Cengage MECHANICS 1

5.54 Physics for IIT-JEE: Mechanics I 9. d. Velocity of police van = 30 x 5 = 25 m/s\" 13. c. Relative velocity of boat with respeet to water is - - 18 3 Vb - Vw = 31 + 4] - (-31 - 4]) = 67 +8] Muzzle speed of the bullet = 150 ms-' 14. a. V, = 251, Vb/, = 25../3] Speed of the bullet w.r.t ground = (150 + 25/3) ms-' N Velocity of thief's ear 5 32 x 5 160 = 192 x - = - - = - m/s 18 3 3 relative velocity of the bullet w.r.t. the thief's car 2Sf3 ----- = ISO + -25 - -160 = 150 - 135 105 m/s -v>, -= 33 3 e 10. c. To find the relative velocity of bird w.r.t. train, superim- \"----'---+ E pose velocity - VT on both the objects. Now as a result oflt, 25 the train is at rest, while bird possesses two velocities - VB Fig. 5.154 Vtowards Nand - T along west. Vb/, = Vb - V, =} Vb = Vb/, + V, N =} Vb = 251 + 25../3) -> IVbl = J25 2 + (25../3)2 = 50 km/h VA = 40 kmlhr e25 I 0 tane = M = M =} = 30 25-v 3 -v 3 90' 15. a. Shortest time = -d = 1/2 I. -= - hr = 10 mm W---t-L=->~~~--E v 36 VT = 40 kmlhr 16. d. Relative velocity of the bird with respect to the train is: Fig. 5.151 VliT = VB + VT = 5 + 10 = 15 m/s [because they arc going in opposite direction] N N --->- ' - - - + •s VB 5 In/sec -> VT = 10 m/sec \", -V>B ~I'-----I~ 150 m ------~.I \"\" -> \" Fig. 5.155 VBT '\" '\" W_------,--9-0'-'--\"\", -----_______ E .. time taken by the bird to cross the train = 5ISO = lOs -> -> Vr vr d 336 17. d. ShOltest possible time: t = - = - = 336 s Fig. 5.152 v1 where, - VB = Velocity of bird - VI' = Velocity of train. 18. b. s.ma = -u;; =../23\" =} a = 60\" e=} = 90° + a = 150° IVii l' I = JIVBI2 + I-VTI2 vSf\\: [By formula, .'. 0 = 90°1 Fig. 5.156 = y'402 + 402 - 40 v'2 kmlhr North-West 19. a, up = vi. UbiI' = -ui \" v 10 Vb = Up + Vb/!' = (v - u) i = v -'- u towards right. e11. b, Finally he will swim along B. tan = - = - = 2 20. c. V, = - 5 .7, vb/' = - 2-J6 I u5 Vb = vbjo + v, = - 2 v'6 I - 5 ) B Vb = /c;--:;;'j2 + 52 = 7 m/s _E v -It- 1/=5 21. a. Let the stone be projected at an angle a to the direction of motion of truck with a speed of v = 20 m/s. A Since the resultant displacement along horizontal is zero, Fig. 5.153 =} 0 = tan-'(2)N of E 12, b. Net velocity of boat in river = y'5 2 - u2 t= Distance I I =} u = 3 kmh-' =} - = Velocity 4 y'52 _ u2

Motion in Two Dimensions 5.55 v\"\"'20mis 24. a. Motion of the person, making an angle (say a) with the downstream. f --» dis --iI> dis d d Fig. 5.157 ! u• The velocity along horizontal = 0 3 Fig. 5.160 4 15+20 cos'\" =0 ::::.} cos a = d The time taken to cross the river = - . - (-D=? . '\" = cos- 1 vsma The distance carried away down stream in the same time = 138°35 is equal to speed X time. d x = (u + vcosa)·····.- (i) 22. d. We know that to cross the river by shortest path: sin a = \": v SIn a v Motion of the person, making a angle with upstream. The h'me taken to cross the\"flver IS equaI to - .d- .. vsma Distance carried away downstream in the same time Fig. 5.158 d x = [u + vcos(180' - a)]-.- But u > v, =? sin\", > 1, which is not possible. .5 vsma 23. b. Speed of train = 108 x 18 = 30 mls X = (U - vcosa) - .d - (1'1') vsma Let llR and II T represent the respective velocities of (ll + d 2 rain and train. vcosa)-.- Now, the relative velocity of rain w.r.t. person (train) is Given, vsma = given by VR.T = VR - VT => VR + (-VT) d (u - vcosa)-.- oRLet and Iff represent the vectors and, respectively, v SIn a in magnitude and direction. :(l:l::+..\"v-c=o=s:a.): = 2 =? 3vcosa = u - (u-vcosa) I v seca (iii) =? u = 3 Vertica! Vertical seca I seca 2:: 1. -- >- -3 =? 3 -, 4 0 Horizontal From . ... v 2: -1 so, 11 cannot less than -1. ..~ I'r equatIOn (111), - \" Horizontal ~ - ,,, vu 3 u 3 0 v, MR 25. a. Change in velocity = f - Vi ~q, ~ \\ T,, VIi . V,., T Fig. 5.159 o OT 2 = OR2 + RI'2 + 20R x RI'cos 1200 2=202 +302 -2 X 20 X 30 X 1 = 400 + 900 - 600 = 700 = ,;7o(j m/s Fig. 5.161 From ~OMR [where M is the foot of perpendicular drawn Its magnitude is :Jv' + v2 - 2vv cos 40° = 2v sin 20° from point 0 on RT] 26. i. a, In the absence of air resistance, the projectile moves OM = 20 cos 30\" = 1Ovr:i with constant horizontal velocity because acceleration due to MR = 20 sin 30\" = 10, MT = 30 - 10 = 20 gravity is totally verticaL eIf be the angle which the apparent velocity makes with ii. c. The vertical component goes on decreasing and eventu- e I'M 20 ally becomes zero. iii. b. vertical, then tan = - - = M iv. c. Horizontal component of velocity remains constant OM 10,,3 throughout the motion, as it is not affected by acceleration due to gravity which is directed vertically downwards e2 = tan-'(-) vr:i

5.56 Physics for IIT·JEE: Mechanics I . u2 u2 sin2 a ,M 27. I. d. R = - and H = ---c:--- :h g 2g P: eFor the maximwn range, = 45° - --8-----,:-H----- - = =u2 sin2 45° ul R H= 2g 4g 4 A !+-RI2-+!C B \" eu2 sin' . 2u sin e Fig. 5.162 n, d, We know that H = and T = . From 2g Ii NowtanO = MC = MP + PC these two equations, AC AC gT2 we get H = -8-' So if T is doubled, H becomes four times. 4=h+H 4= (h+H)2 .•. -u2 Rj2 =} H III b. Here Rmax = h= g =}h=H . .. euZ s1n2 dx dy HeIght H IS gIven by: H = - c : - - 31. c. vx = - = 6 and Vy = - = 8 - lOt 2g dt dt eH is max when = 90° (for a given velocity) put t = 0 (-: we have to find initial velocity) u2 h v, = 8 - 10 x 0 = 8 Hence Hmax = - = -. v =Jv~+v~= -/62 +8' = IOmls 2g 2 iv, a. We know that 32. b. tan 0 = Vy = ~ = ~ orO = tan\"-I ~ Vx 6 3 3 x = (u cos e)t and y = (u sin e)t _ ~ gt 2 Letx2 - x, = (u, cosO, - u2cosfl2)t = X 33. i. c. Velocityat the highest point, Y2 - . 12 (uz . Oz)t + 12 V\" = i(ucose) y, = (u, smOI)t - 'jgt - 'jgt sm Velocity at the starting point = (u I sin 01 - U2 sin 02)t = Y 13,\\. = 1(11 cos 8) + .7 (u sin 8) Y Uj sinOI - uzsin8z l6.vl=lv\" - v,l = I-J(usine)1 =usinfl = = constant, m(say) X /.II COSOI - U2 COS O2 Y=mX ii. d. Speed at the highest point = u cos fI It is the equation of a straight line passing through the Speed at the starting point = u\" Hence, change in speed = (u cos 0 - u) origin. 34. b. Range will be the same, because the sum of two angles is Alternatively: We can think in this way: Relative acceler- 90\". ation of one projectile w.r.t. another projectile will be zero. 35. b. The other angle is 90\" - 30\" = 60\" Hence relative velocity of one projectile w.r.t. another will 36. d. The sum of these two angles is 90\". be constant. If velocity is constant, it indicates straight line e37. a. For maximum range: = 45\", hence tanO = tan45° = 1 motion. v, c. The upward motion is with higher retardation while the downward motion is with lesser acceleration. Further, the 38. b. H\"\"\", = 100 m, R_ m\" = 2 x 200 = 400 m time ofrise is less than the time of return. A part of the kinetic [H':R4 x 100 tane] = -4- tane = 400 = 1 energy is used against friction. =} 0 =450 e\" 28, a, Smce R = 2H or v2 sin 28 v2 sin2 39. c. The time taken to reach the ground depends on the height = 2 x -c:-- g 2g from whieh the bullets are fired when the bullets are fired or 2 sin fI cos 0 = .sin2 0 or tan fI = 2 horizontally. Here height is same for both the bullets and v22 sin 0 cos 0 2vz 2 1 4v2 hence will reach the ground simultaneously. - - - - - = - xv-'s xv-'s =- 40. b. R = h = u' sin 20 When 28 = 90\" g g 5g g g uZ 8in2 u' e e. . 29. d. As In FIg. 5.162, H = R or u22 sin cos () =} - =h = ----- g 2g g eor tan 0 = 4 or = tan-I (4) eu2 sin2 30. a. AC = R/2, PC = H Height H is given by: H = ---co--- 2g We have to find h = MP Whene=90°, H == Hlllax = u2 h - =- From Problem 29 (Fig. 5.162), we know that if H = R, 2g 2 fhen tan 8 = 4, 41. a. We know that

Motion in Two Dimensions 5.57 . 1 gt Z 2usinO 2 x 30 x 1/2 (u S1l1 e)t - = = 2x T=---=- =3s (u cos e)t and y g 10 Then X2 - Xl = (U2COSeZ - u, cose,)t = x Thus, after 1.5 s the body will be at the highest point. Aod yz - .v, = . - I2 +. -21g t 2 So the direction of motion will be horizontal after 1.5 5, the (U2 sm 8z)t -2\"\"t - (U2 sm 82 )1 angle with the horizontal is 0\". = (u, sine, - u2sine2)t = y =53. c. Here angle of projection 45\" y e, -(li, sin U2 sin 82)t -g~. u2 u2 X (U1 cos 81 - Uz cos Rmax = 1.6= = 10 oru =4ms,--1 eU 1sin 1 - Uz sin (h Hence, the distance covered in lOs UI cos 81 - HZ COS = horizontal speed x time = u cos 45° x time = constant, m (say) I =4x v'2 x 1O=20v'2m =} Y = mX, It is the equation of a straight line passing 54. i. a. If h be the maximum height attained by the projectile through the origin. u2 sin' O· u2 sin 20 . e4H 4 x 4 4 8m =-4 then h = or R = - - - 5 2g g 42. a. tanO = - = - - = - =} R 12 3 R 2 sin (I cos 0 \"'R \"'h -= = 4cot8, therefore - =- 2 h (sin 0)/2 .R h .'. percentage increase in R = percentage increase in h = 5% (I[=} u = ,.)2gH = ,.)2 x g x 4 = 5 e ... h u2sin2 g2 g sine 4/5 \\ 2 8'It d. T' = 2g X 4u2 sin2 8 = e4.3. b. t = 2u sin or 2 = 2u sin II. =} u sinO = g 2r' rThus h\"'h = gg '\"T. \"'T = 21 f'h,h I.e., u2sin2 e g2 g [\"'h ]or 100 x _\"'1.-' = -I -- x 100 = -10 = 5% hili = = - = - =5m 2g 2g 2 T 2h 2 44. c. Range will be the same, because the sum of two\"angles is 55. d. Let Ux = 3 mIs, ax = 0 90°. Hence the ratio I: I. v,.=u,.+ayt=O+1 x4=4ms-' 45. d. Acceleration remains constant, and equal to g always. =.;v v?; + v~ = ,,/32 + 42= 5 mls 46. a. R is same for both e and (90 - e). If angle w.r.t. vertical is Angle made by the resultant velocity w,r.t. direction of initial 4(F then w.r.t horizontal direction it will be 90\" - 40\" = 50°. velocity, i.e., x-axis, is R u2 sin 28 I g g 47.d.-= =-cot8 T2 4u2sin20lg2 2 f3 = v)' , 4 tan- ' = - - tan- 3 ei.e., gT2 = 2R tan Vx If T is doubled, then R becomes 4 times. 56. c. U x = u, uy = 0, Vx = U x = U 48. c. We know that at the uppermost point of a projectile the ++Vy = u y ayl = 0 gt = gt vertical component of the velocity becomes zero,' while the Resultant vel: v= ';v'; + v~ = Ju 2 + g2t2 horizontal component remains constant. The acceleration due . . fIh57. I. a. Given h = 490 m, U = IS mis, Apply T = - g to gravity is always vertically downwards. Therefore, at the uppermost point of a projectile, its velocity and the acceler- ation are at an angle of 90°. ii. c. Vy = gT 49. a. For the person to be able to catch the ball, the horizontal Juiii. c. Resultant velocity = 2 + g21'2 component of velocity of the ball should be same as the speed 58. b. h = 150- 27.5 = 122.5 m e ~ ~of the person, i.e., Vo cos = or cos 0 = or 8 = 60° Vg= =(2h- /2 x 122.5 Time taken, T = '1--9.-8- 5 s . fIh50. c. u =180 kmlh =50 mls = 50)2 x 490 = 500 m Now s = u T or 30 = 5u or u = 6 mls 9.8 Honzontal range = u - euZ sin2 u2 sin 28 g 59. d.h = and R = - - - 51. b. The bullets are fired at the same initial speed. Hence 2g g = e eh 45 tan h' 2g 52. a. The time of flight is given by - = - = -- or = 45' R 180 4 60. = =d. Range 150 ut and h = 11050 = 21 x gt2

5.58 Physics for IIT-JEE: Mechanics J , = 2 xiS 30 J3 =--ort=- ort 100 x g 1000 10 150 x 10 __ 500 M -I H: u2 sin2 e J3u_- 150 -_ v~\"ms n = llg t 2x- 10 61. h. The horizontal range is the same for the angles of projection . = Hm -H~ x 100 e and (90\" - e) % decreases III Hm Hm 11 = -2u-si-n e, 2u sin(90\" - e) cos e g t2 = g = 2\" :~)=(1- g xI00=9% tit, = 2usine x 2ucose = -2 [U'Sin2e] = -2R ,. usin8 I using usine 10 II. d. t = -- ,t = -- = -~- = -t g g gg g g g' II where R = u2 sin2e 10 g % decrease in t = t -t' x 100= (1- 10) x 100=9% t II Hence, tit, ex R (as R is constant). e... Ill. 62. c. y, = u2 sin' a u' sin2 (90\" - 0) a:u:.'-c.:o.:s:'.:~ a. Here H' = u2 sin2 = III 11g , y, = 2g 2g 2g 2x- 10 u2 1 2at2 YI + yz = 2g =Using x = ut + where x = H~I' U 0 and 63. h. Given y = 12x - ~x\" Ux = 3 mls a = g - !l.. = 9g. we find t = J2H' /a 10 10' m v = -dy = dx - -23 dx = /200 x u sin a y 1 2d -t xd-t dt 198 g dx It is almost equal to the time of fall in the absence of friction. At x = 0, 1Iy = u y = 12 dt = 12l<, = 12 x 3 = 36 mls R' sin2e g 2X)ay = 66. h. T' = g 4 sin' a = 2: cot e = 5 cot a d (dY ) = d'x - 23: (ddXt + x d . R = 5; hence 5 = 5 cote or a= 45\" dt dt 12 dt' dt2 GIVen T' d 2x = a, = 0, hence 67. h. t = 2u sina 2u sine . but dt' or 2 = or u sine = g -~ = -~u -~ -~ gg a = 2 dx 2x = 2 x 3= 2 mis' = u'sin2 e = -g' g dt 2g - y hm 2g = 2 =5m Ra n g 2uxuy = 2 x 3 x 12 =16 ill eR=-- ay 9/2 68. c. If the ball hits the nth step, then the horizontal distance tra- 3 versed = nh. Here, the velocity along the horizontal direction :\\ Alternatively: We have y = 12x - x 2 When projectile = u. Initial velocity along the vertical direction = O. So again comes to ground, then y = 0 and x = R. nb = ut (i) 3, .:IJI+,:c,.------=-::---l>. U 0=12R-:\\R =}R=16m e a.64. d. Range is same for angles of projection and 90 - R = u' sin 2e hI = :u:..2..:s:.i=n..2..:e:. - --, g 2g e=u2-c=o~s2,Hence and h2 = U2 sin2(90 - e) 2g 2g 4H II 1!.~ = u'sinecose = ~ [u2 Sin2e] = Fig. 5.163 2g 4 g 4 . nh = 0+ -1gt2 (ii) 2 65. i. h. Retardation due to the friction of air = !1l.0.. Hence., in From t = -n. pbuttm.g .t m equatI.On (I..I) v upward motion: (nb)2nh = 21:g x -;; . = g+ g = 1Il0g 2hu' total retardatIOn gl or n = gb' 10

