5.2 Physics for IIT-JEE: Mechanics I y p RELATIVE VELOCITY y It is given by the time rate of change of position of one object w.r.t. another. Relative velocity of a body B with respect to some -; ~) other body A means velocity of B is recorded by an observer rpH sitting on A. Mathematically, Relative velocity of B w.r.t. A: VB/A. = VB - VA .-.\"JV VSA \"'---F-ra-m-c-B---\" Fig. 5.3 Fig. 5.1 Figure 5.] shows a certain instant during the motion, At this' instant, the position vector of B relative to A is rnA. Also, Proof: From Fig. 5.1, PH/A = fB - fA. Differentiating this cqua- tion w.r.t. time, we get the position vectors of particle Pare rpA relative to A and dCI-BfA) -drH- -- d--r-A-- but drjJ rPl? relative to B. From the arrangement of heads and tails • cit dt dt = VB and of those three position vectors, we can relate the vectors with dt rpA = (PH + ':HA. d(rll/,') _. - -{-- = By taldng the tilllederivative of this equation, wecan relate the c t V·Hf A velocities Up/II and Vp/B of particle P relative 10 our ohservers. putting these values we get VB/A = VB - VA. Hence proved. +We get 1)p/A = uP/fJ UB/!\\' - -.Similarly, we can prove that relative veloci~y of A w.r.t. B: VAIB = \" We can understand the concept orrelativc velocity hy a simple VA ._. VB· situation as follows: Let two cars P and Q move with velocities vp and uQ towards north and east, respectively. Graphical Method to Find Relative Velocity ~ -, V/J!A 1'[1 ,, eIX ~ ., Pi~, ·v0, V..l .. VQ <4---\"----------- -- - - - - f > VQ (a) Fig. 5.2 When two bodies' move at an angle f) with each other then their tV relative velocity is given by: \"~r ·~~~'--_17~ H Magnitude:!vlJ/r\\1 = Ivu - vAl = ,,, (v;\\ + vY) ;\"2'~;'~~~~-'~;;~1-8o'-=-e) = /v~ + v~ - 2VAVJjcosf) --} I Direction: tan Q' = VB sin(l80 - 0) ------------. ,-- VI' : VA + vBcos(180-0) t vJJsinfJ (1)) tan a = ---''---- Fig. 5.4 VA -vJJcosO Velocity of c~r P w.r.L Q (i,e., velocity of car P for an ob- Note: server on car Q) is vP/Q, where vp/Q = VI' -- vQ as shown in tan a: =V-JJ Fig. 5A(a), which implies that for an observer on car Q, car P VA appears to move along north-west instead of north. Similarly We can find the velocity of a particle in a frame if we know the for an observer on P, car Q appears to move along south~cast particle's velocity in some other frame and the relative velocity of frames W.Lt. each other. instead of east as shown in Fig. 5.4(h). In Fig. 5.3, two observers are watching a moving particle P A cal' A moves with a velocity of 15 mls from the origins of reference frames A and B, while B moves at and B with a velocity of 20 mls as shown in Fig. 5.5. Find tl'\" a constmlt velocity VB/A relative to A. (The corresponding axes relative velocity of B \\V.l'.t. A and A w,r.t. B. of these three frames remain parallel.)
Motion in Two Dimensions 5.3 15 m/s --2-0 -m/-s » Graphically: VII/A = V/i + (-VA). from Fig. 5.8 A~ NS = M P = 208in30\" = 10 10 -------- -, Fig. 5.5 ~~~ I s A--:.\" :VAc.-~P 10 -., 20 V VOlA Sol. Givc:n: VA.:::;:; 15 mIs, V/I:;:: 20 m/s. IX to M Relative velocity of B W.Lt A: UN/A ='Va - VA' = 20 ~ IS = 5 m/s. 1300 0N Relative velocity of A w.r.t B: J.iA/B = VA - Va Fig. 5.8 = IS .... 20 = ···5 m/s. and ON = OM - NM = 2000s30\" - ]() = 10(~ - I) (Negative sign indicates th(/t this relative velocity ;.'1 in back- ward direction.) IV/I/A I = .jON' + NS' = 10/(~ - 1)' + (I)' Note: If the .uirs. were. moving in.opposite directions .as = 1OIs=2~ m/s shown, then VA = -IS.lIlts, VlJ :;:: 20 mis, Relative velocity 11f B w.r.t. A: villA'\"' Vl1 ~ VA = 20 NS 10 I - (-15) = 35m!s -- = ON Relative velocity of A w.a B: V\"lll = VA\"\" VB =-15 =7f=ltana = 10(~:\"\" I) - 20= -35 mls. (Negative signilldicaiestilat this relalive velocityi\" ill back- Ram was riding in a car traveHing: at a ward direction.) speed of 20 mls towards east. Shyam was standing on tbe !5 !HIs •20 m/s sideway when the car passed bim. .Just before tbe car passed close to ShYaIll (whell it was north-west of him) Ram had </---- threw a banana peel with a velocity of 10 mls relative to him Fig. 5.6 =as Vu\",u\",\". Ibm (-10) i + (-10)]. The banana peel had ve- Imp: 1;11/;\\ and VA/II have equal magnitude but opposite direc- locity compollents 10 rills towards south and 10 mls (owlIrds tions. west in Ram's reference frame. When Ram met Shyam later, Shyam accused that Ram had thrown the banana peel at him. Now the following illustration shows how to find relative ve- Explain this with the help of velocity diagram. locity when two objects move in directions at sornc angie. ~- Two objects A and B lYre moving along VShynm the directions as shmyn in Fig. 5.7. Find the magnitude and Ram the direction of relative velocity of B \\V.r.t. A. (n) 20 m!s P [}----+ \", -, A !() .-- m/s vBanana, Ram VBan~na. Ea1ih Fig. 5.7 Sol. Analytically: s -, VRam. ['Mill VA = 107. U/I = 20cos30\"7 +20sin30\") = 10~7 + 10) (b) 1) Relative velocity of 13 w.r.L A: Fig. 5.9 vIII,' = V/I - V', = 10 (II ~ I) 7+ lO] Sol. The banana peel was thrown from a point in south-west r-----2--- direction relative to car (reference frame of Ram). Shyam is lOy ( 1)'* IVIIIA I = on the ground; velocity of banana pee] relative to him can be )3- + J2 = 10)5 - 2~ mis, considered to be absolute velocity of banana peel. tan a = -.-------1.-0..---.--..;..- = ----I...---. VBanana/Ram = VB,mana - VRain 10()3- 1))3-1 +::::} VBannna = VBamlna/Ram VRain = (~ 10) i + (~ 10) J+ (20) i = (10) i ~ (10) J
5.4 Physics for IIT-JEE: Mechanics I We can derive this result by resolving v into its components. Sum of x-components Vx = vcosO - V i.c., the absolute velocity of banana peel is directcd towards south-east, or towards Shyam as shown by the third figure. The Sum of y-components V:y = v sin e above exanlple shows that two observers moving at constant Resultant velocity = 'V~v} +V} velocity relative to one another find that their velocity measure- ments yield different results. = /Cv cos 0 - I'p'+-(~;sin 0)2 When One Body Moves on the Surface of Other = / + V2 - 2vVcos e; tan a = -1'',-. = vcvosiensI-i v Body vv2 Vx v, In Fig. 5.10 a plank is moving on grouud Change in velocity: with a velocity and a hlock is moving on the plank with ve· locity V2 relative to the plank. What will be the actual velocity Let initial velocity of an object he VI and therefore after some- of block? time velocity of the object becomes V2· v!Then, change in velocity: ~v = V2 - magnitude of cbange in velocity: I~ VI = IV2 - VII and change in magnitude of velocity(or change in speed): IV21 - IV11. -----Jtr> vI Notc: lJe car~fult\"aiajJerageaccelerationis change in velocity (/lot change ill speed)divMed by time taken. Fig. 5.10 ' . . ii2~ ~'ih aar,,=\"-'-t- Sol. From Fig. 5.10 we have vI' = VI and Vb!p = \"2· A particle is moving with a velocity of 5 s it is found to be moving with a velocity of Now apply the formula Vb/p = Vb - vp 12 mls in a direction perpendicular to the original direction. During these 5 s find => + +VI; ::;: vbl p v p :::::} Vb = 1)2 VI 1. change in speed, A block slips along an incline of a wedge. 2. change in velocity and magnitude of change in velocity, to reaction block on tbe wedge it slips backwards. and 3. average acceleration and its magnitude. An observer on the wedge will see the block moving straight Sol. down the incline. Discuss how to find absolute velocity of block w.r.t the wedge. iV'\",.li1 \"\" v v,'\"\"5m/s M Fig. 5.13 Fig. 5.11 1. Change in speed: = j 2 - 5 = 7 m/s vSol. We know that VmjM = lIl - VAl 2. Change in velocity: \"\"\" = V2 - VI = 12] - 5f => vm = Vm/M + VM = -5f +12] Note that a single subscript implies absolute velocity. Abso- Magnitudc of change in velocity = I\"\"vl = JS2 + 122 lute velocity of block' is vector sum of its velocity relative to the wedge and velocity of wedge relative to the ground. = 13 m/s \" \"\"v3. Absolute vclocity of block (ground reference frame) is shown aa\\' = -- = -5f -+5 -12-] = - , +1 -2 c in the vector diagram given in Fig. 5.12. '''''I -- l 5 .1 Magnitude of average acceleration = /aa\\, I v v cos e y I;'= !l~:(¥) 2 = m/s2 - - - \" ' - - e----+ Lx ,,, We can also find magnitude of average acceleration from 'I' v le\"I;\" I = TI\"\"vtl = 513' .2 vsin 8 m/s Fig. 5.12 River-Boot or River-Man Problems Iv\",1 = ';v2 + V2 + 2vV cos(rr =::-0) = ';v2 + 1'2 - 2vV cosO Let a boat travel in still water with a velocity, v. If water also moves in the river, then net velocity of boat (w.r.t. ground) will be different from v.
v+Net velocity of the hoat: Vb = Vb/w w , where Vb/w is ve- Motion in Two Dimensions 5.5 locity of boat in water or w,LL water and Vw is the velocity of x-...c water in river, ~ Special cases: --\"//' 1. If a boat travels downstream we have Vb/w = vi and Vw = ui, d v1---'u,-;. So, Vh = Vb/w + Vw = vi + ul = (v + u) 7 Hence, the net velocity of the boat will he v + u downstream, Fig. 5.17 u =:> cosO =' - - v g 'd 1=---= = Fig. 5.14 vsinO v../I- cos2 0 2. If the boat travels upstream, v 1-- v2 Li d =:> t = r../',,T2 =_ 'u\"2 +---\" Net velocity of boat: Vb = \"sin e] . ----- 2 I _ =\"/,,2 _ 2 Fig. 5.15 v vj u ueMagnitude: Iv\" I = sin = ,,2 We have Vh/w = -v7 and Vw = ui. So u e Vi} = Vb/H) + Vw = --vi + III = -- (v - u)! fJ (-ve sign indicates backward direction.) Hence, the net ve- v locityof the boat will be v - u upstream. 3. If the boat travels at some angle 0 with downstream then, Fig. 5.18 Vw = Ill, 1.j/J/1II = veosal ..+-...vsinO]. Vb = Vb/(II + Vw -- (u + v cosO)! + 'VsinO] v\" = flu + v cos'O]:;-:;:(;; sin ii)2\" = ../u2 + v2 + 2uv cos () A man wishes to cross a river in a boat. d crosses river in minimum time then he takes 10 min Time taken to cross the river: t = --.-', with a drift of 120 m. If he crosses the river taking shortest route, he takes 12.5 min, find the velocity of the boat with v sm fJ respect to water. Drift: x = (u + vcase)1 =:> x = -(u-+-v.c-o's-8-)d- Sol. I, = 10 = '1. =:> d = 10 v. x = ut, =:> 12u = u x lO V eV sm Ie x =:> u = 12 m/min 10 v HI d '(' 1/ 12.5=--~ id \" ,J----'-}t 12 = 12.5 = -../v2 _ u2. =:> 1\"sinO 0J2-122 v =20m/min AI Rain-Man Problems I«I<-----~.I Formula to he applied: vr/m = Vr - tin/; where vr/m is velocity Fens'\" II vrof rain W.Lt. man, is the velocity of rain (w.r.t. ground), and Fig. 5.16 7:\\/1 is the velocity of man (w.r.t. ground). Case (i): Rain isfalling vertically downwards with a velocity V/, To cross the river in shortest time: d du and a man is. running horizontally with a velocity Vm as shown ., in Fig. 5.19. What is the relative velocity afrain w.r.t. man? sin 0 = I =>0 = 90' , tmin = -, x v v ux 1_'v,. x can be calculated from Fig. 5.16. Also: tan a = .- = - vd Fig. 5.19 du =:> X = - v Net velocity of boat (pute = 90'): Vb = ui + VI, Magnitude: IVbl = V1I2 + u2 To cross the river by shortest path: For this x = 0 +=:> U 11 cosO = 0 (Fig. 5.17)
5.6 Physics for IIT-JEE: Mechanics I Magnitude: vr/m = ./v;n + v? and dsirection: tana =!:!!: ~ile man is stationary anci The man-starts ,;i'()ving forward. VI\" the rain is falling at his The relative velocity of rain w.r.t. -, man shifts towards vertical direc- \".. 11m , back to an angle 1> with tion. ,, ~ ~, a the verticaL ~v,. I~:v\",~~I Vd\", I ~I',,,, 1/11 I\\.\\. Fig. 5.20 Velocity diagram Case (ii): If rain is alreacly j(llling (It some angle 0 wilh hori- zontal, then. with what velocity the man should travel so that the rain. appears falling vertically downwards to him? ~ •I~ ~v,. _ __ v'\" Fig. 5.21 man--~'As--thelnmlTl'i111cr i;;:~' 1{-lhe i nCI:e~l.~csi1Ts speed crease his speed, then at further more, then the rain ap- v vmHere,vlII = villi, VI\" = Vr sinei - VI\" cose] a particular value the rain pears to be falling from the for- Now, r/III = VI\" - = (vrsine - vm)l- vrcose] appears to be falling ver- ward direction: Now for rain to appear falling vertically, the horizontal com- tic ally. vponent of r/ m should be zero, Le., r-----\"-~---~- ---~- .= . = -V-II-/ = \"''''Gt t tV,,,, ~v. e eVr S1I1 (-) - sm v,. VII/ 0 ;:;:} and V\"/III VI\" cos 2 Velocity 1 _ vI/!. Diagram V,~ -~ Vm o Fig. 5.23 ~ • In some situations you have to presume the velocities. For \"<---cc_~\"_ _... 1',. cxample, if the problem says that a man can walk at a maximum of g kmph and if it asks you to find the velocity -1'\", on a train, then you have to assume that. the velocily of Fig. 5.22 the man with respect to the surface he is on (in this case the train is 8 kmph). Similarly, the velocity of a bullet is We can illustrate the whole situation by the diagrams given always measured respect to the gun. If the gun is mounted in Fig. 5.23: on a truck, the bullet will have a differcnt velocity. It is quite interesting to notice the steady rainfall sitting in a • Similarly if [ throw a ball, the velocity of the ball is nor- vehicle such as bus, car, etc, While moving on a straight track mally given with respect to the man throwing the ball. If the direction of rainfall changes when the vehicle changes its the man is having some velocity, then you need to add this velocity. That means, the velocity of the rain you observe is the up when you want to rind out the velocity of the ball with velocity of the rain relative to you. Therefore, your observed respect to the ground. velocity of rainfall (both magnitUde and direction of velocity of rninfall) is the velocity of the rain with respect to the vehicle • Therefore, depending on the circumstances, the velocity will have to be computed with respect to the ground or the (you) VII!' It cannot be equal to the velocity when the vehicle object on which the body is moving. becomes stationary 1),1/ = O. If you measure lhe velocily of the • You can consider any of the hodies as observers and any rainfall while the vehicle is stationary, t.hat gives actual velocity of the bodies as observed. However. note that at a mini- of rainfalL mum you will have 3 bodies, 2 observers on different ref- erence frames and one body, which is heing observed. You Problem Solving Tips for Relative Velocity can change onc of the observers as the observed and vice versa and use the relat.ive velocity/relative displacement • If the velocity is mentioned without specifying the frame, equations, You will always get. the same answer. assume it is with respect to the ground. • In many cases, a body travels on water or in air. Depending on the context you will have to figure out whether the velocity is with respect to the water!air or with respect to the ground.
