7.110 Physics for IIT-JEE: Mechanics I 116. Let F, and Fz are applied in such a way that accelerations of both masses mj and 1n2 are same both in mflgnitudc and direction. Then For Problems 111-113 a. - - -m-2g= - - - - A time varying force F = 6t - 2t 2 N at t :::; 0 starts acting on a body of mass 2 kg initially at rest, where t is in seconds. Thc In, force is withdrawn just at the instant when the body comes to rest again. We can see that at t = 0, the force F = O. Now answer b. F, Inl the following. m2 Ill. Find the duration for which the force acts onthe body. c. -FJ +m-lg=F-l +m-!l? ml mj m2 m2 a. 2s b. 3 s c. 3.5 s d. 4.5 s d. F, = F2 In! m2 112. Find the time when the velocity attained by the body is For Problems 117-119 maximum. A mass M is suspended as shown in Fig. 7.517. The system is in equilibrium. Assume pulleys to be massless. K is the force a. 2 s b. 3 s c. 3.5 s d. 4.5 s 113. Mark thc correct statement: constant of the spring. a. The velocity of the body is maximum when the force Lower support acting on the body is maximum for the first time. aiwl,w-- b. The velocity of the body becomes maximum when Fig. 7.517 the force acting on the body becomes zero again. c. When the force becomes zero again, the velocity of the body also becomes zero at that instant. d. All of Ihe above. For Problems 114-116 117. Thc extension produced in the spring is given by For the system shown in Fig. 7.516, there is no friction anywhere. Masses m I and m2 can move up or down in the slots a.4MgjK b. MgjK cut in mass M. Two non-zero horizontal forces FJ and F2 are c,2MgjK d. 3MgjK applied as shown. The pulleys are massless and frictionless. Given In] =I=- m2· 118. The net tension force acting on the lower support is a. Mg b.2Mg c.3Mg d,4Mg 119. If each of the pulleys A and B has mass M, then the net M tension force acting on the lower support (assume pulleys to be frictionless) is a. 2Mg b. 6Mg c. 3Mg d. 4Mg Fig. 7.516 For Problems 120-122 On a stationary block of mass 2 kg, a horizontal force F starts 114. According to the abovc passage, which is correct? acting at. t = 0 whose variation with time is shown in Fig. 7.518. The coefficient of friction between the block and ground is 0.5. a. It is not possible for the entire system 10 be in equi- Now answer the following questions. librium. __F __I. 2 kg b. For some values of F, and F2, it is possible that entire t system is in equilibrium, P\"\" 0.5 c. It is possible Ihal F, and F2 are applied in such a way Fig. 7.518 that m] and 1n2 remain in equilibrium but M does not. 120. The time when the acceleration of the block is zero is d. None of the above. FIN) In,115. Let F, and F2 are applied in such a way that and 20 1n2 do not move w.r.t. 'M. Then what is the magnitude of acceleration of M? Let m J > m2. ----F--~...-I. 2 kg 10 a. +(fIl, 1n2)g (Ill, ~ \"'2)g t 15 M+ml+m2 b. P 0.5 5 '0 I I(S) M (In, - m,)g F, - F, 10 - - - - - - - - - - - d. M c. Fig. 7.519 ·M +m! +m2
Newton's Law of Motion 7.111 a. at 5 s only Columnl.. CO(UnlnlI h. at 10 s only .i.$tring\\Vbreaks 1l.1;;8L=0 c. bothat5sandiOs d. at a time after I = lOs only .. U.SpringX b1'e.#& 121. The velocity of the block when first time its acceleration m. StringY. breUks .iv,.SprirrgZlJreaks becomes zero is a. 12.5 mis h. 25 mls c. 10 m./s d. none of these 2. There is no friction anywhere in the system shown in Pig 7.522. The pulley is light. The wedge is free to move on l22. The velocity of the block at I = 12 s the frictionless surface. A horizontal force F is applied on the system in such a way that m does not slide on M or both move a. 20 mls h. -12 mls c. +6 m/s d. zero J2together with some common acceleration. Given M > ITI. For Problems 123-125 ~-\"\"F A long conveyer belt moves with a constant velocity of 8 m/s. Two blocks A and B each of mass 2 kg arc placed gently on Fig. 7.522 the belt with B on A. Initial velocity of both blocks is zero. Coefficient of friction between A and belt is 0.1. There is no Match the entries of Column J with that of Column II. friction between A and B. Length of A is 4 m. 2kg @] A 2 kg I II •Column Co.b.lmn~-=~~=-~----~~--=---- Fig. 7.520 ..•.. .·.iflF i.Pseqd)i-fOfce acting a.equal t()~ . ()nJnasseen from the . m+M 123. The time when B falls off A. Initially B is on the right ........> ..ffam~of Mis· . end of A. Ignore the dimensions of B is ii'PS~lldo-force ...• acting ·.ml' a.ls b.3s c.2s d.4s h.. gre.a.t.er.t.h.an -~'.n-'.+'-.-M'-... 124. The velocity of A when B falls off A is .. QnM.as seen from the a. 2m/s b. 4m/s c. 6m/s d. 8 mls frameofmis 125. If the coefficient of friction between the block B and belt iii.• NormalMf~e (forO' c.less thanmgsinO ..,;, 45°) between In and • is 0.4, then the separation between 'the two blocks when Mis . . B comes to rest w.r.t. belt is .' Jv:.N9rm'llfor,el1et'l(ee,'.d,gn'at¢rthanl1lgsinO.... groulld ul1gMis L8m ~6m c.2m d. None of these Matching 3. In the system shown in Pig. 7.523, masses of the blocks are Column Type such that when system is released, acceleration of pulley PI is a upwards and acceleration of block 1 is.a! upwards. It is 1. In Pig. 7.521, strings, springs and the pulley are light and found that acceleration of block 3 is same as that of 1 both ideal. The system is in equilibrium with the strings taut. in magnitude and direction. Masses arc equal. Match Column I with Column II. Fig. 7.523 Fig. 7.521 a,Given that > (l > 'Q2, ' Match the . folloWlllg
7.112 Physics for HT-JEE: Mechanics I Columnl '. I ColumnII . hangs from nIl by an inextensible light string. Then match i. Acceler-at-io-n-o-f-2- a .-2-a ,+-·(1-1 -..•- - - -. '. thc entries of Column I with that of Column n ii. Acceleration of 4 ..... b.2a-ai . T~ ~1111 F iii. Acceleration 'of 2 'w.r.t.'-) ,,,' C~ upwards .\" m2 Fig. 7.526 iv.Accelel'ation of 2· w.r.t.4 . '.' d. downwards ....... 4. The coefficient of friction between the block and the surface isii~:,Fort£<;tding'on m2 b.ln2g sec 0 is 0.4 in Fig. 7.524(i-iv). Match Column I with II F ,iii. :Tensionin the -strihg is ~w ~....~.... .F.~.2.5N +C·l11-2 \" -_ - (i) (ii) tnI niz , I,~T I 1.1 '. t~\" iv. -F0rc_e -acting-:on ml by-the d•. (ntJ \"'2)gtan(! (iii) (iv) wire is Fig; 7.524 7. Column I describes the motion of the object and one or more of the entries of Column II may be the cause of motions - - - -C- o- _l u.m\"u I .\" '\" C()lnmn II described in Column I. Match the entries of Column I with the entries of Column II. i., Force. of frictioR is zero in . a.Fig. (i) '.' ii; Force oHriction .is 2,5N in ...... b.Fig:(ii) .... - i_ii.. .Accele'ra.tion of the'blo'ck ..i~s.-z~e.r..o...-i h _'.::.!'iJ[ (iii) . ColnmnI- ColumnII ..~--- _i. An object is m~ving a. Net' force acting bn the iv. Normal force is not equal to 2g in d. Fig. (iv) towards east object must. be towards 5. For the situation shown in Fig. 7.525, in Column I, the state- east ments regarding friction forces are mentioned, while in Col- ------~---+~~~~----'-'- umn II some information related to friction forces arc given. H.. An -6bjecC-is moving b.-At .leasCone forcc\"lTIust Match the entries of Column I with the entries of Column n. towards east' with con- act towal\"ds east stal1t acceleration iii•.An object.is. moving C., No force may aCHowards t()wardseastwith vary- east ing acceleratiOll iV•.. An object is moving d. No force may acton the .. towards east\\vlthcon- object ~__s~ta~'n~tyelocity .~~~~~~~~~__~~~~ Fig. 7.525 Column I .' ColumnII . 8. Column I Column II a. towards i. Total friction {orceon 3kg block is i.. If friction force is less than ap- a. Stalic right . b. towards left plied force then friction may be - -ii.-T-ot-al friction force on 5 kg block is c. zero '. '. ii.\"If'frictiori' force is equal to','the b. Kinetic iii; Friction force on.2kg; block due to d.Jlon-zero force applied. then friction may 3 kg block is be ---- I-----'~-:...:.:.---------- iv. Friction (orce on3kgblockdue to iii. If object is.moving, then friction c. Limiting I 5 kg block is maybe . __________ _ ____ iv. If object is at rest, then friction d. No conclusion e6. A horizontal force F pulls a ring 'of mass m! such that __llC'''Y be____ can be drawn remains constant. with time. The ring is constrained to move 9. The system shown in fig. 7.527 is initially in equilibrium. along a smooth rigid horizontal wire. A bob of mass 1112
Newton's Law of Motion 7.113 Columnii Both the blocks m:c coilllccted by the massless' inelastic string. The, tllagnitude of tension in the string is ii.' F\\'.~ \" ' ',' ~'F1 \" /l17?7I1I1 ,Bo,th the blocks arc coni1ected by the:ma~sless.,inclastic string. The , I1.1a'gnitudc oficnsion in the strillg,is Fig. 7.527 iii ~~>:;;13fEJ~\"F; c. {lllnt2 (,,'F2 _ 1'\\ ). The\"Ihagnitudp'ofhonnal lnl\"+'ln2' m2 m[ r'eaction bcnvecp the ,blocks Column.! .-C-o-l-u=m-n-I:I---,--------1 is ' I i. Just after the Spring). 1\\. Acce)eratesup liV :FI<+--a~]2f2 is cut. theblockD ·ii.Just afterthe Spring 2 h. A,celeratesdown . is cut, the block C <.\" - The'inagnit6d'ci'ofnormal ~~~'~~~~~7~~~~·- reaction betWeen the hlocks'js • iii.. Just after the Spring2 c;Momentarilyat rest . is cut,the block_A_\"__f_,-,-,_-,-c-:-__ 12. The system shown in Fig. 7.529 is initially in equilibrium. Masses of the blocks A, B, C, D, and E arc. respectively, 3 iv•. Justaftct.the SpriJlg d. MoVesulf with aceel- m. 3 In, 2 111, 2 m, and 2 111. Match the conditions in Column connecting. Aand B is eratiOIlg I with the effect in Column II. cut, the block D --~-~---~~ 10. For the situation shown in Fig. 7.528, match the entries of Column I with the entries of Column II. Spring 1 a?f--l>F Fig. 7.528 1 - - -- C O i u m n -'\"CC=oILn:::m::::n'-r;C-C~--'-:-~-C----' Fig. 7.529 Llf F=12.N,then a. There. is relative motion be- i: Aftet',:spring -2 is cut, :tension in a.jncreases . tween A aUd.B string AB. ~\"\"\"\"\"'---7~-'--'--i ii., Afterspring'2 is'cut,\"tensiori'in b. decreases ii. If F = l5N,then b. There is relative motion be- tween Band C stdn.~{,CD\" -----'----\"-'-1 ~~~~~~~f~ iii, After.stringbetwcenC:alldpul- c.rcmain constant iii. If F= 25N, then c. There is .relative motion be- ley\"is,cut;, tensiqn 'in 'strhl.fAB tween C and the ground I' .1________-'I-iv. A.·[(er.string .b.etwe<mC~ndPlll-. d.·.. zero iv.If F=40 N, then d. There is relative motion is not ~. ley cut; tensIOn m strmg CD there atany oLthe \".c:1f-,,_c.e...;..,.......! 1L Column I gives four different situations involving two blocks 13. In Fig. 7.530 block is attached to an unstrctchcd vertical of mass 111-1 and 1112 placed in different ways on smooth hori- spring and released from rest. As a result, the block comes zontal surface as shown. In each of the situations, horizontal down due to its weight, stops momentarily and then bounces forces F! and F~ arc applied on blocks of mass m I and ln2, respectively nnll also 1112 FI < m! F2. Match the statements . back. Finally the block starts (lSeiJIating up and down. Now in Column 1 with tb corresponding results in Column II. match Columt, I with Column II.