Motion in Two Dimensions 5.59 69. b. Here. U x = 6 and u y = 8 is great enough to travel the horizontal distance to the tree R = 2ux uy = 2 x 6 x 8 = 9.6 m . before hitting the ground. (For large u lesser will be the time g 10 70. e. v2 = u2 - 2gh or u2 = v2 + 2gh of motion; so the monkey is hit near its initial position and or U2x +u2y _- 2Vx+ 2Vy+ 2 1g 1 , for smaller u it is hit just before it reaches the floar.) Bullet u; v;= + 2ghor u; = (2)2 + 2 x 10 x 0.4 = 12 will hit the monkey only and only if Y > 0, . H - 2I. gt 2 > 0 I.e., u y = v'I2 = 2.J3 mis, Ux = Vx = 6 mls or H > I 2 or H > I x2 _gt -g 22 2ucos8 g or u > g-(x-·2-+-H--2) = Uo 2H 2H '*atan~u, I 2 = = 2.J3 - a= 30' -- = J 1r--or u > ~ Ux 6 .J3 ycos8 71. c. At the two points of the trajectory during projectile mo- If u < uo, the bullet will hit the ground before reaching tion, the horizontal component of the velocity is same. Then the monkey. u cos 60Q = V cos 4SO '* I 1 ISO 150 x - = v X M or v = M mls 74. a. The motion of the train will affect only the horizontal com- 2 ,,2 ,,2 ponent of the velocity of the ball. Since, vertical component is same for both observers, the h\", will be same, but R will •• • _ u • 600 -_ -15-02.J-3 mls be different. ImtIally . u y - sm 75. d. hz = u2 sin2 82 x 2g = 150 I ISO hi 2g u2 sin2 81 Finally: Vy vsin45°= M x M = - mls sin2 :n:16 . ,,2 ,,2 2 + 150 150.J3 TBut Vy = --2- - lOt uy ayt or = 150 lOt = T(.J3 - I) or t = 7.5(.J3 - I) s 72. a. A bullet fired at angle 45' will fall maximum away, and all 2 other bullets will fall with this bullet fired at 45 0 • Rm\" = ':.. 76. c. Suppose the angle made by the instantaneous velocity with g (~)Maximum area covered = :n:(Rm,,)2 = :n: 2 the horizontal be \". Then 73. d. If there were no gravity the bullet would reach height H vy usin8-gt in the time t taken by it to travel the horizontal distance X, tan\" = - = ------,-\"- i.e., .x Vx u eose H = usmat and x = ucos8t ort = - - u eose Given that: \" =45\", when t =1 8; \" '\" 0\", when t =2 s However, because of gravity the bullet has an accelera- tion g vertically downwards, so in time t the bullet will reach e e -This gives: u cos = u sin g (1) a height u sin8g - 2g = 0 (2) Solving equations (I) and (2), we find ausing = 2g and u cos 8 = g I I Squaring and adding: u = .J5g = 10.J5m/s gt 2 '*77. c. v cos 45\" = u = 18 mls y = u sin 8 x t - - = H - - gt ' v = 18../2 mls 2 2 ....\"\"\"... '\" 18m/s usin (} 45\" '\",,, Fig. 5.165 +---x_ ~=Vertical component: v sin 45° = 18../2 x 18 mls Fig. 5.164 78. b. Vx = u, = 100 mis, Vy = +uy ayt = 0 + 10 x 10 ~This is lower than H by gt2 which is exactly the amount '*Vy 100 8 = 45\" the monkey falls in this time. So the bullet will hit the monkey tan 8 = - = - = I regardless of the initial velocity of the bullet so long as it Vx 100

5.60 Physics for lIT-JEE: Mechanics I 79. a. Ux = 16 cos 60\" = 8 mls 11 =-gt'=- x IOx(])'=5m Time taken to reach the wall = 8/8 = ] s 2, 2 Now U r = ]6 sin 60° = 8v'3 mls 86. c. From Fig. 5.167 II = 8v'3 x 1 - 2I: x 10 x I = 13.86 - 5 = 8,9 m H u' sin2 e/2g sin' e 1 tan¢= -- = = -- = -tantl 80. b. When a projectile is projected at an angle Ii or at an angie R/2 u2 sin 28/2g sin 28 2 (900 - Ii) with the horizontal, the horizontal range remains the same. The horizontal range is maximum when the angle of projection is 45°. 81. d. lOA = IBe (Fig, 5,]66) Fig. 5.167 To find: lOA + tOB + +tOA taB = IBe (taB) = T y 11 87. e. H = eu2 sin2 eu2 sin2 or 80 = - - - o cL.-'----~-L.-x 2g 2 x 10 Fig. 5.166 or u2 sin2 0 = 1600 or u sin 8 = 40 ms- I Horizontal velocity = u cos tlg at 3 x 30 = 90 ms- 1 82. b. Let at any time t, the ball is at height of 15m. usinO = 40 94 orO = tan- 1 (49) ucosO 90 ortane = S•,\\' = u\", t + -2]ay,t 88. b. II = (u sinO)t - ~gt' =} 15 = 4 sin iii - ~gl' d = (ucose)t ort = -d - , u cosO 5I , =} ]5 = 52 x -t - - x lOr 13 2 ~-el2 ~ d 1 usinG x --,- - -g +=} t' - 41 3 = 0 =} (t - 1)(1 - 3) = 0 eh = x uco~e 2 u2 cos2 =} t = 1 s, I = 3 s, required time: 3 - I = 2 s, dg u=-- u'83. i. b. II = 'in' e = (56)' sin'l:I costl 2(dtanO-II) 2g 19,6 e, v'3u 40 x 19,6 1\" 21: (56)2 = or sm 1:1 = 89. c. GIven -2- = u cos = speed at maximum height or 4, 2 or Ii = 0 s sm 1:1 = 30 .. u'll,a,R= -s-in -2t1 (56)' sin 60' cos8 = 2v'\"3 orO = 30° (i) g 9,8 Given that P Hm\" = R (ii) 56 x 56 x v'3 6 r:; Rtane = =10y3m We know Hmax = - 4 - 19,6 84. d. Angle of projection from B is 45°, As the body is R4 4 eP = - = - = - - = 4 v ' 3 v'able to cross the well of diameter 40 m. hence R = - or Hmax tan tan 30° g 90. c. Y = ax - bx2 , for height or y to be maximum: ely = 0 or v =..fiR = ,j10 x 40 = 20 mls - dx On the inclined plane, the retardation is: a - 2bx = 0 or x = -a 2b g sin a = g sin 45°= 1Ov'2 mis' Using v2 - u2 = 2ax )2i.Y,m, (a a' =a (2ab) -b 2b (20)2 - U' = 2 x ( - ~) x 20v'2 = 4h u = 20v'2 ms- I • i,e\" V = 20v'2 mls (dY) eoii. = a = tan where 80 = angle of projection 85. b. Time taken by bullet to reach the target dx .t=.() distance distance 80 = tan- t a. = velocity -uc-os-B , as Ii is very small, cosO = 1 u~nli 2 ,, , distance 400 91. a. tan () = - - = - . The desired equation IS u costl ] TIme = u = 400 = 1 s gx2 lOx' Vertical deflection of bullet Y = x tan e - 2u2 cos2 Ii = x x 2 - -(--\"\"\"'~=)-'-(--1-)-';2 2 '12' +]' - vIS

Motion in Two Dimensions 5.61 or y = 2x - 5x2 '7)or a = 30° + tan\"! ( 92. c. Average velocity = -D-isp'=lac.e-m-en-t 98. a. Horizontal component of velocity. u H = U cos 60° = ~ T!me YI~ '2ut AC = UlJ X I = and Ulx (U;)AB = AC see 300 = (~) = ~ () 99. d. Speed in horizontal direction remains constant during whole journey because there is no acceleration in this di- +RI2+ =rection. SO, VII 5 rus .. ·j Fig. 5.168 In vertical direction: Loss of gravitation potential energy Here, H = max height :::= gain in K E Le., mgh = ~mv~ v2 sin2 e = 2g v2 sin 211 v~ = 2gk = 2 x lOx (70 - 60) = 200 R = range = - - - Hence, the speed witli which he touches the cliff B is: g 2v sin Ii v= Jv~ + v~ = ../25 + 200 - '=../225 15 ms\"! and T = time of flight = - - - g +_ V / 2 100. c. AB =\"yfI2hi V\" - ZV 1 3 cos 8 93. b. .u._' s_in..2..8- = (..u_/2_)'.._sin....3_0.\".. u' K g g 8g U G)~ ~.'. sin2B = orB = sin\"! yg94. c. x = v f2h, 2x = v,)2 (2h) Solve to get: v' = V2u AB g 5)2= 600 x 18 Fig. 5.170 95. b. T = 210s0 = 4 s =? 2u sin Ii = 4 =? u sinO = 20 mls x 2000 = 3333 m = 3.33 km g 10 u2 sin2 e u2 sin 28 u2 sin (2 x 15°) u' 96. d.H = ,R = ._- 101. b. 1.5 = =? - = 3 2g g g g maximum height will be same because acceleration a = g/4 u'R = - sin (2 x 45°) = 3 km g is in horizontal direction (2USinO)'R, =ucosliT+I -aT'=R+1-g- =R+H 2 24 --,- 102. c. x = 6t, Ux = -dx =6 g dt . 2u sin(a - 30°) dy 97. a. tAB = time of flIght of projectile = _ _.c.....=,-_ y = 8t - 512, vy = - = 8- lOt . dt g cos 30° Vr(t~O) = 8 mls Now component of veloeity along the plane becomes zero at point B. u u= Ju; + v~ = \"/6' + 102 = 10m/s 103. b. tan Ii = u,. = -8 = -4 ~ Ux 6 3 104. c. vertical eomponent of both should be same: v,\", = sin 30° =? V2 = Z1 = 0.5 ;;;- A 105. d. Apply equation of trajectory: Fig. 5.169 ('7)../3 g =? Vo = 00 . o= u eos(a - 30°) - g sin 30° x T 0.5 = ....2 tan 30\" - 2 2 2uo eos2 30° ' . 2u sin(a - 30°) or u cos(a - 30°) = g sm 30° x -_.o....=c-':\" 106. b. a, = -u' = -52 = 12.5 m/s2 r2 g cos 30° cot 30° '../3 ortan(a - 30°) = -2- =2-

5.62 Physics for IIT-JEE: Mechanics J 107. a. Since velocity is in tangent direction so its component eAngle of v'i'\" with inclined plane is 60 - = fJ a10ng radial direction is zero. tan 60° tan e 108. e. a, = 2m/s2, v = u +a,t = 1 +2 x 2 =5 mls tan fJ = tan(60 - 0) = -;-c--:c;:cc---;:--;: 1 + tan 60\" x tane -vG - VS/7 = 3-vG 1 +-VG x -vG/7 5 Ja; arnet ace = + = --/12 + 2~ =../5 m/s2 =? fJ=tan.,(3~) 109. c. ac = v2 -7 constant in magnitude if v is constant. - r 3. c.,d. Distance between two buses on road = Vb T 110. c. After releasing the string. centripetal acceleration will be- For A to B direction: distance =:: relative velocity x time, come zero, due to which direction of velocity cannot change VbT = (Vb - velT, =? T, = -V-b T- now and stone flies tangentially. Vb - Vc 111. a. Angular velocity is always directed perpendicular to the For B to A direction: plane of the circular path. Hence, required changc in angle VbI' = (Vb + V,,)12 =? Vb T = 0\", +1'2= - - - Vb Vc 112. b. Angular acceleration and angular velocity are along the 4. b.,e.,d. If they collide. their vertical component of velocities axis of circular path. So they cannot be perpcndicular to each should be same, i.e. other. e e100 sin = 160 sin 30\" =? sin = 4/5 =113. c. W = 300 rpm = 300 x 2JTI60 IOrr radls Their vertical components will always be same. Hori- zontal components: 114. c. r = I km = 1000 m, v =900 kmlh = 900x5/18 = 250 mls 160 cos 30 = 80VS mls a,. = -vr2 = (250)2 2 and -10-0=0 62.5 mls 115. d. a = (W2 - WI)/I = (400 - 100)/5 = 60 rev/min' 100 cos 8 = 100 x 3/5 = 60 m/s They are not same, hence their velocities will not be 60 x 2n 2JT 2 same at any time. So (b) is correct. = = - rad/s (60)2 60 2rr 50 rr 2 /Z a, = ar = 60 x 100 = 60 m/s fof-- x --t>I14-- \"2----Jo.l 14: Xl _I MultipLe CorreCt Answers [Mpe Fig. 5.172 1. a.,c. For angle e in west of north, x = X,. - Xz = 160cos 30\"1 - 100cosei esin 8 = 150 = 0.5, Le., = 30° =? X = (80-vG - 60)1 V,. = Resultant velocity = \";°(1\"'002,,\")-_-:(CS::'\")2 = S-vG ms· 1 Time of flight: T = 2 x 160 x sin 30 == 16 s So, options (a) and (c) are cOITeet g ._ ,_ 10 --v-G1 +,. 1'0, Now I < Tx-+ to collide in air -30}, v\", -2 2. a., d. v, = = 2.1 =? ,/3 < 16 =? X < 1280-vG - 960 80. 3 - 60 Since their'times of flight are the same, they will simul- taneously reach their maximum height. So it is possible to collide at highest point for certain values of x. e 7tane = .5_-vG- = --vG' 5. a.,e.,d. [K.E: + P.E.) = [P.E.] projection point with 35 7 =? = tan·-1 (-vG) W.Ith vertI.CaI ground In all situation K.E. (i.e., speed) at the ground are equal. i.e., option a. is COlTect. h vt, igt?1 . . 0) = + [for (first) particle) h = VI, - ~gli [for (second) particle) Oi) Fig. 5.171 From equation (i) and equation (ii) 12 > It (t2) = maximum, (t,) = minimum i.e., option (c) and (d) are correct.

Motion in Two Dimensions 5.63 6. b.,c. y = 2x implies that the particle is moving in a straight a straight line having negative slope. Hence, option (d) is wrong. line passing through origin. 8. a.,c.,d. Angular speed is constant. Hence, linear speed is also s constant, i.e., magnitude of velocity is constant, but direction is changing. Fig. 5.173 9. b., c., d. If the particle is projected with velocity u at angle B, then equation of its trajectory will be: gx' y=xtanB-2u-2 ~co~s2-B We know slope is given by m = dy dx vx =4 2t, vx=ux+axt, ux =4, ax =-2 gx Now.y = 2x dy =I-d-x Therefore, slope m = tan B - 22 u cos B =} - It implies that the graph between slope and x will be dl 2 dl straight line having negative slope and a non-zero positive intercept on y-axis. But x is directly propOltional to the time I, therefore, the Uy = 2anday =-1 shape of graph between slope and time will be same as that 7. b., c. Since the particle is dropped, it means that the initial ve- of the graph between slope and x. Hence, only option (a) is locity of the particle is equal to zero. But the particle is blown over by a wind with a constant velocity along horizontal di- correct, i.e., option (b), (c), and (d) are incorrect. rection, therefore, the particle has a horizontal component of velocity. Let this component be Vo. Then it may be assumed 10. a.,b. x = a cos(pt), y = b sin(pt) that the particle is projected horizontally from the top of the Equationofpathinx-yplane: [~r + [~r = I tower with velocity Vo. Hence, for the particle, initial velocity u = Vo and angle i.e., the path of the particle is an ellipse. of projection B = 0\". Position vector of a point Pis: We know equation of trajectory is: r= a cos pt! + b sin pI} v=? = p(-a sin ptf + bcos pI}) gx2 aand, = -p'(a cos pt! +bsinpl)) = _p2r ey=xtanB~-2-~- a °also, V· = att = rr/2p 2u 2 cos2 11. a.,b. Time of ascent = 2 + I = 3 s gx2 usinB . Here,y =-2 =} - - = 3 =} usmB =30 g 2vo And tan fJ = U sin B - gI =} tan 30\" = ::30=---...:I::O::x:,::::2 The slope of the trajectory of the particle u cos B u cos B -dy- - -2g-x - - gx dx - 2vo' - v02 =} u cos B = lOy') Hence, the curve between the slope and x will be a From hereu = J(loy')), + 102 = 20y') mls straight line passing through the origin and will have a neg- ative slope. It means that option (b) is correct. 30 And lanB = h = y') =} B = 60° Since. horizontal velocity of the particle remains con- IOv 3 . = dy = -g-t stant, therefore x vol. We get - -v2 (21), dx Vo 0.2 12. a.,b.,c.,d. v = 2t, ac = r = So the graph between m and time I will have the same shape as the graph between m and x. Hence, option (a) is = 2012 = 20 x 22 = 80 mis' wrong. dv 2 - The vertical coinponent of velocity ofthe particle at time at = =2m/s a Ja; atdl I is equal to gl. Hence, at time I, K E ofthe particle, K E = Net acceleration: = 2I m [{gl)' + (vO)2] + > 80 m/s2 It means, the graph between KE and time t should be a Assertion-Reasoning Type parabola having value ~mv5 at t = 0. Therefore, option (c) is correct. 1. d. At the highest point only horizontal component of velocity As the particle falls, its height decreases and K E in- is present, and vertical component is zero. creases. The K E increases linearly with height of its fall or the graph between K E and height of the particle will be 2usinB u2 sin2B, 2. c. T = =} T ex It, R = =} R ex u gg

5.64 Physics for IIT-JEE: Mechanics I ° k8 _4 A -L.._ 3. b. If we cut the string anywhere the bob will follow the parabolic path. Q\"----~R 4. b. Before attaining the maximum height. angle is acute and Fig. 5.175 after this the angle is obtuse. At the highest point it is per- pendicular. towards B (AB is perpendicular to river flow) (Fig. 5.176) 5. d. Projectile motion is a motion with constant acceleration Us = us/? + VR but it is not a straight line motion. A body with a constant magnitude of acceleration may not USR = velocity of swimmer relative\"to river speed up, this is possible in uniform circular motion. vR = velocity of river 6. b. In a non-uniform circular motion, due to change in mag- Ii c nitude of velocity, tangential acceleration arises and due to change in direction of velocity centripetal acceleration is pro- L':-' to Ol e 11 =., 1 km!hr duced. 7. c. In a uniform circular motion, velocity is along tangential ~> A direction and acceleration is always_ towards centre, so angle between velocity vector and acceleration vector is always Vs. R Fig. 5.176 '72f ' But in general, the angle between the velocity and the Hence the net velocity of the swimmer is directed from acceleration can be acute or obtuse also. eA towards C at an angle with the river. 8. a. Since the \"relative acceleration is zero and initial relative e = tan~1 ~ velocity is also zero, so relative velocity at any moment will 2 be zero. 9. a. Time taken is shortest when one aims perpendicular to the eFrom triangle ABC, BC = AB cot = (10)(2) = 20 m flow. .'. the swimmer lands on the other side of the river at a point 10. c. Urjll! = Jv; + v;;/ C which is 20 m from the point B, towards which he was trying to swim. Comprehensive time = Sy/ Vy = 10 m /1 kmlu·-· 1 = 36 s TYRe Method 2: For Problems 1-2 v.> = \",.R + \"R\"S.R = 1] km/hr VR = 21 km/hr 1. b., 2. a. Sol. 0 A represents velocity of man due east. Vs = 1) + 21 (km/hr) = 21 + J(km/hr) o Q represents relative velocity of rain \\V.r.t. man. Actual velocity of rain is represented by 0 R (Fig. 5.174). y t .I. .S. . .]. ...•.,..O....\". ' \". '.A.•. ,I·\" -> 1V.,.8 Q.. R , X• A VR Fig. 5.177 Fig. 5.174 Time to cross river Displacement in y-direction 10 t= =--=36sec foOR = .jOQ2 + RQ2 = + 32 = 3V2km/hr Velocity in y-direction 5/18 e eAlso, tan = 0Q QR = 1 =? = 45' For Problems 5-6 The rain is falling at 45° east of vertical with a velocity S.c., 6. b. 3Y'2 km h~1 ii,. = v,.\", + v\", (Fig. 5.175) Sol. Time taken to cross the river For Problems 3-4 L 400 t=--=-- 3 a., 4. b. Vb .w 10 Sol. Assume that the person starts from point A and tries to swim t =40 s.

