess ore Yaa nati essues eth ye isaffected by adragtorceequalto (5, eae i ‘my? (where, mis mass of the bal, a pete {A ane Siew? visits instantaneous velocity andy ang. gy S vtae-tnt isa constant), Time taken by the erceareen Albena gts Certo ree tote zenith is Given, reistance tere byte wa var ate. Apel BE) aF=-25x107N = te 1 Bi So, deacceleration of bul (eed (oh pest v0 pak PSE? Sigg? Miata i f 4 ar wo A prigh-teg state (o) on i.) fem=209=20%10"ka) ct Sa Lint [2 = Hed tng + a?) sesin'{ Uy 8 ae ee wai) enna? 10-2 fox ims" ands =20em=20 10°? m) sate] pet)Laws of Motion 13 A man(mass = 50g) and his son (mass =204g) are standing on a frictionless surface facing each jother. The man pushes his son, so that he starts moving ata speed of 0.70 ms “ with respect to the man. The speed of the man with respect. to the surface is 12010, 12 Api shit 2) 0.28ms (0) 0.20ms~* (c) 0.47 ms" (4) 0.14 ms" Ans. (6) The gven situation can be shown as. Before colision mns0 i me=20 S = aL 2 An Ty en Son ‘Alter colision Using momentum conservation law TotaimomentUMioe ese =(Total moment ann (rn x0+ (m, xOl=my,em vs Demeuli+ mi ‘Again, rolative veloci But trom figure, relative veloc Wty #07 Fram Eas.{ibana(il we get vi+2.51,=07 = w(5}=07 yal 44 A ballis thrown vertically up (taken {as + Z-axis) from the ground) The correct momentum-height(p-n) diagram is (co) h(a) 12019, 9 Apri site, Ans. 0) Whena tli thrown vertically uward then the acceleration of the bal a acceleration duet gravity (giactng inthe downward direction} Now. using tn equation of motion, av? —29h eed % Aswe know, momentum, p= mv or vepim So, substituting the value of vin Eq). we get or 0 pectoiny a Age kro tat. a the maximum neigt oct fhe bel own wou bowers $0, torte tight when theta isthrown tie reschs ine acm eign ‘Y-vchanges ttm vtoO = p-venanges tom muted Sir. nent recher isi poe, ten 1s hanges omg 0 Aso, changes om Of some aes Thus. these conlton ee ony aise Ine lt grenimapionta 45 Two forces P and 0 of magnitude 2F and 3F, respectively.are at an angle @ with each other. Ifthe force is doubled, then their resultant also gets double(d) Then, the angle eis 12019, 10 Jon Shite) (20° (w)120" (c)30° (aja0° Ans. (0) Resultant force , of any two forces, (ie. PlandF; (ie. O)with an angie between them can be given by vector addition as 45 FF, c0s8 (i) Infirst caseF, =2 andF, =3F FP 49? + 2x2xIF? cost IS? + 12F* cos “ay Insecond case, =2F andF, =6F (Force 0 gets doubled) ang Far, (Given) By putting these values inEq.(ik we get (ar, =(0FF + FP + 2x2x6F? cose = EF 240F? +206 cose ay From Eq. (i)and Eq. i, we get: SOF? + 496? cosB=A0F* + 24F* cos = 12+2hcos@=00rcos0=— 1/2 or @=120" —_reosi20°=— 1/2) 16 A particle of mass mis at rest at the origin at timet =0.Itis subjected toa force F(t)= Fe in the xdirection. Its speed vit)is depicted by which of the following curves? & TaleeE 2012] (al (o) (c) (a vo = Ans. (0 As the force is exponentially decreasing, soits acceleration, ie, rate of ncrease ‘of velocity wil decrease with time. Thus. the graph of velocity willbe an increasing curl with decreasing slope with time.me 47 The figure shows the position-time {x-t)graph of one-dimensional motion of a body of mass 0.4 kg. ‘The magnitude of each impulse is q {AIEEE 2010) For AB, ty ° 368 10 12 14 16 ys (2) O4N-s (b) 08 N-s (c) LNs (a) 0.2N-s Ans. () From the graph. it is astraight ine, 0 Uniform motion, Because of impulse sirection of velacity changes 2s canbe seen from the slope of the graph, theorem 4B A ball of mass 0.2 kgis thrown vertically upwards by applying 2 torce by hardd) f the hand moves (0.2m while applying the force and the ball goes upto 2m height further, hon, for PA Fromabove equations, 2(ng)=22N -tternate Solution Using work-energy ng x2.2+F x02=0 49 A player caught a cricket ball of ‘mass 150 g moving at a rate of 20 mis. If the catching process is completed in 0.15, the force of the blow exerted by the ball on the hand of the player is equal to JEE Main Chapterwise Topiowise Physics | ‘O} ‘Ol Va! +20x02 20 A particle of mass 0.3%g 5 subjected to a force F =~ kx with k=15Nm |, What willbe its iitiat acceleration, if itis released from a point 20 cm away fram the origin? {alee 2005) (a) 3ms? () 1eme* (c) ms? (a) ms? Ans. (d) Oni 292 ea2gx2 Negative sign indicates that acceleration E is always towards the mean pasition. «Initial acceleration, 2A A machine gun fires a bullet of mass 40 gwith a velocity 1200 ms". The rman holding it, can exert a maximum force of 144 Non the gun. How re many bullets can he fire per second = 0 o=109=100ms at the most? (aieee 20047 ‘Then, for PA.FBDOt balls al (bo) 4 F-m=mo (oh2 (3 [isthe force exerted by hand on ball Ans. (d) (go) Force exerted by machine gun on man's hand in firing abullet= Change in momentum per second on a bullet oF rate of change of momentum -(ia)* 1200=48N, ‘000. Wg + We =O =22N Force exerted by man on machine gun Force exerted on man by machine gun 144N Hence, numberof bullets field) = 44 22 A rocket with a lift-off mass 35x10" kgis blasted upwards with find the magnitude of the force. [AIEEE 2006] ‘an initial acceleration of 10 ms“*. Considerg=l0rm/s®, (AIEEE 2006) (a) ISON (oan ‘Then, the initial thrust ofthe blast {a)4N-(B) IBN (0) 20N (0) 22N (e)30N (a) 300n is {AIEEE 20031 Ans. (6) Ans. (6) (a) 35x10°N—(b) 20x10" The stuationis shown nfigue. tint! _Thisisthe question based on mpulse- fq WOXIOEN — (e) L78X108N time, the bal ist then under the ‘momentum theoren. hee detien of force exerted by bana trom P IF -Atl= [Change in momentum + (0) Tonandinentiomato Bietacceleration =F xON=I~P| Hie. ting ores reaper ‘accelerate the rocket, so initial thrust of ‘af bat during PA beams (assumed to ‘asthe ball will top after catching eee be constant) in upward direction and p= nw, =015%20=5.) a . 