IConcept Application EJ(erds~---- Motion in Two Dimensions 5.7 1. State 'Ii'uc or False: 11. A man can swim in still water with a velocity 2.5 km/h, He wants to cross the river flowing at a speed of 1.5 km/h. u. A ball is dropped ,from the window of a train moving He always swims at an angle 120\" with downstream. The width of river is SOO ITI. Find along horizontal rails. The path that ball travels to reach B the ground is parabola. River flow ---+ h. In the above part, the ball appears to move along circu- \\ 120\" lar path for the man in train who dropped the ball, Fig. 5.24 c. In order to cross the river in shortest time a boat needs a. the time in which he will cross the 'river. b. where will he arrive on the opposite bank? to move in the perpendicular direction to the flow of 12. A car with a vertical wind shield moves in a rain storm at the speed of 40 km/h. The rain drops fall vertically with a river. terminal speed of 20 m/s. I<ind the angle with the vertical at which the rain drops strike the wind shield. d. Similar to above, if the river has to be crossed through the shortest path, the boat must move at an angle () < 9(Y' inclined downwards to the flow. e. A man has to hold the umbrella at an angle (VIII)o= tan~l with vertical (where symbols have v, their usual meaning) while moving in rain falling ver- tically in an attempt to save himself from the rain. f. A person aiming to reach exactly opposite point on the bank of a stream is swimming with a speed of 05 m/s at an angle of J20° with the direction of flow of water. MOTION WITH UNIFORM ACCELERATION IN A PLANE The speed of water in the stream is 0,25 m/s. 2. A train 200 m long is moving with a velocity of 72 km/h. Let a particle move in x-y plane with a constant acceleration Find the time taken by t.he train to cross the I km long bridge, a= a.J + a.,.]. 3. Two cars A and B are moving on the straight parallel paths Magnitucl~ of acceleration: (( = I~T'~~~~;} with speeds 36 km/h and 72 km/h, respectively, starting from the same point in the same, direction. After 20 min Let the particle is at point A at t = t\\, where its velocity is hmv much behind is car A from car B? it = urf+Uyl 4. Two trains; 110 m and 90 m long, respectively, are running in opposite directions with velocities 36 km/h and 54 km/h. At t = t2, it reaches B where its veloCity becomes Find the time taken hy the two trains to completely cross each other. v=vxi+vy). 5. Two persons A and 13 arc walking with a speeds 6 km/h Now apply, v= u+ at and 10 km/h, respectively, in the same direction. Find the separation bctween A and B after 3 h. => v) + Vyi = u.J + u y} + (axi + (ly))1 + +;=} Vx = Ux a.,J and Vy :::::~ u y ayt y \"\"\"} A (XI, )'1) \" 6. Two parallel roads run east-west. Car A moves east while 8 (Xl. Y2) car B moves west with speeds 54 km/h and 90 km/h, re- spectively. Find relative velocity of 13 w.r.t. A. and relative \"I, I, -> velocity of ground w.r.t. 13, -> ~r, \"2 7. The relative velocity of A \\V.r.t. B is 6 ms--! due north. o\"\"--------+X What will be the relative velocity of B w.r.t. A? Fig. 5.25 8. A rilan A is going duc east with a speed of 5 m/s. Another man B is going due north with a speed of 7 m/s. Find Here we see that the velocity along x~axis is related to ac- celeration along x-axis only and the velocity along .y-axis to is a. the relative velocity of B w.r.t. A. related to acceleration along y-axis only. h. the relative velocity of A w,r.L 13, It means that acceleration in some direct.ion can affect the 9. A car is going due south with a speed of 15 m/s and a bus velocity in that direction only, not in a perpendicular direction. is going clue north-west with a speed of 10.)2 m/s. What velocity of car, a passenger in bus will record? Displacement 10. A moving sidev·mlk in an airport terminal building moves fZ-f, = \", , t =} r2 = r, + C;V)t at a speed of 1.0 m/s and is 35,0 m long, If a woman steps on at one end and walks at 1.5 m/s relative to the moving r, + u+u,+zz.t sidewalk then find the time that she requires to reach the opposit.e end when (i) she walks in the same direction ---t the sidc\\valk is moving; and (ii) when she walks in the 2 opposite dircctio.n.
5.8 Physics for lIT-JEE: Mechanics I r, + • + 1.. 2 To study the motion of a projectile, we make the following assumptions: ut '2at 1. No frict.ional resistance of air. =} X2 , + , = , + Yt,i + +. uyt,j +I2ax{~,i- + I 2' 2. The effect due to rotation of earlh and curvature of earth is i Ylj xli lay! j ~l:,Ji negligible. 3. g remains constant at all points of the motion of projectile. It + +I does not change with height. From here, we get, X2 = .:1:1 uxt 2o.v{2 and Projectile given angnlar projection: Y = YI +uyt + I, Let a particle is projected at t :::: 0 from a poinCA (origin) with \"2(lyr velocity u at an angle 0 with horizontal. Displacement along x-axis: 5,'.,. = X2. - Xl Displacement along y-axis:-S), =)'2 - Yl = uyt + lIay! 2 Note: Equation ojpath can be obtained by eliminating t from both.tlle above equations. . y ax \"'= 0 ga./~ ~- Ji'rom izere, lve note that the displacement-in (l direction is lIy ' - A\"P(x. y) 11)\" D(Ri2, H) •\"0 covered by velocity andqcceleratioTl in that direction only. ,, ,,,u : t =1 )Ix : Finally, equations of motion for two dimensional motion un- der uniform acceleration arc as follows: (i) ,,,,:y Vx = U x +axt +Vy = Hy uyt (ii) B(R,O) T IX {=o N ,, M x '2Sx I2 I:,1(0.0) 11.,_ x lI!/? = uxt + (iii) axf I, tV2 Sy = uy. t + 2-a'yr (iv) (v) v; = u.~ + 2a,.Sy Fig. 5.27 v.~ = u;. + 2aySy (vi) Ux = u cos 0 -> horizontal component of velocity, eUy = u sin -> vertical component of velocity Projectile Motion The object covers horizontal displacement due to horizontal A projectile is the name given to a body t.hrown with some veloc- velocity and vertical displacement due to vertical velocity. ity at an angle with horizontal and then allowed to move under the aclion of gravity alone, without any external force being Lel the particle reaches at poi I1t P(x, y) at any time' = t. appliedtoiL Velocity of projectile at this point becomes v, say at an angle f3 The path followed hy a projectile is known as its trajectory. with horizontal. Then Vx = vcos/3 and Vy = vsinfJ. +For motion from A to P: apply Vx = Ux ax', we get vx = Ux because ax = O. It means horizontal component of velocity remains constant. It is because there is no force on the object in horizontal direction. So, finally we get: (i) ev cos f3 = 11 cos +Now apply Vy = u y ayt, we get Vy = u y - gf • => Ii sin f3 = u sin 0 - gt (ii) Fig. 5.26 It means vertical component of velocity decreases as the pro- Example: jectile goes up. It is because force of gravity is downward. A stone thrown at some angle At highest point D. velocity is purely horizontal. f3 = 0 stone\"\"\"')' projectile curved path ......,). Tr~~ectory After crossing the highest point, again vertical component A projectile moves under the effect of two velocities: starts increasing in downward direction. 1. A uniform velocity along horizontal djrection, which would not change provided that there is no air resistance. Path of Projectile 2.' A uniformly changing velocity (either increase or decrease) (Equation of trajectory) in the vertical direction due to gravity (the motion is taking place along horizontal) as well as in vertical direction. From A to P: apply Sx = uxt + I, where, 2Clxr, 5\\ = x - 0 = x =? x = U cos Of
Motion in Two Dimensions 5.9 x If = u s_m(e\" -S-in-O) -- -Ig (U-S-i-n-O)2 g 2g t=--- u2 sin 2 0 1 u2 sin2 () u cosO =? H = --------- - -- ------ Now apply Sy = uJ + 1 where Sy =Y- () = Y g 2g 'lay I', v; u;Alternatively: From A lo D, = + 2ays)' y = u sin e( __X_) _~ g (~-)' ucosO 2 cose gx2 (iii) 0' = (u sinO)' - 2gH =} y = x tanG - - - - _- If = -u--2--s-i-n--2---0- 2u 2 cm;2 () 2g It is known as equation of trajectory. It is an equation (~r , (vi) parabola, Hence path (~r a projectile is paraholic. (vii) u~. Equation (iii) can also be written as Also H = ~ x'y =xtan() - -,---,- = x tan 0 - x 2 sin e 2g 2u\" cos- () ( CO:OSinO) cosO • Time of flight and the maximum height attained depend g only upon vertical components of velocities, (I -= x tan e - -x-2-ta-n-e- ~) • H......:;.. it is the maximum vertical height attained by the R object above the point of projectiop. =} y = x tana This is another from of equat.ion (iii) which has important ap- Horizontal Range: (R) plications. It is the horizontal displacement covered between point of pro- jection and point of hiUing. Time of Flight (1) From A to B: S'x = UJ + I t 2 :::::} R = If cosOT \"2Gx It is the total time for which the object remains in flight (air)_ Time of flight consists of two parts: =} R = 2u sinD = 2-u-.\\-.u~y (viii) [{cosO---- (ix) 1. Time from A to D -+ known as time of ascent, let it be II gg 2. Time from D to B --? known as time of descent, let it he tz u2 sin 20 Aloo R = - - - - g It is found that time o1'asccnt= lime of descent::::} Alternatively: Put x = R, y = 0 in equation of trajectory to get R =u-2-s-in- -20_ +First method: from A to D, Vy = U y tty! () = u sin e - gIl g u sinO Maximum Horizontal Range =} I, = - - - We know that R = uZ sin2 (} _R will be maximum if sin 2e = I - g ' g u sinO =} Tj2=-- g 2u sin 0 (iv) =} T=--- (v) g +Second method: from A to B, Sy = u y ! 1? Fig, 5,28 '20yr II=\"2=? 20 IT =? 0 = 4IT ' then R\",,,, = u2 I (x) =? 0 = u sinOI' - 'lgT' =} gT' = 2u sinOI' =} T = -2u -sin-O Two angles of projection for the same horizontal range, pro- vided the two aI1SWCrS arc complementary. g , 2u)' also, 7 = --'- g Maximum Height: (H) u From A to D: Sr = \"-r.t + 1 ~u 'lar-12 Fig, 5,29 _I R, = -u'-si-n -20, R2 = -u-' -s-in-[-2-(-9-0----e-)-i ' gg =? If = u sm 01, - 'l gIl