Motion in Two Dimensions 5.65 y + vw eJVb(absolute) = Vbw + = (2 - 4 sin 8)1 + 4 cos 1~ Vbx = 2 - 4sine, Vby = 4cosB VS•R VI< X• 7. a. For directly opposite point Vln = 0 A sinO = ~ = sin 30' => e = 30' Fig. 5.178 2 B:'---X--.. c Hence to reach the point directly opposite to the starting point he should head the boat at an angle = =f3 (90 + 30) l20with the river flow. L\"-- 400 m =2.8.c.I=l'... => I=_d_= 4 ../3 hr. Vby 4 cos 0 4 cos 30° u = 2 m/s 9. ed. For 1 to be minimum cos = 1 => e = 0' A 4 tmin = - - = 1 hr. Fig. 5.179 4cosO Drifting Be = Uw x 1 10. d. TJ = -2- = 1 h2r, Tz = -1- = - hr Be = 2 x 40 = 80 m. (4-2) (4+2) 3 Method 2: Vb = Vb,w + Vw = 10] + 21 (1+T = _2_hr + _2_hr = ~) hI' = ~hr (4-2) (4+2) 3' 3 i\\ = 21 + 10) mls For Problems 11-12 ,,y 11. c., 12. d. IE C Sol. Taking N as + Y-axis and E as + Axis. Imagine yourself as an observer sitting inside thc ear. You will regard the car as being at rest (at e), Relativc to you, the speed of the motorcyclist is obtained by imposing the reversed velocity of the car on motorcyclist as shown in the Fig, 5,182. L= 400 In North e , ___ _~t:::,, p A ------\" X Fig. 5.180 Time to cross the river (Fig. 5.180) East X-axis L 400 View from inside the car (at C) 1=-=-=40s Fig. 5.182 Vb,w 10 Vme = ~152 + 20z = 25 mls Drifting Be = u x 1 = 2 x 40 = 80 s. (~~)0= tan- J = 53\" withX-axis For Problems 7-10 The motorcyclist appears to move along the line M P 7. a., 8. c., 9. d., 10. d. with speed 25 mis Sol. B is a point directly opposite to the starting point A, The shortest distance = perpendicular distance of M P Let the man heads the boat in a direction making an angle B with the line A B, Here, Vw = 21 from e = d => d = 50 cos 53° => d = 30 m Vbw = -4sinol +4cosO] v in ms-! Time taken to come closest = time taken by motorcyclist 4 2 to reach B =MB 50sin53° 1= - = Ville 25 => t 1.6 s. 2 ______________\"-_..JI. 8 For Problems 13-14 4 13. e., 14. a. Sol. .d O.5km 1 e1 3 . c . I = - - = =--h v sin 3 sin 120'km/h 3../3 Fig. 5.181

5.66 Physics for IIT-JEE: Mechanics I 14. a. x = (u + vCose)1 = (2 + 3cos 120°) 1 20. d. - = _ + 1_ , M S ul Zal 3v 3 1 , + y , = (5iA)1 + I, + \" =--km xi j Z(3i 2j)1 6../3 For Problems 15-16 = 51 + 3 = + 3 Zl2 Zl' 15. b., 16. b. x =} 84 5t Sol. v 31' + 101 - 168 = 0 =} 3t' + 281 - 181 - 168 = 0 15. b. sin 30° =..!!'. =} v, = 1 mls t[31 + 28] - 6[3t + 281 = 0 1=0 v, Y = 12 Y = 6' = 36 m v, 21. c. y = 12 • X = 51 + ~12 dx = Vx = 51 + 3t. dy = Vy = 21 - - elt ell Vm = 0.5 Vx = 5t + 3(6) = 23. Vy = 12 Fig. 5.183 v = Jv~ + v~ = /(23)' + (12)' '\" 26 mls 16. c. v,)m = v,. cos 30\" = 1 x 2../3 = 0.5vr3;; mls For Problems 22-23 For Problems 17-19 22. c., 23. d. 17. c. d., 18. d., 19. d. Sol. eSol. (i) 17. c. Velocity of first body at any instant I. VI = 21 - gl) 22. c. Maximum height: h = u2 sin' - 2g 8 mls --6--r:\"-- 2 mls u22 sin ecos e ,,,, 2h = (ii) g eDividing equations (i) and (ii); tan = 2 =} 8 = tan-, (2) 23. d. tane = I2 15/1 2 L,J Fig. 5.184 Velocity of second body at any instant Fig. 5.185 V, = -87 -gt) u'. . 2gh Since l:it.lv2 Le., 01 x ih = 0 eFrom equatIOn (I) = -'-2- sm (21-gl)(-87- gl)=0 =}-16+g'I'=0 2gh 5gh u = /5gh12 (21N 2 YI!'{l61= =} 4 For Problems 24-25 1= 10 = 0.4 s 24. c., 25. a. Sol. 18. d. .51 = (2 x 0.4)7..52 = (8 x 0.4)(-7) 24. a. Here r will become range .5, = 0.8f and .52 = -3.21 u2 sin 2e =} gr = u2 sin 2e Separation = 51 (-.5,) = 0.87 + (3.21) = 47 r= g 19. d. S- , = 21i, - Z1gl2j'. S- 2 = -81i, - z1gt\" j et Z25. a. h = u sin + 1 (Fig. 5.188) gl2 as 5,1..52 i.e.• 5,.52 = 0 x = ucoset x -161 + -1 x 1 = 0 =} g212 = 16 x 4 1=-- _g214 ucos8 22 8 x 1 x2 =} gl = 4 x 2 =} 1= - = 0.8s n8-- cos ueose -g 10 2 For Problems 20-21 h = u s i + eg x 2 20. d., 21. c. u2 2 Sol. + 2u' sinD cos ex - 2hu2 eos2 e = 0

Motion in Two Dimensions 5.67 32. c. h = v sin el - ~ gl' u when I = 1 s, then h = v sin 0 - ~ g gx4 g g 3g or h = -2- - = 2g - - or h = - I<--x-->l 222 Fig. 5.186 Aliter: ~g12 - vsinOI +h = 0 x = -2u' sin 0 cosO 1\\12 = h h 23g -I-or 1 x 3= g/2 orh = +J4u' sin' 0 cos2 e + 8ghu' cos' e -g =U-CgO- SyOU[/s\"m20 +2gh-usm,O] 2 For Problems 33-35 .33. c., 34. a., 35. d. Sol. [~ 2 2 sin' = -2g v sin' For Problems 26--29 v; _v 0 0]33. c. 26. a., 27. b., . 28. d., 29. c. 2 2g Sol. -v2-si,2n-2 -e 26. a. 1 = mg(3 - I) = 2mg or v2 = v2 sin2 e - '2mv2 y [from conservation of energy] e eorvy2 = or v = .J4g' = 2~ v2 sin2 v sin Vertical component at A = 2~ sin 30\" = ~ 2 orvY=..fi 27. b. This is equal to time of flight 34. a. v\" = (v cos 0)' + ( V.S.fiine)' 2vsinO 2~ 2 or v'z = v2 cos2 e + -v 2 ;s:in:2-e- T=--=-=- - g g~ =, ul + I, we get 0)2 '2ar, 28. d. Usmg S or va = v' ( cos' 0 + -si2n' - -I = ~I - 12 I , ~I - t= 0 sin2 e '2gt or '2l1t - or Vi = V cos' 0 + - - 1= ~±Jg+4X ~g l±v'3 2 g o r t =~- - - ~v235. d. v'cos'O = [cos'O + Sin; 0 ] 2x - 2 or 5 cos' 0 = 2cos2 0 + sin' 0 Neglectm, g -ve tI,me, I = -1-+-v-'3- ~ or 3 cos2 0 = sin' 0 29. C'X=2~COS300[1~] orx=(v'3+3)m or tan' 0 = 3 or tan 0 = v'3 or 0 = 60' Angle of projection with vertical = 90' - 60' = 30', For Problems 30-32 For Problems 36--39 30. c., 31. b., 32. c. 36. b., 37. d.,. 38. a., 39. c. Sol. Sol. 1I - 36. b. As shown in Fig, 5,187 30. c. h = vsinOI - '2l1t2 or '2gt' - vsinOI + h = 0 4~\" 20 mig 30° -vsinO 2vsinO 1\\ + I, = ---,-- or t\\ + I, = --- = T -g g 2 15m =or T(I + 3) s 4 S I+-- 15 m -t>l ~2 sin2 e Fig. 5.187 31. b. hma< = - , - - - -15 = 20 sin 30\"t - ~ 101' 2g =sgg'T=' gg1T'=ggI X4X4=2g [ ','vsinO=gTT]

5.68 Physics for IIT-JEE: Mechanics I =} 5t' - lOt - IS = 0 2vsine 2 x 5 3 =}t=3s =} t = - - - = - - tane = - s gcose 10 4 37. d. R = 20C0830't = 30v'3 m R = uxt + 1 = veost + I. 38. a. Vx = Ux = U cos 30\" = 1Ov'3 mls 2axt' 2gSmet2 u, = uy +ayt = 20sin30' -10 x 3 = -20mls 3 1 . (3)'5 cos 37° x 4+ 2 x 10 x 8m 37\" x 4 = 7156 m v = Jv; + v; = 1Ov'7 mls 39. c. Maximum height above top of tower: 46. b. Vx = U x +axt = 5 cos 37' + gsin37' x ~ ~v)' = u)' +ayt = 58in37' - gcos37' x = -3 mls (20)' sin3 30\" H= =Sm Sg Maximum height attained above ground v = JVx2+ vy2= ( -127)' + 32 -- -52.J13- mls. =1S+H=20m For Problems 40--41 40. c., 41. c. For Problems 47-49 Sol. 47. c., 48. a., 49. b. 2 Sol. l~ 3s40. c. R = ut = uf!J- =} R = 3/ 1 = m v2 =} r = -v2, where r is known 47. c. We know that a, = - 41. c. vx = ux =3 mls r ac as radius of curvature. At the highest point v = ucos e, vy' = u'y+ ,2gh =} vy2 = 0' + 2 x 10 x 1 = 20 eu2 cos2 v=Jv~+u;=mm/s ac=g=>r= g 48. a. Similarly find the velocity and a, at the given point and For Problems 42-43 then find r. ac will be a component of g.L' to velocity at 42. d., 43. a. Sol. that point (Fig. 5.189). e - e42. d. y = x tan 2u2 e082 =} IS = 30 tan 4S\" IO(3? =} u = 1Ov'6 mls g g cos (O!2) 43. a. v, = Ux = 1Ov'6 cos 45\" = 1Ov'3 mls Fig. 5.189 v; = u; + 2ays), = (l0v'6 sin 45\")\" + 2(-10)lS = 0 vcos(012) = ucose ==} v ucosOsec(eI2) a, = g cos(e12). Now find r = -v' So the velocity is 1Ov'3 mls horizontally. a, For Problems 44-46 49. b.h = H12. v cos</> = ucose (i) (ii) 44. c., 45. a., 46. b. v' sin' </> = u2 sin' e _ 2g H 2 Sol. 44. c. 45. a. Fig. 5.188 F l' 4-~--..... v u, ,, ,,III 8 ,,'Ii /I e x Fig. 5.190 Fig. 5.188 Squaring equation (i) and adding in equation (ii) ax =gsinB,a y = -geose =} v2 = u' - gH Sy = 0 =} uyt + -21a.,t2 = 0 u2 sin2 e .I whereH= - - - =} vsmet - 2gC08Bt' = 0 2g ac = gcos¢ = gu cose = -v 2 ,Now r a, v

Motion in Two Dimensions 5.69 Matching VJ = vcos457 - vsin45) Column' Type vJ\"'v = + (-Vi) = 2v sin45(- j) 1. i. -7 b.,d., ii. -+ c., iii. -+ a., iv. -+ c. Distance will be minimum because the man will reach I\"'ii I = 2vsin45 = v.fi from point A to point B directly. . . Total displacement IV ~ d. Average veloCIty = --;::--:-:--- Total time H mRI2H II -7 ~> _) RI2 VI/IW VII/ V(lI Fig. 5.192 e A J(~)Displacement = 2 + H2 Fig. 5.191 e, = -Vw and .v.m.....L..v..w ~ ~ 8111 '';-4' + H2 v'R2 + 4H' Vmw e e-Vav = v sin = -2vsin Hence i. -+ b., d. Time taken is minimum if VII/It) is perpendicular to i\\.o. gg Hence ii. -> C., iv, -+ c. Vav = v e(ii) If Vir/CO < w , then drift or distance is shortest if sin = -VmO.J Hence (1.1..1) -+ a. vw 2. i. -+ a., ii. -> b., iii. -+ c., iv. -+ d. i. -+ a. Range is maximum, when the angle of projection is 45'· (i) v2 v2 H = - sin' 45; H = - 2g 4g Velocity, at half of the maximum height is Vi .\"V = -! ,2 =} ,v 3. i. -+ a.,c., ii. -+ b.,d., iii. -+ a.,c., iv. -+ a.,b.,c.,d. v =- i. In the uniform circular motion, the acceleration and the 42 velocity are perpendicular to each other, but in non uniform ii -+ b. Velocity at the maximum height circular motion the angle between the velocity and the acceleration lies between 0 to 7t12. V ii. In a straight line motion, the. acceleration vector and the Vi = vcos45 =} v ' = - velocity vector are collinear to each other, i.e., the angle .fi between them is either 0 or 1800 • iii. In a projectile motion, the angle (e) between the velocity [because vertical component of velocity is zero at the highest eand the acceleration can vary from 0 < < Jr. point] iv. In space angle between velocity and acceleration may be iii -+. c. Projection velocity OSB SJr· At projection point. Vi = V cos 451 + v sin 45) At the point. when the body strike the ground



MisceLlaneous Assignments and Archives on Chapters 1-5

6.2 Physics for IIT-JEE: Mechanics I from the ceiling. Which curve best represents the position of the bolt as a function of time? EXERCISES a.A h.B e.C d.D Solutions on page 6.14 x 1. For three particles A, B, and C moving along x-axis, x-t Fig. 6.4 5. Fig. 6.5 shows the velocity-displacement curve for an object graph is as shown below (Fig. 6.1). Mark out the correct relationships between their average ve- moving along a straight linc. At which of the points marked, locitics between the points P and Q. is the object speeding up? x a. I b.2 e. I and 3 d. 1,2 and 3 Q \" - \" \"C\" - - - - ._ _ 1 Fig. 6.1 a. vav,A > Vav,B = vav,c h. Vav,A = Vav,R = vuv,c c. Vav,A > Vav,B > vav,c d. Vav,A < Vav,B < vav,c 2. A particle is moving along a straight line whose velocity- 6. A ball is thrown downwards from the edge of a very higb cliff displacement graph is as shown in Fig. 6.2. What is the ac- celeration when the displacement is 3 m? with an initial speed that is greater than the terminal speed. a. 4vS mis' h. 3vS m/s' Mark the correct statement about its acceleration a. It is always acting in the upward direction e. vS m/s' d. 4/,/3 m/s2 h. It is always acting in the downward direction v c. Initially, it is aeting in upward direction and then it be- --\\l4m/s comes zero ,,,' ,,, d. Initially, it is acting in upward direction and then it attains ' a non-zero constant value in the downward direction 160°\\ \"_-'- _-» 7. The particle is moving along a circular path as shown in Fig. 6.6. The instantaneous velocity of the particle is L_ - : : '_ s 3m ,.; = (4m/s)i - (3m/s)] Fig. 6.2 Through which quadrants does the particle move when it 3. The acceleration of an object, starting from rest and moving is travels clockwise and anticlockwise, respectively, around along a straight Jinc is as shown in Fig. 6.3. Other than at t = 0, when is the velocity of the object equal to zero? the circle? r a. Att ~ 3.5 s -ffi-u b. During the interval from I s to 3 s c. At! ~ 5 s d. At no other time on this graph l Fig. 6.6 a. First, First h. First, Second c. First, Third d. Third, First Fig. 6.3 8. A particle is moving with a velocity of 4 rnls along +ve X di- rection, an acceleration of 1 tn/s2 is acted on the particle along 4. An elevator is moving upward with a constant acceleration. The broken curve shows the posit.ion y of the ceiling of the -ve X direction. Find the distance travelled by the particle is elevator as a function of time t. A bolt breaks loose and drops 10 s. a.IOm h.26m e.16m d.8m