7 velocity of ballat be sym, a Reoiwdian Fe ma=35x 10" x1=55x10°NLows of Motion 23 Two forces are such that the sum of their magnitudes is 18 N and. their resultant which has magnitude 12 N, is perpendicular to ‘he smatier force. Then, the ‘magnitudes of the forces are (AlgEE 2002) (W) ISN5N (a)6N.2N The sum of the two forces. AsBe08 “ 2AB Cos0 i) i) Solving Eqs. (iL idanati we get A=5NB=13N Since, 4+8=18 andcose=-4 8 Squaring both sides in Ea, (iL we get lehea + B+ 288 cos On puting cost» in above equation we get tebma? + GF + 200(-4 waaa + 28( 3) nae Bt 2a = Nee BF a? = TW4=(B~A)(8 + 4) On putting 8 + A= 18. we have a-a 18 Solving B-A=8 B+ A= 18 We have, B= 15NA=5N TOPIC 2 Equilibrium of a Particle and Common Forces in Mechanics ‘24 An object of mass mis being ‘moved with a constant velocity under the action of an applied force of 2'N along a frictionless surface with following surface profile. (202%, 1 Sep Shit) The correct applied force versus distance graph will be F tb) UJ ? oN 47 F 2No(--ve) constant hing the upward tation, the force Positive constant ant during te downward motion the force is tegative ‘constant Hence, tha corroct guaph ie tbh ant aN —— 25 A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is. Given, m=8kg, M=16kg] ‘Assume all the surfaces shown in the figure to be frictionless, 12021,1 Sep shit) @ z 2, Ans. (b) Lots draw the free body diagram, (During the upward direction) F =2N-=(+ ve)constant {During the downward direction) Fa2N 4 6 3 2 (alta wi8q tcl (aja Eo tela Ualeg Ans. (d) Here, both block and wedge are moving. Consider the acceleration of the block with respect to the wedge isa, and the acceleration of the wedge isa, ‘Given, mass of the wedge, M= 18g ‘and mass of block. m= 84g Let's draw the free body alagrarn of the wouige, Hl we i x, A Nc0s60"= Ma, MO0.51= 160, ” Ne=32m, Now, draw the free body diagram of the block with respect to the wedge, 08 60°* Poeudo foson=7r | SOS mgsnsee een PN ‘Along the inclined plane. ‘mg sin30° + ma, coss0"= ma, Vex Bgx¥ gx! + 8x¥5 gx 8 =, 9x5 8x9 29 3 ‘The acceleration ofthe block with respect to the wedge is 29/5. 26 Acar is moving on a plane inclined ‘at 30° to the horizontal with an acceleration of 10 ms~ parallel to the plane upwerd. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is (Take,g=10ms~?) (2021, 31 Aug Shift} Ans. (30) Given, Angle of inclination. @=30° Acceleration. = [Acceleration due to gravity. = T0ms ‘According to the question the car and ‘bob is as shown below. Toos a+ Here, Fis the pseudo force acting on the bob when we considered:t from car's frame and is the tension onthe string. Inequiibrium, EF, 20 and ZF, =0 JEE Main Chapterwise Topicwise Physics > Freosi0 macesst ae 1 aot eperon => masin30° + mg = MAC08! [r-using Eq. (ii) cone gt (cosa) 27 Statement | three forces FF, andF, are represented by three sides of a triangle and F,+F, =—F,, then these three forces are concurrent forces and satisfy the condition for equilibrium. ‘Statement IA triangle made up of three forcesF,,F, andFy asits sides taken in the same order, satisfy the condition for transiatory ‘equilibrium. Inthe light of the above statements, choose the most appropriate answer from the options given below. (2024, 51 Aug Shit] {a} Statement is false but statement ‘ste. {(b) Statement {is true but statement is false. (c}_ Both statement !and statement i aretalse (c) Both statement land statement Ans. (6) Three forces, F;andF, are actingon a tbody andi this body isin equilibrium, then resultant of these three forces must be 72F0 .2. Fag =F, + F + = Fi+h=-Fs This situation can be shown graphically py three concurrent forces at 120" with each others. or, by three forces inthe same order ‘lang three sides ofa triangle F Hence, both statement {and statement Hare rue BB The boxes of masses 2kg and 8 kg are connected by a massless string passing over smooth pulleys. Calculate the time taken by box of mass 8 kg to strike the ground starting from rest (Use,g=10m/s®) 12021, 27 Aug Si) (a)0.345 (v0.28 (c)0.255 (a)04s Ans. (6) cording tothe given figure and tree bady diagram masses of two bodies m, ‘oceleration of mass m, Acceleration of mass m Tensor Distance hetween m, and ground =20 om 202m Initia velocity u=0 Equation of mation of 2hg block, 1-2Laws of Motion 49 Equation motion of @ kg back 1 fim ft 34 Consider a frame that is made up 2 Vien ty of two thin massless rods ABand a 5 AC as shown in the figure. A vertical From Eas. (idand (ii) = 1+ Vin 4-4/3 : (i i 2 i se force P of magnitude 100 N is airings applied at pont A of the frame, and 8g 1 ven) Sm 4g-T=20 i) 3 = 139 ting the value ne get ee ee 39-29=20-99=20 = ae will) ee According to question, the coefficient of o,= 8m andoy tpeton between te body and plane ~& therefore, on comparingit with Eq Since. s =ut+ lot? 5 2 (ih we canwete x3 20 og 180 ee 2x4 wo"°*22"" = "=F .559 30 A steel block of 10 kg rests on a horizontal floor as shown. When > three iron cylinders are placed on it = as shown, the block and eylinders go 29 A body of mass mis launched upon down with an acceleration 0.2 m/s: 8 rough inclined plane making an The normal reaction R by the floor, angle of 30° with the horizontal if mass of the iron cylinders are i i ‘Suppose the force is P resolve Thecoeticientoffrcton between equalandof 20g each oN, paraiditothearaapenine x =10m/s? andy, = the body and plane is “ifthe Mokeg=t0m/stendus 602) thetrame: Tester an eat time of ascent is half of the time of resolved component along the descent. The value of xs... cl arm ACis XN. The value of x, to the 12021, 20 July Sita nl de nearest intege Ans. (3) (Given, sin5°)=0.575, Letus assume that fof 819,sinfi0")}=0,939, 7, be the time of ascent of body and, 7 > 0.342] be the time of descent of body. encom? 