7.114 Physics for IIT-JEE: Mechanics I izontal direction. There is no friction between M and the ground, /11 and /1,2 are the coefficients of friction as shown M between the b1ocks. Column I gives the different relations between iJ..j and J.tz, and Column II is regarding the motion Fig. 7.530 of M. Match the columns. , , Column I ColulllnII F ----~--~-+~==~~~~~ i. When the block is at its a.Acceleration iSll! up- maximum downward ward direction ' ' 'displacement position Fig. 7.532 ,(may be, known as extreme position) ii. When the blocki, at its h. Acceleration is 'in I Column I .... --'Column II . .... equi1ibrium,Pbsitio~' downward direction tif Ih= ).<2 = 0 a. may accelerate towards right . iii. When the block iss()me- c. Acceleratipn is zero ii. if ).<1= 1]2,,0 0 .... b.may accelerate towards left.· where' between equilill:- Iii. jfll! >\"2 .....••.. c. dpps not accelerate ..' riumposition and down- wardcxtrelllc position iv. ifll!<1l2 •.' .... d, lilaY or may not accelerate iv. When theblock is, above' d. VelocitYlllity beinup~ 16. The coefficient of friction between the masses 2m and m ward, or in downward is 0.5. All other surfaces are frictionless and pulleys arc equilibrium\" position but direction massless (see Fig. 7.533). Column I gives the different bel()wthe, initial un- values of m I and Column II gives the possible acceleration stretched position of 2m and m. Match the columns. 14. In Fig. 7.531, a block of mass m is released from rest when spring was in its natural length. The pulley also has mass In but it is frictionless. Suppose the value of In is such that finally it is just able t9 lift the block M up after releasing it.. <4-Rod Fig. 7.533 m Column I· Column II .. . '. M ii. In! = ~31ft~.b.~ac~cde~rat~ion~of2~ma~nd~ma~red~iffe~ren~t Fig. 7.531 iii. -1n\\ :::; 4m c~':ac6eleniJiiJn, of 2m, Js: 'gr~a:ter than m', 'i~. m 1=6m . d.acceleration of fn iskssthanO.6g .. '.• Column I Column II 17. When the system shown in Fig. 7.534 is released, A i. The weightofmrequired tojustliftM accelerates downward. ii.Thetel1si()nin the rod, whenm in b.Mg ,,--Ie equilibrium d.2mg [I iii. Thel1ol.mal forceactingonMv.:henf~ A is in equilibrium iv.Thetension in the stringwhendisplace- ment of m.is maximum possible 15. In Fig. 7.532. both the pulleys are massless and frictionless. Match the entries of Column I with that of Column II. A force F (of any possible magnitude) is applied in hor\"
Column I ColumnII ... Newton's Law of Motion 7.115 ioAcceleration orB a. towards left . 7·J D ii.- Acceieration of C b. towards right . A w.r.t. B B . iii. Accele~ation-:- of • I w.\"t C c. at some angle.1i with hori- zontal; 0< Ii <: 90° iv. Acceleration of d. at saine angle.8 withverti- cal, 0 < 11.< 900 • w.r.t A '. Fig. 7.536 18. Column I , .. Column II Coluri1ll I Column II i. '11\\ +hl is . . a, True i; In equilibrium, the I.;I.' l'1 - T,2 'IS • . .' a. T2 . .' body must be station- b. False ary iiCi. l4T, +1·'3,Iis -- -b.-T3- - - - , - - c, Net force acting on the ii, If a body is moving, bodyiscertainly zero iv.T, +his c. ::: 40 N .. then no forceinaybe acting on it d. The body may have some d;>lOON '. iii. If \"body is at rest, it acceleration also Archives Solutions on page 7.169 means no net' force' is acting on. it Fill in the Blanks Type iv. Ifa body isacceler- L A block of mass I kg lies on a horizontal surface in a truck. ated, then direction of The coefficient of static friction between the block and the acceleration '-is always surface is 0,6. If the acceleration of the truck is.5 mIs', the in the direction of net frictional force acting on the block is _...... __\".,_,.. ,...newtollS. force (IIT-JEE, 1984) 19. For the situation shown in Fig. 7.535, in Column I, the state- 2. A uniform rod of length L and density p is being pulled ments regarding friction forces are mentioned, while in Col- along a smooth floor with a horizontal acceleration Q' (see umn II some information related to frict~on forces is given, Fig, 7,537) The magnitude of the stress at the transverse cross-section through the mid-point of the rod is _________. I' ~ 0.2 4kg F~ ]00 N (IIT-JEE, 1993) 2kg I' ~ 0.4 6 kg ,li _.- 0.5 Fig. 7.535 Fig. 7.537 Match the entries of Column I with the entries of Column II. Column I CoJumnII . True or False i. Total-friction force-on a. Towards right 1. When a person walks on a rough surface, the frictional force 4 kg block is I exerted by the surface on the person is opposite to the direc- ii. Total friction force 011 b. Towards left tion of his motion, (I1T-JEE, 1981) 2kg block is '. 2. A simple pendulum with a bob of mass m swings with an m.• Friction force on 6 kg angular amplitude of 40\", When its angular displacement is c. Zero 20\", the tension in the string is greater than mg cos 20°. blockdue to 2 kg block. (IIT-JEE, 1984) I· is .' 3. The pulley arrangements of Fig. 7.538 (a) and (b) arc identi- iv. Total friction force on d.Non-zero caL The mass of the rope is negligible, In (a) the mass m is . 6 kg block is . lifted up by attaching a mass 2m to the other end of the rope. In (b), In is lifted up by pulling the other end of the rope with 20. In Fig. 7,536, the whole system is in equilibrium. Tensions a constant downward force }I' = 2mg. The acceleration of m in different strings are shown. Match the followings is the same in both cases (IIT-JEE, 1984)
7.116 Physics for IIT-JEE: Mechanics I ~nD a. b. r )D--f ~D--f c. d. (b) 7. An insect crawls up a hemispherical surface very slowly Fig. 7.538 (see Fig. 7.539). The coefficient of friction between the 4. Two particles of mass I kg and 3 kg move towards each other insect and the surface is 113. If the line joining the center under their mutual force of attraction. No other force acts of the hemispherical surface to the insect makes an ;mglc on them. When the relative velocity of approach of the two a with the vertical, the maximum possible value of a is particles is 2 mis, their centre of mass has a velocity of 0.5 m/s. When the relative velocity of approach becomes 3 mis, given by (IIT..JEE,2001) the velocity of the centre of mass is 0.75 m/s. \\CL) (IIT-JEE,1989) Fig. 7.539 Multiple Choice Questions with One Correct Answer a. cota=3 b.tana=3 c. seca=3 d.coseca=3 1. A ship of mass 3 x](J1 kg initially at rest, is pulled by 8. The pulleys and strings shown in the Fig. 7.540 arc smooth a force of 5 xl 04 N through a distance due to water is and of negligible mass. For the system to remain in equi- elibrium, the angle should be negligible, the speed of the ship. (IIT-JEE, 1980) (IIT-JEE,2001) a. 1.5 mls b. 60 mls c. 0.1 mls d. 5 mls I I 2. A block of mass 2 kg rests on a rough inclined plane I I making an angle of3()G with the horizontal. The coefficient I I '8 of static friction between the block and the plane is 0.7. The frictional force on the block is a. 9.8 N b.O.7 x 9.8 x,f! N c. 9.8 x,f!N d.O.7 x 9.8 N mm 3. During the peddling of a bicycle, the force of friction Fig, 7.540 exerted by the ground on the two wheels is such that it a. 0° b. 30\" e. 45\" acts. (IIT-JEE, 1990) a. in the backward direction on the front wheel and in 9. A string ofnegligible mass going over a clamped pulley of the forward direction on the rear wheel mass In supports a block of M as shown in the Fig. 7.541. h. in the forward direction on the front wheel and in the The force on the pulley by the clamp is given by backward direction on the rear wheel (IIT-JEE,2001) e. in the backward direction on both the front and the m rear wheels d: in the forward direction ont both the front and the rear wheels M 4. A car is moving in a circular horizontal track of radius 10 til with a constant speed of 10 m/s. A plumb bob is suspended form the roof of the car by a light rigid rod. Fig. 7.541 (lIT-JEE, 1992) a.v2Mg b. v2mg d. .j(M+ml'+M2g a. zero b. 30\" d, 6()o c. .j(M+m)2+m2g 5. A block of mass 0.1 is held against a wall applying a to. What is the maximum value of the force F such that the horizontal force of 5 N on the block. If the coetlicient of block shown in Fig. 7.542, does not move? (IIT-JEE, 2003) friction between the block and the wall is 0.5, the magni- F tude of the friction force acting on the block is (IIT-JEE, 1994) a. 2.5N b. 0.98N c.4.9N d. 0.49N m=.J3kg 6. A small block is short into each of the four tracks as shown below. Each of the tracks rises to the same height. The Fig. 7.542 speed with which the block enters the track is the same a. 20 N b, ION c.12N d. 15 N in all cases. At the highest point of the track, the normal reaction is maximum in (IIT-JEE,2001)
Newton's Law of Motion 7.117 AB 11. Two particles of mass m each arc tied at he ends of a light string of length 2a. The whole system is kept on a friction horizontal surface with the string held tight. so that each mass is at a distance a from the centre P (as shown in the pQ Fig. 7.782). (IIT·JEE,2004) Fa Fx Fig. 7.544 a. b. a. 2U case b. u/cose 2m ~a2 -x2 2mJa2 -x2 c.2U/cose d. U case F ~a2_x2 Fx ~ ~ 2m x 2m a 12. A block of base 10 em x 10 cm and height IS cm is kept 2. A reference frame attached to the earth on an inclined plane. The coefficient of friction between a. is an inerti-al frame by definition. ethem is 3. The inclination of this inclined plane from' h. cannot be an inertial frame because the earth is re-' the horizontal plane is gradually increased from ()O. Then (InJEE, 2009) a. at 0 = 30\", the block will start sliding down the plane volving round the sun. b. the block will remain at rest on the plane up to certain c. is an inertial frame because Newto.n's laws are appli- e and then it will topple cable in this frame. ec. at = 6()\", the block will start sliding down the plane d. cannot be an inertial frame because the earth is rotat- and continue to do so at higher angles ing abut its own axis. (IlT-JEE, ed. at = 60\", the block will start sliding down the plane 1986) e, eand on further increasing it will topple at certain 3. A simple pendulum of length L and mass (bob) M is f oscillating in a plane about a vertical line between angular limit -¢ and +¢. For an angular displacement 8 (101 < ¢). the tension in the string and the velocity of the bob are T and V respectively. The following relations hold good under the above conditions (lIT JEE, 1986) e a. TcosO = Mg w MV' FBD at just toppling condition b. T - Mgcos8 = - - Fig. 7.543 L c. The magnitude of the tangenial acceleration of the Assertion and Reasoning bob?aTI = gsin8. Mark your answer as d. T = Mgcos8. a. If Statement I is true, Statement I.I is true; Statement II is 4. A patticle P is sliding down a frictionless hemispherical howl as shown in Fig. 7.545. It passes the point A at t = O. the correct explanation for Statement I At this instant of time, the horizontal component of its b. If Statement I is true, Statement II is true; Statement II is velocity is v. A bead Q of the same mass as P is ejected from A at 1= 0 along the horizontal string AB, with the not a correct explanation for statement 1 speed v. Friction between the bead and the string may be neglected. Let P and IQ be the respective times taken by c. If Statement I is true; Statement Il is false P and Q to reach the point B. Then (IIT-JEE,1993) d. If statement Tis false; Statement II is true 1. Statement I : A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. (IIT-JEE,2007) Statement II: For every action there is an equal and opposite '0' pC reaction. Fig. 7.545 2. Statement I: It is easier to pull a heavy objecft than to push a.tfl<:tQ h. Ip = IQ it on a level ground. (IIT-JEE,2008) C.tp>IQ Statement II : The magnitude of flrctional force depends on d. II' = length of arc AC B the nature of the two surfaces in contact. length of AB Multiple Choice Questions with One, or More than One Correct Answer 1. In the anangement shown in the Fig. 7.544, the ends P and Q of an unstretchable string ,move downwards with uniform speed U. Pulleys A and B are fixed. (nT-JEE, 1982) Mass M moves upwards with a speed
7.118 Physics for IIT-JEE: Mechanics I ANSWERS AND SOLUTIONS Subjective Type N = F + mg cos 30° +mg sin 30° = /LN = /L(F mg cos 3~') 1. Denote the common magnitude of the maximum acceleration as a. For block A to remain at rest with respect to block B, or, F = mg sin 30° _ mg cos 300 a < f..lsg. The tension in the cord is then /L + + +T = (rnA mB)a /L,g(mA mB) (v'3)= (2) (10)(1/2) _ (2) (10) = (mA + I1IB)(a + /L,g). 0.5 2 This tension is related to the mass me by or, F = 20 - 17.32 or F =2.68 N T = mcCg - a). Solving for a yields 4. The friction force on block A is /Lk WA = (0.30)( 1.40 N) a = me - +/Lk(mA mB) < /L.,g = 0.420 N. This is the magnitude of the friction force that g block A exerts on block B, as well as the tension in the string. rnA +mB +mc The force F must then have the magnitude Solving the inequality for me yields + ++F = ILk(WB WA) /LkWA T = /L,(W/I + 3w,,) + +(rnA mB)(/L, /Lk) = (0.30)(4.20 N + 3(1.40 N)) = 2.52 N me < 1- /L, 2. Draw free body diagram (Fig. 7.546) of block, SFy = 0 Note that the normal force exerted on block B by the table is the sum of the weights of the blocks. 5. Draw FBD of the block with respect to plane (Fig. 7.547). Acceleration of the block up the plane is a= mg cos 37° - mg sin 37\" =g (45-~3) m ,mg cos 370 = 2 mis' Fig. 7.546 .I ApplYlllg, S = 2at2 =} N + F sin 30' = mg cos 37' ! -- ffs _- )2 x I-_Is or, N = ing cos 37° - F sin 30° a2 G) G)=(4)(10) -(10) A Pseudo-force N or. N=27N (i) mao \"\" mg+-+-_ _~< f,n\" = /LN = 0.5 x 27 = 13.5 N G)mgsin45° = (4)(10) = 32N (v'3)and F cos 30' =(10) 2\" =8.66N B mg c Fig. 7.547 Now since mg sin 37° > fmax + F cos 30° 6. Consider the forces on the person as shown in Fig. 7.548: Therefore, the block will slide down and the friction will be maximum. Therefore, the net contact force is y Fe = ../N' + (f,n,,)2 = ../(27)2 + (13.5)2 II Fc=30.2N 3. Since mg sin 30\" > /Lmg cos 30' the block has a tendency to mg slip downwards. Fig. 7.548 Let F be the minimum force applied on it, so that it docs not slip. Then,
Newton's Law of Motion 7.119 LF,,=ma, 10. n -mg =ma x n = L6mg so a = 0,60g = 5,88 m/s2 L y - Yo = 3,0 m, a, = 5,88 m/s2, y vo)' = 0, Vy ='1 Fig. 7.549 v; v6, y= + 2a (y - Yo) gives Vy = 5,0 mls 7. Force offriction between the two will be maximum, Le., I1mg fLmg Retardation of A is aA = - - = Mg m and aceeIerall'On 0 t\" B IS a B = -wn-g = -Mg 2m 2 acceleration of B relative to A is a. L is the lift forcc b. LFy = may Subst'Itutm, g, Ii- = \"2; aHA = 43g' Mg - L = M(gI3) 8. Lct thc tension in the cord attached to block A be TA and the L = 2Mgl3 tension in the cord attached to block C be Tc , The equations c. L - mg = m(gI2), of motion arc then TA - mAg = mAa where m is the mass remaining. L = 2Mgl3 Tc - f1kmBR - TA = trlBa so, m = 4MI9 meg - Tc = mea Mass 5M/9 must be dropped overboard, a. Adding the above givcn three equations to eliminate the 11. m = mass of one link tensions gives Chain Top link + +a(mA me me) = g(me - rnA - MkmB) F~t\\ldent Fstudent solving for me gives a, ++ +mACa g) mn(a Mkg) Fmiddle me = 2ma g-a 3 mg 3 mg ma mg and substitution of numerical values gives Middle link Bottom link 2mg me = 12,9 kg Fig, 7.550 b. TA =mA(g+a)=47,2N The downward forces of magnitude 2ma and rna for the Tc = mcCg - a) = 101 N top and middle links are the reaction forces to the upward force needed to accelerate the links below. 9. a. The only horizontal force on the two-block combination is b. i. The weightof each link is the horizontal component of If, F cos ex, The blocks will mg = (0.300 kg)(9,80 m/s2) = 2,94 N accelerate with F~et 12 N - 3(2,94 N) 3,18 N a = _. a = -3 ill = = :0:,-9c0:0=-k:g- 0,900 kg (ml + m2) b. The normal force between the blocks is mIg + F sin (Y, for = 3.53 m/s2 or 3,5 m/s2 the blocks to move together, the product of this force and Using the free body diagram for the whole chain Its must be greater than the horizontal force that the lower ii. The second link also accelerates at 3,53 m/s\", so block exerts on the upper block. That horizontal force is one of an action-reaction pair; the reaction to this force Fnet = Flop - rna - 2mg = ma accelerates the lower block, Thus, for the blocks to stay together, Ftop = 2ma + 2mg Using the result of part (a), = 2(0.300 kg)(3,53 m/s2) + 2(2,94 N) m, Fcosa =2.12 N +5,88 N = 8,ON S!,,(mlg+Fsin(Y) 1n1 +ln2 Solving the inequality for F gives the desired result.
7.120 Physics for IIT-JEE: Mechanics I 12. a. Let al and a2 be the accelerations of the two men in the over pulley 2, M = mass of the man, m = mass of thc platform upward direction, and T the tension in the rope. Then, (Fig, 7552), l' - Mg = Ma\"and (i) l' -(M+m)g = (M+m)a, (ii) Subtracting equation (ii) from equation (i). we get. or, a2=(~)al-~ M+m M+m a2 < al Hence. the lighter man will reach the pulley first., b. The lighter man ascends a distance h in time t with accel- Fig. 7.552 eration aj. Hence, The. FBDs of man, platform and pulley 2 arc shown in I, (iii) Figs, 7,553(a). (b), (c). and (d), h = -a,t- 2 Let s be the distance travelled by the heavier man in this time f, then s = ~a2t2 = ~ [~al - ~] T, 7i 2 2 M+m M+m N ~~~~2SlI platform 2 T, 2 P s=2(Mt+m) [.Mf(22h- ) -mg ] ID I ::-;-:-;---,- [2Mh - mgt 21 2(M+m) The distance of the second man from the pullcy = h - s = h- I [2Mh - mgt'] N+mg T, T, F (b) (e) (d) 2(M+m) Mg (a) Fig. 7.553 ~_I_-,- [Mh + mh - Mh + ~g2t'] (M +m) ~_2= :-:M(-;-m-'+'--m--:-) [_g + h] Action exerted hy the man on the platform:::; reaction exerted by the platform on the man = N For equilibrium of the man 13. T Mg=7,+N (i) eiliA gsin mngsinO f' me gcosO For equilibrium of the platform (ii) Fig. 7.551 (b) 1', + 72 = N + I11g (a) (i) For equilibrium of pulley 2, (ii) (ili) 1', = 21', (iii) (iv) mAg sin e+ F - l' - f = mAa For equilibrium of point P of the string f = i-',inAgcose F = 1'2 (iv) l' -!nBgsine - f - f' = mBa Adding equations (i) and (ii), we get f' = i-'k N' 1', + 72 + (1'2 + N) = N + mg + I11g t = i-'k(mAg cos e +mBgcosO) 1', +21'2 = (M +m)g Solving above equations, we get Using equation (iii), we get 21'2 + 21'2 = (M + m)g l' = 215 N (M+m) 1'2 = 4 g 14. Let N be thc reaction between man and platform, 1', = tension (60:2()) NF = 1'2 = (M_!m) g = g is string passing over pulley 1, 12 = tension in string passing
Newton's Law of Motion 7.121 = 20 kg wt 16. Let M = mass of the painter = 10 kg Now from equation (i), N = M g T2 = =M mass of the crate 25 kg =Mg-F Let FA be the action force excIted by the painter on the crate, reaction force exerted by the crate on the man is = Mg - (M+m) (3M-m) (v) 4 g= 4g N = FA =450N C= x 6~-20) g The free body diagram of painter is shown in Fig. 7.555(b) = =40 g N 40 kg wt riTT a T /\"'. From equation (v) mg = 3Mg - 4N T Clearly, mg is maximum when N is minimum A N \"-./ MinimumN =0 rT So (mg)m\" = 3Mg = 3 x 60g = 180 kgwt. Ta 15. Let T, be the tension in the rope and T, the tension in the ia tail of monkey A. Let force exerted by monkey A on the rope be F. Let M A and M B be masses of monkeys A and B, respectively. The free body diagrams of monkeys A and B and point P of the string are shown in Figs. 7.554(a), (b), and (e). Mg (M+ m)g (a) (b) (e) T2 Fig. 7.555 jaj A .'. equation of motion of the painter is (i) N + l' - Mg = Ma BP The equation of motion of the whole system is 2T - (M + m)g = (M + m)a (iil M?JK T2 T2 F MUltiplying equation (i) by (ii), we get (iii) (a) MBg (e) (b) 2N + 2T - 2Mg = 2ma Fig. 7.554 Subtracting equation (ii) from equatioii (iii), we get 2N - 2Mg + (M + m)g = (2M - M + m)a Or 2N - (M - m)g = (M + m)a The equations of motion of monkeys A and Bare a = _2N_-:c(:-M,---_m.-:.lcgc M+m 1', - MAg - Ii = MAa (i) (ii) 2 x 450-(100-25) x 10 T2 - MEg = MBa = 100 + 25 Equation of motion of point P is 1', - F = (mass) a 900 - 750 2 a=O=}F=1', (iii) = 125 = 2m/s For maximum tension in tail From equation (ii), we get From equation (ii), we get Ii - MBg = 30 - 2 x 10 2 . . (M +m)(g+a) a= TenSIOn 1 = 2 = 5 m/s ME ·2 then from equation (i), we get = (100 + 25)(10 + 2) = 125 x 6 = 750 N 1', = T2 + MA(g + a) = 30 + 5(10 + 5) = 105 N 2 For minimum tension in tail a = 0 17. a. When the ring slides from A to A' in time interval ,'\"t, the block descends from C to e' as shown in Fig. 7.556(a). .'. from equation (ii), we get A' D is perpendicular dropped from A' on AB. For small displacement A'B = DB. Ii =MBg =2 x 10= 20N Now as the length of the string is constant. Then [rDIn equation (i), we get AB + BC = A'B + Be' T, = MAg + T2 = 5 x 10 + 20 = 70 N or (AD + DB)+ BC = A'B + Be' Therefore, the force exerted by monkey A on the rope AD = Be' - BC ('.·A' B \"\",DB) F = 1', must be between 70 Nand 105 N.