MisceHaneous Assignments and Archives on Chapters 1-5 6.3 9. A body with zero initial velocity moves down an inclined ~ plane from a height 17 and then ascends along the same plane d. The direction of r changes with time; its magnitude may or may not change, depending on the angles of projection with an initial velocity, such that it stops at the same height 17, 17. A point moves such that its displacement as a function of Considering friction to be present, in which case is the time time is given by x 3 = [3 + 1. Its acceleration as a function of of motion longer? time t will be a. Ascent b. Descent 2t 2 a,s2 2t 2t d.-, c. Same in both d. Information insufficient C'x4 x b·x-s X· 10. A car travelling at a constant speed of 20 mls ovettakes an- 18. Two particles are thrown horizontally in the opposite direc- other car which is moving at a'constant acceleration of2 m/s2 tion with velocities u and 2u from the top of a high tower. and it was initially at rest. Assume the length of each car to The time after which their radius of curvature will be mutu- be 5 m, The total distance covered in overtaking is ally perpendicular is a. 394,74 m b. 15.26 m c. 200.00 m d. 186.04 m a. ..fi\":- b.2\":- c. I U d . ~\":- 11. A particle is moving along X-axis whose acceleration is given g g ..fig 2g by a = 3x - 4, where x is the location of the particle. At 19. In Fig. 6.7, the angle of inclination of the inclined plane is 30\". The horizontal velocity Vo so that the particle hits the t = 0, the particle is at rest at x = 34\"' The distance travelled inclined plane perpendicularly is by the particle in 5 s is c. infinite d, None of these Vo a. zero b, := 42 m 12. A particle has been projected with a speed of 20 mls at an angle of 30' with the horizontal. The time taken when the velocity vector becomes perpendicular to the initial velocity vector is a. 4 s b. 2 s c. 3 s d. Not possible in this case 13. At a distance of 500 m from the traffic light, brakes are applied Fig. 6.7 to an automobile moving at a velocity of20 tnls. The position /2 g H ofthe automobile relative to the traffic light 50 s after applying a. VO =V/2-g5H- b. Vo = \\ -7- the brakes. if its acceleration is -0.5 mis', is )g;c. Vo = (iii a.125m b.375m c.400m d.l00m d. VO=YT 14. A parachutistjumps off a plane. He falls freely for sometime, 20. A particle reaches its highest point when it has covered ex- actly one half of its horizontal range. The corresponding point and then opens his parachute. Shortly after his parachute in- on the vertical displacement-time graph is characterised by a. zero slope and zero curvature flates, the parachutist ' h. zero slope and non-zero curvature c. positive slope and zero curvature a. keeps falling but quickly slows down d. none of these b. momentarily stops, then starts falling again. but more slowly c. suddenly shoots upwards, and then starts falling again but more slowly d. suddenly shoots upward, and then statts falling again, 21. Two particles A and B are placed as shown in Fig. 6.8. The particle A, on the top of a tower, is projected horizontally , eventually acquiring the same speed as before the with a velocity u and the particle B is projected along the surface towards the tower, simultaneously. If both the patti- parachute opened cles meet each other. Then the speed of projection of particle B is [Ignore any friction] v;15. An object has velocity w.r.t. ground. An observer, moving with a constant velocity v~ w.r.t. ground, measures the veloc- v;,ity of the object as The magnitudes of three velocities are related by b. Vj ::s V2 + Va +a. Vo 2: Vi V2 d.Allcl~~~ ~~~Vt+~ . 16. Two particles are projected simultaneously from the same +-: B point, with the same speed, in the same vertical plane, and at \"d different angles with the horizontal in a uniform gravitational field acting vertically downwards. A frame of reference is Fig. 6.8 b.dMd)a. g - u fixed to one particle. The position vector of the other particle, 2H b. d)2~ +u d.u 1'.as observed from this frame, is Which of the following statements is correct? 22. Twelve persons are initially at the twelve corners of a regular r~. a. a constant vector polygon of twelve sides of side a. Each person now moves IS ~ with a uniform speed v in such a manner that 1 is always b. r changes in magnitude as well as direction with time -> c. The magnitude of r increases linearly with time; its di- rection does not change

6.4 Physics for IIT-JEE: Mechanics I directed towards 2. 2 towards 3, 3 towards 4 and so on. The e (~)ii. The angle of fall of rain is = tan -I time after which they meet is (Jz)eiii. The angle offal! of rain is = tan -I v~ ~ ~ a, ~ b. -; c. v(i + .J3) a d. v(2 _ .J3) 23. An object is subjected to the acceleration a = 4 + 3v. It is eiv. Velocity or rain is3,,/2kmh- I given above is with verti- given that the displacement S = 0, when v = O. The value of· eal a. Statements (i) and (ii) are correet displacement when v = 2 mls is b. Statements (i) and (iii) are COlTect a. 0.52 m b. 0.26 m c, 0.39 m d. 0.65 m c. Statements (iii) and (iv) are correet d. Statements (ii) and (iv) are correct 24. Two boys P and Q are playing on a river bank. P plans to swim across the river directly and comes back. Q plans to swim downstream by a length equal to the width of the 29. The maximum range of a projectile is 500 m. If the particle river and comes back. Both of them bet each other, claiming is thrown up a plane which is inclined at an angle of 30· with that the boy succeeding in less time will win. Assuming the the same speed, the distance covered by it along the inclined swimming rate of both P and Q to be the same, it can be plane will be concluded that a. 250 m b. 500 m c. 750 m d. 100 m a. P wins 30. Velocity versus displacement graph of a pmticle moving in a b. Q wins straight line is as shown in Fig. 6.10. c. A draw takes place v d. Nothing certain can be stated 25. A projectile is fired with a velocity v at right angle to the eslope which is inclined at an angle with the horizontaL The range of the projectile along the inclined plane is v --+-------------.x Fig. 6.9 v2 sec e Fig. 6.10 2v2 tan () b.--- The acceleration of the particle is g a. constant a, v 2 sin e h. increases linearly with x g c. increases parabolically with x d.-- e2v2 tan (J sec g d: none of these c. --~ 31. The acceleration-time graph of a particle moving in a straight g line is as shown in Fig. 6.11. The velocity of the particle at time t ; 0 is 2 m/s. The velocity after 2 s will be 26. A ball rolls off the top of a stairway horizontally with a ve- locity of 4.5 ms- I . Each step is 0.2 m high and 0.3 m wide. If g is 10 ms- 2, then the ball will strike the edge of nth step where It is equal to c.ll d.12 4 a.9 b.IO 27. A man holds an umbrella at 30· with the vertical to keep himself dry. He, then, runs at a speed of 10 m/s' and finds the raindrops to be hitting vertically. Study the following statements and find the correct options. 2 t (s) i. Velocity of rain w.r.t. earth is 20 ms- l Fig. 6.11 ii. Velocity of rain w.e!. man is 10.J3 ms- I iii. Velocity of rain w.r.l. earth is 30 ms·- I a. 6 mls b. 4 m/s c. 2 mls d. 8 mls iv. Velocity of rain w.r.l. man is 10,,/2 ms 1 32. Velocity versus displacement graph of a patticle moving in a a. Statements 0) and (ii) are correct straight line is shown in Fig. 6.12. CmTesponding accelera- b. Statements (i) and (iii) arc correct tion versus velocity graph will be c. Statements (iii) and Ov) are correct v (mfs) d. Statements Oi) and (iv) are correct to 28. Rain appears to rail vertically on a man walking at 3 kmlh, but when he changes his speed to double, the rain appears to fall at 4SOwith vertical. Study the following statements and 10 dm) find which of them are correct Fig. 6.12 i. Velocity of rain is 2.J3 kmh- I

Miscellaneous Assignments and I'v\"chives on Chapt~rs 1-5 6.5 a (mI,') a em/s2) x ;;::; 0, y ;;::; O. The object is definitely moving towards 0 when 10 - - - --, L~>O,~>O ~~<~~<O 10 v (mls) (a) 10 v (mls) +C. XVx y Vy < 0 +d. xvx y Vy > ,0 a (mls') (b) 10 a (m/s2) 37. Two particles A and B are moving along a straight line, whose position-time graph is as shown in Fig. 6.16 below. Deter- mine the instant (approx) when both particles' arc moving with the same velocity. x B A v (mls) 10 v (m/s) \"'--=''-:-':--~-:'::--'''' I(S) (e) (d) 5 10 11 20 33. Acceleration-velocity graph of a particle moving in a straight Fig. 6,16 line is as shown in Fig. 6.13. Then, the slopc of vclocity- displacement graph a. 17 s b. 12 s c. 6 s d. No where a 38. An object moves along the x-axis. Its x-coordinates is given as a function of time as X = 7t - 3t2 m, where X is in metre and t is in second. Its average speed over the interval t = 0 to t = 4 s is c. -21649 mls 169 a. 5 mls b. -5 mls d'24 mls 39. A cannon fires a projectile as shown in Fig. 6.17. The dashed --f-------------.v line shows the trajectory in the absence of gravity. The points Fig. 6.13 a. increases linearly M, N, 0 and P correspond to time at t = 0, I 3 s, 2 sand b. decreases linearly c. is constant 3 s, respectively. The lengths of X, Y and Z are respectively d. increases parabolically 34. The acceleration of a particle travelling along a straight line M is shown in Fig. 6.14. The maximum speed of the particle is +5 Fig. 6.17 (ms~) t r--~~--~----...,..,-----r--'~ a. 5 m, 10 m, 15 m b.lOm,40m,90m 4 8 12:,, ,,: 16 t (s) c. 5 m, 20 m, 45 m d. 10 m, 20 m, 30 m 40. The speed of a projectile at its highest point is \"1 and at the -5 ~ \"2. Yf'i5:., Fig. 6.14 point half the maximum height is If -\"'- = then the a. 20 ms- i b. 30 ms- 1 c. 40 ms- 1 d. 60 ms- 1 angle of projection is \"2 35. A block is dragged on smooth plane with the help of a rope a. 45\" b. 30° c. 37\" d. 60° which is pulled by velocity v as shown in Fig. 6.15. The horizontal velocity of the block is 41. A particle is projected at an angle of elevation a and after t seconds it appears to have an angle of elevation fJ as seen from the point of projection. The initial velocity will be gt gtcosfJ a. . b. -::-\".-;--'-cc 2 sm(a - fJ) 2 sm(a - fJ) v sin(a - fJ) 2 sin(a - fJ) c. d. --'--7-'- Fig. 6.15 2gt gt cos fJ a. v b. vlsin e c. v sin e d. vlcos e 42. A velley shots are fired simultaneously from the top and bot- 36. An object is moving in the X-Y plane with the position asa tom of a vertical cliff with the elevation\" = 30°, fJ = 60°, function of time given by -; = x(t) 1+ y(t»). Point 0 is at respectively Fig. 6.18. The shots strike an object simultane- ously at the same point. If a = 30.)3 m is the horizontal distance of the objcct from the cliff, then the height h of the cliff is a. 30 m b. 45 m c. 60 m d. 90 m

6.6 Physics for IIT-JEE: Mechanics I i 2. For a particle moving along X -axis, x~t graph is as given in Fig, 6,21. Mark the correct statement(s), h x ~~.'==~. a'\" 30-IT 111 c D EF Fig. 6.18 B A 43. Figure 6.19 show that particle A is projected from point P \" - - - - - - - - -... t with velocity u along the plane and simultaneously another particle B with velocity v at an angle a with vertical. The () particles collide at point Qon the plane, Then Fig. 6.21 p~'aB a. Initial velocity of the particle is zero, AQ b. For BC, the acceleration is +ve and for DE, the aceeler- e\" alion is -ve. c. For EF, the acceleration is +ve, d. Velocity is getting zero, three times in the motion. 3. For a particle moving along X-axis, a scaled x-I graph is shown in Fig, 6,22, Mark the correct statement(s), Fig. 6.19 a. vsin(a - 00) = u b. vcos(a - 00) = u x (m) c. v = u d. None of these to --------- c 44. A platform is moving upwards with an acceleration of 5 mis', At thc moment when its velocity is u = 3 mis, a ball is thrown 1---+--'---'---_1 (s) from it with a speed of 30 mls w,r.t. platform at an angle of 0 =30\" with horizontal w.r,t. platform, The time taken by the ball to return to the platform is -10 a. 2 s b. 3 s c. I s d. 2,5 s A 45. Two balls arc projected from points A and B in a vertical Fig. 6.22 plane as shown in Fig, 6.20, AB is a straight vertical line, The balls can collide in mid-air if vr!v, is equal to a. Speed of the particle is greatest at C, b. Speed of the particle is greatest at B, B~2\", c. Particle is speeding up in region marked CD, \", d. Average velocity is greatest for region AB among marked 8, regions. A 4. Mark the correct statement(s), Fig. 6.20 a. A particle can have zero displacement and non-zero aver- a. cc.o-s O-r dc.o-sO-, sina2 cosO, cos Ot age velocity. h. A particle can have zero displacement and non-zero 1. For a particle moving along X-axis, mark the correct statc- mentes), velocity, a. If x is +vc and is increasing with time, then average ve- c. A particle can have zero acceleration and non-zero veloc- locity of the particle is +vc, b. If x is -ve and becoming +ve after sometime, then velocity ity, of the particle is always +ve, d. A particle can have zero velocity and non-zero accelera- c. If x is ~vc and becoming less ~ve as the time passes, then average velocity of the particle is +ve. tion. d. If x is +vc and is increasing with time, then velocity of the particle is always +ve. S. At time I = 0, a car moving along a straight line has a velocity of 16 mis, It slows down with an acceleration of -0,51 m/s2, where t is in second. Mark the correct statement(s). a. The direction of velocity changes at 1 = 8 s, b. The distance travelled in 4 s is approximately 59 m, c. The distance travelled by the particle in lOs is 94 m, d. The velocity at 14 = 1() s is 9 mis, 6. A ball is thrown upwards into the air with a speed that is grcater than its terminal speed, It lands at the same place from where it was thrown. Mark the correct statement(s), a. It acquires terminal speed before it gets to the highest point of the trajectory, b. Before reaching the highest point of the trajectory, its speed is continuously decreasing,

Miscellaneous Assignments and Archives on Chapters 1-5 6.7 c. During the entire flight, the force of air resistance is great- 11. A particle is moving along a straight line, whose x-I plotis as est just after it is thrown. shown in Fig. 6.24. Which one of the following can represent v-I graph for the particle? d. The magnitude of net force experienced by the ball is max- imum just after it is thrown. x B 7. A particle is moving along X -axis whose position is given by 13 . X = 4 - 9 + -. Mark the COlTect statement(s) in relation to 3 its motion. 3. Direction of motion is not changing at any of the instants. b. Direction of the motion is changing at t ::;;: 3 s. c. For 0 < 1 < 3 s, the particle is slowing down. d. For 0 < I < 3 s, the particle is speeding up. 8. A particle is thrown vertically in upward direction and passes three equally spaced windows of equal heights then , A \" '---,B 1 d. 0 c. ojL.--+:-'D~-'\" Ground V'-E Fig. 6.23 12. An object moves in the XY plane with an acceleration that a. average speed of the particle while passing the windows has +ve Y component. At time I = 0, the object has a velocity satisfies the relation vav! > vav2 > Va\\'3 given by 'J = 31 + 4]. Mark the eorrect statement(s). b. the time taken by the particle to cross the windows satisfies a. Magnitude of the velocity of the particle is continuously the relation tl < t2 < t3 increasing c. the magnitude of the acceleration of the particle while h. X component of the velocity is continuously increasing crossing the windows satisfies the relation ill = il2 f:. Q3 c. Y component of the velocity is continuously increasing d. Y component of the velocity is constant d. the change in the speed of the particle while crossing the windows would satisfy the relation ~Vl < ~V2 < ~V3 13. An object moves with the constant acceleration -:;. Which of the following expressions is/are also constant? 9. Ship A is located 4 km north and 3 km east of ship B. Ship A has a velocity of 20 kmlh towards the south and ship B is b·ld·;1a.dl~I c. moving at 40 kmlh in a direction 370 north of east. Take X- and Y-axes along east and north directions, respectively. v a. Velocity of A relative to B is -321 - 44). b. Position of A relative to B as a function of time is given dl dl by 14. A ball is dropped from a height of 49 m, the wind blows r;B = (3 - 32 I) i + (4 - 44 I)] horizontally and imparts a constant acceleration of 4.90 mis' to the ball. Choose the correct statement(s) where I = 0 when the ships are in position described a. Path of the ball is a straight line. b. Path of the ball is a curved one. above. c. The time taken by the ball to reach the ground is 3.16 s. d. The angle made by the line joining initial and final posi- c. The instant at which their separation is minimum is at tions (on ground after I\" strike) of the ball with horizontal 1= 65 s. is greater than 4SO. d. The least separation between the ships is 4.12 km. 15. An object may have a. varying speed without having varying velocity 10. Mark the correct statement(s). b. varying velocity without having varying speed 3. Average speed of a particle between two instants tl and 12 c. non-zero acceleration without having varying velocity depends only on the position vectors at tl and f2. b. Average velocity of a particle between two instant II and d. non-zero acceleration without having varying speed t2 depends only on position vectors at t] and f2. c. Average acceleration of a particle between two instants I! 16. From the top of a tower of height 200 m, Dall A is projected and 12 depends only on velocities at II and 12. up with a speed of 10 me I and 2 s later, another ball B is d. Average speed depends on mid-instants also. projected vertically down with the same speed. Then

6.10 Physics for IIT-JEE: Mechanics I 24. A body is projected at the angle of 300 and 60° with the same For Problems 32--34 velocity. Their horizontal ranges are R 1 and R2 and maximum Two cars approach each other on the highway. Car A moves towards north at 90 m/s. Car B moves towards south at 70 m/s. heights are HI and Hz, respectively, then a ,II-I > I b.H-I >I c ,II-I < I d .H-I < I 32. The velocity of car A as seen from car B, i.e., VAll is 112 H2 112 H2 •. 140 mls b. 160 mls c. 20 mls d. 40 mls For Problems 25-27 33. The velocity of car B as seen from car A, Le., VBA is A _point moves in x - y plane according to the law x = •. 120 mls b. l30 mls c. 170 mls d. -160 mls a sin wt and y = a - a cos (ot, where a is a positive constant and t is the time, 34. Their velocities relative to car C, which is trave!l1ng north at 100 kmlh, i.e., VAC and vBe arc 25, The magnitude of velocity of the body at any instant of time a, - 10 mis, - 170 mls t is b. -20 mis, -180 mls a. (leu cos wt b. a(i) c. a(osinwt d. None of these c. 30 mis, -140 m/s d. -40 mis, 120 mls 26. The trajectory represents For Problems 35-37 a. x = \"J2a(-J_'-;-_-::--:- Two particles are thrown simultaneously from points A and b. x= ±J2ay (1- Ja) B with velocities u! = 2 InS_oj and U2 = 14 ms-! , respectively, c. x ex y2 as shown in Fig. 6.28. d. None vl /'\" 2m,' 27. The magnitude of acceleration of the body at any instant is a. aw2 b. aU) c. aw2/2 - d. awl2 k-------A ,,,,,,:(0. 20m) For Problems 28-29 20m:,,, \",--14ms' B A helicopter is flying at 200 m and flying at 25 mls at an angle 11m 37\" above the horizontal when a package is dropped from it. 200m x Fig. 6.28 o 35. The relative velocity of B as seen from A is b. 4,,;27 + 3,/3] Fig. 6.27 28. Distance of the point from point 0 where the package •. -8,,;2 i - 6 ,,;2] c. 3,,151 + 2,/3] d. 3,,;21 + 4J3] lands is 36. The direction (angle) ith horizontal at which B will appear a. SO m b.IOOm c. 200 m d. 160 m to move as seen from A. 29. If the helicopter flies at constant velocity, find the .< and y a. 37- b. 53° c. 15- d. 90- coordinates of the location of the helicopter when the package 37. Minimum separation between A and B, lands, a.3m b.6m c.12m d.9m a. 160 m, 320 m For Problems 38-42 b. 100 m, 200 m Two inclined planes OA and OB having inclination (with hor- c. 200 m, 400 m d. SOm, 100m izontal) 300 and 6(Y' respectively, intersect each other at 0 For Problems 30-31 as shown in figure. A particle is projected from point P with velocity u = 10~3 111S I along a direction perpendicular to A particle is moving with a constant acceleration in xy-plane plane ~A. If the particle strikes plane OB perpendicularly at from x = 4 m, y = 3 m and has velocity Q, calculate .1.·17 = 2 m/s 1- 9 m/s The acceleration of the particle is B given by vector; = 4 m/s21+ 3m/52 ]. \"Q 30. The velocity vector at t = 2 s is •. 81 - 3] b.61-7) 60\" c.207-5] o d.107-3] Fig. 6.29 31. The position vector at t = 4 s is •. 441 -9] b. 41- ] 38. the velocity with which particle strikes the plane 08, c.-51+12} d. 407 - 127 a. 15 mls b. 30 mls c. 20 m/s d. 10 mls