12021, 16 March shit ns. From second equation of motion we hire. wiles, ayn THO 22) have Ifthe force Pot magnitude 100 Nis seute toe ‘Ans. (6) ‘esalved paralleta the arms AB and AC 2 Thetforce equationinvertical direcionis _of the frame, the above igure During the time of ascent, the distance f19—R=Mo ‘represented as follows overeat the body, iF A sala ti 20, ‘and during the time of descent, the ‘istance covered by the body, A i f jcon e bid 55 where, M=collective mass of biock and A allthree iron cyinders wa 28in8 ug cosB pein « a = pose ms and = omponent of force along Be and R= normal reaction AC = 100 c0836"N = [esce—n9 cox Force along verticle axis My ~ R= Mo = 100x0.819N=818N = 82N Tg-R=70x02 This isthe require magnitude of the se Ftd ‘solved component along the arm AC. Compare with given in question, =700~ =50 JEE Main Chapterwise Topicwise Physics When BF, =0. then Mg =T costs “ Using Eqs. {andi we get 32 Amass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 46° with the vertical. Then, F equals (Take, g=10ms~? and the rope to be cof Tsing? 1 F Mg" Teosts® Mg = F=Mg=10x10=100N massless) {2020.7 onSNF)_34 Two fixed frictionless inclined (a) 75N_ (0) TON (6) ON (a BON TNane making the angles 30° and ne (2 60° with the vertical are shown in Given situation i as shown below. the figure. Two blocks A and Bare placed on the two planes. What is the relative vertical acceleration of ‘Awith respect to8? [AIEEE 2010) (a) 4.9ms" in horizontal direction (o) 9.8ms" in vertical direction Tos 45° resisé \We resolve tensionT in sting into vertical and horizontal components. For equibrium. and Me Acceleration of system.a = men oe 0, foree acting on mas, me meh Femo= 36 A block is kept ona frictionless inclined surface with angle of inclination a. The incline is given an acceleration a to keep the block stationary. Then, is equal to TAIEEE 2005) a (0) gosee ‘ana (9 (0) gtana Ans. (d) Oniviing Ea by Ea. (HL we get (c) zer0 Inthe trame of wedge, the force diagram Fs taniseorF = Mg =10%10=100N (a) 4 3ms” in vertical direction (of block is shown in figure. From free. Mg a ‘body diagram of wedge 33 Amass of 10kgis suspended vertically by a rope from the roof. ‘When a horizontal force is applied on the mass, the rope deviated at an angle of 45° at the roof point. If the Force applying onthe block where, ais along the inclined plane. Vertical component of accelerations suspended massis at equilibrium, jane ‘the magnitude of the force applied is Relative vertical acceleration of Awith (Take, g =10ms~*) respect to Bis Forbiocktoremsin stationary, (2019, 9 Jan Shit) alsin? 60° si? 30°]=9.= 49s” ie ceaarasaa (a)70N (b}200 N 2 or = g tar (c)100N (a)40N Linvertical direction) oon Ane t ‘35 Ablock of mass mis connected to 7 Tiks masse = 619 end jven systemis follow : ocleecrwes another block of massM by a spring y (massless) of spring constant k The blocks are kept ona smooth im, =4.8kq tied toast hanging over a light frictionless pulley. What is the acceleration of the masses when lft is free to Nig = 10 108 LLetT stension in the rope. For equilitvium condition of the mass, EF, =O{force in xdirection} EF, =01 force indirection) When ZF, =0,then sings “ horizontal plane. Intialy the blocks are at rest and the spring is unstretched. Then, a constant force F starts acting on the block ‘of mass? to pullit. Find the force on move? (g=9.8ms *) (AlgEE 2004) the block of mass m. [AlgEE 2007), me He me a) OE (p) Meme salt) On MF en (m+)Laws of Motion a) 02 (0) 28ms (c) 5ms (a) 48 ms. Ans. (@) ‘elvasing the motion af the system wilbe according to figure mo-T=ma 0 wy iy ‘38 A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the iftis stationary. If the lift moves downward with an acceleration of § ms~?, the reading of the spring balance will be AIEEE 2005) (a) 24N (D)TAN (c) EN (d} 4ON Ans. (a) \n stationary position, spring balance reading=mg=49 8 oskg do Whon ttt moves downward, mg-T=ma Reading of balance T=mg-mo 5198-5) =5x4.8=240N 39 Three forces start acting simultaneously on a particle moving with velocity v. These forces are represented in magnitude and direction by the three sides of a A ABClas shown) The particle will now Move with velocity [AIEEE 2003) ¢ a 8 (a} less than (b) greater thany {c) ivlinthe direction of largest force @C (4) v. remaining unchanged Ans. (d) Resultant force is zero, as three forces ‘acting on the particle can be ‘represented in magnitude ané direction by thee sides ofa triangle in same order. Hence, by Newton's 2nd law fi 2) weve willbesame, 0 A block of mass Mis pulled along a horizontal frictionless surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block's {alese 2003) em em a i), Om tam eM tor 0 oe Ans. (f Let acceleration of system {rope + block} bea alang the direction of applied force. Then, Pp Fretonless suface 51 Draw the FBD of block and cope a shown in figure. whe Toop where, Tis the required parameter Forblock —T=Ma Me tem 7 A tight spring balance hangs from the hook of the other light