5.10 Physics for HT-JEE: Mechanics I 1/' sin(l80 - 20) 1/' sin2e u2 sill 0 = ~-- = ------ 0=} R, = R, 2. The maximum height reached = - - - gg 2g It So the horizontal range is the same whether the angle of (30)' x (sin 30'), 900 c---cc-----c = 11.25 m eprojection is Ii or 90 - 0, tlleir sum is () + 90 ~ = 90\"_ - '--'--- 2xIOx4 It It means that the horizontal range will he the same for those -2g two angles of projection whose Sllm is 90\". - g - -_ 1/ 2 sin 20 (30)' x sin 2(30') e Also, if the angle of projection with the horizontal is 3. Honzontal range = = _. 10 e, e.90 - then with the vertical it is Hence the range will = 9~0-0-x-~~-i = 45J3 m be the same whether the angle of projection f) is with hor- 20 4. The time for which the ball is in air is same as its time of izontal or with vertical. Note: When the range is maximum, the height H reached 2/1 sin Ii 2 x 30 x sin 30\" _ by the projectile is flight = g = 10 --=.1s_ =Hi: _u2_-s_-inc2_(_J A footballcr kid<s the football to check 2g his stamina with velocity 60,fi ms- l at an aJ!gle of 450 , Find i.e., ifaperson-call throw aproiectile to'a maximum distance following after 3 s. Rm:\\x, the maximum'ilelght to which it will i'ise,is (Rmax/4) 1. Velocity of football. and the angle ofpr~iectioll is 450 • 2, Angle made by the velocity with the horizontal, 3, Horizontal and the vertical displacement. atVelocity of Projection ony Time Sol, We have earlier obtained equations 0) and Oi). These are 1. Given /I = 60/2 ms~l, 0 = 45', v cos f? = u cos 0 and v sin f3 = u sin 0 - gt. We know that the horizontal component of velocity re- On squaring and adding these equations: mains the same during projectile and only vertical component changes (Fig. 5.3 I). So we need to tlno the vertical compo- v' = (1/ cos 0)' + (1/ sin IJ _ gl)' ncnt of velocity after 3 s. Hence, we will have both Vx and = j ·\"1- gt 2ugf sin e:;;::;} V vy. We can use v = II~.~ ·;-~i to find out. velocity aftcr 3 s. ----~---- -----~----- 1/ 2 2 -~ Dividing them, we get, tan p = u sin () - (xi) If cos {-} v 2 = u 2 - 2gh ::::} v = j~7r~_ 2gh y Note: We cal! apply the following technique to find the \" \\'\" speed. v Ii \" !+-.\\\" ~I Fig. 5.31 Fig. 5.30 So. U, = /I cos 45\" = 60·/2 x J2I = 60 IllS-I: A batsman kicks a ball at an angle of 300 u r = 1/ sin 4Y = 60 h x ..I/2 = 60 ms- I with an speed of 30 mis, Assume tbat the ball travels in a vertical plane, calculate The value of the vertical component of velocity aftcr L tbe time at which the hall reaches the highest point. 30 s is 2. the maximum height reached. +Vy = I.{y Oyt = u y - gt = 60 - 10 x 3 = 30 IllS--I, 3, the horizontal range of the ball. Vx = 60 1115..... 1 4. the time for which the ball is in the air, JII;v = + \"; = ../3600 + 900 =J4500 = 30../5 111,-1 Sol. Here 8 = 30\", 1/ = 30 m/s 1. The time taken by the ball to reach the highest point is ha1f of 2. The angle made by the direction of t.he movement of a projec- the total time of flight. As time of ascending and descending is the same for a projectile without: air resistance, then the tile or its velocity with horizontal at any time duringjollrney time to reach the highest point is is given by til = ~: = I\",i~ = 30 x sin 30\" 1_5 s tanG =. -v--\"-'-- tan Ii = 30 0=} 0 = tan-I (-21 ) 2 g 10 Vol 60
Motion in Two Dimensions 5.11 3. As horizontal component of velocity remains constant. So S\". = u.vt + 1? to calculate the horizontal distance covered we can Llse the -2Q\"t- relation-Distance:::;: speed x time =? x = vxt Y = -~g (S)2 But vertical component keeps on changing side by side as gravity V = -- ---g_. -x-J keeps on acting in downward direction. So we have to use the - 2u2 . = u.rt ,+I2-Q·?rt- y = -kx2• where k = ~2x2 relatlOn.y So X = 60 x 3 = 180 m and y = 60 x 3 - .2I x g x ? (3)- y = 180 - 45 = 135 m Hence the horizontal displacement is 180 In and vertical dis- placement is 135 tn. Projectile Fired from Some Height 1 + - - - R - - - I > l ( R , h) Sx = R,Sy = -h,ux = ucose,u y = lI-sin8,Qx = O,a}, =-g Fig. 5.34 8 Hence the path of projectile is parabolic. Time of flight: (T) From 0 , u,..t + I to A:.I., = -2a·,.I' 1<I!l'If----R--~ (f(, Ii) => -h = () x I' \" I 01'2 - Fig. 5.32 2\" \"* T = '1r-I-;l;l Horizontal Range: (R) S, = H,I + 'I2\",1-? => 1? Vg=> R = uT => R = u (2h R=ucoset,s.v=uv. t+-G2·rt- Velocity of any Time: Vx = +Ux ax! => - I U sm. 0-t '2Ig?r- = - I We can use these equations to calculate range and time taken. Projectile Given Horizontal Projection -gt Initially: velocity is purely hobzontal and vertically it is zero, => {J=tmf'(-gl/u) (i.) Horizontal velocity remains constant; (ii) vertical velocity increases in downward direction due to gravity. u Particle covers horizontal displacement due to horizontal ve- A ball is thrown horizontally from the locity and vertical displacement due to vertical velocity. top of a tower with velocity of 30 ms-t. During its motion, at a particular point, horizontal and vertical velocities become equal. Determine the time elapsed to reach this point. Sol. Given v = 30 ms·\"! By finding out horizontal and vertical velocities separately and by equating them we can find out the yequired time. Also, remember that the horizontal velocity remains the same and t.he vertical velocity keeps changing during motion. Fig. 5.33 So, horizontnl velocity H,. = V, = 30 ms\" (i) Path of projectile: Vertical velocity v, = u, + \",I = 0 + gl = +10 I Iii) U x = u, Ii)' = 0, ax = 0, a y = - g Comparing equations (i) and (ii); I() t = 30 :::::} t = 3 s. +1 , Projectile from a Moving Body Sy = u.\\'{ '2({xt- Let liS consider a trolley which is moving on a horizontal plane with a constant velocity v. A boy throws a ball from a 1110V- ::::::? x = ut => x t = .- U
5.14 Physics for IIT-JEE: Mechanics I NON-UNIFORM CIRCULAR MOTION v = WI' sin f) Angular acceleration (0:-): Tt is defined as the time rate of change . cl(o.,., and its direction is perpendicular to the plane containing rand titof angular velocity, a = - . Unit of a: rad/s£... w. v w r.By right hand screw rule, we can wyite; = x }-_+_'-_ _.,---.... r ,A rigid body is spinning with an angular x Particle is speeding up velocity of 4 rad/s about an axis parallel to 3) -k passing through the poiut 1+3)-k. Find the velocity of the particle at the point 41-2)+k. 3}-k (1,3, I) 0 \", ,;,': ,I I Fig. 5.44 --)\", \\ I (II~ Sol. Let n. be the unit vector in the direction of 3}-ii. Then, x Particle is slowing down n= 3) --k -(3V)-1--0k-) Fig. 5.46 Relation between linear acceleration and angular accelera~ )32 + (__ 1)2 .'. angular velocity of the particle is tion: . = ,on. = 4 .i -.k) radls (tiOl) /.v = (Dr :::;:} dv ___ :::::.} a( = ar io \",,(3 tit dt '110 ' This acceleration is along tangent, so it is known as tangential The position vector of the point with reference to point acceleration. ci+3}-k)is Centripetal acceleration: There is one more type of accelera- r = (47 - 2) + 1<)-(7 + 3)- k)= 37 - 5) + 2k m tion in any kind of circular motion, which is known as centripetal Hence, linear velocity acceleration. This is given by ~ ~~ 4 •• •• • w 2 ,. ,,2 v =W x r = \",,(3j -k) x (3i -5j+2k) Cl c = =- , '110 r = 4. + 3'.i + . mls -V-1--0.(i 9k) Two particles A and B are moving as shown in Fig. 5.45. At this moment of time, lind the angular Fig. 5.47 This acceleration is direct.ed always towards the centre of the speed of A relative to B. circle. 1t is also known as normal acceleration or radial accel- eration. Sol. We know that, Net acceleration: In Fig. 5.47 a is the net acceleration. efV ABh = V Ay - V By = V A sin A - V Bsin 0 B a -+ a'lMagnitude of net acceleration: = ;a;~ and ,(J) = [V AB],' Direction of net acceleration: - A-B- ' tan f) = ~:::;} e = tan~l (~~,\") = VASineA-VBSinO B ].111 clockW.Ise .dIre.ctIOn. ac ac [ r Analysis of Uniform Circular Motion -----*I~-I VA sin 8,4 In the uniform circular motion, the particle moves in a circular path with constant speed. AB . Let us choose the centre of the circular path as the origin of Fig. 5.45 the reference frame. Point P is an arbitrary point on the path r yJ,whose position vector is = x 7+ where r, the radius of the circular path is related to x and y by follqwing equations: x = r cos 0 and y = r sin 0
Motion in Two Dimensions 5.15 Vp Vp ,. - --x ~ ,P r e 0 0', (a) (b) Fig. 5.48 v +Also, x 2 y2 = r2 ar r = rcosOI +rsinB) o Now, the velocity of particle P is given as (e) (d) \" de dxo dy,. d(rcosO), d(rsinB) }1;g.5.49 v = dl = 'dt' + dt j = dt ' + dt To evaluate the acceleration we translate th~ vectors v p and vQ v e - +dO , dO, to a common origin shown in Fig. 5.49(c). The change in velocity = -r sin i r cos (-) - } dt dt Ll v is also shown. As the time interval !::J.t is made smaller,-points P and Q are found closer together. We can see that angle is But de = (IJ = angular velocity [constant, for uniform circular also the angle between the two velocity vectors shown in figure. dt motion] When becomes so small that v p and v Q arc almost parallel and their difference I'\" v is almost perpendicular to both of them. In vThus = wr(- sin e1+ cos e)) the limit when D.t tends to 0, I'\"v is perpendicular to v. Hence Now vi' = wr'(- esin 0 cos + sin 0 cos B) = v0, i.e., is per- pendicular to r the instantaneous acceleration which is in the same direction as v ! eNow, is along the tangent Ii!I = wr sin' g + cos' = WI' I'\"v, is directed radially towards the centre of the circular path. v = WI' Therefore) a particle moving with constant speed around a circle aNow, acceleration is given as is always accelerated toward the centre [Fig. 5.49(d)]. From Fig. 5.49(d), the acceleration can be easily evaluated. \" e= l ~i ! = (or (- cosOd--ei-,s'inO ddet i') The magnitude of change in velocity !::J. v can be given as elt . (J £It I'\"v = v I'\"e (2) = > Z i = -w'r (cose i + sine J) =[are length radius x angle subtended by arc] a = -(Ji x r = w~(-r) Here Iv pi = IVQ I, and from equation (2), centripetal acceler- The above equation shows that a is directed in the opposite direction of r. Thus, a is always directed towards the centre. ation is a = v I'\"B w or a = v' [as v = Tu'] (3) - --=v D.t I' Magnitude of Zi: a = w2r/cos2 e + sin2 e = (J)2r whose direction is always towards the centre of the circle. .:::} a = (fir CENTRIPETAL ACCELERATION General Methad to find Acceleration in Circular Motion In uniform circular motion the velocity is constantly changing, v w r.We know that = x Differentiating it w.r.t. time t, we will there must be only the centripetal or normal acceleration. The tangential acceleration here is zero as the magnitude of veloc- get the net acceleration. ity remains constant. Let us consider an instant a particle is at position P shown in Fig. 5.49(a), in uniform circular motion a\" = d-i! = -d( w\" x \"1') = \" X dr + -dw x \" it is moving with velocity vl\" In the short interval of time !::J.t, dt dt the particle moves by an angular displacement !::J.() to another W - I' point Q. During the timc interval 1',,1, the velocity changes by an dt elt amount !::J.v = vQ - up. The average acceleration for this small duration is ( .,' dr =1\"1,ddwt _) dt =a a= (I) =} a=ac+at vwhere, Gc = (l) X is called centripetal acceleration because it is directed towards the centre of the circle and its magnitude is a,. = wv = uir = v'! r. It is responsible for changing the di- at a rrection of motion and = x is called tangential accelera- tion because its direction is along the tangent. Its magnitude is a, = ar and it is responsible for changing the speed of the body.