7.122 Physics for IIT-JEE: Mechanics I or AA' cos e = ee' 18. N -A-Ac' os ee' pN 8-- Fig. 7.557 /',/ /',1 N =mg p,N = mru/ i.e., velocity of the ring x cos II = velocity of the block lung = m2 sin Ow' ei.e., velocity of the block e, Vc = v cos (i) b. If a is an initial acceleration of ring, then the acceleration of the block = a cos II. Let T be the tension in the string at this instant. Then the FBD of block is shown in Fig. 7.556(b). /1f5(j)= A A' Lg -- 2sinO 0.1 x 10 ())= 2 x sin 30 B w = I rad/s. c 19. So long as the applied force does not produce the limiting frictional force, the two bodies will have no relative motion, that is. the two will behave like a compact body. Here, the limiting frictional force = 0.25 x 2 x 10 = 5 N. Let F be the force that produces limiting frictional force between the two. Then F - 5 = 2a and 5 = 20a, where a ;::: common acceleration. Hence F ;::: 5.5 N. Here the ap- plied force is only 2 N. Hence the two bodies will behave as _a(a) one body and will have a common acceleration given by N 2 = (20 + 2)a or a = I T, 11 = 0.09 ms'-' Now fr, = 20x 0.09 = 1.8 N =When F 20 N, the applied force exceeds the force required to produce the limiting frictional force. So there will be rela- motion tive motion between the t\"wo. The frictional force = the lim- iting frictional force = 5 N. Mg a,From the free body diagram of the weight and the block, (b) 20 - 5 = 2a, or = 7.5 ms' az = 0.25 ms-2 mg 5 = 20a2 (e) 20. Let X be the leftward displacement of M, and x and y be Fig. 7.556 the leftward and downward displacement of m as shown in Fig. 7.558. Then by constraint relation c. Equation of motion of block M is Mg - T = -Macose (ii) T The free body diagrams of ring is shown in Fig. 7.556(c). Mg Resolving T along horizontal and vertical, we have Tcosll = Ma (iii) M rna mg T+_--1 T=-- cose Equations (ii) and (iii) give Mg cos II 2mg cos 30' +a= = In + 2m cos' 30° In M cos' II Fig. 7.558 = 6.78m/s' x = X =? x =- X => ax ::::: Ax and I,-x+lz +13 -x+14 +y =1, +12+1,+1. where I, , 12, 13, 14 are the instantaneous lengths of the seg- ments of the string.
Newton's Law of Motion 7.123 lx=y=} lx=y =}lax=a,. N = J1Ulx and mg - jkN - T = may andlT-N = MAx = Max Eliminating T, A and N, we get lmg 4mg ax = and a y = M +5m +l\"m ' M +5m +2\"m a= la'+a'= +25Vm5r+ng2jJJn Vx Y m Fig. 7.559 '21. Considering the free body diagram of rn and M and keeping N'=mgcosa in mind the horizontal motion I = rna and F - I = M b, mgsina +\"N' = T where I is the frictional force and a and b are the accelera- 2mgcosa = N tions of m and M, respectively. and mg sina = (2mg cosa + mg cosa) or = :13 tana When there is no sliding, a = b, Hence ,f = mF -m+-M- The mass m will begin to slide when I = Ji;m = kmg or kmg = mF or F = k(m + M)g :I3 =:31 43 = --- m+M Q ,, = tan x 0,25 When the force F is greater than k(m + M)g, the mass 37 m slides over M with acceleration a given by 24, a. When A is projected to the right, B falls freely under grav- fiim = rna or kmg = rna ora = kg ity, The string is tensioned when A and B travel the same The acceleration of M is given by distance, Let t be the time in which they cover the same 1 F - ,/i;m = M b or b = -F-:--k;-m-\"g- distance, Then vt (distance described by A); 2gt2 (dis- M The acceleration of M relative to m is tance covered by B) brei = b - a =, F-kmg - kg F-kg(M+m) u M M or t = 2- From the fonnula S = ut + 21It' g _ 0.1 + 2I bce,t2 _ IF- kg(M + m) , 1- 2 Let v be the initial common velocity, When the string is - Mt tensioned, the two bodies experience the same impulsive force but in opposite directions. change in momentum of A = 2mvf - 2mu 22. a. LetN, andN2 be the reactions of the plane on 1 and2and = 2mv 2mu F be the force of interaction between them. From the free body diagrams of I and 2, we get .. change in momentum of B = m v - mUinitiaJ +m1gsina - k1m]gcosa F = mIG 2u But l-linitial = gt = g- = 2u g m2g sin Q' - k2tn2g cos a - F = m2a Change in momentum of B = mv - m x 2u Eliminating a between these two equations, we get Therefore, by Newton's laws of motion F = -(k'!-'---k-2\")-m'l-m-2-g\"c\"o-s-a- (2mv - 2mu) = -(mv - 2mu) mj +m2 4u b. When the bars just start sliding, a = 0 or3mv=4mvorv= - m,gsina - k,m,gcosa + F = 0 3 b. Let a be the common acceleration. and Jn2g sin a - k2m28 cos Q' - F = 0 mg - T =ma T =2ma m,Adding (m, + In,)g sina = (k,m, + k2m2)g eosak, g or tan a = -' m, -+, -k-2- ' - a=- -- 3 tn] +m2 v v52 - = 221,/i 3 '23. a. Since A tends to slip down, frictional forces act on it from where l' is the distance covered by A' or B with accelera- both sides up the plane tion. Let N be the reaction of the plane on A and N' be the 2u } 2u 2 mutual normal action - reaction between A and B. Now I = I' + ut = l' + ug- =, } 1= 1- -g From the FBD of A Fig, 7.559 N' + mg cos a = N and mg sina = (N + N') From the FBD of B
7.124 Physics for IIT-JEE: Mechanics I = 16u' + 2g1 - 4u2 4u' + 6g1 This gives l' = 4M, M, + Mo(M, + M 2) (v) -9- -3- -3- = --9:---''- .'. acceleration of pulley of A from equation (ii) J4u 2 + 6g1 21' 4M,M, ao= -= (vi) V= 4 M, M, + Mo(M, + M,) 3 Mo 25. The same frictional force is effective on A and B. This force 27. Evidently, the larger block of mass experiences more cen- trifugal force radially outwards, compared to the block of produces retardation on A and acceleration on B till they acquire a common velocity. F = rna = Ma', where a == ab~ smaller block In [M > m and r, > rd. solute retardation of m Figure 7.561 shows their FBD al :::: absolute acceleration of M Relative retardation of m = a - (-a') = a + a' Initial relative velocity = Vo Final relative velocity = 0 .. v6 = 2(a + a') where s =: distance covered by m relative to m or v6 = 2 (~ + ~) s = 2F(m + M)s Fig. 7.561 mM mM Owing to the larger force experienced by block of mass mMv5 M, it tends to fly off radially. or s = :2:-F::(,m-;--+,-\":M-:,), In the situation of limiting equilibrium, we have 26. If ao. a\" and a, are accelerations of Mo. M\" and M\" respec- l' = m,,} /', + I, =? l' + h = M w2r, tively, then [where I, and h are frictional forces for the two blocks and R a, the surface.] T ~/o The above two equations get reduced to +10 (i) T = muir] J.ilmg l' + V2Mg = Mw'r2 (ii) Subtracting equations (i) and (ii), we get IL2Mg = Molr2 - !nU}'1 - J..Ll mg Fig. 7.560 g [gw' = [v,m -i- V2M] =? w = (v,m + V,M)] ,/, -([1 -- a2 Mr2 - rnrl Mr2 - mT] ao= 2 - 28. a. The tension in the cord must be m2g in order that the hanging block moves at constant speed. This tension must 2ao = al - a2 ei) overcome friction and the component of the gravitational For motion of Mo force along the incline, so 21' = Moao (ii) m2g = (m,g sin a + Vkm,g cosa) For motion of mass M 1, and m2 = mlCsin a + J.tk cosa-) M'Ii- l' = M,a, (iii) b. In this case, the frictional force acts in the same direction as the tension on the block of mass m 1, SO, For motion of mass M2, l' - M,g = M,a2 (iv) Substituting values of ao, a\" and a, from equations (ii), (iii), and (iv) in equation 0), we get tn,or, rn2 = ml(sina - fl-k COSO') 41' = 2g _ l' (_1 + _1) c. Similar to the analysis of parls (a) and (b), the largest could be Mo M, M, m! (sin a + /.Ls cos a) m,and the smallest could be i.e, ( -4 + -1 + -I ) 1'=2g Mo M, M,
Newton's Law of Motion 7.125 ODjectiv.e iT~lBe From equations (iv) and (v), we get 1. c. When a body is stationarys its acceleration is zero. It means g L Fnet force acting on the body is zero, i,e., T,=4x-=13N 3 = O. We can 5. c. Let a be the common acceleration of the system also say that all the forces acting balance each other. (Fig. 7.564) 2. a. The water jet striking the block at the rate of I kgls at a Here T= Ma (for block) speed of 5 mls will exert a force on the blank P - T = Ma (for rope) dm o F=v-·-=5xl=5N elt Under the action of this force of 5 N, the block of mass 2 kg m will rnove with an acceleration given by Fig. 7.564 F5 a= -m = - = 2.5 mis' 2 P - Ma =Jna 3. d. When a string is fixed horizontally as shown in Fig. 7.562 or P = (m + M)a or a = P /(m + M) (by clamping its free ends) and loaded at the middle, then for the equilibrium of point P Now MP T=Ma=--. lvl +m 6. d. Let A applies a force R on Ii, T T P at r\";'tl w Fig. 7.565 Fig. 7.562 2TsinO = W W Then B also applies an opposite force R on A as shown Le., T= in Fig. 7.565. 2sine For A: mg -- R = ma Tension in the string will be maximum when sin 0 is mini- =} R = m(g - a) = 0.5110 - 2J = 4 N e emum, Le., = on or Sil1 = 0 and then T = 00, However, 7. h. In Fig. 7.566, the point Ii is in equilibrium underthe action as every string can bear a maximum finite tension (lesser than breaking strength). So this situation cannot be realized of T, P, and Mg. . practically. We conclude that a string can never remain hor- izontal when loaded at the middle howsoever great mass be Here T sin 0 = F the tension applied. or l' = F/ sinO 4. h. For A : T - 2g = 2a (i) ,:8 T +, Tcos8 (ii) ,ITf-si-n8+- IB F For Ii: 1', + 2g -- T = 2a C For C: 2g _. T, = 2a (iii) Adding equations (i) and (ii), we get T, =4a (iv) Fig. 7.563 Fig. 7.566 From equations (iii) and (iv). we get 8. a. T=60N 2g - 4a = 2(1 ~g'r~, +11.\" or a = g/3 T, Fig. 7.567 T=7',+40 60 = T, +40 (v) T, = 20N
7.126 Physics for IlT-JEE: Mechanics I 9. d. Let the tension in the string A Pi and P, PI be T. Consid- 17. b. The acceleration of the body perpendicular to 0 E is ering the force on pulley PI, we get a= F -42=2m/s2 T=WI -= Further, let LAP,PI = 28 m p,.Resolving tensions in horizontal and vertical directions Displacement along OE, SI = vI = 3 x 4 = 12 m and considering the forces on pulley we get Displacement perpendicular to OE ZI 2 I , S2 = zat = x 2 x (4) = 16 m or 2Tcose=W, The resultant displacement or w,2WI cose = orcose = 1/2 /srs = + s~ = v'144 + 256 = v400 = 20 m So e = 60\" LAP2 PI = 2e = 120\". 10. c. Let the tension in the rope be T. Let acceleration of the 18. c. Let l' be the tension in the rope and a the acceleration man and the chair is a upwards. of rope. The absolute acceleration of the man is therefore' For man: l' + 450 - . l10o00a 4 -5g a ) . Equatl.Ons 0f motI.On for mass and man gI.Ve 1000 = ( or l' = 550 + lOOa (i) l' - 100g = 100a (i) (ii) WFor chair: T - 450 - 250 = ( 250) a . (Sg)l' - 60g = 60 4 - a or T = 700 + 25a (ii) Solving equations (i) and (ii), we get, From equations (i) and (ii). we get a = 2 ms~2 T = 1218.75 N \"\" 1219 N 11. c. The spring will exert maximum force when the ball is at its e -19. b. l' sin mg = sin 30\" = rna lowest position. If the ball has descended through a distance e=} l' sin = mg sin 30° + mgl2 (i) x to reach the position. (ii) • l' cose = mg cos 30° I (i) mgx = zKx' or x = 2 mg/K Dividing equation (i) by (ii), we get For the block B to leave contact spring force (ii) Kx = Mg Comparing equations (i) and (ii), m = M 12 12. a. Acceleration of the skates will be in the ratio mgcos 30\" FF Fig. 7.568 4:5 0r5 : 4 Now according to the problem, s = 0+ I t 2 2 -a 2 tane =- 5 v'3 == we get S2 (12 4 20. c. 1'2 cos fJ = mg (i) (ii) iJ'R'CCr-+-R-C R3 7icosfJ = F = mg (iii) 13.a.a= = From equations (i) and (ii), fJ = 45° (iv) mm Ii = =14. b. Change in momentum of one ball 2mu, time taken I s . Total change in momentum B F\"\"mg F,w = - - - - - , = \" ' - - , - - - - - ( Time taken = n (2mu) =2mnu I 15. c. Area under force-time graph is impulse and impulse is mg change in momentum, Fig. 7.569 Area of graph = Change in momentum 1'2 cos fJ + mg = 1', cos\", Z TI 4mu T2 sinfJ = Tl sina =} 1'Fo = 2mu =} Fo = Dividing equation (iv) and (iii) 16. d. From 0 to 1', area is +ve and from l' to 21', area is -yeo tana = 112 Net area is zero. Hence, no change in momentum. =} sin\",=I/v'5
Newton's Law of Motion 7.127 Tl~ 27. a. Acceleration of the system: (1) A /3 T2 mg p Fig. 7.570 a=--- _aM+m The FBD of mass m is shown in Fig. 7.572. RC3O.SRf.f . T2 vI\"i vI's , . R sinf3 From equation (IV), = T, mg => v'sT2 = v\"iT,. Fig. 7.572 21. a. Acceleration att = I s, a, = 23.6 = 1.8 mls. Rsin{3 = ma (2) Rcos{3 = mg From equations (2) and (3), we get (3) 22. d. Tension will be the least during downward acceleration a=gtan{3 fronll = 10 s to t = 12 s. Putting the value of a in equation (I), we get e e23. a. x = 0, till mg sin < fLmg cos and gradually x will in- P = (M + m)g tan {3 ecrease. At angle > tan-'(fL) kx + fLmgcose = mgsine 28. d. Extension in the spring is x = AB - R or mg sin e - fLmg cos e = 2Rcos30' - R = (vS - I)R x= k Spring force Here k = force constant of spring. (vS+ I)mg r;; 24. d. Net pulling force on the system is Mg + mg - mg or F=kx= R x(v3-!)R=2mg simply Mg. Total mass being pulled is M + 2m. Hence, the acceleration of the system as shown in Fig. 7.571 is Mg N a= B M+2m Fig. 7.573 Fig. 7.571 From Fig. 7.573 Now since a < g, there should be an upward force on N = (F + mg) cos 30' = -3-..2/3-m\"g- M so that its acceleration becomes less than g. Hence for any G) G)29. b. Acceleration of the cylinder down the plane is value of M, the spring will be elongated. 25. b. Suppose F = upthrust due to buoyancy a = (gsin 30')(sin300) = 10 = 2.5 m/s2 Then while descending, we find /2V-;;Time taken: (I) = (2i = x 5= 2 s 2.5 Mg - F = Met (i) when ascending. we have F - (M - m)g =.(M - m)\" (ii) 30. b. 31. d. Same solution for both. Solving equations (i) and (ii), we get M, = 100 kg, m2 = 50 kg, a = 5 m/s2 m=[et ~+ gJM T + N - m,g = m,a T - N =mg =m2a Solving these, we get 26. a. Initial force = 2g = 20 N T = 1125 Nand N = 375 N Im.t.Ial acceleratlO. n = Force = 20 T -- Mass 5+1 20 = - m/s2 6 Final force = (load + mass of thread) xg =(2+1)x 1O=30N .. f'inIa fcceIera'tlon = (30; m/s2 mg Fig. 7.574