spring balance and a block of mass M kg hangs from the former one. Then, the true statement about the scale reading is {AIEEE 2003), {a} Both the scales readMtkg each {b) The scale of the lower one reads kg {and of the upper one zero (6) The reading of the two scales can be anything but the sum ofthe readings willbettk (0) Both the scales read? kg Ans. (0) The arrangements shawn n figure. ‘Now, draw the free body diagram of the spring balances and biack , B Uart sping balance ahi spring S216) oalance For equilibrium of block, T= Mg where, T,=Readingof S, For equilibrium of ST; =7; ‘where, T, =Reading of S, For equilibrium of, = Ty Te 1 5 Se |B} Bloc Mg Te 1 Hence. So, both scales read Mka,52 2 A litt is maving down with acceleration a. A man in the lft ‘drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively {ateee 2002) (a9 (0) 9-a.9-0 (e)g-0.9 (a)a.g Ans. ( Apparent weight of bal wi =w — Rlwnere R=ma yormal reaction) {acting upward) nig ~2) Hence, apparent acceleration in the ift isq—a. Now, f themanis standing stationary on the ground, then the ‘apparent acceleration of the aling ball 's9. mgm 3 wen forces F;, F,.F are acting on ‘particle of mass m such that F,and F, are mutually perpendicular, then the particle remains stationary. I the force Fis ‘now removed, then the acceleration ofthe particles (AIEEE 20021 wh 28 @ Sa mor, m Ans. 6) “The particle remains stationary under the acting of three forcesF,,F, andFyit ‘means resultant force is 220 a) ‘since, in second casesFisremoved(in terms of magnitude we are takingnowh the forces acting areF, andthe resultant of which has the magnitude 2s FF, so acceleration of particleisinthe direction opposite to that ofF. A Aight string passing over a smooth light pulley connects two blocks of masses mand mz (vertically). If the acceleration of the system is gB, then the ratio of the masses is {[AleEE 20021 (a8 (0) 9:7 (43 (a) 5:3 Ans. () asthe stringie inextensible, both masses have the same accelerationa iso, the pulley is massless and frictionless, nence the tension at both tends of the strings the same. Suppose, JEE Main Chapterwise Topiowise Physics the mass mis greater than mass m0 the heavier mass m is accelerating downward and the lighter mass m, faccelerating upwards. rte al mg ma Theretare,by Newtons Znaiaw T=mg=me 0 and mg-t=me id ter slving Eas. (bani we get (m-m) og aaa gal [given] im+m) 8 sa {t=mum) = | «go t a7 milemim) - ta Bex ™ Thus, Ea. (iilbecomes tex) ier 8 So. the ratio ofthe masses is@:7. ‘45 Three identical blocks of masses kg are drawn by a force F =10.2N with an acceleration of 0.6 ms~? ona frictionless surface, then what is the tension (in Nin the ‘tring between the blocks Band C? AIEEE 2002] ss (a) 92 (b) 7.8 ge es poe. Te ty F From the figure, F-T=mo i} mo fi and Ea. ibis, 402-1)=2%08 = 1 st02-12=8N gain. trom 6 One end of massless rope, which passes over a massless and frictionless pulley? is tied toa hook Cwhile the other endis free. Mayimum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in ms~*) cana man of 60 kg climb (on the rope? (areEE 2002) ¢, fais (we ee (a 80 Ans. (o) The free body diagram of the person can ve crannas : mo nu eo Lexine person move upwithan acceleration a.then 1 ~809 =800 2 380-609 _, ve value ‘That means. its not possible to climb up conte rope. Evenin this problem, itis not possible to remain at rest on rope. Hence, ne option is core But. f they willask for the acceleration of climbing down. th T OF OnaLaws of Motion TOPIC 3 Distance covered by the blockis same Friction forboth the case, - 90-5 pare 47 When a body slides down trom rest janes WEead along a smooth inclined plane [from Eq.{i)& Ea. ii) ‘making an angle of 30° with the orizontal,it takes time T. When . the same body slides down from the rest along a rough inclined Comparing witha. plane making the same angle and through the same distance, it takes The valueaf hex time eT, where cris a constant — Greater than. The coefficient of 4B The coefficient of static friction friction between the body and the between two blocks is 0.5 and the ip L(#-1 table is smooth. The maximum Tough plane is—-|——z-}wherex —forizantal foree thet eo be applied to move the blocks together is eves (202K1Sepsnitemy (Take, g=10ms~?) Ans. (3) 12021, 26 Aug Shit Let's craw the fee bady diagram when Teuie Ta] w-05 ody slides down on smooth surface ag Lr a TT Ans. (15) tay Given, coetfcient of static fiction Se : y=a5 Value of acceleration due to gravity, For smoath surface, a=l0ms ‘mo=mg sing0® Fer compete system ta move together, 9sin30= 9/2 Pees Distance covered by the block on the ‘smooth surfacein timeT, saute lot? 2 Here, mis total mass, Fa(le2o From free bady diagram of 1kg block @ vty = - ths ‘Now. let's craw the free body diagram when body slides down on rough surface i) ur oe Ng wmg cos 20° Balance frees in horizontal direction Faye mta ‘i OX Balance forces in vertical direction, rans cnc remg | Put value of Ring. i) For rough surface ma =mg sin30!