5.16 Physics for IIT-JEE: Mechanics I Case I: When at = 0, a = Q c = (fir; in this case motion is rlv d called uniform circular motion. 2. The tangential acceleration: at = -dl = -(krt) = kr rlt Case II:. When at '# 0, Q c = w2r, in this case motion is called 3. The resultant acceleration non-uniform circular motion. Kinematical equations in circular motion for constant a: Clrcs = ;;;i~~~i = jek2-;ti)i + (kr)2 = kr Jk2 f4 + i +1. W2 = (/)j at 4. ~ a. = k 2rr2 = kt 2 -- 2. e = Wj t + 2I at-, From Fig, 5,50: tan a =-'- a, kr 3. ())~ = (OT + 2aO =} a = tan-I (kl'l Tangential acceleration of a particle RELATIVE ANGULAR VELOCITY moving in a cirenlar path of radius 5 em is 2 m/s'. Angular Angular velocity is deHncd with respect to the point from which velocity of the particle increases from 10 radls to 20 radls the position vector of the moving particle is drawn. Let a particle during some time. Find v.P is moving with velocity Its position vector w,r.t a point A is 1. this duration of time, and 2. the number of revolutions completed during this time. r, rsay. Take some fixed line as a reference line and let makes Sol. r = 5 em =0,05 m, al =2 mis', 'Ut = J0 radls, an angle 0 with this line as shown in Fig. 5.51 (a). Then angular velocity of point P w,r.t. an observer on point A can be defined 0)2 = 20 radls dO Vi. vsina +1. W, = 'Ot al =} 20 = 10 + 40t =} t= 0,25 s r , .as (u = -dt. It is also .g. iven as co = ~..- = Angular velocity of a given particle can be different about different. points. For example in the Fig. 5.51(b), a particle P is moving on the circumference of a circle. 2. e = 'Ut t + _I at' = !O x ,I. + -I x 40 x (1- )2 = -15 rad 2 42 44 \\' ('os (X ,& -\"15~ = 0,6 -, Number of rcvo1utlOns = - = I' 27l 4 x 27l A particle moves on a circle of ra- (» dius r with centripetal acceleration as function of time as A () ReCline a, = k'r (2, where k is a positive constant. Find the follow- L.L-_ _ _ _ _ ing quantities as function of time at an instant: (al (\\1) Fig. 5.51 1. the speed of the particle, Here the angular velocity of the particle w,r,t, 0 and A will 2. the tangential acceleration. of the particle, 3. the resultant acceleration, and be different and given ,~s 4. angle made by the resultant with tangential direction. da. df3 (OPjO = . - and {j)PjA = - dt dt Sol. Definition 1. In the problem it is given that ac = k2r t 2 Relative angular velocity of a particle B with respect to another moving particle A is the angular velocity of the position vector of B with respect to A, This means it is the rate 'at which position vector of B with respect to A rotates at that insth.nt. Relative velocity of B w.r.t. A perpendicular to line AB ,, COB/II = \" Scpcration between A and B ,, 0, ,, --- -- Fig. 5.50 Since ac = -v-2 ::::} -v2 = k2 rt 2 r r ::::} v~ = k2 r2 t 2 taking square root on both sides. We get, l! = k r t Fig. 5.52
where, (VB/I1)J.\" = vi sin f}2 - VJ sin OJ, so we get Motion in Two Dimensions 5.17 WB/A = V2 sin fh ~ Vj sin 01 that circular arc which fits at that particular point on thc curve as shown in Fig. 5.55. r ~., , ' ----il Important Points 11----, ,I (/i· \\ I RI C w o particles are movmg on the same circle or dIfferent (a) (b) coplanar concentric circles in same direction with differ- Fig. 5.55 ent uniform angular speed (VA and (VB, respectively, then '*= = -.V2 v2 the angular velocity of B relative to A (for both the Figs We know that, a\" -R R ac This is the expression 5.53(a) and (b) below) for an observer at the centcr will for radius of curvature, dO .-----11 Concept AppLica'tjon Exercisl:l53 [1----, be (OB7\" = (l)ll - (VA = ...\"..- dt B , ~\"\"'\" 1. Calculate the angular velocity of (a) (b) a. second's hand of a clock, b. minute's hand of a clock, and Fig. 5.53 c. hour hand of a clock. So the time taken by one to complete one revolution 2. A stone tied to the cnd of 2 m long'string is whirled in a arou nd () W.r.t. the other is horizontal circle with constant speed. If the stone makes 10 revolutions in 20 s, calculate the magnitude and the -~,- = ~ direction of acceleration. (\"2 - \"'I 1'1 - 3. A motor car is travelling at 30 m/s on a circular road of ?:':. 7, 'l' = where \"'I = radius 500 m. It starts increasing its speed at the rate of = 2][ and 21118--- 2 . What is its acceleration at this moment? ~ W~I 1'1 4. A body is moving in a circle of radius 100 em with a time period of 2 s. Find the accclerafion. 2rr =(v'} - - 5. Calculate lhe centripetal acceleration of the moon moving \" 1'2 around the earth in a circular orbit in terms of its time period T and radius of the orbit R. If two particles are moving on two different concentric circles with different velocities then angular velocity of B relative to A as observed by A will depend on their positions and velocities. Consider the case when A and 15 arc closest to each other moving in same direction as shown in Fig. 5.54. In this situation v,,1Vrc! = IUB - = VB - VA and = rll - rA so, (ORA = (vrelh = -VB.----VA- here (vreJ) I.:::::: RelatI.ve vc- .- ._\"\"-\"._- rrc1 rn - r.1 loeity perpendicular to position vector +_ \"B ~ ,, A bird is fiying dne east witb a velocity of 4 ms-I. The wind starts to blow with a velocity'of 3 ms-I VIJ <--,-..... due north. What is tbe magnitude of relative velocity of bird w.r.t. wind? Find out its direction also. ..,,, 'vA , A t\", ,,rB ,,,, Sol. Method 1: Whenever two bodies are moving at right angle • to each other, then their relative velocity is obtained by taking rA square root of the sum of squares of respective velocit.ies. • N .,, 0 \".w-----+;:--':2·b~r'-_ E . ~ ---~ Fig. 5.54 l_~~~ _ _ _ _ _,_ _ _~___ _ _ _ _ _ _ _ -1 Radius oj Curvature s Any curved path can be aSSllme to be made of infinite circular Fig. 5.56 arcs. Radius of curvature at a point is defined as the radius of
5.18 Physics for IIT-JEE: Mechanics I Here velocity of bird = Vb =4 ms- 1 towards east. Distance covered 10 Velocity of wind = Vw = 3 ms- 1 towards north. .',thctimctaken= Relative velocity -- 20 -- O.Ss. N ~ Rain is falling vertically with a speed of 12 ms- I. A cyclist is moving east to west with a speed of W------~----~---.E 12./3ms-I. In order to protect himself from rain at what angle he should hold his umbrella? s Sol. Method 1: In the case of rain falling vertically with a veloc- Fig. 5.57 eity of tan = Vr,! and a person (cyclist, bikers, etc.) is moving Velocity of bird w.r.!. wind = Vbw =? Vbr Vb,w = Vb - Vw = Vb + (-vw) horizontally with a speed Vm , the person can protect himself Jvt v;,So, Vbw = + = ,/(3)2 + (4)2 = v'25 = 5 ms- 1• from rain by keeping umbrella in the direction of relative veloc- ity of rain W.r.t. person vrp. If e is the angle that Vrp makes with vertical or rain. . -.'. velocity of rain w.r.L cyclist . v rp = Vr - vp To find the direction we need to find the angle fJ, the angle between the relative velocity and the velocity of bird. tan fJ = 3 sin 90\" = 3 -> 4 4 + 3cos90\" IV VI' E [.'. e = 90\", angle between Vb and vw .] Method 2: Vb,w = Vb - Vw N ~v,. M Fig. 5.59 s (v,')Here,vr =12ms-1 andv,,=12yh3,tan8= - Fig. 5.58 11, = Vb + (- v;o) = 41 + (- 3}) (m/s) = 41 - 3 J (m/s) ande = tan-! (./3) = 60° So the cyclist has to hold the umbrella at an angle 60' to the IVb.wl = ,/(4)2 + (3)2 = 5 m/s verticaL Method 2: lt p,,\"oo = -12./3 1 (m/s) (Fig. 5.60) Vrain = -12] (m/8); Vrain,person = Vrain - vpcrso!l = [(-12}) - (-12./31)] (m/s) = (l2~31- 12))m/s Hence, the direction of orientation of umbrella with vertical is Here, the direction of the relative velocity of the bird is tane = 12./3 =./3 => e= 60' n)ItanfJl=~ => fJ=tan- 1 12 y; Hence, the relative velocity of the bird with respect to wind is NOlih (~)5 mis and the direction is tan- 1 from east towards south. -> -> IIMp\"1 Assume two cars A and Beach 5 m long. Vr,b Vrain e Car A is travefling at 84 kmlh and overtakes another car B -> ,,, • - - --x which is traveling at low speed of 12 kmlh. Find out the time East taken for overtaking. VpCrsOll ,, Sol. To analyse the motion in case of overta)<ings we need rel- Fig. 5.60 ativc vclocity of object which overtakes w.r.!. the other object. Therefore, we need to find relative velocity of car A w.r.t. car B iii!ll!nlI!IJI An aeroplane pilot wishes to fly due west. which is A wind of 100 kmlh is blowing toward the south (Fig. 5.61). 84 - 12 = 72 km/h = 20 ms- 1 1. If the airspeed of the plane (its speed in still air) is 300 Total relative distance covered with this velocity = sum of kmlh, ill which direction should the pilot head? lengths of car A and car B = 5 + 5 = 10 m.
Motion in Two Dimensions 5.19 2. What is the speed of the plane over the ground? Illustrate . Displacement with a vector form. 2. TIme taken to cross the river = ' Velocity of boat W.r.t river Sol. Given, Velocity of boat w.r.1. river is used since it is the velocity Velocity of air with respect to ground \"AIG = 100 kmlhr with which the river is crossed. =} 8~0 = 200 s vVelocity of plane with respect to air PIA = 300 kmlhr So the boat will cross in 200 s. ->+---~---- N 3. Desired position on other side is A, bUldue to current of river VPIA boat is drifted to position B. To find out this drift we need -VPfG time taken in all to cross the river (200 s) and speed of current (2 ms- 1). E So the distance AB = Time taken x speed of current =200 x 2=400m -> Hence, the boat is drifted by 400 m away from position A. Vl'fG Fig. 5.61 1. In which direction should the motorhoat given in Example 5.5 head iu order to reach a point on the opposite hank 1. As the plane is to move towards west, due to air in 'south directly east from the starting point? The boat's speed direction, air will try to drift the plane in south direction. relative to the water remains 4 mls. Hence, the plane has to make an angle 8 towards north-west, south west direction, in 6rder to reach at point on west. 2. What is the velocity of the hoat relative to the earth? VPjA = VI'/G - VA/G and VPjA sine = VAG 3. How much time is required to cross the river? ~ A river flows dne south with a speed of2.0 mls. A.man steers a motorhoat across the river; his velocity Sol. As the boat has to reach exact opposite end to the point of relative to the water is 4 mls due east. The river is 800 ill estmi, has to start (velocity 4 m/s) at an angle aiming somewhat wide. upstream. Taking into count the pus~ven by the current. 1. What is his velocity (magnitude direction) relative to the earth? v,,, nVelocity of boatw.r.t river Vb\" = 0 A = 4 mls 2. How much time is required to cross the river? Velocity of river w.r.t earth = = 2 mls 3. How far south of his starting point will he reach the op~ Velocity of boat W.r.1. earth Vb, mls = OS =? posite bank? Sol. Velocity of river (i.e., spced of river w.r.1. earth) ii,., = 2 mls Velocity of boat w.r.t. 'river VbI' = 4 m/s Width of the river = 800 m ,,,, 4 m/s East e , , ,I I I I j 2 m/s ,, South -> C ---------S Fig. 5.63 V'1' 1. To find the direction of boat in which boat has to go we need to find angle 8. ,South I AB 2 Fromrt/'\"OBA,sin8= OA = 4 = 2 Fig. 5,62 8 = 30\" Velocity of t'JOat w.r.t. earth Vbe =? mis Hence, the motorboat has to head at 30\" north to east. According to the given statement the diagram will be as given 2. To find the velocity of boat w.r.1. earth, we can usc pythagor- in Fig. 5.62. ous theorem again, 1. When two vectors arc acting at an angle 0[90°, their resultant Using rl. /'\" 0 BA. we have can be obtained by pythagorous theorem, Vb2 e- -vb2 r -ve2 Vb, = /VE, + v~, = vl6 + 4 = v'2O = 4.6 mls 3. Time taken to cross the river Width of river To find direction, we have Velocity of boat W.r.t earth tanO = -Ure = 2 I .- = - =} Vbr 4 2
5.20 Physics for IIT-JEE: Mechanics I V!)(' is. taken in this because V/,e is along 0 B (the line of the angle between velocity of rain and velocity of rain W.1'.t. the 800 400 v'3 person. movement) = - - = - - - s Values of VI' and vrP canbe obtained by using simple trigono- 2v'3 3 metric relations. A person walks up a stationary escalator 15c1. Speed of rain drops W,Lt earth = ii, = in t1 second. If he remains stationary on the escalator, then it can take him up in tz second. If the length of the escalator CB ., => CB is L, then From rt '\" OCM, OC = si1160' OC=-- sin 60\" 1. Determine the speed of man with respect to the escalator. 20 40 40)3 -1 2. Determine the speed of the escalator. = --=-=---illS v'3/2)3 3 3. How much time would it take him to walk up the moving OB2. Speed of rain \\V.r.t. the person tir ? = escalator? From rt '\" OCM, CO'BB = cot 6,0.'' Sol. 1. As the escalator is stationary. so the distance covered in 11 => OB = CBcot60' = 20 = 20v'3 l - - ms second is L which is the length of the escalator. - L v'3 3 Speed of the man w,r.t the escalator Vme = ~ A boat heading due north crosses a wide Il river with a speed of 12 kmlh relative to the water. The water 2. When the man is stationary, by taking man as reference point in the river has a uniform speed of 5 kmlh due east relative to the eartb. Determine the velocity of the boat relative to an the distance covered by the escalator is L in time fz. observer standing on either bank and the direction of boat. Sol. Imagine a situation in your mind of a boat moving across L tile river. The boat is heading north, which means it wants to go Speed of escalator Vc = - straight, where the CUlTent pushes the boat along the direction of current, Le., east. We are given 12 +3. Speed of man W,f.t. the ground VIII = Ville Vc Velocity of boat relative to the river VI,,· = 10 km/h _ + [!.. +!..] _ +~ ~ -I L [11 12] Velocity of river = velocity of river relative to earth t1 t2 t] tz tjtz = v,., = 5 km/h => L=v\", [11~+J1.2 [ ~]t1 + t2 is the time taken by the man to walk up the moving escalator. A person standing on a road has to hold w-t-,==========:VI'£!: ======= =: N his umhrella at 600 with the vertical to keep the rain away. - - - - - - -.;r---~ He throws the umbrella and starts running at 20 m5-1• He finds that rain drops are falling on him vertically. Find the s speed of the rain drops with respect to 1. the road, and Fig. 5.65 2. the moving person. Velocity of the river can be taken as relative to the earth as the velocity measured has only earth as reference. We have to find Sol. Given 0 = 600 and velocity of person out the velocity of the boat relative to an observer standing on U\" p = A =-c--'oO 20 rn-s1. the bank. Since the observer is stationary with respect to earth, This velocity is the ,same as the velocity of person W.r.t. so the velocity of boat relative to observer will be same as the velocity of boat relative to earth, ground. First of all let's see how the diagram works out. Let us suppose due to push of current the boat gets drifted by It,.p = DB = velocity of rain w.r.t. the person. an angle 0 from the straight line path (see Fig, 5,65), c As seen from Fig. 5.65, velocities in situation from a right angled triangle and we havc the valucs of two sides. Therefore, Fig. 5.64 the third side can be calculated which represent.s the desired velocity. oRV,, = DC = velocity of rain W.Lt earth v,.p is along as 1. From pythagorous theorCll1, bl;E = /VEr-+ V;E\" a person has to hold umbrella at an angle with vertical which is = )(12)' + (5)' = ,,1144 + 25 = 13 km/h
2. To find out the direction, we need to find the angle 0 through Motion in Two Dimensions 5.21 which boat has deviated. ' :N tan {J = v/\"(' :::} 0 = tan·\"] (vr(') (2)= tan-'\\ NE 12 Vhr VI);' Hence, the boat is moving with a velocity 13 km/h in the f;)direction tan I ( east of north relative to earth. Ifthe boat or the preceding example trav- , els with the speed of 13 km/h relative to the river and is to 'S travel due north as shown in Ilig. 5.66, what should its angle of direction be? Fig. 5.67 Sol. Given VIn- = 10 km/h. The flags will flutter in the resultant direction of winds coming As the boat has to move due north, so it needs to start at an angle 0 move upward direction of the river. from two sides (i) along N. E direction (ii) in backward direction This is necessary because the boat during the motion will be due to forward motion. We can say the flag will flutter along vwc, drifted downwards due to the push of current. relative velocity of wind W.r.t. the car. 1\\\"C = Vw + (-vc) [as velocity of air due to car is opposite to motion of carl Here IVwl = 41.4 km/h, I - vpl = 40 km/h eAngle between i\\v and V!, is 13Y\\ Le., = 135\". Then direc- tion of the resultant is given by Fig. 5.66 tan fJ = 41.4 sin 135\" 40 + 41.4cos 135\" V!U' = velocity of boat w.r,t. earth is along hypotenuse = 13 km/h 41.4 sin (90\" + 45') Vre = velocity of river w.LL earth is along perpendicular 40 + 4J .4cos(90\" + 45\") = 5 km/h 1 ;nI 41.4X72 11/)(, = velocity of boat \\V.r.t. earth is along base =: ? 41.4 x VB,. = VUe - VI'I\" => lJ!lr = 13 km/hr 40+41.4 (-sin45\") = 40-41.4- ..Ifi VBc = VBI' + UrI' => Vrc = 5 km/hr , Using PythagorLls theorem we have, vEl' = VEe + v;e 10 vie = V~r - VI~{! => _ ri\"___--:; == => VVbe - Vhr Vre 40 - 10 3 Vbe = J(lW -- (5)2 = ,jT(;9-=- 25 = ,fi44 = 12 m/s G)=> fJ = tan- I Now to find the right direction of movement of boat so that it .'. angle W.r.t. east direction f3 = tan- 1 ~ - 45° 3 egoes straight in north direction, the angle needs to be obtained tant) =!!!!.- => e = tan:- l (.1!!.:::) = Jan-- I (~) Two roads intersect at right angle, one goes along x-axis another along y-axis. At any instant two Vhl' V'ne J 2 . cars A and B are moving along y and x directions, respec- tively meets at intersection. Draw the direction of motion Hence the boat has to start at an angle tan,l ( ]5 ) in order of 2 1. car A as seen from car B, and 2. car B as seen from car A. to move due north, A political party has to start its procession y in an area where wind is blowing at a speed of 41.4 kmih and V8 party flags on the vehicles arc fluttering along north-east l-~------_x direction. If the procession starts with a speed of 40 kmih towards north, find the direction of flags on the vehicles. Fig. 5.68 Sol. When the procession is stationary, the flags flutter along the -north-cast direction. It means wind is Howing along the north- east direction. The Hags will start fluttering along the direction of relative velocity of wind w.r.t.. procession. As soon as procession gets into motion (see Fig. 5.67). As th~ car will move along north with velocity vc, the air will How in opposite direction to that of car and will influence the direction of fluttering of car.