7.128 Physics for IIT-JEE: Mechanics! 32. d. As shown in Fig. 7.575 (a) and (b) T eos45\" = Ina (1) I, + 12 = C Teos 45° < T rll, rlI2 -+-=0 rlt rlt 1B mg (b) (a) Fig. 7.575 Mg - T cos45\" = ma (2) From equations (1) and (2), we get Fig. 7.577 T = mg/..fi - VI cose l + V2 cos (h = 0 33. d. From constraint relations, we can see that the acceleration VI cos 82 of block B in upward direction is = ~aB = ( ae aA ) with proper signs. V2 COS f)j CSo, 12t = 37. b. an = 2 m/s2 (t) an = -2 ) 1.5 - 6t For the following questions (Fig. 7.578) assume or dV/i = 1.5-6t Fig. 7.578 dt I, + 12 = C J J\"8 , 13 + /4 + 15 + 16 = C or rive = (1,5 - 6t)dt I;' + l~ = 0 o0 L~ + LX + 19 = 0 or Vn = 1.5t - 3t2' - ap - ap + 12 = (} m vB=OMt=l~s ap - all - all - all = 0 34~ c. When mass 1112 moves downwards, the centre of mass of ap = 6 m/s2 system (m! + m2) moves downwards. It means the acceler- ation is found in centre of mass in the downward direction during motion of m2. This is possible only when net down- ward force is greater than that of upward force. Mathematically, In 1g + m2g > N N<In,g+ ln 2g 35. b. Length of the string (Fig. 7.576) is I \"'XA +2XB +xe c A Xc B ap = 3 all =} all = 2 m/s2 Fig. 7.576 38. a. -bi - 4b] (Fig. 7.579) Differentiating twice w.r.t. time, we get II d2xA 2r12xB d2xe -f- i2 o = rlt2 + dt2 + dt 2 or, 0 = -aA + 2aB + ae 13 aA - ac -f-' 14 _Yis aB = ~B 2 b 36. e. From Fig. 7.577 Fig. 7.579
Newton's Law of Motion 7.129 Acceleration of A in horizontal direction = the acccler- F ation of B = b in rightwards A Acceleration of A in vertical direction = the acceleration TT of A with respect to b in upwards direction = a = 4b. BC --------~--------- Hence the net acceleration of A = bi + 4b]. (-x, 0) (x, 0) =39. a. Acceleration of C in horizontal direction acceleration Fig. 7.582 of A = a (right direction) = ,,1 ma = TsinO = _F_ x sinO Acceleration of C in vertical downward direction 2cosO =-2(a+b») Hence, net acceleration of C = a i - 2 (a + b») (see Fig. 7.580) F F OB F --,0x===· - tan 0 = - - - /02--= x = x X2) 2 2 OA 2 Fx So, a = - x [\"=== 2m )(a2 _ X2) Fig. 7.580 42. c. In a given system, g From length constraint = I, +h+i3 +/4 =C m! +m2 8 mj - m2 1 1~/+I~+I~+I~=O = °- °-a - b + a - b + c = m,+m2 8 c=2a+2b 8m! - 8m2 = tn, +m2 Prom wedge constraints, acceleration of C towards right siue 7m, = 9m2 ml 9 is a. Acceleration of C w.r.t. ground = a i - 2 (a + b») -=- 40. c. In Fig. 7.581 givcn (V = lOm/s) 43. b. ab,! = ab - a, = (g - a) + V.I\" = v sin 45° iib = g.j. IOm/s 44. c. T = N sinO and N = mgcosO e . e 2T = mg cos S10 = mg8.m 211 V:v = V cos 45° Fig. 7.581 ...N cos (900 - e) N sine Vg x2 .~,----+- V', .= ~ .- f- T (900 ~ 8) fi i 10 10 V., = -.fi - -.fi x2 V, = - .1f0i mls and Vy = .1f0i m/s N sin (900 \"\" fl) tooV = 2 + 2100 = lOy~2+ 2 = 10 mls Fig. 7.583 45. d. a-:> F = -10 A (m/s2 ) - = m j Displacement in y-direction 41. c. From Fig. 7.582, F = 2T cosO y = ut + 1 2at2 or T = FI(2cosO) 0=4 x I - :12 x 10 x (2 The force responsible for motion of masses on X-axis is TsinO.
7.130 Physics forIIT-JEE: Mechanics I 4 50. b. m' = -3 x 10 = 1 kg t= - s 30 5 T=20=lx4=>T=24N 46. c. At equation 2T cose = Mg 'T = 2mMg m+M , T ', T 2mg =--m-;2mg e',e 1+- Mg M Fig. 7.584 Mg M Hence the total downward force is 4mg (Fig, 7,588), cose= -- =- 2mg 2m 2T cose < 1 M TT - <1 2m m M <2m 47. b. From Fig, 7,585 M Case M I Fig. 7.588 Fig. 7.585 51. b. In the given system, Mg- B = Ma B-(M-m)g=(M-m)a (m, - m2)g g From equations (i) and (ii), we get a= ::::: m! +mz 8 mj - m2 1 = - -+ mz 8 ml 8m, -8m2 =m, +m2 7m, \",9m2 (i) mj 9 (ii) 52, c, 2Mg Fig. 7.589 m= a => T = mg -rna = e48. b, Force exerted by block on inclined plane M g cos 53. a, As F2 and F3 are mutually perpendicular, their resultant Mgcos& JF{+ Fl- Fig. 7.586 As particle is stationary under F\" F2, F3; therefore, 49. a, As shown in Fig, 7,587 fFf+ FJ must be equal and opposite to F, ' a = 32 - 30 =4 m/s2 3 J- - -When Fl is removed! resultant force is Fi + Ff· ~~T~.......•. Therefore, acceleration of the particle 1+-+1 JF{+FJ F, IOcm = =- mm Fig. 7.587 54. d. For a: (Fig, 7,590) 5g - T = 5(2C) ForC: 21'Sg = SC => C = !L = ~ ms-2 14 7
Newton's Law of Motion 7.131 60; c. From Fig. 7.593, Fig. 7.590 FF 2F 55. c. Charge in the length of the string should be zero. ~ N Mg mg Fig. 7.593 Fig. 7.591 2F-N - Mg = Ma Hence, 2F-mg+N =ma IXI = Ixl 4F - (M + m)g = (M + m)a IXI = displacement of M with respect to ground 4F - (M +m) Ixl = displacement of m with respect to M. a= M+m 61. b. The situation bet\"re breaking the string 20 cos 60° + 20 cos 60' is shown in Fig. 7.594. dP •,20 cos 60~ + 20 cos 60° 56.d.AsF=- A '\" f Jdt fdt dp = /:'1' = Area under F -t graph Fig. 7.594 /:'1' = 21: x 5 x 4 - I x 5 = 5 kg m/s 57. c. T, (Hanged part) = 3T; (Sliding part) . . x = 3x' =} 3 x 0.6 = 1.8 m/s From Fig. 7.594, 58. a. VA = 2 m/s (towards right) 20 cos 60\" + 20 cos 60° = mg VPI = \"V2A = 1 m/s (upwards) , mg = 20N 62. d. The acceleration of block-rope system is VB = 2 m/s (towards left) F a = -..:..:- 13 A (M+m) =whcre M = mass of the block and m mass of the rope So, the tension in the middle of the rope will be T = {M + (m/2))a = M + (m/2) F , M+m Given that m = M /2 .. T= [M ++((MMI/24)) ] F = 5F M 6 m 63. d. The reading of the spring scale is the normal reaction be- Fig. 7.592 tween the man and the spring scale (see Fig. 7.595). Now 2VP2 = VB + VPI r VB + VPI 2 + I ' ~l;ft ~~a VP2 = 2 = -2\" = 1.5m/s ~ ~ 59. b . a = -.3'.:1_'_-,.:m=g,::s=in::.:.O Initially, N',\"\" mg N'+ma=mg m Nb mg ·- ma 3 x 250 - 100 x 10 x sin 30° 100 = 2.5 m/s2 Fig. 7.595
7.132 Physics for IIT-JEE: Mechanics I As the rcading decreases, it means that the normal reac- 68. d. Let upthrust = F (ii) tion is also decreasing. Firstly, the lift must be moving up- Mg-F=Ma (ii) wards with a constant velocity and then decelerated to rest. F = (Mg - Ma) 64. b. F 1,fl fl Balloon m Mg Fig. 7.599 Fig. 7.596 Let In should be removed, then tan;'! =\"152 10 F - (M - m)g = (M - Ill)a cos;'! = -- Solving equations (i) and (ii), we get 13 m =2 M-a- T, cos;'! + T2 cos;'! = mg (i) (ii) g+a 1', sin;'! = 12 sin;'! 69. c. By virtual work method: T,=72=1' 2YXA = T X XRA 2xA =xBA 2T cos;'! = mg X/JA =2 x 5 = 10 T = _.,,!_L =* aj) = aHA + aA J102 + 52 = .1100 + 25 -.J12s 2 cos;'! 5 65. b. To move lip with an acceleration a the monkey will push the rope downwards with a force of 40 a + +~11al1 = mg 40amax ; 600 = 400 40a 40200 2 a max = = 5 m/s So, the rope will break if the monkey climbs up with an acceleration of 6 m/s2. 66. c. From the Fig. 7.597 60\" '0 T P 30N Fig. 7.597 1'eos60'=30N=* T=60N 70. c. From Fig. 7.60 I 72 cos 0 = mg T sin 60° ;::;; 1'2 = W 12sinO = mg w = 60 v\":l = 30vr3;; N T 67. b. From Fig. 7.598 a A F 2 Tease F f3 f3 ee '1-B'''---1> F'\"\" mg mg T, TT Fig. 7.601 m 1', sin a = mg.J2 sin 4S G Fig. 7.598 COSet 2mg T=F Tl sina = m,g;:;:;} sina = m,g 2TcosO = Mg I 2 F=~ tana = ~ cosa = -.J5 2eose 2 To make the rope straight, 0 = 90\" F should be infinite, 2 T, = mg.J5 1'1 I< =2mg v5
Newton's law of Motion 7.133 · T, vis Here, weight of the hanging part will be balanced by the T2 = V2 =} V2T,= vlsT2 frictional force acting on the upper part, Le., 71. a. For vertical equilibrium of the block (Fig. 7.602) mlg = J-Lm2g Solvc to get (m,IM) x 100=20%. Fsin8 75. c. Frictional force: R---~ F = \"R = 0.5 x mg = 0.5 x 60 = 30 N ,8 F Now F=T,=T2 cos45\" \" ~+--, or 30 = T2 cos 45° Fcos8 j1R mg and W = T2 sin 45\" Fig. 7.602 Solving. them we get W= 30 N. R = Feose +mg (i) 76. c. Acceleration of the suitcase till the slipping continues is While for horizontal: a= -,{m-ax (ii) m FsinO = ,\"R a = vmg = Vg = 0.5 x 10 = 5 m/s2 From equations (i) and (ii). we get m F sinO = II(F cosO + mg) Slipping will continue till its velocity also becomes 3 m/s. e ._.or F = ~,,_m~g~~ +v = II al (sin \"COS IJ) or 3 = 0 + 51 or I = 0.6 s In this time, the displacement of the suitcase 72. a. Force of Ifiction, F = \"R = \"mg I, I ., Reta!'datt·on due to f.n.ct.Ion: F =-c \"-m\"-g- = /J-g 8, = 'lal- = 'l x 5 x (0.6) =O.9m m rn and the displacement of the belt, Given II = 10 mis, s = 50 m, v = 0, a = -\"g 8, = vt = 3 x 0.6 = 1.8 m Now v2 = u2 + 2as Displacement of the suitcase with respect to'lhe belt or 02 = 10' + 2(-10,,)(50) s,-s2=O.9m 73. b. or Ii = 0.1 77.' a. Normal reaction on the block from the wal!: R=F=ION R + I' sin 60\" = Mg Weight of the block will be balanced by the frictional R = Mg - I'sin60' force Frictional' force W = Frictional force = \"R = O.2x 10 = 2 N. 78. a. For upward accelerat.ion of M I F=\"R = ,,[Mg - I' sin 60\"J M,g 2: M,gsinO + VM,gcosO Rp =} (M')m;\" = M,(sinO + Veose) For downward acceleration: F (M,gsin 0 - VM,gcoslJ) 2: M,g Mg =} (M2)m\" = M, (sin e - V cos 0) Fig. 7.603 The body just moves when 79. a. During downward motion: I' cos 60\" = F F '= mg sin 0 - \"mg cos 0 or~> HI x P~]or I'cos60\" = ,,[Mg - Psin60\"J During upward motion: 10- 2F = mg sine + vmgcose or P '= 5.36 N Solving above two equations, we get V = (tan e)/3 80. 3. The string is under tension, hence there is limiting friction between the block and the plane (Pig. 7.604). 74. a. Let the total mass of the chain is M and the mass ofhanging +=} VN 50 cos 45\" = ISO sin 4SO (i) part is m, . Then, the mass of the part placed on the table will bem2 = M -nIl. L:F,.=O
7.134 Physics for IIT-JEE: Mechanics I N celeration a. Then for whole system: y x F - JL2(M + m)g = (M + m)a (i) LL45°~_ __ (ii) V For block of mass m: Mg~ 150N /1 = ma or lJ...jmg = ma Fig. 7.604 or From equations (i) and (ii), we get => N = 50 sin 45\" + 150 cos 45\" (ii) F = (M + m)g(JLl + JL2) Solving equations (i) and (ii). we get JL = 1/2 85. c. Maximum frictional force on block B 81. c. The minimum force required to just move a body will be f, = JL,mg. After the motion is started, the friction will I:=JLmBg =0.4 x 3x 10= l2N become kinetic. So the force which is responsible for the Hence, maximum acceleration = = 4 ms-2 increase in the velocity of the block is Hence, maximum force F = (rnA + m,,)a = (6 + 3) x 4 = 36 N F = (JL, - ti.lmg Aliter: We can also apply the forrmila discussed in the pre- vious problem by putting JL2 = 0 and JL, = 0.4. = (0.8 - 0.6) x 4 x 10 = 8 N 86. b. For the equilibrium of block of mass M,: So a= F 8 Frictional force, f = tension in the string, T (i) -= (ii) m -4 = 2m/s2 where T = f = ,\"(m + M,)g e82. a. If the plane makes an angle with horizontal, then For the equilibrium of block of mass M2: tane = 8/15. IfR is the normal reaction 'T = M2g C~)R = 170gcosB = 170 x 10 x From equations (i) and (ii), we get = 1500N JL(m + M,)g = M2g Force of friction on A = 1500 x 0.2 = 300 N m= -M-2 M , Force of friction on B = 1500 x 0.4 = 600 N Cunsidering the two blocks as a system, the net force JL parallel to the plane is 87. a. Maximum frictional force between the blocks is = 2 x 170gsine - 300 - 600 fmax = Ilmg = 1600-900=700N So maximum acceleration :. acceleration = -700 = 35 - = amax = fmax/ m = f-Lg 340 17 mjs2 88. b. Horizontal acceleration of the system is Consider the motion of A alone. a= 2m F = F +m +2m 5m 170g sine - 300 - P = 170 x 1357 (where P is pull on the bar). P=500-350= 150N -----0-. a 83. a. For first half, acceleration = g sin 1> Fig. 7.606 .'. velocity after travelling half distance: (i) Let N be the normal reaction between Band C. Free v2 = 2(g sin 1»1 body diagram of C (Fig. 7.606) gives N = 2ma = 2 -F 5 Smooth Now B will not slide downward if JLN ?: mBg Rough F?:(~F)Or 8 JL ?: mg or }JL mg Fig. 7.605 5 -mg For second half, acceleration = g(sin 1> - JLk cos 1» or Fmin = 2JL So 02 = v2 + 2g(sin 1> - JLk cos 1>)1 (ii) 89. a. j, = JLMg, f = F = Fot Solving equations (1) and (ii), we get JLk = 2 tan 1>. Motion will start when f = j, 84. d. Here, the force applied should be such that frictional force = ==} FoT JLMg => T iL Mg F\" acting on the upper block of m should not be more than the limiting friction (= JL,mg). Let the system moves with ac-
Newton's Law of Motion 7,135 90. d. Maximum friction that can be obtained between A and B 94. c. In the free body diagram of m [Fig. 7.609(a)] (i) T = mg is II = f.l.mAg = (0.3)(100)(10) = 300 N T and maximum friction between B and ground is j, = f.I.(mA + mB)g = (0.3)(100 + 140)(10) = 720 N Drawing free body diagrams of A, B, and C in limiting case ~ T T ~ Fig. 7.607 (a) N (b) Equilibrium of A gives Fig. 7.609 TI = II = 300N (1) In the free body diagram of M [Fig. 7.609(b)]: Equilibrium of B gives ~N=3T 00 2T, + II + j, = T, ~ T+~=N N = (m + M) g O~ or T2 = 2(300) + 300 + 720 From equation (iii) f.I.,(m + M) g = 3 mg = 1620 N and equilibrium of C gives me g = T, From equation (ii) or lOme = 1620 or me = 162kg (2) f.I.,M 1/3 x 8 91. d. If the blocks move together, m = (3 - f.l.2), = -=(3:'-_-'1:-7/3:::-) = 1 kg F 10 5 _, 95. c, As shown in the free body diagram of 1 kg and 2 kg blocks a= =-=-ms (Fig. 7.610) rnA +mB 6 3 IB (frictional force on B) $ mg = mfiF = 20 N fllNI rnA +ma 3 N, iBmex = f.l.mAg = 0.4 x 2 x 10 = 8 N ~I~I~M\"g II As f B mex > J/l, Hence the blocks will not be separated +-- F 92. c. Minimum effort is required by pulling a block at the angle of friction. fl2N2 N2 93. d. In the free body diagram of B [Fig. 7.608(a)] Fig. 7.610 Minimum force required to pull block M -+ F=f.l.INI +f.I.,N2 = f.l.1 mg+f.l.2(Mg + Nil I = f.l.1 mg + f.l.2(Mg + mg) = 0.1 x 10 + 0.2 (20 + 10) = 7 N T mg /!!;s (a) (b) => tJ) = - aD Fig. 7.608 N=mBa (i) f =msg 2s - 2s gsine + f.l.gcose 2 gsine - f.l.gcose f.l.N = mBg (ii) e -4g sin 4f.1.g cose = g sine + f.l.g case From equations 0) and (ii). a = f = 20 mis' 3gsine = 5f.1.gcose f.I. In the free body diagram of the bob [Fig. 7.608(b)] 33 f.I.=-tane=- eT sin = rna (iii) 97.c. N=Mg-Fsin¢5 5 Tease = mg (iv) From Fig. 7.611 From equations (iii) and (iv), we get Fcos¢ - f.I.(Mg - Fsin¢) a tane = - '* e = tan\"1(2). a= g M