—n mgcosso# - aarp Putte value ofainEa.fi we get Distance covered by the blockon the Fa3x5=I6N rough surface ntimea, Thus the masimum horizontal free eaten required to move block together is 5 sutra _ Hi : 49 Two blocks A andBof masses 520+ ligsins0t—p o sino" ae Cero im, =1kg and mp =3kg are kept on =2()-Smtar? i) the table as shown in figure, The t Coefficient of friction between A 53 and Bis 0.2 and between Band the surface of the table is also 0.2. The maximum force F that can be applied on Bharizontally, so that the block Adoes not slide over the block Bis [Take, g=10m/s?] [201,10 Apri! Shitty Lal abr (NN (DIN (18N (4) 40N Ans. (0) ‘Acceleration of system of blocks A and eis Net force Total mass +m, where, =triction between Band the mM Substituting the above values in Eq.) wea i) Ove to acceleration of block 8, apseudo force Facts on A This force "is given by Fame where, ais acceleration of Aand 8 caused by net torce acting on B. For itt slide over &: pseudo force on ie F’ must be greater than fiction between A ana 8 = moze We consider timiting case, =ma=nimg > a=4g=02x10=2ms? fi) Potting the value ofa from Ea. iiinto Egil, we get Fo solid cylinder of mass mis wrapped with an inextensibie light String and, is placed on a rough inclined plane as shown in the figure. The frictional force acting54 JEE Main Chapterwise Topicwise Physics Ans. (5) Ans. (30) jeen the cylinder and the inclined plane is Given, mass ofthe body. m= hg The free body diagram forthe wooden backis shown below CCoottoient of static friction, gt = WS ae an cts draw the tree body diagram of the ff = af al | sin@ = —T | y Fane . + CY | en L co] a : teenie ho uta, Inthe xdirection, the summation ofall the forces isto be zero, (The coefficient of static friction, ™? c.|80.4) (202418 Moreh ShiR-I]_——_Using the condition of the equilibrium ne ailme wisme (2 (wd Ine-direction T205R 4 F cost=f=O=HN “a Inthe y-direction, , the summation of all ee Iny-direction the orcesis tobe zero Letra ee boy daa! ee wo paper “ Nemg-F snd = 05R4R-00=0 = R=60N Hence the roxmalforee on the wooden Substituting te vale ofWin Ea Hesee. thane met ei Usama =e are 7-=08160)=30N Sanaa Jinx” perce amar! tension in the rope, so that wooden Substituting the vauesin the above eee block will not move is SON. ‘equation, we get 1 = 4" 53 Twoblocks(m=0.5kgandM =4.5 4g) Fells = FaEN are arranged on a horizontal re 1 ) frictionless table as shown in \ Ws figure. The coefficient of static Using the condition ofthe equilibrium of the olinder: inthe direction of inclined plane, T+f—mg sin60"=0 = inthe perpendicular drectionof inclined A boy of mass 4 kgis standing on 2 Hence, the body move by applying ion between the two blocks is minimum possible force of BN. So, the 3/7. Then, the maximum horizontal whist wade’. force that can be applied on the larger block so that the blocks move together iS... Ne i sande W-mgcossot=0 Ta et igen (Rout sth erent mn Name the wood and the floor is 0.5, the eke, gam") twine “nen ema —— txert onthe rope, so thatthe piece ol, wo fabnmg cost = fading of wood does not move from its u i) place is .. N. (Round off to. [200,17 Moreh Shit] a = the nearest integer) ‘him. : _ (Take,g=10ms™) » 51 A body of mass 1kg rests ona (2021, 17 Moreh Shift) ‘When both the blocks move together 2s horzontal oor with which thas @ asam tenasceratonot hs coefficient of static friction 1/V/3. It is desired to make the body move by applying the minimum po ssible force F newton. The value of F will ti) o= mer Frictional force on mass. tema i be came rom Eas.Jandl we get (Round off tothe nearest integer) Frensham woe (Teke,g=10ms"*) (sen) [2021, 17 Moreh Shift]Laws of Motion © slipping. sump {ew being the tic Friction) 54 A block of mass m slides along a floor, while a force of magnitude F is applied toit at an angle@as shown in figure. The coefficient of kinetic friction is, . Then, the block’ acceleration ais given by (gis acceleration due to gravity) (2021, 16 March Shift) nF ‘The lagram andthe required components ot force on given black are shown below Fsne tn UF tn cose _ fr “| te From the above diagram, N=mg-Fsing iv where, N-=narmal force: and Fcost-f, =ma = F cos. where. f, =kinetic friction force. From q.(iband Ea.{il we get F cost —p,lma~F sin = os Eon-u,(9-L sn) oO) Thisis the required acceleration of the Dock. ‘As shown in the figure, a block of mass v3 kgis kept on a horizontal rough surtace of coefficient of friction 1/ 3y3. The critical force to be applied on the vertical surface {as shown at an angle 60° with horizontal such that it does not ‘move, will beSx. The value of x wll 3 be... ; Ans. (3.33) Given, mass of black, m Coetticiont of fretion According to diagram, LetF be the force applied on the body, we the weight =a) Nbe the normal reaction, Friction forcef =yN For no movement of body along X-axis, net force along X-axis should be zero, fF, be the net force along y-axis then it willalso be zero because body isnot accelerating ata. sin80°+ mg = nates ios ) 2 Similarly. F, =F cos60°—yN From Eq.(il we get ~ & aa(Sevms} Plz FLW FF 0 Ses TEs = Fen Gren, Fade = x2Be333 3 ‘An inclined plane is bent in such a way thatthe vertical cross-section is given by y=* where, y isin 4 vertical and x in horizontal direction. If the upper surface of this curved plane is rough with coetticient of friction, 0.5, the ‘maximum height in om at which a stationary block will not slip, downward is... M. (2021, 24 Feb shift) Ans. (25) The graph forgiven equation is shown below yin) x(n) ‘Atmaximum height, the slope of tangent drawn, onde eX [oye 42 4 = os=% (= tare) 3 a 57 The coefficient of stati friction between a wooden block of mass 0.5 kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall Will DE Ne (Take, g=10 ms~?) (2021, 24 Feb shit] Ans. (25) Given, coefficient of tate friction, wy =02 Force 2702 \Various forces acting on block below Frictional force < mg 3Nx0255 = NS Magnitude of horizontal force, FeN=25N,56 JEE Main Chapterwise Topicwise Physics a catroughwthecooticent at Yoga rn at hao Ls fhetiong. Retauncthemetiock — TSR es comes to rest as it reaches the mgs useoney | 5 uss bottom (point A) of the inclined = oe plane. If C= 2AC, the coefficient 4 of friction is given byt = k tan®. Consider a uniform cubical box of + 0y=545VSH N19 = 10ms *) Let s be the distance between A anato) side aonarough floor that istobe _—_‘Then. thevalue of kis... - moved by applying minimum [2020.2 Gap ene] Fromthird equation of motion, possible force Fatapointbabove ANS. 3) Ye-0=208 its centre of mass(see figure}. If Ditterent forces acting on the incined = =, “ the coefficient of frictionis=0.4 —_‘Paneareshown below 2a, 5+ 65H) the maximum possible value ot _Fewmaros ® {Using Eat When the block comes backtoits initial a nme mosn® S16 position(e, romB tol its velo before moving is smoot, m9 2 So, the acceleration of the block while iaza 7 oonsnnen—ychand Sache om a ese ‘a, = gsin30°-pg coss0° against friction. ‘Again, using third equation of motion, we ; ain (tJ -a-me Some ‘toppling. - . &, Coe eee BW avlocksttsmovingypon in {Using Ea. ill Equating Eqs tidandtiv we get 4. 4 e+ Svan) 5S) = 45-55 coefficient of kinetic friction = 20-20V5p =5+5v5u between the block and the inclined = plane i close to the nearest wit velocity“. The value ofthe = Fee (3) 2 Using resut of Eat we et ) +n Integer £0118 an ‘ot0,3 ep shin so, i Ans. (246) Ley be the coetcant of kinetic - 3481000 =346 thelonbetweentheblockandthe gy . fomes pene 1 An insect is atthe bottom of a ‘The free-body diagram of the given hemispherical ditch of radius 1m. It uations erow belo. Craw up the ditch but starts es slipping ater itis at height hfrom ASS the bottom. If the coefficient of - ge theton between the ground end ‘Asmail block starts slipping down P the insect is 0.75, then his{Take, from a point Bon an inclined plane Yo, 1oms~*) 2020, 6 Sep Shift-t AB which s making an angle® ie ! Mien. with the herizontal section BCis (e)08om (4) 0.80m ‘smooth and the remaining sectionLaws of Motion Ans. (a) LLethbe maximum height up to which insect crawls up the ditch The free body ‘iagramis shown, Resolving the components of force slong tangential and radial direction " mash ng cose For balancing, mg cos8 = N 193i fy HN = gsind=ymg cose £08 yng cost” mg S = cos=4 O75=5 > coeds (From Pythagoras theorem) From diagram % ort 62 A block of mass § kg is(i) pushed in case (A) and (i) pulled in case (B), bya force F =20N, making an angle of 30° withthe horizontal, as shown in the figures. The coefficient of friction between the block, the floor is =02. The difference between the accelerations of the block, in case (8)and case (A) will be (Take, g=10ms) 12019, 12 api shite a N @ (a}0.4ms? ——(b)3.2ms"* (c}oams® (gamer? Ans. () Case | Block is pushed over eurtace Jao Free body diagram of bio " ~e Ichis Foun 0 mg F F sina Inthis case, normal reaction, Ne=mg + FsinB0° =5x10+20%/<60N {Given, m= 5kg,F=20N] Force of function, =n 2x60 [:=0.2) =2N So, net force causing acceleration(a,)is Fea = M0, =F cosS0°—f Case Block's puled over the surtace Finger i =e Free body diagram of boc okie, Pan. }—* F cos sor io Net torce causing acceler Ft =F cosso*— uN rationis =F coss0* = Frat =F 20850" img -F sin30°) HWaceelerationis nowa,, then a= fat F coss0°. ulm 57 vs ' x ~02(5x 10-20% OF: ( nm) 0.8ms —+ Fcos3yr 63 Ablock of mass 10 kgis Kept on a rough inclined plane as shown in the figure. A force of 3 N is applied ‘on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force F, ‘such that the block does not move downward ?(Take, g=10 ms~2) Me (30 25N an tan Ane. Free body diagram. forthe given figure is astoliows, Forthe black tobe in equlibrium i.e. so that it does not mave downward, then %,=0 34.Mg sind ~F - or 34-Mgsino=F + As. frictional force, f 3+ Mg sino. “0 Similarly x, Mg cos8+ R=0 or Mg cos8=R i) Substituting thevalue of ® from Eq.i)to Fa.(i. we get 34 Mg sind=F + u(Mg cosa.) Here, M=10ka, and W=0658 Substituting these valves is a. (Hi. weget 34+{10 «10 sine} = (08x 10x 10 08 45" 1, 100_ 605, 40 3420 /2=31.8N of F=32N 84 Ablock kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2.N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10'N. The coefficient of static friction between the block and the plane is. (Take, g=10m/s*) (2019, #2 Jon Shit : Ans. (b) Block does not mave uptoa maximum ‘applied force of 2N down the inclined plane. So, equating forces, wehave 2+ mgsind=f or 2+ mgsind=nmg cos ll Similarly, black also does not move upto ‘amaximum applied force of 1) Nup the plane. 10N Now, equating forces, we have ‘mgsind+ f =10N or mgsin8+ wme cos = 10 ov, saving Eqs.tilandid we get si JEE Main Chapterwise Topicwise Physics uid (iv) mg sind and wimg cos8 = Dividing, Eqs. (i and\iv)we get 5 cot e 2 Stand _3tand0" a 2 Two masses m,=5kg and m, =10 kg connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.18. The minimum weight m that should be put on top (of m, to stop the motion is (DEE Main 2018) mg (ay18.5 kg (0)273k9 (c)433k9 (a)10.3kg Ans. (b) Motion stops when pull due tom, force of friction between mand, and surface. = mgsulm + mig = Bx 10S 0I8{10-4 mx 10 = m2 28359 Here, nearest value is 27.3kg 80. May #27349) {66 A point particle of mass m, moves along the uniformly rough track POR as shown in the figure. The coefficient of friction between the particle and the rough track equals 11. The particle is released, from rest, from the point P and it comes torest at a point R. The energies, lost by the ball, over the parts, PO ‘and OR, of the track, are equal to each other, and no energy is lost when