5.22 Physics for IIT-JEE: Mechanics I Direction of mation of A as seen from B Sol. Direction of motion of car A as seen from car B VA,S = VA - VB = VA + (-VB) Here car B will consider as rest and car A will be observed. y Location of B, considered as rest ~ Fig. 5.72 t--V_B ::..-.------. x , y This is the direction in which iiliWH1II11 Consider the situation giveu in Fig. 5.73, ,, car A will appear to move as two cars are .moving along road 1 and road 2. Draw the seen from car B, direction of the motion of -4 t,V-)A+(~--V>B}:' _____ -) 1. car B as seen from car A, and VAB= VA 2. car A as seen from car B. , :~ y , VB --~-~~-~::..-.-----.x 0\"-\"\"\" Situation of car B R~:~> at rest Fig. 5.69 \"Road 1 lIi€imClm Two roads one along y-axis and another ~/ \" -4 ealong a direction at angle with x-axis are as shown in VA / ' \", VB Fig. 5.70. Two cars A and B are moving along the roads. ,, ,A, , 8, e, ,B,, , , x Consider the sitnation of the diagram. Draw the direction of Fig. 5.73 1. car R as seen from car A, and 2. car A as seen from car B. Sol. 1. Direction of motion of B as seen from A Fig. 5.74 y y Direction of motion ofB , as seen from A ~ / Road VB / e-~~----~~~~---x car A ,,/ carB \" Fig. 5.70 Sol. 1. Direction of motion of B as seen from A. Fig. 5.74 +VB, A = VB - VA =} VB,A = VB (-VA) Vli.A = VB - VA =} VB + (-VA) y 2. Direction of motion of A as seen from B (Fig. 5.75) /'Dircction of y -) \" motion of B ._) Direction of motion of A VB I as seen from ~ VA as rest from B __ _Ae+_ _ _- e \\'/-\"point A. ,, , Car B considered ,,,i:dCJ:._ _ _ _ x Fig. 5.75 Location of /\"/T /\" / car B VA.B = VA - VB =} VA + (-VB) :~ ~ ! I A, considered I~ Consider two cities P and Q hetween which consistent bus service is available in both directions on rest Fig. 5.71 2. Direction of motion of A as seen from B. VA.B = VA - VB =} VA,8 = VA + (-VB)
Motion in Two Dimensions 5.23 every x minutes. A morning jogger is jogging towards Q and R = =2u-2=s=in:a:.:c:o.s a (iil g from P with a speed of 10 kmlh. Every 18 mins a hus crosses (i) this jogger in its own direction of motion and every 6 mins Dividing equations (i) and (ii), (ii) another bus crosses in opposite direction. What is the time period between two consecutive buses and also find the speed. Hmax li 2 51n2a g I of buses? R= x = -tana Sol. Analysis ofproblem: The equations of motion are applica- 2g 2u2 sinacosa 4 ble in any frame of reference. Attaching the frame of reference at jogger, The velocity of bus is now required w,r,t. jogger. Suppose R speed of buses s ~ v km/h Hmax = \"4tana. Distance between two buses on road: s = xv The relative Hence proved. • • •~10km/h u2 sin2 a' I' J Q 3. Hmax = 2g , and 2u sina T = ----~ g Fig. 5.76 .'. squaring equation (ii) and dividing it by equation (i), velocity of bus w,r,t. the jogger, for the buses moving from P to T2 4u2sin2 a 2g 8 g Q = (v - lO)km/h vx Hmax g2 X u2sin2 a So we have 18 = -v .--10 =} vx = ISv - 180 (i) gT 2 = SHmax +Similarly. for the buses moving from Q to P ~ (1! 10) km/h Hence proved, So we have 6 = ~ =} vx = 6v + 60 (il) v+ 10 From a point on the ground at a distance, Solving equations (i) and (il). we find a from the foot of a pole. a ball is thrown, at an angle of 45°, 1! = 20 km/h and x = 9 min, which just touches the top of pole and strikes the ground at The horizontal range of a projectile is a distance of b, on the other side ofit. Find the height of the pole. times its maximum height. Find the angle of projection. ~ Sol. If u and a be the initial velocity of projection and angle of o 4--a--++--b---J!r- projection, respectively, then Fig. 5.77 u2 sin 2 a -2g-The maximum height attained = and horizontal 2u 2 sin a cos ex Sol, Let h be the height of the pole, range = Using equation, we have y = x tan a (1 - ~) g Since top of pole, lie on curve (1), 2u 2 sin (X cos a According to the problem, = ~--- g =2J3 ,,2?sin2,-,) =}tana ( -J23 ) ( -,1,230) a a_) a[a a]h = tan 45\" (I _ _ ( -g =}a=tan-·! = + I> - = ~ ,a+I> a+b a+b For a projectile, show that (~ A particle is projected over a triangle ~)(1) gT2 = 2R tan a; (2) Hnmx = ( tan a; and (3) from one extremity of its horizontal base. Grazing over the =gT2 8Hm\"\" where the symbols have their usual meanings. vertex, it falls on the ot,herextremity of the base. Ifa and f:J be Sol, ethe base angles of the triangle and the angle of projection, prove that tan e ~ tan a + tan f:J. 2u sina 1. We know that T = ----, and (i) Sol, Let ABC be the triangle with base BC, Let h be the height g orthe vertex A, above BC, If AM be the perpendicular drawn on 2u2 sin a cos a R=--- (ii) base Be from vertex A, then tana = h h - and tan,B = -, where g ab Squaring eqnation (i) and dividing it by equation (ii) we get BM = a and CM = b, T2 4u 2 sill a g 2 Since A (a, h) lies on the trajectory of the projectile; =x =-tana y=xtan()(I-~) R 2u 2 sinacosa g (i) gT2 = 2Rtana Therefore it should satisfy equation (i) Hence proved. u2 Sil1 2 a a (I - _a_) ai.e,. h = tane (i) a+b [e, range R = + b] 2. Again, we know that Hmax = -~- 2g
5.24 Physics for IIT·JEE: Mechanics I [_b ]~ =tanG = tan a a2 + ah + b' -- a2 -- ab] [ a a +b a2 + ah + 7 b' (~~I:) h h tan O! = -a2-+-a;b-+-b-2 a = tan _I [a 2 +ab+b2 ] +- tan8=h ab ab = ba Hence proved, tan () = tan a + tan f3 Hence proved. A hall is thrown from the top of a building ~A 45 m high a speed 20 mls above the horizontal at an ti . o hi, l angle of 30°. Find B~~C 1. the time taken hy the ball to reach the ground, and M 2. the speed of ball just before it touches the ground. ~ (l .. Ii\\! b~ Sol. Given v =20 m/s, G = 3(F, H =45 m Fig. 5.78 A particle projected at a definite angle 1. As the ball has been projected at an angle of 30e above the horizontal, so first of all we need to analyse the velocity \" to the horizontal, passes throngh points (a, b) and (b, a), horizontally and vertically. This will be useful while using r~ferred to horizontal and vertic.al axes through the point of distance-time relation in horizontal and vertical directions, projection. Show that: Vxl = veos30c' = 20 x \"vS2 = IOvh3 m/s; 1. the hOr.Izontall'ange R = .a2-+- -ab-+- -b2, and 1 a+b U.ri = lJsin30° = 20 X -2 = 1O m/s 2. the angle of projection a is given hy It will be easy for us to use distance - time relation in vertical [a2 2 (il as it will involve less calculation tan-1 +ab+b ] . ab In y direction: --45 = lOt -- ~ x gt 2 =? t 2 -- 2t -- 9 = 0 Sol. 1. The equation to the trajectory of the particle: which On solving gives t = 1 +/10 s (positive value), (other Y = .x tana (1 - RX) value is I - Jlo s, a negative value of time which is not Since the particle passes through the points with coordinates acceptable). 2. v,'f = 10 - 10 x 4 = - 30 mis, (a, b) and (h, a) they should satisfy the first equation of curvc (1 - %)b = a tan 0: and (ii) (I -t)a = btana (iii) j;;-'\"jV~fV{ = + v;f = (lOvS)' = 20/3 m/s Solving equation 0) from equations (ii) and (iii) y ~ = (1-~) tanaand (iv) I(0,0) v;=20m/s (v) a R x j; = (I -t) tana 0, ,,'- 30.0°\"\"\\ Dividing equation (iv) by (v) we get (to eliminate tan a) \\ \\ (1 -%)(1h' -~} =? Ii - _I,,' =a2 _ a3 45m \\ RR \\ a' \\ \\ \\ ,,\\ \\ =? 2.[(/3 _ h31= a' _ b2 1 ,\\ \\ xf= ?1 ):r'\" ··45.0 m; I+---xr~ R + +a3 -- b3 (a - b)(a 2 ab b2 ) Fig. 5.79 =? R = . - . - = '-\"-'~c---c--'- a'-b' (a-h)(a+b) a2 + ab + b2 a+h 1. With what velocity Vo should a hall he projected horizon- tally from the top ofa tower so thatthc horizontal distance Hence proved Ie. a # hi. on the ground is qH, where H is the height of the tower. Z. Substituting the expression [or R in equation (ii) 2. Also determine the speed of the ball when it reaches the ground. ~ = tan a [I .. :;,al((;;!~b']
Motion in Two Dimensions 5.25 1rH ,,,,,,,,,,,,, Now analyzing the motion in x and y directions, we have 4-llI1~ Fig. 5.80 Ux = U u y = 0 ax = -gsine ay = -gcose Here we can usc t.he following formula v = u + at in x-direction. As we have values of initial velocity, final veloc- ity, and acceleration we can find t. +Vx = Lt.,\" axt At position B, VX = 0, as final velocity is equal to Sol. 1. Given Horizontal distance = ry H, where H = height of y-component of velocity. u the tower. O=u-gsinet :::::} I = - - which is required thne To flnd out the sufficient velocity as asked, we have to usc the to travel. g sine formula. Distance::: velocity x time. Here the time involved wiIl be time of flight as we arc considering projectile motion which Method 2 (using vectors) is given by T = -V1g2H' As u and v both are perpendicular to each other, we can use the orthogonality property of dot product. This means that if two vectors are perpendicular to each other their dot. product is zero, t.o find out time of travel to desired position, f2H So, UV = 0 =} li(u +at) = 0 = vOVIiSo, voT = I)H =} ryH =} uu+uat=O u2 + u g cos (90\" + 0) t = 0 2. During projectile motion horizontal component of velocity [:, Angle between u and g is 90\" + 0 as from figure.] remains same and its vertical component keeps on changing u2 + ug (- sin OJ ' t = 0 So t = u is the desired -- under the effect of gravity. So horizontal speed Vx = vo, vertical g sinO speed Vy = .j2gH. time, /~6 + 2gH 1)g2H~+2gH To find out the velocity we can use the same relation as used jv; vi= =Total speed =+ 2 in this question. But as at final position (considered) only y ': VO=I)/g; component of velocity is present. So we need to use the same rclation in y-direction. vy=uy+afil ::::::} v=o-gcoset; eA particle is projected at angle with = -g cos e- ,Lt - = -u col e is the . at position B, eV the horizontal. Calculate the time when it is moving perpen- g sm velocit.y dicular to initial direction. Also calculate the velocity at this position. At what angle should a hall be projected ,, , up an plane with a velocity Vo so that it may hit the : : /'''''4- Initial direction incline normally. The angle of the inclined plane with the :: / horizontal is el. [i~ v v'l \":, Fig. 5.81 \"\\'\\ 110 sine 1'0 Sol. Method 1 (using properties of projectile motion): ,, As we have to calculate the time between two positions A and 13 where final direction of movement is perpendicular to initial \\,,, direction of movement. So for our own comfortability we can o, choose initial direction of motion as x-axis. Also let's assume velocity at position B to be v. \"\"\" U, ,x k\" :a\\ 'gsina t ~ \", , , , , , <, ,, , , , .g -- g cosa v Fig. 5.83 Sol. As the ball has to hit the inclined plane normally, so in that position the x-component of velocity will be zero and the velocity will have y-component only. . The ball will hit the incline normally if its parallel component of velocity reduces to zero during the time of flight. By analyzing this motion along incline, I.e., x-direction +Vx = U x alt, Herevx =0, Ux = lJaCosG,ax = ~gsina (i) Va cos e O=vocose-(gsina)T =} 1'= gsina Also the displacement of the particle in y-direction will be Fig. 5.82 zero. Using