7.136 Physics for IIT-J.EE: Mechanics I From Fig. 7.613 Fsin 11 r l - + - + F cos ¢ N Mg Fig. 7.611 fIs98. c. From s = ut + _2I at2 = 0 + -2Iat2• t = - a For smooth plane a = g sin 0 Mg For rough plane, a' = g(sin 0 - Il cos 0) Fig. 7.613 -nt-n ~t'= 2s Equations of motion are y.' g(sine - f-LcostJ) - - T - fl + mg sin 0 = ma ~ (I) .. n2g(sinO -ILeOSO) = gsinO Mg sin 0 - T - h = Ma (2) When 0 = 45°, sinO = cosO = 1/../2 Solving equations (I) and (2), we get T = 0 (1 -Solving, we get Il = ~2 ) options (a), (b), and (d) are wrong. 99. b. Retardation of train = 20/4 = 5 m/s2 M It acts in the backward direction. Fictitious force on suit- 102•. a. Mass per unit length of chain = (; case = 5 m Newton, where m is the mass of suitcase. Mass . suspended part = (M; y of the It acts in the forward direction. Due to this force, the For equilibrium of chain, the weight of suspended part suitcase has a tendency to slide forward. If suitcase is not to must be balanced by force of fi'iction on the portion on the slide, then S m = force f of friction table. or Sm = Ilmg or Il = 1S0 = 0.5 M(6-y)g M Il . = -yg 66 100. b. For a frictionless surface, 2I: (6 - y) = Y =? (6 - y) '= 2y a = gsin4S0 = g/../2 .. 1= Ig X 2 fIs(1) 3y = 6 =? Y = 2 m t2 2- .x.-/2- In the presence of friction, ut + _I at' 0 + -I at2, t 103. b. From s = = = -- 2 2 a a = ..L _ Ilg = ..L (I - Il) For smooth plane, a = g sin 0 ../2 ../2 ../2 For rough plane, 0' = g(sin 0 - Il cos 0) .. 1=~X[Jz(l-fl)Jxti (2) I 2s ____ t' = / . . Yg(sin 0 - Il cos 0) Dividing eq. (2) by eq. (I), we get (2 1l=3/4 n)g~~e= lit = tiI = ( I - I l ) _ c \" 101. c. NI = mgcose and fl = wngcosO .. 1l2g(sinO -Ilcose) = gsinO From Fig. 7.612 When 0 = 4S\", sinO = cosO = 1/../2 (I - ':2)Solving, we get Il = 104. b. Fig. 7.614 EC= vf2 +N 2=m{g+a) fN mg m(g+a)sinB Fig. 7.612 meg + a) cos f) N2 = Mgcose and.f2 = IlMgeose Fig. 7.614
Newton', law of Motion 7.137 105. b. Net forte applied by block on the inclined plane is equal 11 1 to the weight of the body. K' K (I - 113) 106. d. During downward motion: F = mgsinB - Jl.mgcose K'= 2K 3 During upward motion: 115. b. The free body diagram of m in frame of wedge, e e+2F = mg sin {.Ung cos Solving above two equations, we get: I1 N /1= -tanO=- 33 107. d. j, = /1mg (frictional force on mass m) ~ mg So, maximum acceleration Fig. 7.615 f,· /lmg N = mgcosCi - ma sina = - = ~- =fi8 Now f = fiN = macosa +mg sina mm 108. d. Suppose due to the force R on 8, both blocks A and B acosa+ sina move together. fi= In this case, g COSet - a sin ex F = +(rnA m/3)a = (2 + 5)a or a = I'/7 a + tana 5 12 Now force on A = rna = (2F/7) fi= For no relative motion between A and B, 2F/7 must not g - a tan 0: exceed the limiting force of friction between A and B. The 116. d. The minimum value of I' required to be applied on the limiting force of frict.ion between A and B is given by blocks to move is 2 x (2 + 4) x 10 ~ 12 N. Since the applied fimAg = 0.8 x 2 x g force is less than minimum value of force required to moye the blocks together, the bloeks will remain stationary. 117. b. Velocity of liquid through inclined limbs ~ 2v: 2F Rate of change of momentum of the liquid - = 0.8 x 2 x g or F ~ 56 N. 7 (¥)2= pAv' + 2 [PA cos 60.J = ~PAv2 109. b. The frictional force on the block A that represents the carl back is given by rna. 118. c. In equilibrium, there will not be any friction between the cylinder and the wcdge. If 0 be the required angle The upward frictional force, F ~ fima Cylinder A: mg sin 60\" ~ kx cos (600 - 0) (= mg cos 30°) For block A to be stationary. g -::= 20 Cylinder B: mg sin 30° Ii ::;:::} amin = m/s2 fLlna :::: rng or a ::: - = kx cos(30\" + 0) = kx sin(60° - 0) Ii fi cot 30° = cot(60\" - 0) +110. b. Frictional force ~ fiR = fi(mg Q cos 8) and horizontal 09 0~30°. epush ~ P - Q sin 119. b. x 2 + y2 = /2 For equilibrium we have, On differentiating, we get eli(mg + Q cos 0) = P - Q sin xV,+yV,.=O P - QsinO 09 Vy = Y component of velocity of B which is along the rod, i.e., BA (Fig. 7.616) fi=~--- mg+ Qcose 111. c. Given horizontal force F ~ 2S N and the coefficient of friction between block and wall (fi) ~ o. L We know that at equilibrium horizontal forec provides the normal reaction to the block against the wall. Therefore, nor- mal reaction to the block (R) ~ F ~ 25 N. We also know that weight of the block (W) ~ Frictional force ~ fiR ~ 0.4 x 25 ~ ION. 112. d. As there is no relative slipping between the block and Fig. 7.616 cube, the friction force is zero. ~y =. y 3 m/s dm Iv,::1IV.'! = 113. d. Force = V~ .2 ell Hence, VII cos 30\" = r;; m/s 114. c. I = 1 + -I + I + ... 00 '13 - - ~ K' K 3K· 9K
7.138 Physics for IIT-JEE: Mechanics I VB = 1 m/s 123. h. The free body diagram of 10 kg 120. d. Free body diagrams (Fig. 7.617): N' =10 kg ~ N' ... block 10 kg will slip, ~ •F !LN' = 0.3 x lOx 10 = 30 N Friction =' 30 N Fig. 7.617 124. c. Equation of motion for A (Fig. 7.619) Equations of motion: Kx =ma => kx a=- m F + x-direction) aB = (in For B, F - T = ma' - M F-kx a\" = !... (in - x-direction) =} a' = - - - m Relative acceleration of A w.r.t. B: m =} The relative acceleration == ar = la' - aI = F-2kx --- m m kx kxkx m 0-----0- ~ ~F AB (along -x-direction) Fig. 7,619 Initial relative velocity of A w.r.t. B 125, h. As the springs are fixed to the horizontal and have the same UA,B = Vo natural length. Hence, if one spring is compressed, the other will be expanded. Hence, the compression will be negative Final relative velocity of A w.r.t. B = 0 The free body diagram of m, [Fig. 7.620(a)] Using v' = u2 + 2as T + F2 = 80 Nand F2 = 70 x 0.5 = 35 N o- ,_ 2 F (m + M) S - Vo mM S= Mmv2 T = 80 - 35 = 45 N 0 TT 2F(m +M) 121. a. +V2cosa VI cos a = v\\ V2 = v\\ [ 2 Sin2(a/2)] COSa' 122. h. Choosing the positive X-Y axis as shown in Fig. 7.618. the momentum of the bead at A is p, = +mv. The momentum of the bead at B is Pi = -mv. A (a) rl/2g=80N (b) rIIjg\"\"20N FBD ofmz FBD ofmt ••_--'.l_._\"y___...Jt. --x +x B Fig. 7.620 Fig. 7.618 FBD of In, Fig. 7.620 T+F, +mg Therefore, the magnitude of the change in momentum between A and B is or 45 = F, +20 IC.p = p,. - p, = -2mv F, =25N i.e. IC.p = 2m v along positive X -axis. 25 25 The time interval taken by the bead to travel from A to X, = - = - = 0.5 m B is k, 50 rrd/2 rrd .': compression in first spring = -0.5 m M=--=- 126. h. Equation of motion for M: v 2v Since lvt is stationary Therefore, the average force exerted by the bead on the wire T - Mg = 0 is => T = Mg F\" = IC.p = (2mv;'fd) = 4mv'. Since the boy moves up with an acceleration a Fig. 7.621. M 2v rrd
Newton's law of Motion 7.139 TT TT T T 111g sin 30° \"--'--_ _ _ _-' Fig. 7.623 T T Mg 130. C. l\"t = 2 sin wI - 1 (a) (b) Fig. 7.621 (e) j,na, = 0.25 x 10 x 1 = 2.5 N As maximum value of the force applied on the block is T - rng = rna less than 1m,,' hence, the block will not move. =} T=rn(g+a) 131. d. As there is no relative slipping between the block and Equating egs. (i) and (ii), we obtain cube, the friction force is zero. Mg=m(g+a) tim 132. d. Force = V - dt 133. b. (~)=} a = g, the block M can be lifted Tn--9.r N, m -I Nt T'b 127. d. Figure 7.622 T N~ & [0-..fl- ./ I A .rz. A TB .ft mg mg T-fi =ma ij-ii=ma (a) (b) Fig. 7.624 • /1 134. a. We draw axes for each block along the incline and the normal to the incline. The component of the acceleration for each block are as shown in Figs. 7.625 and 7.626, where a is accelenition of wedge. .h=ma Mdg-T=MDa mt Fig. 7.622 Solving we get, o g (sin 0) a sin e g (sinO) o =} I, = 2rna;MDg = (MD + 3rn)a It It and I, = f1.N = f1.(2mg) +=} MDg = (MD 3rn)f1.g Fig. 7.625 Fig. 7.626 3mf1. It is obvious that vertical component of acceleration is MD=-- larger for block in Fig. 7.626. t -f1. •. _ F-JI _ F-f1.m,g _ mls2 .. T, > T2 128. e. a, - --- - - 10 135. b. The free body diagrams of two large blocks are given in tn! mj a2 = -F + f1.'\"2g = -1 m/s2 Fig. 7.627 and Fig. 7.628. m2 :. s I a\",,1 2 = I 1 0- (- 1 )) 12 =2: 2:[ =} 1=2s (i) Fig. 7.627 Fig. 7.628 129. d. For S, (see Fig. 7.623) From FBD it is obvious that the net force on each block rng - 2T = ma (ii) is zero in the horizontal direction. For A, al = a2 = 0 mg T = - - = 2rna 2 Solving a = 0
7.140 Physics for IIT-JEE: Mechanics I 136. a. . = _1at' =} a should be same in both the cases, 142. b . .ft = Itmg, friction will provide the necessary centripetal SIIlCD, h 2 force (Fig. 7.633). f = I/lIiJ'r because hand t are same in both the cases as given. I :0 II =} moir:o Itmg =} It 2: gw2r 2' x 50/100=10 =} It 2: 0.2 •f mg mg Fig. 7.629 Fig. 7.630 In Fig. 7.629 Fl - mg = ma Fig. 7.633 => F! = mg+ma 143. c. (b) and (d) are the standard statements related to Newton's second and third law, respectively. In Fig. 7.630 21', - mg = ma For option (c), pseudo force is an imaginary force, New- mg+ma ton's third law tells us about the material interaction forces. FOl' option (d), although Newton's lirst and second laws =} F, = -\"-...,.-- are valid in inertial frames only as they are concerned with 2 the motion of body (directly or indirectly) but Newton's third law is valid in all frames . Fl > F'z 144. d. 137. a. For a chain to move with a constant speed P needs to be 145. d. Due to the malfunctioning of engine, the process of rocket equal to the frictional force on the chain. As the length of the chain on the rough surface increases, the friction force fusion stops and hence net force experienced by the spacecraft becomes zero. Afterwards the spacecraft continues to move .It: = J1.kN also increases. with a constant speed. 138. b. 1.8t -Itk 15 = 1.5(1.2t - 2.4) r . . .1----..j~~ 1.8 x 2.25 Fig. 7.631 146. d. Spring balance reads the tension in the string connected to its hook side. As the spring balance is light, the tension in the string on its either side is the same. Now the only thing that For T = 2.85 S remains to be found is the tension in the string which could Itk = 0.24 be found easily by using Newton's second law. 139. a. tan e = v'/ Rg 147. d. If we consider the situation shown (a) in Fig. 7.634 the FBD would be as shown in Fig. 7.634(b). 140. a. Vmax = v2b IIII\\~\\\\~\\I\\I\\\\ \\111\\111\\1\\\\\\\\1\\1\\1\\\\\\ 17=- tal (al Fig. 7.634 Rg Rg [tanO + It] 141. b. K = 10' N/cIn = 104 N/m. Let the ball moves a distance x away from the centre as shown in Fig, 7,632. Fig. 7.632 In this situation, the option (b) seems to be correct, but in other case like if this system is in a life moving with some kx = mw'(O.l + x) acceleration, then N oF mg. So, (d) is the correct option. 104x = 90 (10')' x (0.1 + x) 148. a. Let us flrst assume that the 4 kg block is moving down, -- X then different forces acting on the two blocks would be like 1000 as shown in the Fig. 7.635. [Normal to the incline forces are not shown in the Fig. 7.635.1 Solve to get: x '\" 10\" In To have the motion, the frietional force .f should be equal to the limiting value. Le., fl, = \"\"smg cos 37 4 =0.27x IOgx-=2.16g 5
Newton's Law of Motion 7.141 component of gravity force along the hill is balanced by lim\" iting the frictional force. emg sin = f-L.I'mg cos () '* e e= 11m\" 1(II,,) :::0: 3T where is the new slope angle of hill. 152. a,The FBD of the block is shown in the Fig. 7.638 N = 80cos3T=64N Fig. 7.635 80 sin 3T Here, the 4 kg block is not able to pull the 10 kg block +-4-N up the incline as 4g < 109 sin 37 + fL, so the system won't 80 cos 31\" move in the direction that we assumed. So, if there is, a chance 4g Fig. 7,638 of motion of system it can only move down the incline, and So, /L = 0.2 x 64 = 32 N the system will move only if the net pulling force down the As 4g < 80 sin 37°, so the frictionali'orce will act down\" incline is greater than zero. For down the incline motion, the wards. FBD is as shown in the Fig. 7.636, Net applied force in upward direction (excluding friction force) is 109 sin 80 sin 3T - 40 = 48 - -40 = 8 N 4g As 1~1pp!icd in vertical direction is < /L so block won't Fig. 7.636 move in vertical direction and value of static friction force is, For a to be nOll-zero, i.e\" +ve f = 8N, 109 sin 37 > fL +4g which is not, so the system is neither moving down the incline, 153. d. The FBDofblockfrom the lift frame is shown in Fig. 7.639, nor up the incline and so the system remains at rest. From the given data, as meg + ao) sin e > 2 mao cos e. 2maocosO 149. c. Let the weight of each block be W (Fig. 7.637). y~ -...lstcasc '\\\\' \\\\\\\\ \\ WN, m(g + ao) I;jrl-N, \"\" W Fig. 7.639 , \" N)=N2+W'\"'2N So, the friction force acts upwards, ~2 f = m (g + aD) sin 0 - 2mao cosO WN) 9g 4mg mg Fig. 7.637 ----= 10 5 10 N3 eN = meg + aD) cos 0 - 2mao sin = 9mg So, N, = N, = 5 2 150, c. Here the frictional force would be responsible to cause the acceleration of truck. Here the maximum frictional force can As.h = 1l.,N = S18mOg = 92m5g > f \"M2g- . be f = II x where M -+ Mass of enttre truck. TIli'S I.S 1hc net force 'actmg on tyre, so Ma = \"I-l-M'2\"g- so the static friction. a= 0.6 x 10 =3 2 0Reaction force, - - 2....- mls R = y ~f\"-+ 2N = -mg .. + 92 =-m-\"g-v::'I-3- 54 2 151. c. As the sand particles arc sliding down the slope of the 154, c, As the block is in equilibrium, a gravity free hall, its FBD hill gets reduced. The sand particles stop coming down when would be as shown in Fig. 7.640.