particle changes direction from PO to OR. The values of the coefficient of friction and the distance a= OR) are respectively close to [DEE Moin 2016) aa ram tl < vores —+0 soe (e)azena6sm (002 an055m {c)0.29and3.5m (0)0.29and6.5m Ans. (0) Energy lost over path PO =p mgcox4 OBE mG——F Eneray lost over path OF =n max ile. wmg cos30°x4=ymgx (+0=30°) x 45m From0 toR energy lossis halt of the total energy loss. Vemagh = 20 eum The values ofthe coefficient of friction and the distance x= ORlare 0.28 and &7 Given in the figure are two blocks A and B of weight 20 N and 100 N respectively. These are being pressed against a wall bya force F ‘as shown in figure. Ifthe coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall in block Bis (JE Main 2015) oy 20N 100 (a)i00N (o)80N Ans. (¢) Key idea tn vertical rection weights rebalanced by frictional forces. Consider F80 of block Aand Bas shown indiagram below. By 1 T 20N 7, 100N ‘asthe blocks are in equitrium. balance forces are in horizontal and vertical direction. [e)0N (@)150NLaws of Motion For tho system of boc he(A + A N Forbiook ay !ONand tor block fh to 1006 1208 88 A block of mass mis placed ona surface with a vertical Cross-section given by y=x° /6. If the coefficient of friction is 0.5, the maximum height above the ground: at which the block can be placed without slipping is (Ee Main 20% ‘im awi2m tedtm (a)! Kalam (oem fehtm ta! Ans. (0) block of mass m is placed on a surface with a vertical eross-section, then f *) tana dh LE oo At limiting equilibrium, we get W=tan0, O5=¥ 12 = Pale yet! Now, putting the value of riny = we get When x21 ot Go. the maximum elgh above ha Ground anich ine ocKeanbe paced suet ppg im 69 The minimum force required to start pushing a body up a rough (frictional coefficient) inclined plane is F, while the minimum, force needed to prevent it from sliding down is F>, If the inclined plane makes an angle @ from the horizontal such that tan=2y, then theratioF,/F,/s (are 20) (aye be)? das Ans. (d) = mg (sind cos as body justin Position to move up, tretion force downward) natin ost boy i toad down, tet A Fy int yee Fin pect Lond Ay ton en 70 Consider a car moving on a straight ‘oad with a spoed of 100 ms '. The distance at which car can be stopped. is [py 0.5] ware 2008) (0) 800m (0) 1000 (6) 100m (0) 00m Ans. (b) From Newton's equations, we have Vu! =a Given, ved Lear is stopped} otandation As fiction provide the o=ta.v = 1001s M00¥ = 2 gs 1001 100 100 «100, 240610 5x2 000m 71 Asmooth block is released at rest ‘ona 46° incline and then slides a distance d. The time taken to slide isntimes as much to slide on ough incline than on a smooth incline. The coefficient of friction is. (on, 7 (c)py e hy r) a sai ‘ vinloytt 0 } > nas 7 ih ret 6 68 tin y= 48 Wy go Frome i a). 06 get lost lat ae + attentingt —tetpane 48-9 C080, a ase on PON yg contin aint ” @ ' o wet 72 The upper half of an inclined plane with inclination 6s perfectly ‘smooth, while the lower half is rough. A body starting trom rest at the top will again come to rest at the bottom, if the coefficient of friction for the lower half is given by (wieee 2005) (a) 20in@ (b) 2009 ; {c)2tan@ (a) tang . Ane. (¢ According to work nary theorem, Workdone = Change kn enagy Weak =0 + Work one by ion + Wor done by very =o > sum cand! + ntsnge or } conga sin PF conga sing or weztang 73 A block rests on a rough inclined plane making an angle of 50° with the horizontal, The coefficient of Static fiction between the block ‘and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block (in kg) is(g = 10 m/s”) {tet 2004) (0) 4.0 (25 (a) 20 (ch 16 Ans. (o) Lot mass of the block bem.TOPIC 4 » \ Dynamics of Circular Motion nen aa EN gn 76 Apatcl of mass mis suspended aa odie _— from a ceiling through a string of Frictional force in rest position length L. The particle moves in a Fm mg sins0' horizontal circle of radius r such {thiisstatictictional force andmaybe that r=. The speed of particle Sos tan te king iorlrce] a Pada wile aor 260 98 tolyro oie 2x0 a omen me 0 KO teaig ‘ay z FA Ahorizontal force of 10 N is Ane. (a) NT A ex Cer ad ct horton ee stationary against a wal The en cootficient of friction between the ydyagnahanetatil, wile block and the wall is 0.2. The weight of the block is [AIEEE 2003), f suspended froma ceiling is shown, 10N. a) 20N {b) SON (c) 1OON (a) 2 Ans. (d) Let Re the normal contact force by wall cn the Block, i 40 i 108 A Divide Ea. (iby Ea. ib tand=v 19 w = v= yearend i) Now. fom figure, we can write few and (,=aR LE Ase aes 2x10=2N sine ST = ma Abs 5 A marble block of mass 2 kg ving Subatuting te va fn a. ‘on ice when given a velocity of 8 a oe ms" is stopped by friction in 10s. ve Veg tant = Vig Then, the coetticient of friction is ‘Thus, the value of speed of particle is [AIEEE 2003] vi. (2) 0.02 (8) 0.03 (c) 006 (41 0.01 77 Consider a binary star system of star A and star B with masses m, retardation wil bey g.From equation of rel mig revolting va chalet oct poicker aan of radiir, andr, respectively. fT, “ Betewent ‘andT» are the time period of star A : ye cone ‘and star 8 respectively, then ‘66 (2021, 20 ly Sit) Ans. (c) Let coetficiet of friction bep. then JEE Main Chapterwise Topicwise Physics ole (i Wit =Ty Aci >Tylitma> 4) (lg >Tylts> fg) Ans. (0) ‘As per question, abinary star system of two stars Aand B with masses m, and m, are revolving ina circular orbit of radi ‘andy respectively means both stars Aand B wilhave same angular velocity because to remain perfectly aign w.rt. each other they need to cover equal angular Aisplacement in equa time intervals. ‘so, we know that Time perioa,T =2® Tan, are the time periods of stars ‘and B respectively, therefore, we can Trott “0 on “From Eqs.