5.26 Physics for IIT-JEE: Mechanics I + 1, --,-,--> .'-,.,.--- + mQVQ '2(lyt- JJlpVp mQVQ After collision y = uyt =:? eThis gives T = 2-v-o-s'in~- Before collision g cos a: (ii) From equations (i) and Oi), we have eVo cos () 21)0 sin case 2 sin B Also, v2 - U' = 2as - (. .4j93 )' = 2(-9.8) X s g sina gcosa sina cosa S = 61.25 '* 2tan8tana = I =} tanB = [~cota] The collision takes place at the maximum height where the velocities of both the particles will be in the horizontal direction. a)=} e = tan-l (~cot which is the required angle of Applying conservation of linear momentum in the horizontal direction with the information that P retraces its path therefore projection. its momentum will be same in magnitude but diffei'ent in direc- Particles P and Q of mass 20 gm and 40 tion. Momentum of system before collision = Momentum of sys- gm, are sinmltaneously projected from points tem after collision. A and B on the ground. The initial velocities of P and Q / ! 2rnpvp -mQvQ mpvp-mQvQ=-mpvp +mQuQ :::::} vQ = - make 450 and 1350 angles, respectively, with the horizontal AB as shown in the Fig. 5.84. Each particle has an initial rnQ speed of 49 m/s. The separation A B is 245 m. Both particles = 2 x 0.02 x (49/,;'2) - 0.040(49/,;'2) travel in the same vertical plane and undergo a collision. Af- 0.040 I]= ~ [0,04 _ = ~ x 0 = 0 ter the collision, P retraces its path. Determine the position of Q when it hits the ground. How much time after the col- ,;'2 0.040 .,fi lision does the particle Q take to reach the ground? Take New path of P after collision, Considering vertical motion of g = 9.8 m/s 2. (I1T-JEE,1982) Q, U V = O. Sy = - 61.25, a y = -9.8, Iy = ? ~w 2:I , 1 2 S=UI+Zal = x(-9.8)xI =(-61.25):.1=3.53s / Fig. 5.84 Considering horizontal motion of Q. Sol. Tn p = 20 g, In Q = 40 g, The horizontal velocities of both the VbSince the = 0, therefore the particle 0 falls down vertically so it falls down on the mid point of abo lll!:!!l1t!'!11,m A body falling freely from a given height particles arc same and since both are projected simultaneously, II hits an inclined plane in its path at-a height h. As a re- these particle will meet exactly in the middle of A B (horizon- sult of tliis impact the direction of the velocity of the body tally). becomes horizontal. For wbat value of (h / 1I) the body will To find the vertical velocity at the time of collision let us take maximum time to reach the ground? (IIT-JEE,1986) consider the motion of P in vertical and horizontal directions. Sol. For A to B: u = 0, s = -(H - h). a = -1(,1 =? y4'i9 m/s ,r24m9 /s '*S = U I + 21: a l 2 12 -(H-h)=2:(-g)1 49 m/s 1 = [2(lI - h)f' pQ g 49 m/s A 49!VI- m/s 49l{:2\" m/s B TA T 141.----245 m-----J<.I II-h II Fig. S.85 Horizontal direction, S, = 122.5, I, =? Vx = 49/,;'2, jrh+I\"TTTTT\"ri'-r-r-n-rr;Br}7-!,.-rl '* T '*disp Vel. = time 49 122.5 (122.5),;'2 ,;'2 = Ix = 49 49 (122.5) ,;'2 c 49 Vertical direction v.}, =?, u.v = -,f,i ty. Fig. 5.86 av = - 9.8 m/s2 For B to C vertical motion: u y = 0, Sy = -h, a y = - g IffhS = ut + _Ia1 2 ::::} -11 = __1at2 :::::} [/ = -. 2 2g
Total time of fall T = I + I' = [2(Hg-h)J'/2 + [2ghJ'/2 Motion in Two Dimensions 5.27 For finding the maximum time llsing the concept of differen- 10 cos60° .. dT hatlOn: - = 0 Vc=lOm/s 10 sin60\" VA = 10 tnJs dh A 30m .yt__________________ Q 20m PM BED I,••- - - - d -----+.1 Fig. 5.88 ~. 1\\vo towers AB and CD are situated at 2 t = -_ s a distance d apart as shown in Fig. 5.87. AB is 20 m high and CD is 30 m high from the ground. An object of mass ):1 m is thrown from the top of A B horizontally with a velocity of 10 mls towards CD. Simultaneously another object of 13 [) = PM + M Q = 101 + 51 = 15 I = 15 2 mass 2 m is thrown from the top of CD at an angle 60° X . ;;; to the horizontal towards AB with the same magnitude of \",3 initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick = 1OV3 = 17.32 m to each other. (i) Calculate the distance d between the towers. (ii) Find the position where the objects hit the ground. (ii) Applying conservation oflinear momentum (during collision (IIT-JEE, 1994) of the masses at M) in the horizontal direction c m x 10 - 2m lOcos 60° = 3m x Vx :::::} 10nl - lOm = 31lt x Vx ::::::? Vx = 0 Since the horizontal momentum comes out to be zero, the combination of masses will drop vertically downwards and fall 2 atEx13E=PM=IOI=IOx ;;;=11.547m \",3 1\\vo guns, situated on the top of a hill of A 30111 10 Ill, fire one shot each with the same speed at some 20m internal of time. One gun fires horizontally and other fires upward at an angle of 600 with the horizontal. The shots B D collide in air at a point P. Find (i) the time-internal between Fig. 5.87 the fringes, and (il) the coordinates of the point P. Take origin of the coordinate system at the foot of the hill right Sot. (i) Let I be the time taken for collision. For mass m thrown below the muzzle and trajectories in x-y plane. horizontally from A for horizontalmotion. (IIT-JEE, 1996) PM=101 (I) Sol. For bullet A: Let I be the time taken by bullet A to For vertical motion u y = 0, Sy = y, a y = g (2) reach P. I 'iat2. Sy = uyl + '1i ayl 2 12 (3) Vertical motion: 11 y = O. S = ul + Sy = 10 - y. =} Y = 'ilil (4) a,. = 10 m/s2, +ty = t, Vy = u y Qyt = gt 10 - y = 51 2 (i) For mass 2 m thrown from C. for horizontal motion fy = t =QM [lOcos 60°]1 =} QM = 51 Horizontal motion: x = 5v'1 I (ii) For vertical motion: u,. = 10 sin 600 = 5V3 For bullet B. Let (t + I') to be the time taken by bullet B to =} Vy = 5V3 + gl (5) reach P. . 1Y + 10, S = ut + -2a l2 (6) Vertical motion t +u,. = + 5V3 sin 60\" = + 7.5 a.,. = \"g. S. = S = ul + 1 'J .. 2 -at\" j. -ay = -10 m/s 2 )' _ 10 = 7.5 (I + I') - 5 (t + 1')2 1 (iii) =} y+lO=5V31+'i liI2 +S,. = -(10 - y) = y - 10 Iy = 1 t' Iy = I. From equations (ii) and (vi) Horizontal motion x = (5V3 cos 60\")(1 + I') I + + 1 'i8t2 10 = 5V31 'i1i12 =} 5V3 1+ 5v'1 I' = 2x (iv)
5.28 Physics for IIT-JEE: Mechanics I ground also. Y-direction motion (taking relative terms w.r.t. box) y u y = +usina,tly = -gcos8Sy = o(activity is taken till the Bullet B time the particle comes back to the box) ty = t. v S = III + ~(lI' =} 0 = (usina)1 - ~gcosO x 12 60° V\" SJ] 22 1 Bullet A 2u sina IO-y =}1=Oorl= gcos8 X-direction motion (Taking relative terms w.r.t. box) 10m J....................... Ux =+ucosa y 1 p +I 2 ........... (x,y) S = ul 'lUI y 2u sin a u2 sin 2a Ix = 2u sin et x eax = 0 :::::} Sr = UCOSet X - - \" - = . g cos (J g cos 0 --- g cos () x 2. For the observer (on ground) to see the horizontal displace- Fig. 5.89 ment to be zero, the distance travelled by the box in time Substituting the value of x from equation (ii) in equation (iv), ( -2U.S-in-a) should be equal to the range of the particle. Let g cos8 we get the speed of the box at the lime of projection of particle be U. 5../3 1+ 5../3 t' = 10../3 I =} I = I' Then for the motion afbox with respect to ground (Fig. 5.91). Putting I = I' in equation (iii) y - 10 = 151201' (v) Adding equations (i) and (v) 0 - 151 - 1512 =} I = I s (ii) Putting I = I in equation (ii), we get x = 5../3 Putting I = I in equation (i), we get)' = 5 Therefore, the coordinates of point Pare (5../3, 5) in meters. A large, heavy box is sliding without fric- tion down a smooth plane of inclination O. From a point P on the bottom 01' the box, a particle is projected inside the hox. The initial speed 01' the particle with respect to the box Fig. 5.91 is u, and the direction of projection makes an angle Cl with the bottom as shown in Fig. 5.90. (lIT-JEE, 1998) +Ux =U, S:::::: ut 2I at 2 ,ax = -gsm. 0 S _ -l~~.~in2 2ex _u 2 sin 2ex x - gcos8 gcos8 =-u ( -211 -Sin-o:) _. -Igsm. O (2.U_S-in-a)2 2 , gcos8 gcos() Fig. 5.90 Iy = -2u-.sin-a- on so.lvmg, we get U = -u -co_st.a_+-0-) g cose cose 1. Find the distance along the bottom orthe box between the point of projection P and the point Q where the particle An object A is kept fixed at the point lands. (Assume that the particle does not hit any other surface of the box. Neglect air resistance). = =x 3 m and y 1.25 m on a plank P raised above the =ground. At time t 0 the plank starts moving along the + x 2. If the horizontal displacement of the particle as seen by an observer on the ground is 0, find the speed of the box direction with an acceleration 1.5 m/s 2. At the same instant a with respect to the ground at the instant when particle stone is projected from the origin with a velocity u as shown was projected. in Fig. 5.92. A stationary person on the ground observes the stone hitting the object during its downward motion at an Sol. 1. II is the relative velocity of the particle with respect to angle of 45\" to the horizontal. All the motions are in the x- y the box. Resolve u. plane. Find and the time after which the stone hits the object. Ux is the relative velocity of particle with respect to the box (IIT·JEIC, 2000) in ,\"\"(-direction. Sol. Let i be the time after which the stone hits the objcctanclO be Uy is the relative velocity with respect to the box in y-direction. Since there is no velocity of the box in the y-direction, there- the a.ng1c which the velocity vector ii makes with the horizontal. fore this is the vertical velocity of the particle with respect to According to question, we have following three conditions. (i) Vertical displacement of stone is 1.25 Itl. Therefore,
Motion in Two Dimensions 5.29 y A 1.25 m p u x Q 3.0 m Fig. 5.92 2. Usm\" g SillC rule - ,V-B- - = VT =? Vu =2 m/s 0_,- - sm 135\" sm IS\" 8 u 1.25 m v Igt -1/ s.m8= IJ~\\' CtD5 A ~ \"-'-'-_-------'------..I'V\"-,-,-/~l-'-, 60\" u cosO 0 \" \" \" -4-5\"' - \" ' - - - - - \" Fig. 5.93 Fig. 5.95 1.25 = (u sin II) t - :I2 gt 2, where Ii = 10 m/s2 A particle A moves along a circle ofradius R = 50 em so that its radius vector r relative to point 0 rotates +or (u sin 0) t = 1.25 5 t 2 (i) with the constant angular velocity w = 0.40 radls (Fig. 5.96). (ii) Horizontal displacement of stone = 3 + displacement of ob- ject A. Thcre!(lfe, (u cos II) t = 3+ 1 where a = 1.5 m/s2 _at' 2 or (/I cos e)t = 3 + 0,75t' (ii) (iii) Horizontal component ofvdocity (of stone) = vertical com- ponent (because velocity vector is inclined) at 4Y with horizon- tal), Therefore, (/I cos 0) = gt - (u sin 0) (iii) Fig. 5.96 (The right-hand side is written gt - u sin f) because the stone Find the modulns of the velocity of the particle and direc- tion of its total acceleration. is in its downward motion, Therefore, gt > u sine. In upward Sol. Consider X and Y axes as shown in Fig. 5.97. Using sine law in triangle CAO, we get motion /.l sin () > gt). Multiplying equation (iii) with t we can \"~., write, (ueosil)t +(usinO)t = IOt 2 (iv) Now, equations (i) - (ii) - (i) gives 4,25 t' - 4,25 = () or t = 1 s Substituting t = 1 s in equations (i) and (ii) we get, u sin Fig. 5.97 II = 6,25 mls or u\" = 6.25 m/s and u cos II = 3.75 mis, Urlor Ux ::= 3.75 mIs, therefore, U = u)} + or ii = (3,751 + 6,25)] m/s On a frictionless horizontal surface, as- r Rr R -:-..,-- = - - or ------ = --:,r = 2Rcosll sumed to he the x-y plane, a small trolley A is moving along sin (n - 20) sin II 2 sin II cos () sin Ii a straight line parallel to the y-axis (see Fig. 5.94) with a con- Now r = r cos oi + r sin II] = 2R cos' oi + 2R cos II sin e] stant velocity of (~ - I) mls. At a particular instant, when _ df de, dll , Now v = --- = -4RcosOsinO·-i +2Rcos211-j the line 0 A makes an angle of 45' with the x-axis, a ball dt cit dt is thrown along the surface from tbe origin O. Its velocity = - 2R sin 211wl + 2R cos 28,,)] makes an angle 1> with the x-axis and it hits the trolley. Ivl = 2,,) R 1. The motion of the hall is observed from the frame of the _ eli! = dO, dO, Further a = - -4Rcos28,,)-i -4Rwsin20--j etrolley. Calculate the angle made by the velocity vector of dt dt tit the ball with the x-axis in this frame. 2. Find the speed of the hall with respect to the surface, = -4R,v' cos 2IJi - 4R,o' sin 20) if 1> =4813. (IIT·JEE,2002) 101 =4Rw2 VSol. 1. From the Fig. 5,95 IIT makes an angle of 45' with the x-axis.