7.142 Physics for IIT-JEE: Mechanics I Fig. 7.640 157. b. If the wedge moves leftward by x, then thc block moves down the wedge by 4x, i.c., W.r.t. wedge the block comes As the block is at rest, the friction is static in nature and down by 4x. its value is equal to the applied force (F), i.e., I = F. a+----,-------- As gravity is not present, and no other force or compo- a 4a nent of any force is acting on the block in the vertical direction a hence nonnal contact force between the table and the block Fig. 7.644 would be zero. IPReaction force, R = So the acceleration of block w.nt. wedge = 4a along thc + N' = --/ F' + 0' = F incline plane of wedge (Fig. 7.6;44). 155. a. Free body diagram of various components are shown in Acceleration of wedge with respect to the ground is a, along left. So acceleration of block with respect to the ground Fig.7.64!. is vector sum of the two vectors shown in the figure. i.e.,laBGI = #+c'4a'\"jT+TX-;;--x4a x'-c-os-(;--n---a-::) = (--/17 - 8 cosa) a mis' 158. d. Using the constraint theory (Fig. 7.645) II + 2/, + IJ = constant. =} VI + 2v, + V3 = 0 Fig. 7.641 The adjacent for horizontal equilibrium, F = Nt of 15 kg block and Nt = N,for25kgblockasshowninFig. 7.642. i.e., Nt = N2 = F ILl (limiting friction force) = !\"NI = !\"F 25g Ii 15g Fig. 7.645 Fig. 7.642 Take downward as +ve and upward as -yeo So, +12 + 2(-4) + V3 = 0 VJ = velocity of pulley P = -4 m/s = 4 mls in upward direction 156. d. The direction of acccleration of B is along the fixed incline, VB = Vp - VAP = -4 - (-3) = -7 mls and that of A is along horizontal towards the left. C',,--------Ia A i.e., block B is moving up with a speed of? m/s. \",, ,' tlh=if 159. c. As the eraser is at rest w.r.t. board, friction between two is , static in nature. \\\\,, ,, For Figs. 7.646 (a) and (b), the friction force is same as ___()~I that of gravity force as shown in Fig. 7.646(a). a B Fig. 7.643 From Fig. 7.643, acceleration of B is represented by ,V Mg (b) AB while its horizontal and vertical components are shown (a) by AD and 0 B, respectively. Fig. 7.646 DC,Acceleration of A is represented by DC = ea sin a cot
Newton's Law of Motion 7.143 For (c), f = F2 + Mg > Mg [shown in Fig. 7.646(b)] V = -I[40t - 40 - 5t 21 For (d), f = F2 - Mg as angle by which arm is tilted is m P = mv = 40t - 40 - 5t' very small, so F2 would be small. Hence the correct graph is (c). 160. ~ =4,, o~ = F, - 4 }0 164. a. Let m starts moving down and the extension produced in - =1 spring is x at any time as shown in Fig. 7.649. Value of ;; b.u +2},a m required to move the block m is: Let at any time, the coordinates are (x, y) X- 2= u x' + I2 lax! =} x - 2 = 4t + 2I t' and Y - 3 = 2t - -14t2 2 =} y - 3 = 2t - 2t 2 When y ~ 3 m, t ~ 0, 1 s at t ~ s, x - 2 = 4 x 1 + 2I =} x ~6.5 m 161. a. fl, = \"'lmAg = 0.3 x 300 = 90 N Kx = \",mgcosli +mgsine \"'2fl2 = (mA + mB) g = 0.2(300 + 100) = 80 N =} kx = 4 + 3 = mg 0.5mg- mg- \"'3f l3 = (rnA + mB + me) g 55 For minimum M, it will stop after producing extension =0.1(300+ 100+200) = 60N in the spring x, Mgx = -I kx2 1 2 =} Mg = 2kx fl, p 2 2\"I m fl2 ,--J---\"---1, =} Mg = mg =? M = fI) 165. d. Acceleration of both m will be zero after releasing. But Fig. 7.647 M will accelerate down. So the spring will get elongated for any value of M. 162. a. 2T - (M + m)g = (M + m)a 166. c. If we take two points I and 2 on a string near pulley P as T-mg+N =ma (1) shown in Fig. 7.650, then velocities of both points I and 2 will be same. Hence P does not rotate but only translates. (2) :=T T 2 Fig. 7.650 Mg mg N 167. c. In equilibrium (Fig. 7.651) mg T=mg Fig. 7.648 N =3mg From equations (1) and (2), we get: f =21' = 2mg In limiting case f < fmax N=(m~M)(g+a»o 2mg < \",N As m > M. thus if T increases, a increases and if a 2mg s: 3\",mg rn->3_2 increases then N increases (see Fig. 7.648). T N 163. c. From 0 to 2 s: at any time t, F = JOt 1\"=} a = Flm = 10tim m 2T mg 2m =} dv = l'lO- dt t T=mg o 0m 2mg Momentum: P = mv= 5t 2 at t ~ 2 s, P ~ 5(2), ~ 20 kg mis, v ~ 201m from 2 to 4 s: F = 40 - lOt Fig. 7.651 \" l'dv = 40-lOt elt 120/m 5. 3N = 6a (Fig. 7.652) 2m 168. c. F - N sin 37\" = 6a =} F - (1)
7.144 Physics for IIT-JEE: Mechanics I -a T N sin 37° F--1L__~_kg_'~~~701~~--+a Fig. 7.652 II fi 54N (2) mg cos 37° N, = 4g - N cos 37\" = 40 - (3) Fig. 7.654 N sin 37' - f = 4a (4) .fFrom equations (1) and (3): F - = lOa (5) =} F - ItNz = lOa =} F - It [40 - 4;}= lOa Put the value of N from equation (1) in equation (5) and From free body diagram of B SF - 60 N\\ = mg cos 37') . also put the value of It to get: a = ~-. T = mg sin 37\" + fl Now to start the motion: a>O (where It is coefficient of friction) =} F>I2N From the free body diagram of A So the minimum force F to just start the motion is 12 N. Now the maximum F will be when Nz just becomes O. Nz = Nl + mg cos 37° = 2mg cos 37<' =Then from equation (2): N 50 N. mg sin 37\" = fl + h =From equations (1) and (4) we get F 75 N. h = /LNZ =} It = I (by putting N2 = 0) If we apply F > 75 N. then B will staIt sliding up on A. -4 but we do not want this. 172. c. The free body diagrams of two blocks are shown in 169. d. T = F/2 = 62 N. This tension is not sufficient to lift 7 kg Fig. 7.655. Under the assumption that both the blocks are a,block as in Fig. 7.653. Hence its acceleration = O. moving together, For 5 kg block: T - 5g = 5a, F + 2g sin 37\" + 3g sin 37\" - fl - .fz = 5a =} a, = 2.4 mis' where fl = It x 3g cos 37\" and h = It x 2g cos 37\" F h TT .NI / 2g sin 37° / 3 kg Fig. 7.653 m,a,+m,az 5x2.4+7xO 1 m/s2 3 ,gsin 37\" +m, m,acm = = 5+7 = Fig. 7.655 170. a. Centre of mass will not shift in the horizontal direction. Let vS m moves a distance x on wedge in the downward =} a = 46 mis' - direction. .J2 ill will also move on the other side in downward 5 + 3g f, For 3 kg block, N sin 37\" - = 3a direction by a distance of x. Then mlX1 = m2X2 =}N=I2N vSmx cos 45\" = .J2mx cos e 173. c. Friction between 2 kg and 8 kg blocks is kinetic in nature e=} cosO = vS/2 =} = 30° =(see Fig. 7.656), so F It x 2g = 0.3 x 2 x 10 = 6 N 171. a. The free body diagrams of two blocks are as shown in Fig. 7.654. For 2 kg block, 10 - 6 = 2a, For 8 kg block. 6 = 8a2
Newton's Law of Motion 7.145 dy x dx dt 2~h2 + x 2 tit 3m------. Fig. 7.656 (I,=} (II = 2 m/s2. = ~m/s2 Acceleration of 2 kg block relative to 8 kg block is, = at =\"45 2 h m/s arel -a2 US'l11g the equatI.On 0 fmo'tIOn, 3 = \"I2 x 45t2 =} t =2.19s 174. c. At any instant, the velocity of two wedges would be of same magnitude but in the opposite directions. This cali be concluded from the conservation of momentum or by sym- Fig. 7.657 metry. v,178. d. Let at any time, their velocities are and v, respectively, , = 34\" Vm ethen VI = V2 cos From constramt theory, v M tie 1)2 sin ~ From energy conservation, e - eDifferentiating: al = a2 cos tit + __~Mv2 x 2 m v2 0 = mgh Hence none of them is correct. III - 22 Nt)t~: Optio# .(a) is ,orrect.initially, because initilltly _ _V,w = _. 32mg.h - vz=Ol . . .. 32M +9m g So the velocity with which wedges recedes away form 179. c. tan 0 = - =} a = gcote a each other is 2VM = Vj3322mMgh+ x4 9m 175. d. Before cutting the string the tension in string joining m4 and ground is, Fig. 7.658 and the spring force in the spring joining m3 and m4 is 180. c. Let a plank move up by x, then pulley 2 will move down by x as shown in the Fig. 7.659. Let end of the string C moves T +m4g. down by a distance y. As the string is cut, the spring forces doesn't change 2 x~ instantly, so just aftercntting the string the equilibrium of mI. f112, andm3 would be maintained butm4 accelerates in upward direction with acceleration given by, a = T +m4g - m4g ----''----''- j v;v v;176. d. Speed: =+ dv 2vx dvx +2v,d.v-v· I, - xi ,-...Ll+A_.L..-, Rate of change of speed: _ = tit dt dt 2j~,~ + v~ - 3x2+4xl -2 m/s2 - ~32 +4' - y~ c 177. c. From Fig. 7.657 (I) Fig. 7.659 L = 2 h - 2y + ~x2 + h' Let Initial1ength of string passing over pulley 2: Differentiating the equation +)'1 A2
7.146 Physics for IIT-JEE: Mechanics I After displacements x and y, mentioned above, the 185, e, T = 2ma lengths become (2) T NT + +(I, - 2x) (12 y - x) Equating (I) and (2): y = 3x length of string that slips through A y +x = 4x and through B: y = 3x ReqUI.red ratio;:;;: -4x = 4- 2mg 3x 3 Fig. 7.662 181. b. For equilibrium mg - T = rna f fe e{Ldmg cos \"= dmg sin f for {LAd! g eose \"= !..dlgsine a = gl3 f f{L dl cose \"= dl sine 186. d. a, = (m, - m,)g and a2 = (m2 m,) g m,'---''----'= (- sine = ~f' cose = ~~) (ma,, + m2) Q,Hence - = I f for {L dx \"= dy or {L! \"= h a, m2)+( I -ml 182. a. From Fig. 7.660 - <1 N F a2 m 2mlm2g As T, = (m, + m2) and T2 = m28 Hence T, 2m, 2 = T, (m, + m,) (1 +m-2) ml Hence, TT~ will depend upon the values of in, and m,. mg sin () mgcos 8 Fig. 7.660 So the relation of N,IN2 will be same as T,. T2 N = mgcose, f = mgsine 187. a. Figure 7.663 Net force applied by M on m (or m on M) mgl2 F= ,fN' + l' = j(mg'cose)' + (mg sine)' = mg N 183. d. As the block docs not slip on prism (Fig. 7.661), the combined acceleration.of the prism is a = g sin e. •N mg , mg sin () Fig. 7.663 i'+--H-L----\" . mg - 2mg: = 2mg: (down) Dnving force = mg Max,.mum restl.llg force = {Lmg = 2mg: (up) 188. h. For constant acceleration if the initial velocity makes an 8 angle with acceleration then the path will be parabolic. Fig. 7.661 189. d. From constraint, the velocity of both the block and the wedge should be same in a direction perpendicular to the emg sin is the pseudo force on m. inclined plane as shown in Fig. 7.664. (aAlL = (aBh, aAX = 15, QAY = 15 N + mgsine sine = mg or N = mgcos'e And for no slipping: mg sin ecos e :s {LN mg sin ecos e :s {Lmg cos' e or {L \"= tan e 184. b. Since m is in equilibrium w.r.t. the observer, so the accel- or aB = -5 mls or a~ = - 51 eration of m should also be a2. So the net friction force (as there is no other horizontal force on m) acting on m should 190. a. In this ease, the spring force is initially 0 be = mass x acceleration = ma,. FBD of A and B (Fig. 7.665)
Newton's law of Motion 7,147. N ,, Fig. 7,66.8 - \\ -... an Nc o s 0 - F = m a =m F- Fig, 7.664 2m m 2m 60 mg 2mg =Solving, we get N 3 F 194. c. For a block to be stationary T = 800 N (Fig. 7.669) ~ Fig. 7.665 Tension in the string and the spring will be zero just after mg sin 8\"\"' 800 N the release. 191. c, Fig, 7.669 If a man moves up by acceleration a. Sol. (Check Fig. 7.666 in the frame of the car). T -mg =mG T 8000 - 5000 = 50a =} a = 6 m/s2 195, b, As in Fig. 7.670 1,+12+Z,=C ilia a mg Fig. 7.666 Applying Newton's law pellJendieular to string I 2m!s e ea mg sin = ma cos 0 ::::} tan = - g Applying Newton's law along string Fig. 7.670 T - mg cos e - ma sin-O = ma d-l,+ -dl2+ -dl', +t-it=. 0 J::::::} + +T = In g2 a 2 ma elt elt elt tit =192. b. Aceelerati~n of the box 10 m/s2 -v-v+0+v+2=0 Inside the box forces are acting on the bob (see v = 2 mls Fig. 7.667). 196. d. As in Fig. 7.671 t VB.I = VBg - ~g VB.! = 4 m/s T 'G)----+. ilia lJ0 4 mist ,,,,: (Pseudo force) Fig. 7.671 ---------! 4 m/s = VBg - 2 mis, VBg = 6 m/s t mg 197. d. Metbod·l Fig. 7.667 As the cylinder will remain in contact with wedge A T = J (mg')+' (ma l' = 10../2 N Fig. 7.672: 193. d. Sol. Acceleration of two mass systems is a = -F leftward. FBD of block A (Fig. 7.668) 2m
7.148 Physics for IIT-JEE: Mechanics I M g sin\", = f (i) B By applying Newton's second law to the man along the u A-~'\" 2u incline, will be 30' Mgsina+!=ma (ii) Fig. 7.672 (1 ~)a = g sin a + down the incline III As it also_ remain in contact with wedge B 200. b. From length constraint on AB (Fig. 7.676) u sin 30° = v:~, cos 30° - Vr sin 30° V,. = Vx sin 30° + -u s-in-30-° --- . cos 30' cos 30° V,. = 3u tan 30' = ,fl u v=JVi+v~2=../7u Method-2 mg In the frame of A (Fig. 7.673) Fig. 7.676 3U~' A cos 45' = h cos 45' J?~)/7// T sin 45' = mea) mg - T sin 45° = mb r:), :2g Fig. 7.673 211la = mga = 3u sin 30° = V:V cos 30° T mg or T= mg V, = 3u tan 30' =,flu ..fi ..fi = 2 and V, = 2u =} V = [Vi + V,~= V7 u 201. d. As in Fig. 7.677 198. a. As in Fig. 7.674 eT sin = mao + mg sin a u cos 45° = v cos 60° mgsin a o mg cos 0: A! It Fig. 7.677 Fig. 7.674 T cosO = mgcosa ·or v=..fiu etan = a_o.-+.-g\"-s-i-l1'-O-:- 199. b. Free body diagram of man and plank is given gcosa below (Fig. 7.675) 202. d. As in Fig. 7.678 / Fig. 7.678 Mgsin a I, +I,+h =c Fig. 7.675 I; + I; + I; = 0 For the plank to be at rest, applying Newton's second - VB + VA - VB + VA - VB = 0 law to the plank along the incline will be 3VB = 2VA JaB = 2aA Applying Newton's law on A and B F -2T = 2maA 3F = (6aA + 8aB)m