Miscellaneous Assignments and Archives on Chapters 1-5 6.11 39. time of flight of the particle 9. Statement I: Speed is always positive, while velocity may be positive or negative. a.8s b.6s c.4s d.2s Statement II: Speed is the magnitute of velocity and mag- nitude is always positive. 40. the vertical height h of P from 0, 10. Statement I: The v - t graph perpendicular to time axis is a.IOm b.5m c.15m d.20m not possible in practice. Statement II: Infinite acceleration cannot be realized in prac- 41. the maximum height attained by the particle (from the line tice. 0) ll. Statement I: Plotting the acceleration-time graph from a given position-time graph of a particle moving along a a. 20.5 m b.5m c. 16.25 m d. 11.25 m straight line is possible. Statement II: From the position-time graph, only the sign 42. the distance PQ, of acceleration can be determined but no information can be concluded about the magnitude of acceleration. a.20m h.!Om c.5 m d.2.5 m 12. Statement I: For uniformly accelerated motion started from Assertion~Reasoning Solutions on page 6. 24 rest, the displacement versus time graph is a straight line. Statement II: For uniformly accelerated motion, the velocity Type in equal intervals of time changes by the same amount. In the following questions, each question contains STATE- 13. Statement I: An iron ball and a wooden ball are both released MENT I (Assertion) and STATEMENT 11 (Reason). Each qucs- from the same height. In the presence of a medium both the tion has 4 choices (a), (b), (c), and (d) out of which only one is balls reach the ground with different velocities and different correct. times. Statement II: Both the balls reach the ground simultane- (a) Statement J is True, Statement 11 is True; Statement II is a ously. COITect explanation for Statement I. 14. Statement I: If velocity-time graph is a straight line parallel (b) Statement I is True, Statement II is True; Statcment II is to the time axis, then the acceleration of the body is zero. NOT a correct explanation t()r Statement l. Statement II: Acceleration is equal to the rate of change of velocity (constant). (c) Statement I is True, Statement II is False. (d) Statement I is False, Statement II is True. 15. Statement I: The sum and difference of two non-zero vec- tors will be equal in magnitUde, when the two vectors are 1. Statement I: The dimensional formula for gravitational po- perpendicular to each other. tential is [L'T-'). Statement II: If the above two vectors are perpendicular. Statement II: The gravitational potential is the potential en- then their dot product is zero. ergy per unit charge. 16. Statement I: The direction of velocity vector is always along 2. Statement I: The equation representing the distance tra- the tangent to the path, therefore its magnitude may be given by its slope. versed in •h second, Sn = U + I (2n -- 1) is numerically Statement II: The slope of the tangent to the path only mea- 2a sures the direction of velocity at that point. n! 17. Statement I: The relative velocity of A w.r.t. B is greater and dimensionally correct. than the velocity of either, when they are moving in opposite Statement II: Dimensions of both sides are not matching. - v;,.directions. 3. Statement I: Least count of all screw-based instruments is samc. Statement II: Relative velocity of A w.r.t. B = v~ Statement II: Least count of all screw-based instruments is found using the ratio pitch per division of circular scale. 18. Statement I: The relative velocity between any two bodies is equal to the sum of the velocities of the two bodies. 4. Statement I: Backlash error can be minimized by turning Statement II: Sometimes, the relative velocity between two the screw in one direction only when the fine adjustment is bodies is equal to the difference in the velocities of the two. done. Statement II: Backlash error is because of the weal' and tear or loose fittings in screws. 5. Statement I: Screw gauge with a pitch of 0.5 mm is more accurate than 1 mm for same number of circular scale divi- sions. Statement II: Higher pitch can make an accurate device. 6. Statement I: Pendulum of a clock is made of alloys and not MatcHing , Solutions on page 6.24 of pure metals. Column Type Statement II: Usc of alloys makes the pendulum look good. 7. Statement I: The speed of a body may be negative. 1. For a particle moving along X-axis, if the aeceleration (con- Statement II: When a body reversc!'-%its direction, its velocity stant) is acting along -ve X-axis, then match the entries of Column I with tlle entries of Column II becomes negative of the previous velocity. 8. Statement I: Speed and velocity are different physical quan- tities. StutementII: Both speed and velocity have same unit (rn/s).

6.12 Physics for IIT-JEE: Mechanics I .' Columnl (;olUiunII 4. The trajectories of the motion of three particles are shown in Fig. 6.32. Match the entries of Column I with the entries of 'j;tnldal y~logity:> 0 '••. a.Rartick ..ll1aymove in. +.vc Column II \" X,directibn' With increasing y speed b.particlernaymovein' +ve . g .direction with decreasing speed t. Partide maYlnove in c-ve X.dire.c.ti.?l1with. iricreasing Fig. 6.32 speed . d.pa,tic1e may move itl __ve . Columil I .ColulllilI-I -____'4 i. Time. of flighl,.is !castror X ·direction with decreasing; ... A speed 2. The velocity-time graph of a particle moving along X-axis ii.Vertical cO(nponent onhe ve.- b. B is shown in Fig. 6.30. Match the entries of Column I with the locity isgrcatesc:t,:f0.:cl:.·--'-'--+-'---'---'-~--1 entries of Column II. iii•. HorizOlitalcomponentofthe e.C vc B ·.velocity is greafestfor A )<'----\\;=--_ iv,. Launchsp~ed is least for d; Noappropriate' L.:~-C~_,_~_~-,~=~~=-=matehgiven S. The path of a projectile is represented by y = Px - QX\" E Column I .Columu II i. Range Fig. 6.30 ~~~~~T----' I).p!Q ii. MaxiillUm.helght b. P . ~~~~~~~------~'~~~-'- - - m.. 1'iJne offlight --,-ColllmriI_ ColumuI! .... •••.... ...• .•..•... . ...••... ..! li.F()rAB,tlie\\pfUJi- a.moving .inCrv. X .direction iv. taJ~gent oflhe angle of projection is d. withanincre~singspeed . .'. ble is. u.iRiJr BC; the par~- b.movlng in. +veX-diret:tlon 6. A body is projected with a velocity of 60 mls at 30\" to hori- zontal I. . cn'ds . ••.•. wjthadecreasingsjleed v\" .. )(..HURiwC1;l;•\\tie.ll!R..c•.moYing.1n. ... ,d.ir,e.c\\iQil. .... ,Coltimril .......... Omimull , __:\", . tiele!s ..\" with. an increasing speed •.••.• •'j. Initlalvelocity vector· Ja. 6001 +40 i~.J'orD1iTtiiepar- d;mdVing hl ...veX~direction t •.•..• tiele i\". . '. .\\Yithadesreasing speed .••. .' ii. Velocity aftclr 3sec .•.•. b.3001 +10] 3. Fig. 6.31 shows the position-time graph of particles moving iii.. pispJacement after' 2 sec'· ..... 0 \" ' \" e.• 30 3i+30j along a straight line. Match the entries of Column I with the entries of Column II , iy.Vel{)city al\"ter2 sec '.' •• '. d.3001 .. x Archives Fill in the Blanks Type 1. A particle moves in a circle of radius R. In half the period of Fig. 6.31 revolution, its displacement is ___\",__and distance covered is _____. (lIT-JEE, 1983) ........• 'C;pl\\JIllIlT '. '. ColuUlUII ..... ........ ..•.... . 2. Four persons K, L, M, N are initially at the four corners {'T~eparticle.AiS ..•..... . a.Nce]etating ...•..•.••. '...•.. of a square of side d, Each person now moves with a uni- 'il;Thepat\\icIeB is . . •.... .b. Decelerating form speed v in such a way that K always moves directly 'Ji!;Thep(\\rticl~Cls •·...·.··c.Speeding up towards L, L directly towards M, M directly towards N, and '. N directly towards K. The four persons will meet at a time i,,:Thep(\\rticJ~Dis .•. d. SloWing down '. \".. _____. (IIT,,JEE, 1984)

3. Spotlight S rotates in a horizontal plane with constant angular Miscellaneous Assignments and Archives on Chapters 1-5 6.13 A velocity orO.l radian/second. The spot oflight P moves along the wall at a distance of 3 m. The velocity of the spot P when e = 450 is m/s. (IIT-JEE, 1987) s 3m p B Fig. 6.34 Fig. 6.33 a. 3.14 mls b. 2.0 mls c, 1.0 mls d, Zero. 4. The trajectory of a projectile in a vertieal plane is y = 4. A ball is dropped vertically from a height d above the ground. ax - bx2, where a, b are constants, and x and yare respec- tively the horizontal and vertical distances of the projectile In hits the ground and bounces up vertically to a height d /2. from the point of projection. The maximum height attained is . and the angle of projection from the horizontal Neglecting subsequent motion and air resistance, its velocity' is (lIT-JEE, 1997) v varies with the height h above the ground as (lIT-JEE, 2000) True or False v v a. t Two balls of different masses are thrown vertieally upward h. b with the same speed. They pass through the point of pro- jection in their downward motion with the same speed (Ne- glectcd air resistance). (IIT-JEE,1983) 2. A projectile fired from the ground follows a parabolic path. r W-Cl~\" dr d The speed of the projectile is minimum at the top of its path. hh (IIT-JEE, 1984) 3. Two identical trains arc moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails. (IIT-JEE, 1985) 4. An electric line of forces in the x-y plane is given by the 5. A particle starts sliding down a frictionless inclined plane. If equation x 2 + y2 = I. A particle with unit positive charge, S\" is the distance travelled by it from time 1 = n - I sec to = =initially at the point x I, y 0 in the x-y plane, will move t = n sec, the ratio S,,/ Sn+! is (IIT-JEE,2004) along the circular line of force. (IIT·JEE, 1988) 2n-l 2n+l 2n 2n+1 a. - - h . - - c. - - d . - - 2n+1 2n 2n+1 2n-1 Single Correct Answers Type 6. A particle starts from rest. Its acceleration (a) versus time (I) 1. A river is flowing from west to east at a speed of 5 m per is as shown in Fig. 6.35. The maximum speed of the particle minute. A man on the south bank of the river, capable of swimming at 10m per minute in still water, wants to sY'irn will be (IIT-JEE, 2004) across the river in the shortest time. He should swim in a Acce!eration (m/s2) 10 direction a. due north b. 30\" east of north c. 30° west of north d. 60° east of north (IIT-JEE, 1983) 2. A boat whieh has a speed of 5 km/h in still water crosses a 11 Time (sec) river of width I km along the shortest possible path in 15 min. The velocity of the river water in km/h is Fig. 6.35 a.1 h.3 c.4 d,.j4T (IIT.JEE, 1988) a. 11 0 mls h. 55 mls c. 550 m/s d, 660 mls 3. In 1.0 s, a particle goes from point A to point S, moving in a 7. The velocity displaeement graph of a particle moving along semicircle of radius 1.0 m (Fig. 6.34) . The magnitude of the a straight line is shown in (Fig. 6.36) average velocity is (IIT-JEE,1999) The most suitable acceleration-displacement graph (Fig. 6.37) will be (IIT-JEE, 2005)

6.14 Physics for IIT-JEE: Mechanics I ANSWERS AND SOLUTIONS Objective Type v Vo '-----'+-+x 1. b. We. have Vav = -D::\".i-,sPc.I_a_ee_m_en..,t Xo Time interval Fig. 6.36 = Slope of chord on x-t graph. a Here, slope of chord between P and Q for all three par- .. ~ X ticles is same, so average velocity of all the three particles V would be the same. 2. a. a = v dv = 4 x tan 60\" = 4vi3 !Ii/s3. ds dv 3. d. We know that a = - dt f f=} dv = adt Fig. 6.37 R.H.S. of the above expression represents area under acceleration-time graph. Multiple Correct Answers Type Area of acceleration-time graph is not getting zero over 1. A particle is moving eastwards with a velocity of 5 mls in the interval for which graph is given, hence velocity is not 10 s, the velocity changes to 5 mls nOlthwards. The average getting zero gain over the interval. acceleration in this time is ·4, b. At the instant, when bolt starts falling it acquires the ve- locity of the lift. a. zero VBL =OandvLG = v b. 1/v'2 m/s2 towards north-west vBG = 0 + v = v in upward direction c. 1/2 m/s2 towards north-west d. 1/2 m/s2 towards north and aBC = g in downward direction So, displacement of the bolt w.r.t. ground is given by the equation, YBO = vt - 'Iigt2 (IIT-JEE,1982) 2. A particle of mass m moved on the x-axis as follows: it. starts which is a parabolic equation but this is valid only for the from rest t = 0 from the point x ::: 0, and comes to rest at t = 1 at the point x = 1, No other infonnation is available about its time interval which it takes to the floor of elevator. motion at intermediate times (0 < t < I). If a denotes the After that displacement-time graph of bolt and elevator would be same under the assumption that bolt sticks to the instantaneous acceleration of the particle, then: floor of elevator after striking it. a. (X cannot remain positive for all t in the internal 0 ::::: t ::: 1. dv . . b. I'\" I cannot exceed 2 at any point in its path 5. a. From a = v - , we can find the SIgn of acceleratIOn at c. lal must be,::: 4 at some point or points in its path ds d. a must change sign during the motion, but no other asser- various points. V is positive for all three points 1,2 and 3. tion can be made with the information given dv is positive for point I, zero for point2 , negative for point (IIT-JEE, 1993) - ds 3. So, only for point I, velocity and acceleration have same 3. The coordinates of a particle moving in a plane are given by sign, so the object is speeding up at point I only. x(t) = a cos(pt) and yet) = b sin(pt) where a, b « a) and p are positive constants of appropriate dimensions. Then 6. c. As the speed with which the ball has been thrown in the a. the path of the particle is an ellipse b, the velocity and acceleration of the particle are normal to downward direction is greater than the terminal speed. each other at t = Jf/(2p) c. the acceleration of the particle is always directed towards a focus d. the distance travelled by tile particle in time interval t = 0 Fig. 6.38 to t = Jf /(21') is a (IIT-JEE, 1999)

MisceUaneous Assignments and Archives on Chapters 1-5 6.15 So, }~csislivc > mg From the diagram. we get And hence initially net force acting on the projectile is 20 I = 10+ 21: x 212 =} in upward direction and hence the acceleration also is in the I' + 10 - 20 I = 0 same direction. 1 = 0.513 s, 19,487 s Out of these two. I, = 0.513 s conesponds to the situation But this acceleration is opposite to velocity, so speed =when overtaking has been completed and I, 19.487 s eor- decreases and hence Frcsistive, At a particular instant, the responds to the same situation as shown in Fig, 6.41, but for projectile acquires the terminal speed and at this instant tj < t < t2 the separation between two cars first increases and then decreases and then B overtakes A. Fresistivc = mg and hence a = 0 and the particle moves with Total road distance used = 5 + 201, = 15.26 m. constant speed, i.e., the terminal. 11. a. Att = 0, v = o. x = 4/3 7. c. r r 4x3 ~CD'-\\...<-- ......•.............. -<v\\c+rvyq10j--/+x a = -3- - 4 = 0 . x As both the particle velocity and acceleration arc zero ;:/it at t = 0, so it will always remain at rest and hence distance 1 travelled at any time interval would be zero, Clockwise Anti-clockwise 12. d.ll = 20 cos 30 1+20sin30j Speed ~)- -)- Let v is perpendicular to u at time t Fig. 6.39 y 8. h. In this case, velocity and acceleration are acting in 20 opposite directions and velocity is getting zcro ,at t ::;: 4 s. ., Distance travelled in lOs = 2 xldisplacement in 4 sl- displacement in 10 s ~~3-0-0 ------~-v--+x Required distance = 2 x 8 - (-10) = 26 m. Fig. 6.42 9. h. Now. -;; =20eos301 +(20sin30-gl») =.._)- -,+ V . U = 0 =} 1 = 4 s. Here. time of flight T 2 s. So, it is not possible at any instant. 13. d. Here. the automobile is taking less than 1 min to stop. So Descend Ascend first find that time in whieh the vehicle stops. Fig. 6.40 -~~ ~)- -'>' From v = u + a 1 =} 0 = 20 - 0.51 =} 1 = 40 s. The distante covered by the vehicle in 40 s would be It is clear that Vj < V2 and from equation of motion, we s = 20 x 40 - 1 400 = 400 m. have -2 x 0.5 X (40.)2 = 800 - h Required distance = 500 - 400 = 100 m. sin (J = 14. c. As the parachute inflates, a large impulsive air drag acts where tJ and t2 are the times tak~n for ascend and descend, respectively. in the upward direction. as a result the parachutist suddenly shoots upwards and then under the presence of air drag force As VI < V2 sotl > f2· and gravity force, he will start falling again but more slow,ly, 10. b. The situation is as shown in Fig. 6.4 L +-+ -~ -+-+ -+ 4 15. d. V2 = VI - VO, V2 Vo = Vj From the law of triangle, we can draw the velocity vector diagram, 1'<---- 20 f r;-]~ 20 Ill/s r:l~ 20 m/s ~~ r:Dl---l>-2 m/s2 1,;l---Jt.2 m/s2 L.!!J 11 0 L£.J I' -I Fig. 6.43 .lx2xt2 From triangle property. sum of two sides 2: third side 2 So, v! :S Vo + V2 Au,,\"' 0 Att, overtaking overtaking ,~tmis finishes Fig. 6.41