5.30 Physics for lJT-JEE: Mechanics I EXERCISES SO/UtiO/IS on JUlge:5.47:' 7. A ship is sailing due north at a speed of 1.25 m/s. The current is taking it towards east at the rate of 2 rnls and a sailor is 1. Two piers, A and B are located on a river. B is 1,500 m climbing a vertical pole in the ship at the rate of 0.25 m/s. downward from A, Two ffiends C and [) must make round Find the magnitude of the velocity of the sailor with respect trips from pier A to pier B and return. J) rows a boat at to the ground. a constant speed of 4.00 kmlh relative to the water and C walks on the shore at a constant spced of 4.00 km/h. The S. A bomber plane moves due cast at 100 krnJh over a town T velocity of the river is 3.5 km/h in the direction from A to B. at a certain instant of time. Six J;l1inutes later an interceptor Find the time takcn by C and D to make the round trips. plane sets otT flying due north east from the station S which is 40 km south of T. If both maintain their courses, find the 2. A boat takes 2 h to travel 8 km and back in still water. If the velocity with which the interceptor plane must fly in order to velocity of the water is 4 kmlh, then what is the time takcn just overtake the bomber. for going upstream of 8 km and then coming back. 9. Velocity of a swimmer v in still water is less than the velocity 3. A railroad flatcar is traveling to the right at a speed of 13.0 of water u in a river. Show that the swimmer must aim himself m/s relative to an observer standing on the ground (see Fig. 5.98). Someone is riding a motor scooter on the flatcar. Cor- at an angle cos \"-1 ~ with upstream in order to cross the river responding to the relative velocities 18 m/s to the right, 3 m/s to the left, and 0 m/s of motor scooter relative to ground, u find the relative velocities (magnitude and direction) of motor along the shortest possible path. Find the drifting(distance scooter relative to the flatcar. moved along the direction of stream in crossing the river) of the swimmer alo1)g this shortest possible path. mis 10. A man wants to reach point B on the opposite bank of a river flowing at a speed as shown in the Fig. 5.100. ,,,,--~:-- B ,,: -)011 iA 45° Fig. 5.100 Fig. 5.98 What minimum speed relative to water should the man have so that he can reach directly to point B? In which direction 4. A boy is cycling with a speed of20 kmlh in a direction making should he swim? an angle 30° north of cast (see Fig. 5.99). 11. A steamer plies between two stations A and' B on opposite N banks of a river, always following the path AB. The distance between stations A and B is 1,200 m. The velocity of the w ---lL.l.llL-E current is ,JT7 ms\"~ 1 and is constant over the entire width of s the river. The line AB makes an angle 600 with the direction Fig. 5.99 of the flow. The steamer takes 5 min to cover the distance AB and back. Then find Find the velocity of the second boy moving towards nOlth so a. the velocity of steamer with respect to water, and that to him the first boy appears to be moving towards east. b. in what direction the steamer should move with respect to 5. A boat wants to cross a river as soon as possible. In doing so it takes 4 s less than if it travels by shortest path. Find the line AB. width of the river if velocity of water in river is 8 ms·- 1, and 12. A boat is rowed with constant velocity u starting from p'oint A boat can travel in still water with a velocity of 17 ms- l 6. To a man going with a speed of 5 mls rain appears to be faU on the bank of a river of width d. Water in the river flows at a vertically. If the actual speed of the rain is 10 mis, then what constant speed flU. The boat always points towards a point 0 is the angle made by rain with the vertical? on the other bank which is directly opposite to A (Fig. 5.101). Find the path equation r = f(e) for the boat. 0:,,,,,,, 0 r u A:, Fig. 5.101
Motion in Two Dimensions 5.31 13. A ship A streams due north at 16 km/hr ancl a ship B due 19. A ball is thrown upwards with a velocity of 30 ms- l at an west at 12 km/h. At a certain instant B is 10 km north cast of A. angle of 600 to the horizontal. Find its velocity after 2 s. a. Find the magnitude of velocity of A relative to B. 20. a. A ball is thrown with a velocity of 10 m/s at an angle with b. Find the nearest distance of approach of ships. horizontal. Find its maximum range. 14. Two particles start moving simultaneously with constant. ve- b. A cricketer can throw a ball to a maximum horizontal locities uland Uz as shown in Fig. 5,102. First particle starts from A along AO and second starts from 0 along OM. Find distance of 100 m. How high can he throw the same ball? the shortest distance between them during their motion. c. Find the minimum velocity with which the horizontal N M range is 160 m. A 21. Prove that a gun will shoot three times as high when its angle '1 \"I of elevation is 60° as when it is 30\", but cover the same e \"2 horizontal range. Take g = 10 ms·-2. () 22. A football is kicked by a player with a speed of 20 ms- I at an angle of projection 45(}. A receiver on the goal line 25 JTI Fig. 5.102 away from the player in the direction of the kick runs the same instant to meet the ball. What must be his speed, ifhe is 15. The front windscreen ,of a car is inclined at an angle 60(1 to catch the hall before it hits the ground? Ignore the height with the vertical. Hailstones fall vertically downwards with a of the receiver. speed of 5../3 m/s. Find the speecl of the car so that hailstones are bounced back by the screen in vertically upward direction 23. A stone is thrown from the top of tower at an angle of 30° up with respect to car. Assumeelastic collision of hailstones with with the horizontal with a velocity of 16 ms-I After 4 s of wind screen. flight it strikes the ground. Calculate the height of the tower from the ground and the horizontal range of the stone. 16. State True or False: 24. A ball is thrown downwards at an angle of 30' to the horizon- a. The speed of the projectile is minimum at its highest po- sition. tal with a velocity of 10 ms-- 1 from the top of a tower which 30 m high. How long will it take before striking the ground'? b. The normal acceleration of the. projectile at the highest 25. A bomb is fired horizontally with a velocity of20 ms· l from the top of a tower which 40 m high. After how much time position is equal to g. and at what horizontal distance from the tower will the bomb strike the ground? Take g = 9.8 ms 2 c. For a given speed, the time of flight docs not depend on the angles of projection. 26. a. An aircraft is flying horizontally with a velocity of 540 d. The greatest height to which a man can throw a stone is H, kmh- l at a height of2,000 m from ground. When it is ver- The greatest. distance upto which he call throw the stone tically above a point A on the ground, a body is dropped is fi12. from it. The body reaches the ground at a point B. Calcu- latedistanceAB. Takeg= 10ms,-2. e. A body feels weightlessness during projectile motion. f. The instantaneous velocity of a particle is always tangen- b. A stone is dropped from a window of a bus moving at with tial to the tn~jectory of the particle. avelocity 54 kmh l. The height of the window is 245 em. g. The instantaneous magnitude of velocity is equal to the Find the distance along the track which the stone covers before striking the ground. Take g = 10 ms-- 2. slope of the tangent drawn at the trajectory of the particle 27. A ball is thrown horizontally from the top of a tower and at that instant. strikes the ground in 3 s at an angle of 30G with the verticaL 17. A hit baseballicaves the bat with a velocity of 100.)2 m/s at 45\" abc)Vc the horizontal. The ball hits the top of a screen at a. Find the height of the tower. the 300 m mark and bounces into the crowd for a home run. b. Find the speed with which the body was projected. How high above the ground is the top of the screen? (Neglect 28. A particle is projected from point A to hit an apple as shown air resistance.) in Fig. 5.103. The particle is directly aimed at the apple. Show that particle will not hit the apple. Now show that if the string 18. A ball is projeetecl from ground with a velocity of 40 ms- l with which the apple is hung is cut at the time of firing the at an angle of 30c with the horizonU1L Determine particle, then particle will hit the apple. a. the tillle of flight, h. the horizontal range, and Fig. 5.103 c. the maximum height. d. At what time is the maximum height attained? c. What is the magnitude and direction of velocity at the maximum height? f. What is the magnitude and dIrection of velocity after 1 s of throwing the balP
5.32 Physics for lIT-JEE: Mechanics I 29. A fighter plane flying horizontally at an altitude of 1.5 km time, One gun fires horizontally and the other fires upwards with a speed of 720 kmh 1 passes directly overhead an an- at an angle of 60\" with the horizontal (Fig. 5.105). The shots tieraft gun. At what angle from the vertical should the gun be fired for the shell with muzzle speed 400 ms- 1 to hit the .collide in air at a point P. Find Vo . plane? // 30. A target is fixed on the top of a tower which is 13 m high. A P(x,y) r-=_-' person standing at a distance of 50 m from the pole is capable P(x,y) of projecting a stone with a velocity 1() .Jif ms·-I. If he wants (0,0) ' - -x-. .-. -- - to strike the target in the shortest possible time, at what angle should he project the stone? Fig. 5.105 31. A stone is projected from the point on the ground in such a. the time interval between the firings, and a direction so as to hit a bird on the top of a telegraph post b. the coordinates of point P. Take the origin of coordinate of height h and then attain the maximum height 3hJ2 above the ground. If at the instant of projection the bird were to fly system at the foot of the hill right below the muzzle and away horizontally with uniform speed, find the ratio between trajectories in the ~y-planc. horizontal velocities of the bird and stone if the stone still 39. A small spherc is projected with a velocity of3 mls in a direc- hits the bird while descending. tion 60° from the horizontal y-axis, on the smooth inclined plane (Fig. 5.106). The motion of sphere takes place in the 32. A ball rolls to the top of a stairway with a horizontal velocity x-y plane. of magnitude 1.8 ms-'. The steps arc 0.20 m high and 0.20 m wide. Which step will the ball hit first? (g = 10 ms-2) x 33. A machine gun is mounted on the top of a tower which is 100 30\" m high. At what angle should the gun be inclined to cover a maximum range of firing on lhe ground below? The muzzle Fig. 5.106 speed of bullet is 150 ms- I Take Ii = 10 ms-· 2 Calculate the magnitude v of its vel~)City after 2 s. 34. Figure 5. 104 shows an elevator cabin, which is moving down- 40. A gun is fired from a moving platform and ranges of the shot wards with constant acceleration a. A particle is projected from the corner A, directly towards diagonally opposite cor- are observed to be R, and R, when the platform is moving ner C. Then prove that forwards and backwards, respectively, with velocity vp. Find C~B-----~-~~/ the elevation of the gun ex in terms of the given quantities. 41. Tangential acceleration changes the speed of the particles / whereas the normal acceleration changes its direction. (True or False) ILIA:. . . A: ~_/_~_/_~_/_- -1 42. Calculate the centripetal acceleration ofa point on the equator D of earth due to the rotation of earth about its own axis. Radius of earth = 6,400 km. Fig. 5.104 43. A cyclist is riding with a speed of27 kmh· 1• As he approaches a circular turn on the road of radius 80 m, he applies brakes a. particle will hit C only when a = g and reduces his speed at the constant rate of 0.5 ms~2. What b. particle will hit the wall CD if a<g is the magnitude and direction of the net acceleration of the cyclist on the circular turn? c. particle will hit the roof Be if a > g 44. An electric fan has blades of length 30 cm as measured from the axis of rotation. If the fan is rotating at 1,200 rpm, find 35. A ball is thrown with a velocity whose horizontal eomponent the acceleration of a point on the tip of the blade. is 12 O1s-- 1from a point 15 In above the ground and 6 m away 45. A particle starts from rest and moves in a circular motion from a vertical wall 18.75 m high in such a way to just clear with constant angular acceleration of 2 rad/s2 . Find the wall. At what time will the ball reach the.ground?(take a. angular velocity, and Ii = 10 me') b. angular displacement of the particle .after 4 s. c. The number of revolutions completed by the particle dur- 36. A particle is projected up an inclined plane of inclination f! at an elevation a to the horizon. Show that ing these 4 s. d. If the radius of the circle is 10 cm, find the magnitude and a. tana = cotf! + 2tanf!, if the particle strikes the plane at direction of net acceleration of the particle at the end of right angles. 4 s. b. tana = 2tanf! if the particle strikes the plane hOl;izontally. 37. Two parallel straight lines are inclined to the horizon at an anglc a. A particle is projected from a point mid way between them so as to graze one of the lines and strike the other at the eright angle. Show that if is the angle between the direction of eprojection and either of the lines, then tan = (,f2 - I) cot a. 38. Two guns situated on top of a hill of height 10 m fire one shot eaeh with the same speed s-J3 m/s at some interval of
Motion in Two Dimensions 5.33 Objective Type Solutions on page 5.53 a. 12km b. 18 km 1. Rain is falling vertically downwards with a speed of e. 30 km d. 25 km 4 kmh- 1, A girl moves on a straight road with a velocity of 3 kmh- I The apparcnt velocity of rain with respect to 7. Rain is falling vertically with a velocity of 25 ms- I. A woman rides a bicycle with a speed of 10 ms- I in the the girl is north to south direction. What is the direction (angle with a. 3 kmh- I b. 4 kmh- I vertical) in which she should hold her umbrella to safe c. 5 kmh- I d.7kmh- 1 herself from the rain? 2. Ship A is travelling with a velocity of 5 kmh- I due east. a, tan-I(OA) b. tan- I(l) The second ship is heading 30\" east of north. What should c. tan-I(~) d. tan- I(2.6) be the speed of second ship if it is to remain always due 8. A small body is dropped from a rising balloon. A person A stands on ground, while another person B is on the nOlth with respect to the first ship? balloon. Choose the correct statement: Immediately, after the body is released. a. 10 kmh- I b. 9 kmh- I a. A and B, both feel that the body is coming (going) c. 8 kmh- I d. 7 kmh-I down, 3. A man swims from a point A on one bank of a river of b. A and B, both feel that the body is going up. width IOO m. When he swims perpendicular to the water c. A feels that the body is corning down, while B feels current, he reaches the other bank 50 m downstream. The that the body is going up. angle to the bank at which he should swim, to reach the d, A feels that the body is going up, while B feels that the body is going down. directly opposite point B on the other bank is a. 10° upstream b. 20° upstream c. 30° upstream d. 60° upstream 9. A policeman moving on a highway with a speed of 30 kmh- I fires a bullet at a thief's car speeding away in the B same direction with a speed of 192 kmh-I. If the muzzle tI speed of the bullet is 150 ms-- I with what speed does the 100m *A bullet hit the thief's car? Fig. 5.107 a. 120 mls b. 90 m/s e. 125 mls d. 105 m/s 4. A boat is sent across (perpendicular) a river with a ve- 10. A bird is flying towards nOith with a velocity 40 kmlh and locity of 8 kmh- I. If the resultant velocity of the boat is a train is moving with a velocity 40 km/h towards east. What is the velocity of the bird noted by a man in the 10 kmh -I , the river is flowing with a velocity train? a. 6 kmh- I b. 8 kmh-· I a, 40.j2 km/h N -E c. 10 kmh- I d. 128 kmh- I 5. A man is walking on a road with a velocity 3 kmh- I Sud- b. 40.j2 knilh S-E denly rain starts falling. The velocity of rain is 10 kmh- I c. 40.j2 km/h N-W d. 40.j2 krnlh S-W in vertically downward ~irection. The relative velocity of the rain w.r.t. man is 11. A river is flowing from west to east at a speed of 5 m/min. a. .j7kmh-1 b. .jj3 kmh- I A man on the south bank of the river, capable of swimming c. 13 kmh- I d. vfI69 kmh- I at 10 mlmin in still water, wants to swim across the river 6. Two trains having constant speeds of 40 kmlh and in the shortest time. Finally he will swim in a direction 60 km/h, respectively are heading towards each other on the same straight track (Fig. 5.108). a. tan-I (2) EofN b, tan- I(2) NofE c. 30\" E ofN 40 km/hr 60 kmJhr 12. A boat which has a speed of 5 km/h in still water crosses a river of width I km along the shortest possible path in 60km 15 min. The velocity of the river water in km per hour is Fig. 5.108 a. I b. 3 c.4 d. AT A bird that can fly with a constant speed 0[30 kmlh, flies off from one train when they are 60 km apart and heads 13. A boat is moving with a velocity 31 + 4] with respect to directly for the other train. On reaching the other train, it flies back directly to the first and so forth. What is the total ground. The water in the river is moving with a velocity distance traveled by the bird before the two trains crash? -31 - 4] with respect to ground. The relative velocity of the boat with respect to water is a, 8J b. -67 - 8J