Newton's Law of Motion 7.149 3F T cos 60° Teos 60° all = N sin 60\" T 17 m ----+ 203. a. a: Acceleration of the block A downwards w.r.t ground (Fig. 7.679). Fig, 7.681 -I--->-c Equation of motion: Fig. 7.679 Fortn:mg.jT3 +mA· x :2 - 2T =111- 3 A (i) (ii) b: Acceleration of the block B w.r.t. inclined plane. .j3 . I (iii) e: Acceleration of the block C W.r.t. ground right side. N + r n A2- = m g -2 ~b + c; :acceleration of B w.r.t. ground. For M: T +N .j3 = MA - Applying Newton's law on a system along the horizontal 2 direction From equatI.Ons (.I), (I\"I),and (I\"I.I) A,= -3.-j3-g me + m(c - b cos 0) = 0 (i) ., 23 Applying Newton's law on (A + B) along the inclined 207. a. Let ao be the acceleration of chosen non-inertial frame of reference w.r.t. some inertial frame of reference and ~i~ be the plane, acceleration of the object in a non-inertial frame (Fig. 7.682). 2mg sin () = m(b - c cos 8) + n,w sin 0 (ii) 2g sin 0 = b - c cos 0 + a sin e From the wedge constraint between A and B a=bsinO (iii) From equations (i), (ii), and (iii) 4g sin 0 uU eb = .. 1 -I-- ~ sin 2 Fig. 7.682 204. c. aB,g = .)h2 + e2 + 2bc cos(ISO - 0) a:;For to be non-zero, the net force acting on the object + ( -bC2OS-O)2 + bbcosO (-cosO) (including pseudo force) must be non·zero. =b e1 +co-s-2 -cos2 0= 'b2 1~+ 3 sin2 e 208. a. Velocity of object w.r.t. non· inertial Ii'ame is constant and hence w.r.t. some inertial frame of reference it changes, hence 4 it is accelerating. So the net force acting on the object must be non-zero. 2g sin e = .)1+ 3sin' e 209. c. Net force without friction on system is 7 N in right side so first maximum friction will come on 3 kg block (Fig. 7.683). So 12 = I N,13 = 6 N, T = 2 N bcose T-1 f--IN~T 205. a. From equation (i) (Question 203) c = 2 J kg 8 206. b. If the initial acceleration of M towards right is A then we /;nax = 2 fmax = 6 can show that the acceleration of In W.r.t. M down the incline is :~2 ~~8 3A a = A(I +cosO) = - Fig. 7.683 2 FED of block m (w.r.t. M) (Fig. 7.680) N T _u210. b. As in Fig. 7.684 the mass of the rope: m = 4 x 1.5 = 6 kg Acceleration: a = 12/6 = 2 m/s2 mA ....-~A / 4m L-~(~1)__~____~(~2)____~~ 12N , mg cos 60° I- 1.6 m -I mg sin 60° Fig, 7.684 Fig. 7.680 Mass of part I as in Fig. 7.685: 1/1, = 1.6x 1.5 = 2,4 kg FED of M (Fig. 7.681) T =m,a = 2,4x 2=4.8N
_a7.150 Physics for IIT-JEE: Mechanics I . + 1 2 Usmg \"lat S = ut , we g et !Ill 100 = (1/2) x 0.6 x t2 or (= 18.3 s Fig. 7.685 Multiple Correct Answers Type 211. b. Force Metbod: 1. a., b., c., d. From P to Q: v2 = 02 + 2a,s, a. As in Fig. 7.688 g' =g +a where al = g sin 30°, Sl = 1 m = g + 3g =4g and 02 = v2 + 2a2s2 where a2 = g sin 30° - f1.g cos 30°, S2:::;;: 2 m solve to get\"' = J'i/2 Work-Energy Method: (Fig. 7.686) N f II p !I1a N ~gSin8 ml-.....,..R mg cos e mg mg cos (J Fig. 7.688 Fig. 7.686 mg' = 4mg =4 W In terms ofenergy considerations you can summarize the b. Think of Newton's third law afmotion. c. mg < fms whole proc.ess as a loss in the gravitational potential energy or mg < fl-sR of the block which is equal to \\\"-Iork done against friction, or or mg < Ilcs 111Cl or g < lisa (mg) x (3 sin 300) = (\",mg cos O)x2 g =} 3 x -1= } xv-'\"x'i 2=} J'i or fl-sa > g or J.Ls > - 2 2 1-'= - a 2 replace \"'., by I-' d. The jumping away of the man involves upward accelera- 212. c. As in Fig. 7.687 tion. 2. b., d. The horizontal forces on the man must balance, i.e., W w the forces exerted by the tWo walls on him must be equal. COSet = - or The vertical forces can balance even if the forces of fric- OA= tion on the two walls are unequal. The torques due to the OA cosa forces of friction about his centre of mass must balance. This requires friction on both the walls. ~?l 3. a., b., c. If the tendency of relative motion along the common o I~'~~::-;w;:-:::==:;j'1 tangent does not exist, then the component of contact force along the common tangent will be zero. Fig. 7.687 e4. a., c. Using the constrained equation, V2 cos = VI. Acceleration of body relative to the incline: (g + a) sin Ol On differentiation (l2 cos 0 = a I 2)So OA = \"1l(g +a)s.ma,r ( from S = ut + \"1lat 5. a., d. As in Fig. 7.689 or W = 1 + a)sina' (2 -2(g -- cosO! 2 W ] 1/2 T2 t= [ cosa(g+a)sina TI 4 W ] 1/2 [ 4 W ] 1/2 = [ (2cosasina)(g+a) = (g +a)sin2a 5 kg (i) Fig. 7.689 213. b. The force of 100 N acts on both the boats Tz cos 60\" = TJ cos 30°, and 2S0a, =100 and SOOa2 =100 or al = 0.4 me 2 and a2 = 0.2 ms-2 +Relative acceleration: a = aj 0-2 :::;;: 0.6 ms-1
T2 sin 60' + T1 sin 30\" = 5g Newton's law of Motion 7,151 From equations (i) and (ii) (ii) R 1', = 25 Nand T2 = 25-J3 N TM 6. a., c. Mg-1'=Ma 0) T =ma (ii) Mg Solving equations (i) and (ii) Fig, 7,692 Mg a= M'g = arM + M') (M+m) Mig N +a = cc--~-,- Mg (M M') Fig, 7,690 FBD of man (Fig. 7.690) masinO = mgcos8 a =gcote Mg-N=Ma M'g N= Mmg g eote = (M + M') (M+m) ma sin () 7. c., d, Figure 7.691 mgcos e emgsin O+macos ~L J!_________ r Fig. 7.693 cote M + coteM' = M' x = 3 em----Jr> I M' = Meote (I - cote) T = Ma = Mgeote Fig. 7.691 1'= Mg +(y - h) v'~x';-+'-'h\"'2 = I tan e 9. a., d. Figure 7.694 dy + x dx = 0 dt v'x 2 + h2 dt dy x dx = dt v'X2 + h2 dt --T-s-\"in\"e dy 3 mg - = --VA (i) Fig. 7.694 dt 5 3 IUBI = SVA d2y h2 T cos eo = mg (i) (ii) -=VA (x 2 + h2)3/2 T sin 80 = mao dt' Dividing equation (ii) by 0) 16 a aB = VA (5)3 16 tan 00 = - as = -VA g (ii) 125 eo = 30\" 8. a., d. Figures 7.692 and 7.693 0) l' _ mg _ 2mg M'g- T = M'a (ii) - cos 30° - -J3' 1'=Ma
7.152 Physics for IIT-JEE: Mechanics I 10. b., d. Acceleration of M. a = (;) FA = -FB Hence aA ::: aB also - mA mB I = ~ F t' 18. a., b., d. 2M ~ t = J2Ml ~~ d p F Newton's second law is, F = - , which itself explains 11. b., c. Here F > fL.,mg (1 + : ) dt the validity of the given statements. Form 19. b., d. In a tug of war, the FBD of the teams are as shown in the Fig. 7.696. From the FBD. it is clear that the team wins on which horizontal force exerted by ground is morc. F - fLkmg = ma ~2 T T~\"'X2 ForM RX fLkmg = MA R, ~ A = 0.4 m/s2 Fig. 7.696 12. a., b. For the two values of F. i.e.. F = 150 Nand F = 120 20. a., b., d. Under the action of two forces if the body is acccler- N; the tcnsions in the string are T = \"F2 = 75 Nand 60 N. In ating, it means a net force is acting on body and it can never attain a constant velocity or speed. [Provided initial velocity the first case, the accelerations of the two masses are equal is either zero or its_direction is same as that of acceleration]. and opposite, \\yhilc in the second the accelerations is zero, If the two forces are equal then for the present situation 13. a.,b.,c. As the acceleration of A and B arc different, it means they can't act along the same line, but if forces arc unequal that there is relative motion bctvveen A and B, The free body then they may act along the same line. diagram of A and B can be drawn as (Fig. 7.695) 21. a., b., c. For (i): Consider a block at rest on a rough surface and no force (horizontal) is acling on it (Fig. 7.697). Now A I-H [B] ·f onfriction force it would be zero. r • ----+ (1,4, Fig. 7.697 Fig. 7.695 For (ii): Consider a heavy block, under the application For A. F - J = MaA = 50x 3 of small force F which is not sufficient to cause its motion, For B, J = maB = 20 x 2 so friction force is static in nature and block doesn't move. =}J=40N, F=190N f±L:,F '\\:\"0: ,,\" \"'\" 14. a., b. Because mg acts downwards which makes sliding along 4 to be easiest and along I to be the most difficult. Fig. 7.698 15. a.,b.,c. From the FBD of A it is clear that friction opposes its For (iii): Refer to Concepts and Formulae. motion. For (iv): Friction force and normal force always act per- Also friction always opposes the relative motion. pendicular to each other. Since velocities of A and Ii are different, hence there 22. a., b., c. If the block is at rest, then the force applied has to is relative motion between them. So there is kinetic friction be greater than the limiting frictional force for its motion to between the two blocks which is IIJllAg begin (Fig. 7.699). 16. a., c., d. In the first case, m will remain at rest. f~7N F \"'~~ aM= - Fig. 7.699' M .h = I\".,mg = 0.25 x 3g = 7.5 N < F,ppl;,d In'the second case, both will accelerate: So, the friction is static in nature and its value would be equal to the applied force. i.e., 7 N. If the body is ini- F tially moving, then the kinetic friction is present Uk = fLkmg am =aM = -M-+m- mF In the second case, force on m = mam = M+m 17. a., b., c., d. When friction between the blocks becomes zero, the relative sliding between the blocks will be stopped hence VA = VB· Also when the friction becomes zero, only force to move the blocks are FA and FB
Newton's law of Motion 7.153 = 6 N), acting opposite to the direction of motion. If the ap- ------Jt> Direction of plied force is along direction of motion. then the situation f: ~\"\"\"\"motion would be as shown in Fig. 7.700. Fig. 7,704 Fig. 7.700 a= -F -- -.fk =k-t --.-fk mm As F > JIr., the block is accelerated with an acceleration dv kt - .fk -=a= [F -of a = .fk] = ~ m/s2 and hence its speed is contin- elt In m3 2kt 2 - .fkt uously increasing, V= If the applied forcc is opposite to the direction of mo- m tion then the block is under deceleration of a = _ [ F : .fk ] -kt 3- -.fl-t2 = _ ~ m/s2 and hence after some time the block stop and \"d-s =v= til ,*.1'= ~_ _2_ dt m kinetic3 friction vanishes but applied force continues to act 26. a., c. Figure 7.705 (Fig. 7.701). mgcos 31\"\" fk:JEt::F 4-7''''-+ ma (inertial force) ,,,,, mg mg cos 37° Fig. 7.701 Fig. 7.705 But as F < .fL, the block remains at rest and the fric- tional force acquires the value equal to the applied force, Le., friction is static in nature. 23. a., c. First of all draw FBD of P3. Let the tension in three strings be T\" 'r\" and T3, respectively, (Fig. 7.702). T'~P,T' Balancing forces perpendicular to the incline N I = mg cos 37° + nIa sin 37° 1\"' Nl = 4 + 3 -mg ·-ma 55 T, and along the incline, mg sin 37° - ma cos 37° = mb\\ Fig. 7.702 '*2T, - T, = 0 x a T, = 0 br = 34 -g- --Q 55 '*Now draw FBD of P4 and Ps (Fig. 7.7(3) 21', - 7, = 0 1', = 0 21'2-1'.1=0,* 72=T3=0 ilia sin 37° -a+ ma mgcos 3]0 mg c s 37° mg sin 37° Fig, 7.703 mg Similarly for the acceleration draw the FBD of P6 and Fig. 7.706 P7 and get the values of acceleration. 24. a., b., c. From the Fig. 7.704 it is clear that the object slows 43 Similarly for this case (Fig. 7.706) get N, = -mg - -Ina Fdown and comes to rest. At this instant is still acting in 5 5 the same direction and is greater than .fL so the block starts 34 accelerating in the opposite direction of its initial motion. and b2 = -5g+-5Q 25. b., c., d. For some time, the block won't move due to the frictional force. When F > fL, the motion of block starts. 43 N2 = 5-mg - 5-ma
7.154 Physics for IlT-JEE: Mechanics I 109 - T2 = lOa, T2 - TI - / = 3a a TI - / = 2a, where / = n.3 x 2g = 6 N From the above equations: 109 - 2/ = 15a =} 10 x 10 - 2 x 6 = 15a =} a = 88/15 m/s2 7, = 109 - lOa = 10 x 10 - 10 x 1858 = 41.3 N mg (g + a) sin 37° ma TI = / + 2a = 6 + 2 x 1885 = 17.7 N inK mg(g+a) 0537° Clearly T, > 1'1 Fig. 7.707 b. This is correct because of the greater mass of 3 kg. Since acceleration is same for both, Similarly for this case (Fig. 7.707) get c. This is incolTcct because the net force acting on 10 kg mass N, = 4Smg + 4Sma is greater due to its larger mass, not due to its acceleration 3 3 downward. 2mg -mg -5g 5-G al = =g 31. a., c. m h3. = + a, mg+mg -mg = = g/2 Similarly for this case (Fig. 7.708) get 2m 44 a3 = 2mg -mg = g/3 N4 = -mg - ·-ma 3m 55 33 clearlyal > a2 > a3 and b4 = -g- -a 32. a., b., c. Figure 7.709 5 5 3 - /1 3 - 0.5 x 0.2g al=--= =IOm/s 2 0.2 0.2 '\"\"4 ilia ---.al 3N ~ -~ }; mg(g---a) in37° mg(g-a) os 37° Fig. 7.709 mg 3 - .fz 3 - 0.5 x 0.51! 2 a 1 a2 = - - - = mls . 0.5 0.5 relative acceleration = 10 + 1 = 11 m/s2 22 = 21: x 11 x [2 =} [ = 2 s Fig. 7.708 33. b., c. Figure 7.710 27. a., d. If initially the acceleration of A is greater than B. then there will be an extension and if that of B is greater than A ~-~J, _ _ _ _+. F~' lOll N then there will be compression in the spring. Otherwise the length of spring will remain same. Fig. 7.710 28. b., d. Since the apparent weight is increasing, hence accel- F = lOal + 40a2 eration of the lift should be upwards. This is possible in case =} 100 = 10al + 40 x 2 =} al = 2 mis' of (b) and (d). 29. b., c.,d. Acceleration of particle w.r.t. frame Sj: Acceleration of particle w.r.t. frame S2: --..(11 where til and Pi are unit vectors in any directions. Now the I10 kg relative acceleration of frames: --I asz - as! = 2(n - fit). Fig. 7.711 Its magnitude can have any value between 0 to 4 mfs2. So acceleration of A must be 2 m/s2 for the given con- depending upon the directions of mand ii. ditions to be satisfied (Fig. 7.711). 30. a., b., c. F = lOal = 10 x 2 = 20 N a. Let the acceleratioll of each block be a. J</I =} 20<f.l m Ag =} 20 < f.l x 109 =} f.l:O: 0.2 Hence f.l can be greater or equal to 0.2