(VJand\ill we can say tly and 0 ‘The normal reaction N for a vehicle ‘of 800 kg mass, negotiating a turn ‘on a 30° banked road at maximum possible speed without skidding is 108 kg-m/s*, (Take... =02) (2021, 20 duly Shit} (o102 (WI72 (e124 (6698 Ans. (0) The given situation can be represented as Equating forces perpendicular tothe inclined plane, N= mg c0s30° + = N=mg.coss0"Laws of Motion 3 Eauatingtorces ong ihe cine plane, img sins0°- y= sso ti On dividing Eg. Hoy Eq. il, we get =10.44x 10°V ‘Therefore, N = 10.2 10% kg-mis? ‘Amodern grand-prix v racing car of mass mis travelling on a fiat track in a circular are of radius R witha speed. If the coefficient of static friction between the tyres and the track is then the ‘magnitude of negative lift f, acting downwards on the car is (Assume forces on the four tyres. are identical andg = acceleration due to gravity) (2024, 17 March shift-1] ¢ é wrf-25+0) tf 5] a ¥ tofrgig} refers) Ans. (6) We know that, static friction force, fun vere, Is the coefficient of static friction and is the normal force acting 1e body. AS the caris traveling ona Circular track so centripetal force tsalso actingon it me Intimiting condition, ‘The magnitude of negative it f, acting downwarascn the car (=mq-Ne=n a we = Kh rod] 80 Statement | A cyclist is moving on an unbanked road with a speed of Tkmh”" and takes a sharp circular turn along a path of radius of 2m without reducing the spee(d) The Static friction coefficient is 0.2. The cyclist will not slip and pass the curve (g=9.8 m/s?) ‘Statement lf the road is banked at an angle of 45°, cyclist can cross the curve of 2m radius with the speed of 18.5kmh™ without slipping. Inthe light of the above statements, choose the correct, answer from the options below, [2021.16 March Shitt-i] {a} Statement is alse and statement ‘strue (0) Statement |istn 's false, (c} Both statemer are false. (a) Both statement | and statement It are true. Ans. () The maximum speed of cyclist on turn of ‘unbenked road without slipping is given Vax = Vi and statement jatement SOTTO =2me-! fsm=02{9\ver Given, speed= Theh 70 = 194 ms) a As given speedis lesser than vy 50 the cyclist willnot sip. Therefore, Statement lis tue, As per Statement Il, angle of banking onus (oR tare 61 and 5m b4 1000 S600 -aven speed= AS. Vpn SVVny, 50 the CYCISt sip. --Statement i v= T85km/h= isalso true Hence. option {alis the correct 81 A block of 200 g mass moves with a uniform speed in a horizontal Circular groove, with vertical side walls of radius 20 cm. Ifthe block takes 40's to complete one round, the normal force by the side walls of the groove is (2021, 16 March Shitt-t] (a)0.0314N ——(b}9.859%¢10°7 N (c)6.28%10"°N —(d}9.969%10-* N Ans. (4) ‘The normal force bythe sce walls of the groove willbe equal to the centripetal force actingon it. where, 1=200m=02m m=200g =200% 1049 mr _Pmx0.2 and v=rw. 2s a) ‘Suostitutng the given values in Ea. {ik weget t2n0x10"(2402) a2 = 9.859% 10"%N 82 A particle is moving with uniform ‘speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to. Its time period of revolution will be given by [26 Feb 2021 shit-1) (air eR? (or (cir ne? alr = Re Ans. (a) Biven, radius of circle= R Central fictitious force's. Fx 162 et Te the time period of revolution, mabe the mass and angular velocity of Earth 1 F = mai R= ‘A smooth wire of Jength 2nr is bent intoa circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating at with angular speed @ about the vertical diameter AB, as shownin figure, the bead is at rest with respect to the circular ring at position P as shown. Then, the value of «” is equal to (2019, 12 Apri shite) Oe (o)29n6-v3) (gN8Vr (d)2g/r Ans. (6) Key Idea For revolution in a ccclar path, there should bea force which balances the necessary centripetal force. Let = normal reaction of wire loop acting towards centre ae ns ‘Then, companentN cos@balances weight of bead. ~ Ncos®= mg a) ‘and.componentNsin@ provides necessary centripetal put on the bead. JEE Main Chapterwise Topicwise Physics = Nin (eo From Eqs.(ibandiih we have tana =‘ ii) 29 Now. from geometry of figure Livh foe “Ty Ay ra @* putts voue in Ei wooet @= 78 wir A A particle is moving with a uniform speed in acircular orbit of radius R in acentral force inversely proportional tothe nth power of R. If the period of rotation of the particle isT, then JE Main 2018) (a) T= R2? foranyn (re Rene (TR Ans. () {BB An annular ring with inner and outer radii and Ris rolling rout slipping with a uniform ‘angular spee(d) The ratio of the forces experienced by the two particles situated on the inner and outer parts of the ring, F,/F is thiecé 20051 to & wo) (®) R, R, a, tor on & y R, Ans. (d) Since.¢is constant, 80 no net force or torque is acting on rin, The force experienced by any particle is only along radial direction or we can say the centripetal force. (oh? 7 The force experienced by inmer part Fm, Sha the force experienced by outer part Fp=mafR, =» f= 8 shh {BE The minimum velocity (in ms") with which 2 car driver must traverse a flat curve of radius 150 m ‘and coefficient of friction 0.6 to avoid skiddingis {AIEEE 2002) (a) 60 (130 (o) 15 (25 Ans. (b) As centrifugal force is balanced by the ‘centripetal force i, frictional force. Using the relation OY ayR Remo = Meumg or ¥ - ¥ =08% 150% 10 or v=soms"