6.16 Physics for IIT-JEE: Mechanics I 16. c. =} U+V=dj2~' .. V=d{fd-U. y R =22. d. Here n 12t = ---==-a--~2rr- v v-vcos- v a fJ n ~I\"-'-'-----+x A A Fig. 6.44 A is fixed and B moves in the direction as shown in Fig. 6.44. v 5 Direction ofr remains same and magnitude changes with . 3 4 time linearly. Fig. 6.46 +17. b. x 3 = 13 1 =} 3 x 2 dx = 3t 2 - dt dx t2 (1) a 2a t = ---.::~C- dt = x 2 - v(2 - J)' v - v c o s2r-r 12 23. b. a = dv dv dS = 4 + 3v = - - dt dS dl f dS=f~ 4+3v s= 3v - 94 10g,(4+3v)+C 18. a. The radius of curvatures will be mutually perpendicular = = 94 At S 0, v 0, we get, C = log, 4 only when the velocity vectors will be mutually perpendicular S= .':' +-4lo g. 4( -+4-3v ) 3 9 ' i.e., after the time t = y'z':'.. Putting v = -2 mis, we get g 24 4 19. a. 0 = u cos 30° - g sin 300 t (1) S= 3+ 9 10g, 10 =0.26m. u cos 30° (2) 24. a. Let d be the river width and u and v the speeds of the water t= current for P; time taken g sin 30° ~- H cos 30° =' -u sin 300 t - g cos 300 t2 By equatIOn (I) and (2), we get 2d (1) I, = [1H = :2 + CO; a ] v = t~H . ./v2 _ u2 {a = 300 J and for Q; time taken t2 = d [_1_ + _1_] (2) v+u v-u 20. b. The vertical displacement vs. time graph has the same form Dividing (I) by (2), we get y * j ('fJ= 1 - 2 <1 .' .t, < 12 25. c.ax = -gsine;a, = -gcose Fig, 6.45 .v v x as the trajectory of the particle. At the topmost point the slope is zero, but the curvature is non-zero (Fig. 6.45). R 21. a. Time taken by particles to collide e t = y(g2H then uy(g2H + V Y(g2H = d Fig. 6.47 =U x = O;u y v

Miscellaneous Assignments and Archives on Chapters 1-5 6.17 Along . Sy = Uyt + 'I2ayt 2 Vre = - -VIII- = -1=0 20ms-1 , y-axIS, sin 30° 1/2 I O=VI- '2gCOset2 2v Vrm = VTe cos 30° = 20 x .J3 = 1 0.3Jm3 -s1. t=--- 2 gcose .. Sx = u.rt + I 2 28, c. Along x-aXIS, \"2a,tt p R =0 x t _ ~gsine (~)2 2 gcose O __-~r-r-\"\"R -2vtanesecO R = -----g----- 26. a, 0.2n = I gt 2 Q - Fig. 6.50 2 t=tx;2n oQCase I: Let 0 P = 31 be the velocity of man. be the velocity of rain. /2O.3n =4t x 0.2n PQ is the velocity of rain relative to man. 0.3n =4.5 g oRCase II: = 6 i is the new velocity of man RQ = new velocity of rain relative to man n=9 Now OQ2 + op2 + PQ2 i.e., OQ2 = 32 + 32 (:.PQ = PR =6-3 =3) i.e., OQ = 3J2 kmh- 1 0.2 n , , , ,,,,,, and t,an . _ PQ _ '32 -_I,.I.e., e -_4\"5 0 - OQ - e29. b. For the maximum range = 45\", u2 . u2 sin 28 u2 u2 R = = - sin 90\" = -or 500 = gg gg The distance covered along the inclined plane can be obtained using the equation 0.3 n v2 _ u2 = 2as Fig. 6.48 or 0 - u2 = 2(-g sin 30\")s 27. a. v;\" = Velocity of the man or S = -u2 = 500 m.· g 30. b. From the graph, velocity--displacement equation can be written as v = Vo +- ax (1) Fig. 6.49 Here Va anda are positive e constants. V:e = Velocity of rain w.r.t. earth Differentiating (1) with respect to x, we get dv v~;!! = Velocity of rain w.r.t. man - = ex = constant dx Velocity of man Iv\", I = 10 m8- 1 Acceleration of the particle can be written as Using sin 30° = !!.::!.. v\" a = dv = (uo + \"x)\" vd-x a - x equation is a linear equation. Thus, acceleration increases linearly with x. 31. a. Area under a-t graph gives the change in velocity (dv = adt)

6.18 Physics for IlT-JEE: Mechanics I vf-v;= 2I :x2x4=4ms-i . Distance travelled = 0' vf = Vi + 4 = 2 + 4 = 6 mls = 49 49 20 = 169 = 169 ms-' 12 + 12 + dv ds dv 6 x 4 24 32. a.a=--=-v 0' 169 1 6 9 , ds dt ds Av. spced = - = - - = - ms Here -a = -dv = I s- 1 t 6 x 4 24 V ds Accelctation versus velocity graph will be a straight line I passing through origin with slope I s-'. 39. c. In general, h = 2: gt' 33. c. Acceleration-velocity equation from thc graph is /C\"\"\" f a = kv, where k = slope of given line. This can also be written as - --- _/- - - - - - - -- dv dv Fig. 6.53 v- =kv =} - =k x= 1 5m, y = 1 20m, z = 1 = 45m _g(1)2 = _g(2)2 = _g(3)2 ds dO' 22 2 i.e., slope of velocity-displacement cquation is same as slope of acceleration~vclocity equation which is a constant. =34. b. Maximum speed is at t 8 s lHII (mfs 2) 40. d. v, = v,cosfJ = ucose () c \" <\\ g --- 5 Fig. 6.51 Fig. 6.54 v at t = 8 s is given by area under curve DABe. vi vi -0= (v2sinfJ)2 - 2g(HI2) =} = gH Max. speed = ~(AB + DC) x BC \"!.PIAlsu V2 = V'5 = 1 x (12) x 5 = 30 ms-'. j2 H TFrom above VI = 2: 3g ' V2 = VISgil 35. a. Let the velocity of the block is u, then velocity component eu2 sin2 . ') 2gH cos2 0 vfH = along the string should be same. 2g =} sm\" 0 = ~-c;-- usine = v =} u = v/sine. 2 ~H , ~Hx3 r36. c. = xi + y} v,=} tanO=--2- =} tan\"O= 2 \" dr dx, dy, , , gH v= -dt -dtl -ti1t' .VyJ = + = Vxl + tan e =.J3 =} 0= 60\". .I r vIf and are in opposite direction, definitely the particle 41. b. -Y = tanfJ x u sm at - 2:.g~ = tan fJ will be moving towards origin D, for this =} U cos Cit r.v +< 0 =? XVx yVy < O. I ___ _ 37. b. At 12 s, slope of B is samc as that of A. 38. d. x = 7t - 3t2 =} dx v = - = 7 - 6t. dt Let us find the time when the velocity becomes zero. Putting v = 0 =} 7 - 6t = 0 =} t = 716 s Fig. 6.55 7 7' 49 usina - tan.Bucosa =\"g2t XU_7/6 ,) = 7 x (5 - 3 X 62 = 12 m 20m -49m gt cos fJ 12 U= Fig. 6.52 att=O--+x=O 2sin(a-fJ) att = 45, --+ X = 7 x 4 - 3(4)2 = -20 m ==42. C. u! COSa' U2COSf3 30.J3 = u~ sin 2fJ 30.J3 = u~.J3 2g g =} ,,~= 600 =} \"2 = 10.)6

Miscellaneous Assignments and Archives on Chapters 1-5 6.19 f f\" I 0.5 t 2 dv = - 0.5 t dt =?v = 16--- (1Ov'6)2U2 sin 60° 2 v'3/2 t2 = g = t2 = Fs 16 0 =? 10 Direction of velocity changes at the moment when it 1Ov'6cos 60' r;; becomes zero momentarily. U1 = = IOv2 0.5 t' cos 30° O=16--- =?t=8s -h = IOhsin30\"Fs - ~10 (Fsr 2 -h = 30 - 90 =? h = 60 m. dx 0.5 t 2 - =V= 16- --. dt 2 43. a. If the particles collide at Q, it means they travel same Let us consider that at t = 0, particle is at x = 0 displacement along the plane in same time. So their velocity components along the plane should be same fx dx = (16t _ 0.~t3); 0.5 t 3 x=16t- 6 o =Distance travelled IDisplacement! for t :'0 8 s. So. the distance travelled in 4 s, Q 0.5 X 43 x=16x4- 6 ::e59m. Fig. 6.56 Distance travelled in lOs v cos(f3 + eo) = u and f3 = 90° - a = IDisplacement in 8 sl x 2 - Displacement in 10 s Solve to get: v sin(a - eo) = u = 85.33 x 2 - 76.55 = 94 m v(t = 10 s) = -9 m/s 44. a. Acceleration w.r.t. platform g + a 6. b., c., d. Initially. the FBD of ball would be as shown in Fig. 6.57. . 2usine So tIme taken: T = - - - v~v g+a 1ng 45. d. If the particles collide in mid-air, they travel same dis- Fig. 6.57 placement in horizontal direction. So their velocity compo- Here! direction of velocity and acceleration is opposite, nents along horizontal should be same: VI cos 81 = V2 cos 82. so particle is slowing down and moving in upward direction, at one instant its velocity becomes zero which would be the Multiple Correct highest point of trajectory. Then it starts falling down, for its Answers Type descent FBD would be as shown in Fig. 6.58. 1. a., c., d. Based on thought. ~\" 2. a., c., d. Velocity is given by the slope of x-t graph. Here v~ the slope is zero at A and at peak of region CD and the bottommost point of Electric Field. mg For AB and EF, the acceleration is positive. For these regions the graph is concave up. Fig. 6.58 For BC and DE, the acceleration is zero as here the velocity is constant. During its descent. the ball will acquire the terminal For CD, the acceleration is negative, as the graph is con- speed (assuming the ball is not striking ground before ac- cave down for this region. quiring the terminal speed). 3. b., c., d. Point of steepest slope corresponds to the maximum As work has been done against the resistive force, so speed of throwing> speed with which the particle lands. speed. Particle will speed up when direction of the accel- Hence, force of air friction is greatest (when speed is eration and velocity is the same. For region AB, both the greatest) just after it is thrown. acceleration and velocity are positive while for CD both are 7. b., c. The particle's velocity is getting zero at t ::::: 3 s, where negative, so particle is speeding up in these regions. it changes its direction of motion. Average velocity = Slope of ehord on x-t graph. Which is maximum for AB. 4. b., c., d. Based on thought. 5. a., b., c. This is the example of non-uniform acceleration a = dv = -0.5 t dt

6.20 Physics for IIT-JEE: Mechanics I For 0 < I < 3 S, V is negative, a is positive, so particle For 0 to A velocity and acceleration are positive so particle is speeding up. Acceleration can be constant or varying so is slowing down. v-I curve can be straight line or curved. For A to B: Constant velocity, at A and B discontinuity occurs For t > 3, both V and a are positive, so the particle is due to large change in velocity, in a very s~all time. For B to D: Velocity is negative and acceleration is positive, speeding up. so pmtic1e is slowing down. For D to E: Constant velocity. 8. a., b., d. As the particle is going up, it is slowing down, i,e., J12. a., c. Along X-axis, Vx ::::: 3 speed is decreasing and hence we can say that time taken Along Y-axis, Vy = 4 + ayt, I-; I = v.~ + v;. by the particle to cover equal distances is increasing as the particle is going up. Hence, 11 < 12 < 13. As Vav = Distance , we have Vav ex time time + +~)- '\" ~ ~ dvx + -dv\\',~} + -dukz .. So, vavj > vav2 > va\\'3 13. b. Let a = a,l Q)'j a,k = Acceleration throughout the motion remains same from dt dt dt equation, As (; is constant, so its magnitude as well as direction ..-.y ---+ is not changing, but v x , Vy and Vz all can be varying (any combination of these if ax, ay or a,. ::::: 0, then corresponding v = u +at, L),V ex t. So, .6.Vj < .6.v2 < .6.v3. 9. a., b. v';w = - 20 1 velocity component may be constant). 1; I= v= Jv.~ + v.~ + v; ,----- ~Ii\" C' 2() klnil) .~ Vl;Hl w---<3p7° \"-'--i-----E B WhiCth is,a variable quantity. . dd vt = ~)- I a I which would be a constant. s t Fig. 6.59 -. V~B = v~w v~w = -321 -44) [a variable quantity] . f f f~B dC; Iv) = .'!.. [vxi + vYl +vJ:.]· It d(~li) = - 32 dt ; - 44 dt ] cit dt j v; + v; + v~ la variable quantity]. 14. a., c., d. u., = u)' = 0, ax = 4.9 m/s2, a\\. = 9.8 m/s2 31+4J 00 1r~H = (3 - 32t)1 + (4 - 44t) T Their relative separation is given by Ir~B I 49m y x = Ir;ll 1= )(3 - 32t)2 + (4 - 44t)2 -''-......JL-_ _ _=o--'I>____ Ground Fig. 6.60 dx 12 I2 For least separation, -- = 0. So, X = \"2a.rf , and y = layt elt Distance travelled --::::;.---:.- .,---==10. b., c., d. v\". = Tune 1l1terval -, _.)- ~> = r2 - r[ y ay 2 V2 - V j ---c> =--- So, path of the ball is a straight line. = ---,VllV Let t be the time taken by the ball to reach ground, then t2 - t1 49 = '12ayt2. 11. b., c., d. At 0, velocity is non-zero as the slope of x-t graph at t ::::: 0 is non-zero, so option (a) is wrong.

Miscellaneous Assignments and Archives on Chapters 1-5 6.21 t' = 10 0=} t = 3,16s, 4. a. I etane = ay = 2 0=} h = -o,t' ax = tan- 1(2), 2\" 15. b., d. A body moving on a circular path with uniform speed i 150111 Floor must have varying velocity as well as acceleration. 200m 16. a., c., d. Total displacement travelled by ball A in 2 s, ~ lOath Floor 11 i S = ut - - gt' = 10 x 2 - - x 10 X (2)2 = 0 400 m 22 1 Hcnce both A and B will reach the ground simultane- Ground floor ously and strikc with the same Velocity, Fig. 6.62 17. b., c. , - - - - - ~ ;,\\30\" - - - 200 = 2I: g X (6.39)' , ,/~\\)30\" ' ,,\".. I \\ ,,,,,, ,,,,,,, 0=} g = 9.798 = 9,80 mls2 , ~~ -30-\"-------,-------3-0\"- ~ 5. c. Lettime taken by rocketeer in catching the boy at the top of loath floor be t, then time from which boy is in free fall motion is (t + 5) s, Fig. 6.61 200 = I g (t + 5)' = vot + -gt 2 - 22 I 2usine 2xlOx- where Vo is the initial downward speed of the rocketeer. T= = 0=} t = 1.33 s g 2 = 1 s. which gives Vo = 143.7 m/s. 10 6. b. Initially, the rocketeer is under free fall and acceleration 18. b., c. dx = t 2 (i), .Y = 1t3 dy t2 (ii) is along vertical downward, direction, i.e., along positive -dt --::::} = Y-axis, so y-t graph is concave up but when he uses his jet 23 pack the acceleration starts acting in upward direction and dt 2 hence y-t graph would be concave down. t= 1, v, = I, Vy = 2I: 0=} • = ' + L 7. c. From solution 2, velocity of rocketeer at the time of catch- v i 2:J, ing the boy is d2x d'y (iv) v = Vo + gt = 157 m/s dt 2 = 2t (iii) dt' = t For safe landing, final velocity = 0 =} 0 = v2 - 2 x a x s[s = 400 m] 0=} a'\" 3g aAtt = 1 s, a, = 2 and a y = 1 0=} = 21 + ], So, the acceleration produced by jet pack alone would ~ Gorilj:ireliimsive be (3g + g) mis' in upward direction, as due to gravity. ac- .J:~Ile' . ' celeration g is acting in the downward direction, \" '> ' 8. b. Initially, the rocketeer's velocity increases with time lin- For Problems 1-3 early having acceleration g and then decreases due to the use of jet pack with an acceleration (acting upwards) of::o 5g. 1. b., 2. d., 3. c. So. the most suitable option is (b), Sol. For Problems 9-13 9. a., 10. c., 11. d., 12. b., 13. c. 1. b. For 0--5 s, initially, for both the particles, velocity and ac- Sol. celeration both are positive, so both the particles arc speeding up. But approximately from 2.3 s onwards, direction of ac- 9. a. Initial velocity of ball W,r.t. ground, celeration reverses for both the particles, so the particles slow down, '\"'> --J 2. d. Make a chord to both the particles on x-t graph, so that VC:G = IS 10 = 25 ms chord from 0 to f(s) becomes a tangent at t s. + + t--Jo ~,. 3. c. At t = 0, slope IA is steeper than slope IB so, VBe = VBE VA > VB tV~BE = IS 111S-1 , a~BG = 10 ms ---, .t.. There is no time for which tangent drawn to both curves have same slope. -,...\" ~ -.~ ,,' For 5-15 S, Va\\', A :::;:: 0 but for B it is non-zero. aBE =aBG - a =IO-(-5)=15ms~.j. All the things can be directly concluded trom the graph. For Problems 4-8 EG 12 4. a., 5. c., 6. b., 7. c., 8. b. 2at Sol. -> = 2m -t. Using s = ut + W.r.t. .- 2 SBE elevator frame =15t-21: xI5/ 2 o=}/=2.13s,

6.22 Physics for IIT-JEE: Mechanics I Sol. gx2 16. c.y=xtan&- 2 2 ,Putx=30m,y=15m 2u cos II We get u = 10.;6 mis, 17. b. Vx = Ux = ucosB, v~ = u; + 2aysy, Sy = 15 m and ay= -g, Apply v = Jv'x + v~ 50 In For Problems 18-21 Ground _1\"-______ 18. a., 19. b., 20. b., 21. d. Fig. 6.63 Sol. , u21 18. a. smj3 = - = - = - or j3 = 300 v42 = =Required angle 90\" + 30\" 1200 , 10. c. The height of the ball from ground at the time of d projection is 52 m, Maximum height from point of projection is 19. b. t = r.J=v;;2=-==ui2i 20. b. For the shortest time, one should head the boat perpendic- ~~~ 2 2 ular to flow h= = (25) = 31.25m 21. d. t = 2 + -2- = 4 -- - h, 2aBG 2 x 10 v-u v+u 3 Required height = 52 + 31.25 = 83,25 m, For Problems 22-24 11. d. Displacement of floor of elevator in 2,13 s is 22. b., 23. d., 24. d. SI = 10 t + 21: x 5 x t' Sol. dy dx clx dx 22. b. y = ax - bx' or - = a - - 2bx- = (a - 2bx)- \" dt dt dt dt Initially x = 0, v)' = avx SI =32,64 m So, displacement of ball w.r,t, ground (Fig, 6,67) -d'\"y = (a - 2bx£)x- + -dx [ -2bdX-] -':>- Sm = -2 32,64 = 30,64 m + +-~ -,l- dt 2 dt 2 elt dt SBG = SBE 12. b. From the diagram above (Fig, 6,63), it is very clear that =(a-2bxd)'x--2b (d- x)2 dt' dt distance travelled by the ball is H = [h - (SI - 2)J = 2h - SI + 2 = 31.86 m, 13. c. Displacement of the ball w,r,t. elevator is given by S = 15t - 21: x 15 t 2 cl'y ,l'r\" att = 0, -dt' = '-(X x = 0, -dt' =0 '* Vfa=-(X Here S is nothing but separation between the floor of -2b (clx)2. = -2b(v )2 or V = 2b elevator and ball. \" x dt x For s to be maX,Imum, -ds = 0 dt ::::} v == Jv.~ + v~ = vxv'f+Q2 = (X (1 +a2) which gives t = 1 s 2b So, the maximum value of s is '*23. d. V = al + bx], Vx = a or clx = adt x = at bat2 (15 x 1 - ~ x 15 X (2)') = 7.5 m. v\", = bx = bat or y = -2- '2;;'* '*(X)2Y = ba For Problems 14-15 y ex x 2 -+ parabolic, 14. a., 15. c., 24. d. Ranges . be same, HI = -sin,2-81, WIll - H2 sin &, Sol. , Displacement r~ For Problems 25-27 14. a. Av. ve1De1ty :::::: 25. b., 26. b., 27. c. Time 2:rrr/4 wheret =-- Sol. v I\", - I15. c. Average accelerat\"IOn:::::: Changte in vel = 25. b. x=asin(ut,y=a-acoswt VI dx d y , --t- v\", = -dt = awcoswt, Vy = -dt = awsmwt where VI = 2,51, ii2 = -2,5], jNow v = v; + v;' = aw. For Problems 16-17 16. c., 17. b.