Newton's Law of Motion 7.155 A.ssertion-Reasoning 7. d. According to Newton's third law of motion aetion and Type reaction are equal and opposite. 1. b. dp 8. a. In the direction of normal reaction net acceleration is zero. Hence the forces in this direction will be halanced. Hence i. F \"\" - (Newton's second law) N =mgcos8. dt 9. b. By the definition of inertial and non-inertial frame. ii. Newton's third law. e10. c, Coefficiei,t of friction iJ. = tan(e). The value of tan way 2. a, Figure 7.712 exceed unity. .\\' r /:;p IIIgsin n 11. a. F = - , If /:;t is more, then F will be less. /:;t F 12. d. al = g - M F ::;:;} (/1 > (/2 (/2 = g - - Fig. 7.712 m i=mgsine M = mgcosB 13. d. While a running boy pushes the ground in backward di- (i) rectjon and the available friction pushed him in forward di- (ii) rection. R =IN'+.f2 =mg 14. a. In equilibrium, net force on the body is 0, therefore. its acceleration (a) is O. If the body is at rest it will remain at 3. b. Force needed when the breaks are applied rest. If the body is moving with a constant speed along a straight line, it will continue to do so. II mv2 =ma=-- 15. d. Inertia is the property by virtue of which the body is unable d to change by itself not only the state of rest but also the state (v: initial speed, d: distance from wall) of motion. when the turn is taken 16. d. Due to change in normal reaction pulling is easier. h=ma= dmv2 17. a. Contact force is sum of friction and normal reaction. ... brakes must be applied. 18. b. Statie frictional force is self adjusting. 't. u. Figure 7.7 13 19. d. Figure 7.715 T o e T mg Fig. 7.713 Fig. 7.715 21' sine = mg FBD of A (Fjg. 7.716) e e7. = -m-g > mg f\"or sm G--PI < -I,.I.e., < \"0\" e2sin 2 Fig. 7.716 _1 , 5. d. Here the assertion is based on the idea that if you substitute may. = 0 in F' = (because there happens to be a particle on which net force F = 0) You get a= 0 : : :;} v= constant => such a particle will move with constant speed along a fixed direction which is 20. For pulling condition (Fig. 7.717): Newton's first law. But the point is, you cannot employ N + Fsine =mg (i) P' = mii, without tlrst ascertaining that it is valid in this form N=Mg-Fsine . (i.e. without pseudo forces). And that ean be done only by ap- plying Newton's first law and checking the behavior of your . F sin () frame against the description laid down in the first law. F 6. b. Figure 7.714 fLJ211- F cos (! Fig. 7.714 mg (ii) Fig. 7.717 For pushing condition Fig. 7.718: N=Fsin8+Mg
7.156 Physics for lIT-JEE: Mechanics I Feos e (EomIlreHensive [MIle mg For Problems 1-3 Fsi11 0 1. b., 2. a., 3. d. Fig. 7.718 Sol. Before hurning Be, the free-body diagrams are shown in Fig, 7,720, 21. e. Here the acceleration of both will be same, but their masses Fig. 7,720 are different. Hence, the net force acting on each o[them will not be samc. 1'2 = Tl + m2g (1) 22. h. During the static friction there is no slipping between the kx=T2 =mjg (2) two bodies. But during kinetic friction bodies slip due to which heat is produced at the cost of mechanical energy. where x is the extension in the spring. Just after burning, TJ will 23. a. Statement II is correct, as it represents Newton's second become zero, but 1'2 will remain same, ~, T2 - ti12g = J112Cl -;. dp (m, - 1n2)g a= -.law as F = ---, from this only we can say that for greater dt 111-2 value of d P , force applied has to be more, As T2 remains same, acceleration of block A will still remain dt zero. 24. a. Once the ski is in motion it melts the snow below it and For Problems 4--6 hence skiing can be performed. To make skiing easier, wax has been put on bottom surface of ski as wax is water repellent 4. b., 5. a., 6. d. and hence reduces the friction between the ski and film of water. Sol. T,1 = 201, 1', = 1'2 = 101 25. b. The acceleration of a particle as seen from an inertial '*For A to lose contact: 101 = Ig 1 = I s frame is zero if the net force acting on the particle is zero. '*For l! to lose contact: 101 = 2g 1 = 2 s 26. d. Reference frame attached to earth is not an inertial frame For C to lose contact: 20t = 3g =} I = 1.5 s of reference because earth is revolving about the sun, as well as it is rotating about its own axis. aA = 1', - 19 Velocity of A when B loses contact 27. d. When a body is at rest the static friction may be less than '-'---\"-, the limiting friction, /2 /2I 28. d, Thc FBD of block A in Fig, 7,719 is V, = aAdl = (lOt _ g)dt = 5 m/s '---+--~N IOx2-IO at t = 2 s, all = 0, (/A = - - - - 1 - - = 10 mg aAiB = aA - an = 10 - 0 = 10 mis' Fig, 7.719 For Problems 7-9 The force exerted by Ii on A is N (normal reaction), 7. b., 8. b., 9. e, The forces acting on A arc N (horizontal) and mg (weight Sol. downwards), 7. b. W' = meg + a) =} 240 = m(IO + 2) Hence statement I is false. 29. d, If the lift is retarding while it moves upward, the man shall =} m = 20 kg, True weight = mg = 20 x 10 = 200 N feel lesser weight as compared to when lift was at rest. Hence statemcnt-I is false and statement-II is true. 8. b, W' = meg - a) =} 160 = 20(10 - a) 30. d. Even a small force will change the state, hence stalemcnt- I is wrong. Statement-II is the statement of second law of =} a = 2 mis' motion. 9~ c. zero, a = g in free fall. For Problems 10-13 10. b., 11. c., 12. b., 13. d. Sol. (1) n-7-1g sinB - T = mla (2) eT - In2g sin = 111.2a Solve to get a and T, N] = fUl.RCOSO
Newton's Law of Motion 7.157 -+\",T ~ (~r~- T ___ +--N -c Fig. 7.721 Fig. 7.724 = =N2 NI +m2gcose (m, +m2)gcos(i al = 8g\" and Q2 = 2g\" For Problems 14-16 al +a2 =85g 14. a., 15. b., 16. c. Thus. the acceleration of A is ~ in horizontal direction and 5: Sol. in the vertical direction. =14. a. Area under F -/ graph change in rnomcntum Acceleration of B is ~ in the horizontal direction (leftwards) and acceleration of C is ~ in the horizontal direction (rightwards), I 3 200 =} 2:Fo(6 x 10-) = 1000[40+20] =} Fo = 4, 000 N For Problems 21-23 15. b. F\" =T.ot..a.~l_c~ha.~n~g~e~i~n~m~o~m~cn~t~u~m 21. d., 22. b., 23. c. Sol. Figure 7,725 tllne taken 16. c. Area under F -/ graph = change in momentum =} I, 200 + 20] 2:(Fo)(4 x 10) = 1000[1' =} v = 20 m/s F For Problems 17-20 17. a., 18. c., 19. b., 20. a. Sol. Let. acceleration of block C is al (rightwards) and the ac- AB celeration of block B is Q2 (leftwards). Theu. acceleration of A will be (al + a2) downwards and a, rightwards. Free body diagram of A is shown in Fig, 7,722, 7{) 7() TT Fig. 7.722 Fig. 7.725 Using I: Fx = max and I: Fy = rna)\" we get Let To = tension in the string passing over A . N = 4m(al). and (1) T = tension in the string passing over B (ii) 4mg - T = 4m (al + a2) 2To = F and 2T = To =} T = FI4 Free body diagram of B (showing horizontal forces only) is When F =600N T = F/4 = 150N As T < Mg and T > mg. M will remain stationary on the floor, whereas m will move. Acceleration of m, shown in Fig, 7,723, T - mg 150 - 100 I2 G= m = 10 =5rns +---({2 For Problems 24-26 ~T 24. b., 25. c., 26. d\" Fig. 7.723 Sol. For upper block a max = Vg = 4 m/s2 and fmax = 40 N Using I: Fx = ma\" we get (l)When F=30N T = 3ma2 (iii) As F < fmax So both blocks will move together, Free body diagram of C (showing horizontal forces only) is F 30 6 2 shown in Fig, 7,724, \" a = --- = - = - m/s Using I: f'x = rna\" we get M +m 35 7 (2) When F = 250 N T - N = 8mal (iv) For upper block We have four unknowns T, N. aI, and a2. Solving these four equations, we get
7.158 Physics for IIT-JEE: Mechanics J '* '*250 - 40 = lOa T - I\"Mg = Ma (ii) 210 = lOa a = 21 m/s2 Putting the value of l' from eg. Oi) into 0) a= 40 = -8 m/s2 (m - fLM) g = (M + /11) a + mb - 25 5 m(g - b) -1\"Mg For Problems 27-29 ~ =a 27. d., 28. d., 29. d. (M +m) Sol. From the constraint relations we can see that T = fLMg + -M-m\"g -- -M-m-h--'-fL-M\"2-g 3TXB = 2TXA M+m '*3 3 fLM2g + fLMmg + Mmg - Mmb - fLM2g aA = 2aB XA = 2XB M+m So let aR = a then aA = 1.5a T= Mmg(u+ I -b) Writing equation of motion: M+m From block A, For Problems 34-35 21' = 70aA = 105a = 3 x 35a 34, c., 35. b., 35a = 321' (i) (ii) Sol. a W =60N, 1'sin8 = W, From block B, = =so, l' (60 N)/ sin45\", or, l' 85 N 300 - 31' = 35aB = 35a b, F, = F2 = 85 N eos 45° = 60 N Solving eqs. (i) and (ii) we get For Problems (36-38) 2T 36, d\" 37, a., 38. e. 300-3T =- 3 Sol. Let the tension in the cord attached to block A be T, and '*,*900-9T=2T 900=l1T the tension in the cord attached to block C be Tc, The equations of motion are then T = 900 N T\\ - mAg = mAa; T2 - J.tkmBg - T\\ = mBa meg - T2 = mea II 180 m/s2 and aB = 120 m/s2 77 77aA = a. Adding these three equations to eliminate the tensions give For Problems 30-33 almA + mB + mel = g(mc - mA - fLkmB) 30. a., 31. d., 32. d., 33, c. Sol. Case I solving for me gives As the monkey doesn't move with respect to the rope, it means me = mAra + + mB(a + fLkg) that the acceleration of the block or the rope and the monkey is g-a same. So equations of motion ilre and substitution of numerical values gives l' - fLMg = Ma (i) me = 12.9 kg = 13 ~-T=m0-M W b. T, = mA (g + a) = 48 N Putting the value T from e'ls. 0) in (ii) mg - fLMg = Ma +ma 1'2 = mc(g -a) = 102N '* (/11 - fLM)g a For Problems 39-40 (M +m) 39, d., 40. e, T = fLMg + M (m - fLM)g) Sol. For an angle of 45.0\", the tensions in the horizontal and the (M+m) vertical wires will be the same. a. The tension in the ve,tieal wire will be equal to :hc weight . Mmg - fLM2g w = 12.0 N; this must be the tension in the horizontal wire, and =fLMg+ M+m hence the friction force on block A is also 12.0 N. h. The maximum frictional force is /LM2g + I\"Mmg + Mmg - fLM2g fL.,WA = (0.25)(60.0 N) = IS N = M+rn this will be tension in both horizontal and vertical Palts of the wire, so maximum weight is 15 N, Mmg (fL + 1) For Problems 41-43 M+m 41. b., 42. c., 43. d., 44. b., 45, e\" 46. d. Sol. Case II The monkey moves downward with respect to the rope 41. b, ill = 0.25 x 4 = IN, il2 = 0.25 x (4 + 8) = 3 N with an acceleration b, therefore; its abso1ute acceleration is F = il2 = 3 N for eonstant velocity a + b where a is the acceleration of the rope. Therefore, the equations of motion are m ~-T=m0+M
Newton's law of Motion 7.159 42. c. F = fll + ii2 = 1 + 3 = 4 N f = 15acos30' = 15 x 4u5Y(23: 43. d. F = III + ii2 + T. T = fli For A not to slide on B: .f =:5 ii '* F = 2jil + 112 = 5 N '* 15 x 45 x /3 (265 x 15) For Problems 44·46 -1-I- 22 44. d., 45. c., 46. d. Y2::5ft Sol. '* 9.J3 44. b. As in Fig. 7.726 I~?: 53 = 0.294 For Problems 51-52 T ...- mg sin 45° - ill = ma, 51. b\" 52. a, 2mg sin 45\" - T - fl2 = 2ma T Sol. Assuming that the system moves together and there is no sliding, therefore, acceleration of the syst.em a = ~~'\"\" (5 + 10) F (i) a= ~ 15 FBD of III (Fig. 7.728) Fig. 7.726 ~\"g1 where iii = /.\"\"'8 cos 45\" = 2mgeos4SO/3 F and If2 = /.L22mg cos 45° mf Now we get: a = - glC; N 9'12 Fig. 7,728 This is negative which is not possible. Hence a = 0 FBD of M: f - F = lila = 10 (~) 45. C• .Ii, < 2mg sin 45\". hence friction only will not be able to prevent slipping of 2 m mass. So on 2 In mass friction will he maximum i.e. i1'2' 2~ f=F(I+~) (iil :::::> T = --m,g 46. d. T = mg sin 45' + .fl If there is no sliding F :5 I\"s N .3 FG] '*:5 0.4 x 10 x 10 '* mg fl = 3~ For Problems 47-48 F = 24 N 47. c., 48. a. 15. F 24 2 From equation (I), a = = 15 = 1.6 m/s Sol. f i f47. c. Net force F = + Fi = v'4T N For Problems 53-56 ii = 0.4 x 109 = 40 N, fk = 0.3 x 109 = 30 N 53. c., 54. b., 55, d., 56. a, Net force is less than .Ii, Hence Sol. For upper block Required li'iction force =applied force =v'4T N ama;>; = fJ,g = 4 m/s2 and fmax = 40 N 48, a. Now applied force will be equal to maximum ii'iction (I) When F = 30 N force, i.e., ,)52 + ,,2 = ii = 40 As F < fma;>; '* a = ,)1575 N So both the blocks will move together. For Problems 49-50 a - F 30 6 2 = M -+m- = -= - 49. a\" 50. b. 35 7 m/s Sol. Figure 7.727 (2) When F = 250 N '*500 - 558 sin 30' = 55a a = 1451 m/s2 For the upper block 250 - 40 = lOa '* '*210 = lOa ~ t / 'v a :;in 3(Y\" a = 21 m/s2 \",,, A A 40 8 , I! a = - = - mis' / 15 g I 25 5 Lacos::l0° For Problems 57-59 Fig. 7,727 57, c., 58. a\" 59. d. Sol. N - 15g = 15asin30' 57, c. 0.5 t = f'mg (1 + ~.) N = 15 [10 + 45 X ~] = 265 ><.2.5, II 2 22 f' = 0.2
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