Miscellaneous Assignments and Archives on Chapters 1-5 6.23 .x y 32. ,b. VAB = VA - VB = 90 + 70 = 160 m/s 26. b. sm\",t =-, cos\",t = 1 - - aa 33. d. VBA = VB - VA = 160 mls Now sin2(JJt +cos2wt =1 34. a,vAC =VA Vc =90-IOO=-IOm/s VBe = VB - Vc = -70 - 100 = -170 m/s =H = ±/2ay (I - :J For Problems 35-37 27. c. Cl-,; = -ddv-l,; = 2 sm. (J)t, a .v = -ddvly = 2 cos (J)t 35. a., 36. a., 37. d. -(l(J) a(J) Sol. a = V/ax2 + ay2 = aui 35. a. VB, A = VB - VA For Problems 28-29 = - 14cos 45'7 + 14 sin 45°] - (2 cos 45° 7) - 2 sin 45] 28. d., 29. a. 16 + 12 -sv'2; - 6v'2j. --; -j Sol. = v'2 v'2 = vy=15mJs 14 2 v ~,\" 25 m/s 72-72 3 1'/'\" 20 m/s 142 = 4 200m v'2+v'2 ~'I----- • -----I>. p Fig. 6.64 [mill 42 (E) 28. d. X - XI) = V,I (i) z y- YO = v,.,1 - -21gt 2 -200= 15t-5+2 X t=Ss (ii) ¢ R X= 20 X 8 = 160 m, 27 m I 29. a x = 20 x 8 = 160 III 41 (E) Fig. 6.66 Y = 200 + 15 x 8 = 320 m, 4tan <p = 3 =} <p = 370 . For Problems 30-31 37. d. LQPR = ° 43 = 2x7 =} x = 481 30. d., 31. a. Sol. 57, 30. d. v = it + at 81 45 += (27 - 9])+ (47 3]) x 2 = 10 7 - 3], Z P =4- - 9 = -4m 31. a. S= ut + 1 4:5. {min 4 45 2ut2 For c,PZX. sm 53° = Z P' Imin = x = 9 m For Problems 38-42 =(2i\" -9j\")4+ 21(4;A +3J'\\) x 16 38. d., 39. d., 40. b., 41. c., 42. a. = 81 - 36] + 327 + 24] = 40 i - 12] Sol. Two given planes are mutually perpendicular and the particle is projected perpendicularly from plane OA. It means .\" is u parallel to plane 0 B. Si =447 -9]. At the instant of collision of the particle with OB, its ve- locity is perpendicular to 0 B or velocity component parallel For Problems 32-34 32. b., 33. d., 34. a. to OBis zero. Sol. Firstconsidering motion of particle parallel to plane 08, - - - - _ - . . VA ~ 90 mls u = 1O.j3ms-', acceleration = -gsin60° ------.111 VIJ = -70 mls ------1>. Vc~ 100 mls = -5.J3 ms-2 Fig. 6.65 v = 0, t =? s ==? Using, V = u + at, t = 2 s s = ut + I, -at- 2 Vc = 100 km/hr or OQ = 10.J3 III

6.24 Physics for IIT-JEE: Mechanics I Now considering motion of the particle normal to plane 13. c. Due to air resistance, the accelerations of both the balls OB. will be different. Hence, they will reach at different times Initial velocity = D. acceleration = g cos 60\" = 5 O1s- 2 and with different velocities. I = 2 s, v = 'I S = PO = ? 14. a. Slope of v-t graph is zero, hence acceleration will be zero. Using, v = u + at, v = 10 ms- 1 15. b·lii + iii = Iii -iii s = ul + I, or PO = 10 01 + + +-4 _+ -4 -> -ar = =2 =} A2 B2 2 A, B = A2 B2 - 2 A, B -~ - ... h PO sin 30° 10 x sin 30\"= 5 01 -=} 4A,B=0 Inclination of point P with the vertical is 30°, thcrcf()fe =} A-LB, its vertical component is, 16. d. The direction of velocity vector is always along the u cos 30() = 15 ms~'1 (upward) tangent to the path but its slope will not give the magnitude of velocity, Magnitude of velocity is given by the slope of Considering vertically upward motion of the particle position-time graph. from P, initial velocity = 15 ms·-- 1, acceleration = -g = -10 ms-2, v = 0, s = H = ? 17. h. When objects move in opposite directions, their velocities are added while calculating relative velocity w.r.t. each other. Using v2 = u2 + 2as, H = 11.25 m 18. d. Depends upon the angle between the veloeities of two Maximum height reached by particle above bodies, 0=h+H=16.2Sm Distance PQ = J(PO), + (OQ)' = J(lO)' + (lO~)2 = 20m :'Mafcl1ii'iir '\" : l\\ssertion-,Reasoningi ' i'oo!uml;ryper ' ~'~, , 0.' \" 'liype , \" ,,' 1. i. -+ h., c., ii. -+ c., iii. -7- h., c., iv. -+ b., c. If initial velocity and acceleration are in opposite directions, 1. c. The gravitational potential is the potential energy per unit velocity reaches zero and then increases in opposite direction. mass. In B, initial velocity and acceleration are in same di- 2. d. S\" = u + !-a (2n - I) is numerically correct but dimen- rection so velocity increases continuously and particle movc 2 along the direction of acceleration. sionally incorrect, because dimensions of both sides are not matching, In C, x > 0 but velocity can be along cither positive or 3. d. Least count depends upon the scale, 4. a. Backlash error is caused due to wear and tear or loose negative X -axis. fittings in screws and can be minimized by turning the screw in one direction only. 2. i. -+ a., ii. -+ a., iii. -+ h., iv. -+ c. Slope of velocity- 5. c. Least count:::;: pitch/no. of circular scale divisions. time graph gives acceleration. If direction of acceleration and Less is the pitch, less is least count so more is the accu- velocity is the same, then partiele is speeding up, otherwise racy. 6. c. Alloys have least variation in length with temperature. slowing down. Particle moves in the direction of velocity. 7. d. Speed of a body is always +ive, its velocity may be +ive or -ive. 3. i. -+ b.,d., ii. -+ a.,c., iii. -+ a.,c., iv. -+ h.,d. 8. b. Speed, nd velocity are different physical quantities, Spced Particle is accelerating when hoth velocity and accel- is a scalar quantity and velocity is a vector quantity. eration arc having the same direction (sign), and decelerat- 9. a. Speed of a body is always +ive, its velocity may be +ivc ing when velocity and acceleration have opposite directions or -ive. (sign), 10. a. If v-I graph is perpendicular to time axis, its slope will be infinite which will indicate infinite acceleration which is not 4. i. -+ d., ii. -7- d., iii. -+ c., iv. -7- a. possible in practice. Here, maximum height for all the particles is same. 11. c. Slope of position-time graph will give the velocity and H= eu 2 sin2 u~ from here we can nnc! both direction and magnitude of ac- =~ celeration. 2g 2g 12. d. For uniformly accelerated motion started from rest, the So, all three particles have same Uy. displacement versus time graph is parabolic. And for uniformly accelerated motion, the velocity in 2u sine 2u\" equal intervals of time chang.;s by the same amount. T= - - = - ' 2g g So, all three particles have the same time period. Range (R) is maximum for C R = horizontal component of velocity x T So, horizontal component of velocity is greater for C. u= Vlu 2 + u2 x y Ux is least for A and ltv is same for all, so u is least for A.

Miscellaneous Assignments and Archives on Chapters 1-5 6.25 5. i. -4- a., ii. ~ c., iii. -?- d., iv. ~ b. 2. The velocity of K throughout the motion towards the centre Y = Px - Qx2 of the square is v cos 45\" and the displacement covered by this velocity will be K 0 (Fig. 6.68). Equation of trajectory of the projectile motion is given KO (hdI2) d < gx 2 t= v cos 45° = vlh = v by Y = xtan8 - 2u 2 cos2 8 < On comparing, tan e = P So, iv. ---* b. N --- ,,, M J22< u2 cgos2 8 = Q; ucose = g Q Range R = 2u2 • 2g = P - smecose = --P Q g g2Q So i. ---+ a. eMaximum height H = ~2 sin2 2g 1 g sin2 8 I p2 Fig. 6.68 = 2g -2Q- c-o-s2-8 = 4Q p2 - 4Q2 Alternatively: So, ii. -+ c. --+ -> -+ = -+ +(- --+ along the Im• e VKN VN VK JTime of flight T = -2u-sin-e = ~ = VK - VN) gp KN=O+[-(-v»)=v g g 2Q Time taken for K and N to meet will be = dI v. < J= ~g p 3: v = r{J) = 3 tan 45° x 0.1 = 0.3 m/s So, iii. ---* d. 4. Givcn that y = ax - bx2 6.· i. -+ c., ii. --+ d., iii. --+ a., iv. --)- b. Velocity u = ucose i + u sine} 2}2 i~, L r;;' , = 60 x +60 x = 30y3i +30} Displacement after 2 s. x = cos eti a y = (u sin lit - 1 gt 2)}, , x = 60 x 2~'x r2i ;= 6;0y'3i e ::2 1 y= ( 60x:I:2x2-::I2x1Ox4)\"}=40j Fig. 6.69 Displacement + y = 60 ~ i + 40 } Comparing it with the equation of a projectile, we get Velocity after 2 s, Vx = Ux = U cos ai y = x tan II - -=gx~2 - ~, 30yr2;;i' 2u 2 cos2 8 2i = 60 x = =>tanll =a (i) e -Vy = (u sin gt) ,j = (60Ix ::2 - 10 x 2) = 10 , and g = b (ii) j 2u 2 cos2 II v = vxi + vy} = 30~i + 10]. From (ii), g 2 g(a2 + I) (iii) 2b cos II 2b Archives ~][ .,' cos II = a 2 + 1 Fill in the Blanks Type , ., eu2 sin2 MaxImum heIght attamed H = (iv) 1. Displacement = 2r 2g Distance = 7T r JbFrom (ii), = u2 cos2 II i7) u2 sin2 II = 2 + I) g (v) g(a 2b displacement 2b Fig. 6.67 From (iv) and (v), we get H = ga2 = a2 2b x 2g 4b

6.26 Physics for IlT-JEE: Mechanics I Alternatively: -----j---fv~-- --- . dy = a _ 2hx = 0 (For maximum height) ,, -(i-r':~~l~ -- dx a : .d X= - x ax - bx2h 0' Substituting the value of in y = 2 to find max- --- - - - --\"--v, ------''- imum height H=a(;b)-b(::2) = ~; - :; = :;. Fig. 6.71 True or False => The swimmer should swim due north. 1. True. When the two balls are thrown vertically upward with 2. b. Vertical displacement = 1 km the same speed then f=1-5=-h1 r + t t'\" t'\" 60 4 1 - + Uj = U Vb cosO = 1/4 =4 km/h s, = 0;U2 = 0 C V,. B al = g; S2 = 0 (as ball reaches back to the same point) lkm VI =?;a2 = -g Vb v2 - u2 = 2as; Vl == ? A VI = J -2gs + u2 = u. Thus. we find thatg is indepen- dent of mass. 2. True. Vy Fig. 6;72 44 sinO = -3 5 cosO = - = - Vb 5 v, 1C By velocity triangle ABC, sine = -'- Vb V,- 3 => V,. 3 => V, = 3 kmlh. = Vb 5 5 5 Fig. 6.70 3. b. I Average velocity I = IDisplacement I As shown in the Fig. 6.74, the velocity at 1 and 3, i.e., at any arbitrary points before and after the topmost point is '---'c:::;-_ _ greater than Vx ' Alternatively: Time T.E. = P.E. + K.E. Di~plncemel1t T.E. = Constant Fig. 6.73 At P, K.E. is minimum and P,E. is maximum 21' 1 Since K.E. is minimum velocity is also minimum at P, = - = 2 x - = 2m/s. the topmost point. f1 3. False. The pressure exerted will be different because one train is moving in the direction of earth's rotation and the other in the opposite direction. 4. :False. For a particle to move in a circular motion, we need· a centripetal force which is not available. The statement is false. Single Correct Answers Type 4. a. Before hitting the gronnd, the velocity v is given by v2 = 2gd (quadratic eqnation and hence parabolic path) 1. a. Time taken to cross the river t = d Downwards dirt.:ction means negative velocity. After colli- FQr time to be minimum eVs cos sion, the direction becomes positive and velocity decreases. cos e == max ef;)(~)Furthcr, v,2 = 2g x =~ = gd; => 0 = 0' orv = v,~.

Miscellaneous Assignments and Archives on Chapters 1-5 6.27 As the direction is reversed and speed is decreased and -ax2Z+ - =)'2 hence the graph (a) represents thcse conditions correctly. b2 I 5. a. S\" = a I); S,,+I = a 2 n + 1) Path of the particle is an ellipse. 2(2n - 2( Hence option (a) is correct. S\" 2n - I y S,,+I = 2n + I 6. b. Till 11 s, acceleration is positive, so velocity will go on --1:--o-~+--l~ X increasing up to II s and maximum velocity will happen at 11 s. Fig. 6.75 The area under acce1eration-time graph gives change in velocity. Since particle starts with u = 0, therefore change From the given equations, we find in velocity = vf Vj:::::; Vmax - 0 = area under a-t graph or Vm\" = 2I x 10 x II = 55 mls 7. a. v= - Vo x+ Vo, a= vdx = (_ Vo x+ vo) (-VO) ddxt = Vx = - . ap sm pI ~ dv ~ ~ d2x dt 2 => a=(VO)2x_ V6 = ax = - ap2 cos pt Xo XO tiy . -dt = V.y = bp eos pt Multiple Correct Answers Type d 2y dt 2 -) ~ .....,. .....,. -+ and = ay = _bp2 sin pt . ~ Vr-Vi v/+(-v;) v 1. b. AverageacceleratlOn a = - ' - - = -'--'-''-= t tt 7f 7f -;. -) To find the resultant of v/ and - Vi, we draw the follow- At time t = - or pt = 21' 2 ing (Fig. 6.74) y 2p N Vy W- ~Ell)~ 5 ~/s b X a 0 ax s Fig. 6.76 Fig. 6.74 ax and Vy become zero (because cos 7f12 :;:; 0). only Vx and ay are left (Fig. 6.76). Or we can say that velocity is along negative x-axis and . lv~/ I = I- I~ acceleration along y-axis. Smce, Vi Hence, at t = ~, velocity and acceleration of the par- -+ -~-). 2p v is directed in between vf and - Vi. ticle are normal to each other. So option (b) is also cor- Therefore, ; is directed t9wards N-W +~ AA rect. At t = t, position of the particle r (I) = xi yj = -, 5v'2 I a cos pt; + b sin pt] and acceleration of the particle is a = --=-. (;(I)=aJ +ay] = -p2[aeospti +bsinpt]J 10 v'2 = _1'2 [xi + y] = _p2 7(I)J 2. a., d. a cannot remain positive for all t in the interval 0 :S t :S I. This is because since the body starts from rest, it Therefore. acceleration of the particle is always direeted will first accelerate, finally it stops therefore a will become towards origin. negative. Therefore a will change its direction. Hence. (a) Hence, option (e) is also correct. and (d) are the correet options. 7f 3.3., b.,c. cos (pI) = x (i) x = a cos pt =} (ii) At t = 0, particle is at (a. 0) and at t = - , particle is a 2p y = b sin pt =} sin (pt) = ~ at (0, b). Therefore, the distance covered is one-fourth of the Squaring and adding (i) and (ii), we get elliptical path and not a. Hence, option (d) is wrong.



Newton's Laws of Motion 7.1

7.2 Physics for IIT-JEE: Mechanics I INTRODUCTION in Fig. 7.2(b), the wagon moves. When a football is kicked, as in Fig. 7.2(c), it is both deformed and set in motion. These examples In kinematics we dealt with motion of particles based on the def- show the results of a class of forces called contact forces. That initions of position, velocity, and acceleration without analysing is, these forces represent the result of physical contact between its cause of motion. We would like to be able to answer general the two objects. questions related to that cause of motion such as \"What mech- anism causes change in motion?\" and \"why do some objects There exist other forces that do not involve physical contact accelerate at higher rates than others?\" In this section we shall between the two objects. These forces, known as field forces, can discuss the causes of the chang~ in motion of particles using the act through empty space. The gravitational force between two concepts of force and mass. We will discuss the three fundamen- objects that causes the free-fall acceleration described in chap- tal laws of motion which are based on experimental observations tcrs 2 and 3 is an example of this type of force and is illustrated and wcre formulatcd by Sir Isaac Newton. in Fig. 7.2(d). This gravitational forcc kceps objects bound to the Earth and gives rise to what we commonly call the weight THE CONCEPT OF FORCE of an object. The planets of our solar system are bound to the SUll under the action of gravitational forces. Another common How a body moves? It is determined by the interaction of the example of a field force is the electric force that one electric body with its environment. These interactions arc called forces. charge exerts on another electric charge, as in Fig. 7.2(e). These The concept of force gives us a quantitative description of the charges might be an electron and proton forming a hydrogen interaction between two bodies or betwecn a body and its en- atom. A third example of a field force is the force that a bar vironment. As a result of everyday experiences, everybody has magnet exerts on a piece of iron, as shown in Fig. 7.2(0. a basic understanding of concept of force. When you push or pull an object you exert a force on it. You exert a force when Some examples of forces applied to various objects. In each you throw or kick aball. In these examples, the word 'forct;' case, a force is exerted on the particle or object within the is associated with the result of muscular activity and with some boxed area. The environment external to the boxed area pro- change in the state of motion of an object.'_ Force does not always vides this force. cause an object to move, however. Contact Forces· Field Forces For example, as you sit reading a book, the gravitational force acts on your body and yet you remain stationary. You can push ,r8+--,--~ on a heavy block of stone and yet fail to move it (Fig. 7.1). , ,I -m • Produces or tries to produce motion in . a body at rest. 1_ _ _ _ .J • Stops or tries to stop a moving body. (a) (d) • Changes or tries to change thc (b) (e) direction of motion ofa body. r- - - - -- - - --I • Produces a change in the shape of a body. :\"tIlton ~ ~1ff!i11;';\"==='=1Is Fig. 7.1 ,,-----------, (0 ClASSIFICATION OF FORCES Fig. 7.2 Based on the nature of the interaction bctween two bodies, forces may be broadly classified as under: NEWTON'S LAWS OF MOTION 1. Contact Forces: Tension, normal rcaction, friction, etc. The entire classical mechanics is based upon Newton's laws of Forces that act between bodies in contact. motion. They, in fact, are simply known as Laws of motion. These laws provide the basis for understanding the effect that 2. Field Forces (Non-Contact Forces): Weight, elcctrostatic forces have on an objcct. forces, etc. Forces that act between bodies scparated by a distance without any actual contact. Newton's First Law of Motion Since we're going to encounter these forces in our analysis, According to this law: A body continues to be in its state of rest we will briefly discuss each force and how it acts between two or uniform motion along a straight line, unless it is acted upon bodies, its nature, etc., and how we are going to take it into by some extcmal force to change the state. account. We are going to discuss some special forces. If you pull on a spring, as in Fig. 7 .2(a) , the spring stretchcs. If the spring is calibrated, the distance it stretches can be used to measure the strength of the force. If a child pulls on a wagon, as

Newton's Laws of Motion 7.3 Every body c'ontinues to be in its state of rest, or of uniform .1 inertia of R('st~ It i~ the inability of a body to motion in a straight line, unless it is compelled to change that r ~s mc,.os a body at ,'cst \"\"\",,;os at rest and state by forces impressed thereon. This law, also called the law of changc,by itself. it~ state nfrcst - inertia, is really a statement about reference frames, in the sense that it defines the kind 'of reference frame in which the laws of . cannol stmi movl11g on ItS own. Newtonian Mechanics hold. If you lind that the net force on a body is zero, it is possible for you to find a reference frame in Inertia of Direetion:- I! is the inability of a which the body has no acceleration. bodY 10 eil(l!\\%C by itself its direction '01' -, ~ I 1110110n,i.(\",,<) ody continues to nwv.: along Type,> of 11l.:r!iQ ---->- the sal1lt~ strail;ht line llnless compelled . => L F, = 0, L F, = 0, L F, = O. by some cxtel~llal klfCC to change iL lncrtia or Motion:- It is the inability of a hody to change by itself its stat.: of uniform H1otion,i.c., a body in uniform '-----> motion can neither accdcralC' nor retard on its own and come to rest. Fig, 7.3 Newlon's Firsl Law DL'fin('s: In,,\"\" Newton's Second Law of Motion [ lorce According to this law: The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the force applied. Force When an unbalanced force is applied on a body, the momen- According to Newton's Hrst law of motion, a body continues to tum of the body changes; the rate of change of momentum with be in a state of rest or of uniform motion along a straight line, unless it is acted upon by an external force to change the state. respect to time is defined as the net external force acting on This means force applied on a body alone can change its state of rest or state of uniform motion along a straight line. Hence we ~ define force as an eXlernal effort in the form of push or pull which moves or tries to move a body at rest, stops or tries to stop a body - dp ,4 ~ in motion, changes or tries to change the direction of motion of the body. i.e. Fext = ; where linear momentum P = m. V, a body. Hence Newton's first law of motion defines force. - Inertia dt ~ We observe in rouline Ihat an object lying anywhere keeps on lying there only unless some one moves it. For example, a chair, m. ::; mass of the body, V ::; instantaneous velocity a tahle, a hed, etc. cannot change their positions on their own, i.e., a body at rest cannot start moving on its own, The reverse ~ is also true, though it is slightly difficult to perceive. If there were no forces of friction and air resistance, etc., a body moving ~ _ d ( -» _ dm -, d v uniformly along a straight line shall never stop, on its own. Fcx! - - m v - - v +m-- According to Newton's first law of rnotion, a body continues dt dt dt to be in state of rest or of uniform motion along a straight line, unless it is acted upon by an external force to change its state. dm This means a body, on its own, cannot change its state of rest If m. ::; constant, i.e., - = 0 or state of uniform motion along a straight line. This inability of a body to change by itself its state of rest or state of uniform dt motion along a straight line is called inertia of the body. Hence Newton's first law defines inertia and is rightly called the law of ma=? j.~xt = inertia. If mass of a body is constant, the acceleration of the body is' Quantitatively, inertia of a body is measured by the mass of the hody. Heavier the body, greater is the force required to change inversely proportional to its mass and directly proportional to its state and hence greater is its inertia. The reverse is also true (Fig. 7.3). L~ 4 This inherent property of all bodies by virtue of which they the resultant force acting on it, i.e., F = m a. This vector cannot change their state of rest or of uniform motion along a straight line on their own is called Inertia. It depends on the mass equation is equivalent to three algebraic equations, of the body. Ll~ = max (i) (ii) L Fr = mar (iii) L:r\"\"'z = maz Points to Remember 1. If j,: = 0, second law gives;; = 0, therefore it is consis- tent with the first law. 2. The second law of motion is a vector law. It means what~ ever be the direction of the instantaneous velocity of a particle, if any net extemal force is acting on it, this force will change only that component of the velocity which is in the direction of the force. 3. Strictly speaking, this law applies to particles, i.e., point masses only. However, with the introduction of the con- cept of center of mass, this law can now be applied in case of extended bodies or a system of point masses

7.4 Physics for IIT-JEE: Mechanics I also. You will study this in the chapter on systems of ~ particles and rotational motion. We can write it as FBA = • This is a local law. This means that it applies to a Action: tyre pushes on road Reaction: road pushes on tyre particle at a particular instant without taking into con- Action: rocket pushes on gas sideration any part history of the particle or its motion. Reaction: gas pushes on rocke ~ Action: man pulls on spring -;. dp Reaction: spring pulls on man -,)0 • As F = - , where P denotes momentum, slope of a dt momentum versus time graph gives force. In Fig. 7 A, tan a gives the force at t = 11 and tan f3 gives the force at t = f2. t p Fig. 7.4 Action: earth pulls on ball Reaction: ball pulls on earth ~ The force F in the law stands for the net external force. F.Any intcmal forces in the system are not included in Newton's Third Law of Motion To every action there is always an opposed equal reaction. It doesn't matter which force we call action and which we call According to this law: To every action, there is always an equal reaction. The important thing is that they are co-pairs of a single and opposite reaction, Le., the forces of action and reaction are interaction, and that neither of forces exists without the other. always equal and opposite. When you walk, you interact with the floor. You push against the floor, and the floor pushes against you. The pair of forces To every action there is an equal and opposite reaction. When- occurs at the same time (they are simultaneous). Likewise, the ever a body exerts a force on another body, the second body als.o tyres of a car push against the road whilc thc road pushes back exerts a force on the first that is equal in magnitude, is in the on the tyres-the tyres and road simultaneously push against opposite direction and has the same line of action, acting simul- each other. In swimming, you interact with the water, pushing taneously. the water backward, while the water simultaneously pushes you forward-you and the water push against each other. The reac- Two point masses act on each other with forces which are tion forces are what account for our motion in these examples. always equal in magnitude and oppositely directed along the These forces depend on friction; a person or car on ice, for ex- straight line connecting these points. ample, may be unable to exert the action force to produce the needed reaction force. Neither force exists without the other. Any of the two forces making action-reaction pair can be called action, and the other reaction. IMPULSE If two bodies are in contact with each other, the action and re- According to Newton's second law, the momentum of a particle action forces are the contact forces. But Newton's third law also applies to long range forces that do not require physical contact, changes if the net force acts on the particle. Knowing that the such as the force of gravitational attraction or electromagnetic interaction between two charged bodies. change in momentum caused by a force is useful in solving To every action there is always opposed an equal reaction; some types of problems. To build a better understanding of this or, the mutual actions oftwo bodies upon each other are always directed to contrary parts. L~ or important concept, let us assume that a net force F acts on To every action there is always an equal and opposite reaction. a particle and that this force may vary with time. According to It appears to be the-easiest law to remember, but what students do not appreciate is that ,although action and its reaction are Newton's second raw, always equal and opposite, yet they never cancel each other out in case of any of the particles or bodies of the system. Why? -+ -+ --+ --+ Because they always act on different bodies. Forces always occur in pairs. If body A exerts a force F on L;F=dP/dtordP=L;Fdt (i) body B, thcn B will exert cqual and opposite force on body A. We can integrate f this expression to find the change in the inomentum of a particle when the force acts over some time inter- val. (If the momentum of the particle when the force acts over

Newton's Laws of Motion 7.5 some time intervaL) If the momentum of the particle changes t ~~ of-----;r-;\"\".,.-, t from p, at time ti, to Pf at time tf, integrating equation (i) gives o2 F ---c>- ---c>- If ---c>- 00 \".\";:' ~P=~-~=JEF~ I; To evaluate the integral, we need to know how the net force varies with time. The quantity on the right side ofthis equation () Fis a vector called the impulse of the net force L acting on a r --i> particle over the time interval !.':J.t = tf - ti. Fig. 7.6 fIf (iii) Note: Area WIder force F versus time t graph gives total change ilunomentum, i.e.', impulse llatcheltarecrinjigure.' ; = I: Fdt I f12 12 I; -, J '\" Fd{\", dp '\" Total challge ill P = Impulse. From its definition, we see that impulse I is a vector quantity having a magnitude equal to the area under the force-time curve Ii II as described in Fig. 7.5. It is assumed the force varies in time' in the general manner shown in the figure and is nonzero in the R~I, In a particular crash test, a car of mass time interval ~t = tf - tj. The direction of the impulse vector 1500 kg collides with a wall as sh~wn in Fig. 7.7. The ini- is the same as the direction of the change in moment.um. Impulse has the dimensions of momentlIm, that is, M LIT. Impulse is v:tial and final velocities of the car arc = -15.0 i mls and not a property of a particle; rather, it is a measure of the degree to which an external force changes the particle's momentum, Vi = 5.00 i mis, respectively. If the collision lasts 0.1.50 s, Equation (ii) is an important statement known as the impulse- find the impulse cansed by the collision and the average force momentum theorem. exerted on the car. The change in the momentum of a particle is equal to the I3efore impulse of the net force acting on the particle. -- 15.0 l1l!s (iv) .....- - - - - A large force acting for a short time to produce a finite change in momentum is called an impulsive force, In the history of sci- After ence, impulsive forces were put in a conceptually different cat- egory from ordinary forces. Now, there is no such distinction. +2,60 m/s Impulsive force is like any other force with the only difference that it is large and acts for a short time. Even if the net force ---j)- is small and acts for long time, we can still calculate the im- pulse imparted by it and equate it with the total change in the momentum of the body on which it has acted. FF I; If Ir Fig. 7.7 (a) (b) Sol. The collision time is short, so we can imagine the car being (a) A net force acting on a particle may vary with time. The impulse is brought to r,est very rapidly and then moving back in the opposite the area under the curve ofthe magnitude ofthe net force versus time. direction with a reduced speed. (b) The average force (horizontal dashed line) gives the same impulse Let us assume that the force exerted by the wall on the car to the particle in the time interval I1t as the time-varying force described is large compared with other forces on the car (such as friction in part (a). The area of the rectangle is the same as the area under the and air resistance), Furthermore, the gravitational force and the curve. normal force exerted by the road on the car are perpendicular to the motion and therefore do not affect the horizontal momentum, Fig. 7.5 Therefore, we categorise the problem as one in which we can apply the impulse approximation in the horizontal direction. Ifwe plot a graph between average force and time (Fig. 7.6), the area under the curve will give impulse imparted during the Now to evaluate the initial and final momenta of the car; time interval under consideration. j;, = In ,!; = (1500 kg)( -15.0; m/s) = -2.25 x 104 ; kg m/s i;~ = In v~ = (1500 kg)(S.OO; m/s) = 0.75 x 104 1 kg mls

7.6 Physics for IIT-JEE: Mechanics I Impulse on the car: The transfer of momentum to the wall [!>.P,]wxll =2!>.mvcos$ and [!>.P,·]wxll =0 ---4 ___ _+ _} If M is the direction of strike. then the force exerted by the jet on the wall is, I =J',P = Pf- Pi F= [!>.P,Jwxll = -2!->.-mv-co-sO=. (!->.-m) 2vcosO = 0.75 X )04; kgm/s - (-2.25 x 104 1kg· m/s) !>.t !>.t !>.t = 3.00 x 104 7 kg mls = P ( ~~) 2v cos e The average force exerted by the wall on the car: F = 2pv'cosO • _ J',p _ 3.00 X 104 ; mls _ 2 00 X 105 ' N F.wg- J',t - 0.150 s -. I Some Examples of Impulse: Force Exerted by Liquid Jet on Wall Liquid Jet Striking a Vertical Wall Normally Wind with a velocity oflOO km h- 1 blows Consider a liquid jet of area A that strikes the vertical wall with . normally against one wall of house with an area of 108 m2. a velocity v. The liquid after striking the wall moves parallel to Calculate the force exerted on the wall if the air moves the wall. parallel to the wall after striking it and has a density of 1.2 kg m-3• Fig. 7.8 Consider small element of liquid of mass Ilm. The change in Sol. Let us first deduce a general formula. Change in velocity of air along the normal to the wall its momentum after strike is, = v - 0 = v. (Velocity along thc normal aftcr the air strikes the .[J',Pxljet = 0 - J',mv = - J',mv wall is zero because it moves parallel to it.) +~ Amount of air striking the wall per second = A x v, But, [J', P,ljot [J',Pwxlll = 0 where A == area of wall So the transfer of momentum to the wall [J', P, ]wxll = J',mv If J',t is the duration of strike, then the force exerted by jet on Mass of air striking the wall per second == A x v x p the wall where p == density of air. .', change in momentum per second == (Avp)v = Av2 p F = [J',P,]wxll = J',m~ = (t;m) v; By laws of motion the wall exerts this force and the air exerts !>.t !>.t. !>.t the same force on the walL but!>.m = p!>. V = (p1~) v = p ( ~~) v; force exerted by the air on the wall == A 1,)2 P = 108 x ( -1-0306(X)(1i-0-0--0)2 x 5 1.2 = 10 N ~-~ncePt Application Exercise 7.11-----, p is density of the liquid. 1. Explain why? If the rate off low of liquid is !>.V = av, F = pav'. a. A horse cannot pull a cart and run in empty space. -- b. Passengers are thrown forward from their seats when a !>.t speeding bus stops ~uddenly, Liquid Jet Striking the Wall at the Same Angle c. A cricketer moves his hands backwards when holding eNow consider the jet strikes the wall at an angle with the a catch. normal and rebounds with the same angle as in Fig. 7.9. 2. As shown in Fig. 7.10, two identical balls strike a rigid wall with equal speeds but at different angles of incidence. They are reflected back without any loss in the speed. l' v .. - ....--+-- (, ) Fig. 7.9

F Newton's Laws of Motion 7.7 e y m\" \" 60.0° (b) ----------:::. ------x Fig. 7.10 \"60.0° a. Determine the direction of force exerted by each ball Fig. 7.14 on the waIL b. Determine the ratio of impulse imparted by the two 8. The magnitude of the net force exerted in the x-direction on a 2.50 kg particle varies with time as shown in Fig. 7.15. balls on the wall in both cases, Find 3. Fig, 7, II shows the position-time graph of a particle of a. the impulse of the force, b. the final velocity the particle attains if it is originally at mass 0,04 kg, Suggest a suitable physical context for this rest, motion. What is the time between the two consecutive c. its final velocity if its original velocity is - 2.0 mls. and d. the average force exerted on the particle for the time \",=\",mlv N\\impulses received by the particle'? What is the magnitude interval between 0 and 5.00 s. [(._---.J_-\"_'-L.::i_~L,,_*--l,,_~--+t t (s) F(N) () 2 4 6 8 10 12 14 16 Fig. 7.11 2 4. A rubber ball of mass 50 g falls from a height of 5 m and o 2 } 4 5 ! (s) rebounds to a height of 1,25 m, Find the impulse and the average force between the ball and the ground if the time Fig. 7.15 for which they are in contact is 0.1 s. FREE BODY DIAGRAMS T In frec body diagrams (FBD) the object of interest is isolated f5.0m from its surroundings and the interactions between the object and the surroundings are represented in terms of forces. 1 125 m ~ After knowing the nature ofdifferent forces, let us draw a \"free Ground body diagram\". The phrase itself reveals that we must mentally Fig. 7.12 free (isolate) the bodies (or particles) in the systcm and then consider all the force acting on all the particles of the system. 5. Water falls without splashing at a ratc of 0,250 Lis from a height of 2,60 m into a 0,750 kg bucket on a scale, If The Common Forces Encountered in Mechanics and Rep- the bucket is originally empty, what does the scale read resentation of these Forces Through Free Body DiagrQ;';s 3.00 s aftcr water starts to accumulate in it? 1. Weight 2. Normal Force 6. Aliquid of density p is flowing with a speed V through a pipe of cross-sectional area A. The pipe is bent in the 3. Tension 4. Frictional Force shape of a right angle as shown in Fig. 7.13. What force should be eXCl1ed on the pipe at the corner to keep it fixed? 5. Elastic Spring Force Fig. 7.13 Weight 7. A 3.00 kg steel ball strikes the wall with a specd of It is the gravitational force with which the earth pulls an object. 10.0 mls at an anglc of 60.0' with thc surface. It bounces The weight of an object can be written as mg, where m is the off with the same speed and angle (Fig. 7.14). If the ball mass of the object and g is the acceleration due to gravity, which is in contact with the wall for 0.200 s) what is the average is always directed towards the centre of the earth. For a small force exerted by the wall on the ball? stretch of the earth's surface which can be considered to be fiat, the acceleration due to gravity g can be taken to be uniform,