5.34 Physics for IIT-JEE: Mechanics I c. 6; + 8) d. 5~ d. None of these 14. A car is moving towards cast with a speed of25 km/h. To 21. A truck is moving with a constant velocity of 54 km/h. the driver of lhe car, a bus appears to move towards north In which direction (angle with the direction of motion of with a speed of 25V3 km/h. What is the actual velocity truck) should a stone be projected up with a velocity of of the bus? 20 mis, from the floor of the truck, so as to appear at right a. 50 krn/h. 30\" cast of north angles to the truck, for a person standing on the earth? b. 50 krn/h, 30\" north of east c. 25 krn/h, 30\" east of north d. 25 km/h, 30\" north of east 15. A swimmer wishes to cross a 500 m 'river flowing at 22. A river flows with a speed more than the maximum speed 5 km/hr. His speed with respect to water is 3 km/hr. The with which a person can swim in the stil1 water. He intends to cross the river by shortest possible path (i.e., he wants shortest possible time to cross the river is to reach the point on the opposite hank which directly opposite to the starting point). Which of the following is a. ]() mill h. 20 min correct? c. 6 min d. 7.5 min a. He should start normal to the river bank. 16. A train of 150 lTI length is going towards north direction h. He should start in such a way that, he moves normal at a speed of 10 m/s. A parrot flies at a speed of 5 m/s to the bank, relative to the bank. towards south direction parallcl to the railway track. The c. He should start in a particular (calculated) direction time taken by the parrot to cross the train is equal to making an obtuse angle with the direction of water current. a. 12 s b. 8 s c. 15 s d. 10 s d. The man cannot cross the river, in that way. 17. A boy can swim in still water at 1 m/s. He swims across a river flowing at 0.6 mls which is 336 111 wide. If he travels in shortest possible time, then what time he takes to cross the river. b. 420 s 23. Rain, driven by the wind, falls on a railway compartment a. 250 s with a velocity of20 mis, at an angle oDO\" to the vertical. The train moves, along the direction of wind flow, at a c. 340 s d. 336 s speed of 108 km/hr. Determine the apparent velocity of rain for a person sitting in the train. 18. A man can swim in still water with a speed of 2 m/s. If he wants to cross a river of water current speed ..j3 m/s a. 20J? m/s h. 10J? m/s c. 15J? m/s d. lOJ? km/h along the shortest possible path, then in which direction should he swim? 24. The ratio of the distance carried away by the water current, a. At an angle 12(f to the water current. downstream, in crossing a river, by a person, making same b. At an angle 1500 to the water current. anglc with downstream and upstream are, respectively, as c. At an angle 90° to the water curr~nt. 2 : 1. The ratio of the speed of person to the water current d. None of these cannot be less than 19. A plank is moving on a ground with a velocity v and a a. 1/3 b. 4/5 c. 1/5 d. 4/3 block is moving on the plank with a velocity u as shown in figure. What is the velocity of block with respect to 25. A particle is moving in a circle of radius r centred at 0 ground? with a constant speed v. What is the change in the velocity in moving from Ata B(LAGB = 40\")') a. v - U ,towards right b. 11 - u towards left a. 2vsin20\" b. 4vsin4()\" c. u towards right c. 2vsin40° d. vsin2()C d. None of these 26. i. If air resistance is not considered in a projectile lTI()tion, the horizontal motion takes place with _Pla_nk v I a. constant. velocity b. constant acceleration Fig. 5.109 c. constant retardation d. variable velocity 20. A car is going in south with a speed of 5 m/s. To a man sitting in car a bus appears to move towards west with a ii. When a projectile is· fired at an angle 8ith horizontal with speed onv'6 m/s. What is the actual speed of the bus? velocity u, then its vertical component a.4ms·,-1 a. remains the same
Motion in Two Dimensions 5.35 h. goes on increasing with the height gin of the coordinate axes. The x and y components of its c. first decreases and then increases with the height d. first increases then decreases with the height displacement arc given by x = 6t and y = St - St 2 What is the velocity of projection? iii. Range of a projectile is R, when the angle of projection is a. 6 mls b. 8 mls c. 10 m/s d. 14 mls 30\". Then, the value of the other angle of projection for the 32. In the above problem, what is the angle of projection with same range is horizontal? a. tan-' (1/4) a. 45\" b. 60' c. SO' d.40' b. tan\"\" (4/3) iv. A ball is thrown upwards and it returns to ground describing c. tan-' (314) d. tan '(1/2) a parabolic path. Which of the following quantities remains constant throughout the motion? e33. i. A particle is fired with velocity u making angle with the a. kinetic energy of the ball b. speed of the ball horizontal. What is the change in velocity when it is at the c. horizontal component of velocity d. vertical component of velocity highest point? d. (ucos e- u) a. ucose b. u c. usine ii. In first part of the problem 33, change in speed is a.liCOSe h.u c.usine d.(ucosO-u) 27. i. If R is the maximum horizontal range of a projectile, then 34. A ball is thrown upwards at an angle of 60\" to the hori- the greatest height attained by it is zontal. It falls on the ground at a distance of 90 m. If the c. R b.2R c. RI2 d. RI4 hall is thrown with the same initial velocity at an angle 30\", it will fall on the ground at a distance of ii. If the time of flight of a projectile is doubled, what happens to the maximum height attained? a. 120 m b. 90111 a. halved b. remains unchanged c. 60111 d. 30 m c. doubled d. become four times 35. Range of a projectile is R, when the angle of projection is 30° . Then, the value of the other angle of projection for iii. A person can throw a stone to a maximum horizontal distance the same range is of II. The greatest height to which he can throw the stone is a.4Y' b. 60\" a. h b. hl2 c. 2h d. 311 c. 50\" d. 'i0\" iv. The path uf one projectile as seen by an observer 011 another projectile is a/an 36. A projectile will cover same horizontal distance when the a. straight line h. parabola c. ellipse d. eircle initial angles oCprojcction are v. A body is projected at 30\" with the horizontal. The air ofIers a. 20°,60° b. 20\", 50\" resistance in proportion to the velocity of the body. Which of the following statements is correct? c. 20\",40\" d. 2(F,70\" a. The trajectory is a symmetrical parabola. b. The time of rise to the maximum height. is equal to the 37. javelin is thrown at an angle () with the horizontal and the time of return to the ground. c. The velocity at the highest point is directed along the hor- erange is maximum. The value of tan is izontaL d. The sum of the kinetic and potential energies remains con- a. b. V3 stant. 1 d. 2 c. V3 28. A particle is projected with a velocity v so that its range on a 38. A shot is fired from a point at a distance of 200 m from the foot of a tower 100 m high so that it just passes over horizontal plane is twice the greatest height attained. 1f g is it horizontally. The direction of shot with horizontal is acceleration due to gravity, then its range is a. 30G h. 45\" c. 60\" d. 70\" 4v2 4g 4v3 4v 39. Two bullets are fircd horizontally with different velocities d. 5g2 lfom the same height. Which will reach the ground first? a. 5g h. Sv2 c. 5g2 a. slower one h. faster one 29. During a projectile motion if the maximum height equals c. both will reach simultaneously the horizontal range, then the angle of projection with the d. it cannot be predicted horizontal is a. tan-'(I) b. tan-'(2) c.tan\"-'O) d. tan-'(4) 30. A particle is projected from ground at some angle with the horizontaL Let P he the point at maximum height If. At what 40. A person can throw a stone to a maximum distance of h metre. The maximum distance to which he can throw the height above the point P the particle should be aimed to have stone is a. h b. 1112 range equal to maximum height? a. H b. 2H c. HI2 d. 3H 31. The point fi'om where a ball is projected is taken as the ori- c. 2h d. 3h
5.36 Physics for IIT-JEE: Mechanics I 41. The path of one projectile as seen by an observer on an- 51. A gun fires two bullets at 60° and 30° with the horizontal. other projectile is alan The bullets strike at some horizontal distance, The ratio of maximum height for the two bullets is in the ratio a. straight line b. parabola c. ellipse d. circle a.2:1 b.3:1 42. The maximum height reached by projectile is 4 m. The c. 4; 1 d. 1 : 1 horizontal range is 12 m. Velocity of projection in ms~l 52. A body is projected at an anglc of 30\" with the horizontal is (g is acceleration due to gravity) and with a speed of 30 ms- '. What is the angle with the a. 5.../g/2 b. 3.../g/2 horizontal after 1.5 s? (g = 10 ms- 2 ) c. ~h/2 d. SI h/2 a. 0\" b. 30° c. 60° d. 90° 3 53. A grasshopper can jump maximum distance 1.6 m. It 43. A ball thrown by one player reaches the other in 2 s. The spends negligible time on the ground, How far can it go in 10 s? maximum height attained by the ball above the point of projection will be about a. 5J2m b. lOJ2m ~~J2m a. 2.5 m b. 5 m ~WJ2m c. 7.5 m d. 10 m 54 i. The maximum height attained by a projectile is increased 44. Galileo writes that for angles of projection of a projectile by 5%. Keeping the angle of projection constant, what is the at angles (45 + a) and (45 - a), the horizontal ranges percentage increase in horizontal range? described by the projectile are in the ratio of a.5% b.lO% c.15% d.20% a.2:1 b.l:2 ii. The maximum height attained by a projectile is increased by ~ 1:1 ~ 2:3 10 %. Keeping the angle of projection constant, what is the percentage increase in the time of flight? 45. At the top of the trajectory of a projectile, the acceleration a.5% b. 10% c.20% d.40% is 55. A body has an initial velocity of 3 mls and has an ac- a. maximum b. minimum celeration of 1 m/s2 DOImal to the direction of the initial velocity, Then its velocity 4 s after the start is c. zero d. Ii 46. A projectile is thrown at an angle of 40\" with the horizontal a. 7 mls along the direction of initial velocity. and its range is RI_ Anolher projectile is thrown at an angle 40\" with the vertical and its range is R2. What is the b. 7 mls along the normal to the direction of initial ve- relation between R 1 and R2? locity. a. R, = R2 c. 7 m/s mid-way between the two directions. b. R, = 2R2 d. 5 mls at an angle tan--'(4/3) with the direction of initial velocity. c. 2R, = R2 47. A pr(jjectile has a time offtight T and range R. Ifthe time 56. When a particle is thrown horizontally, the resultant ve- locity of the projectile at any time t is given by of flight is doubled, what happens to the range? a. RI4 b. RI2 a. gt b. 1 g t 2 - c. 2R d. 4R C. ,ju2 + g't2 2 48. At the uppermost point of a projectile, its velocity and d. ,j~u'''_-g02t 2 acceleration are at an angle of 57 i. A hiker stands on the edge of a cliff 490 m above the ground d. 1800 c. 90\" and throws a stone horizontally with a speed of 15 ms-' . The 49. A ball is thrown from a point with a speed Vo at an ae.gle of time taken by the stone to reach the ground is (g = 9.8 m/s') projection (). From the same point and at the same instant a. lOs b.5s c.12s d. ISs a person starts running with a constant speed vol2 to catch ii. In the first part of question 57, the vertical component of the the ball, Will the person be able to catch the ball? If yes, what should be the angle of projection? velocity on hitting the ground is a. 79 mls b. 89 mls c. 98 mls d. 108 mls a. Yes, 60\" b. Yes, 30° iii. In the first part of question 57, the speed with which stone c. No hits the ground is 50. An airplane moving horizontally with a speed of 180 kmlh a. 89.14 m/s b. 79.14 mls drops a food packet while flying at a height of 490 m. The b.99.14m/s d.109m/s horizontal range of the food packet is (g = 9.8 m/s2) 58. Two tall buildings are 30 m apart. The speed with which a.180m b.980m a ball must be thrown horizontally from a window 150 m above the ground in one building so that it enters a window c. 500 m d. 670 m 27.5 m from the ground in the other building is
Motion in Two Dimensions 5.37 a. 2 m/s h. 6 mls ii. In the first part of question 65, the time taken to reach the c. 4 mls d. 8 mls maximum height will be decreased by 59. A shell fired from the ground is just able to cross hori- a.19% b.5% zontally the top of a wall 90 m away and 45 In high. The direction of projection of the shell will be c.1O% d. none of these a. 2S\" h. 30\" iii. In the second part of questibn 65, the time taken to return to the ground from t.he maximum height a. is almost same as in the absence of friction c. 60' d.45' b. decreases by I % c. increases by 1 % 60. Two paper scrccns A and B are separated by 150 m. A d. increases by II % bullet pierces A and B. The hole in B is IS em below the hole in A. If the hullet is travelling horizontally at the time 66. At what angle with the horizontal should a ballbe thrown of hitting A, then the velocity of the bullet at A is (g = 10 ms-') so that the range R is related to the time of flight as R = 51'2 (Take g = 10 ms-2) a: IOOv'3 mls b. 200v'3 mls a. 30\" b. 45° c. 300v'3 mls d. 500v'3 mls c. 60\" d. 90\" 61. A projectile can have the same range R for two angles 67. A ball thrown by one player reaches the other in 2 s. The of projection. It t, and 12 be the times of flight in the two cases, then what is the product of two times of flight? maximum height attained by the ball above the point of projection will be about a. tttz ex R2 a. 2.5 m b. 5 m b. l,t2 ex R c. 7.5 m d. 10 m C. t,t2 ex IIi 68. A ball rolls off to the top of a staircase with a horizontal velocity 11 m/s. If the steps are h metre high and b metre 1 wide, the ball will hit the edge of the nth step, if d. l,t2 ex R2 62. A ball is thrown at different angles with the same speed u 2ltu 2illl' and from the same point and it has the same range in both a.I1=- b,n=--- the cases. If y, and Y2 be tlle heights attained in the two gb\" gb cases, then YI + Y2 is equal to I\",z c . n =2-ghh-u2-2 d , n =g-h2 a. g 69. A projectile is projected with initial velocity (61 + 8]) c. d. u' m/s. If g = 10 ms\"-\"2, then horizontal range is 2g 4g a. 4.8 m b. 9.6 m c.19.2m d. 14.0m 63. The equation of motion of a projectile is 70. At a height 0.4 m from the ground, the velocity of a pro- Y= 12x - 3, vjectile in vector form is = (61 + 2}) m/s. The angle of -x- 4 projection is The horizontal component of velocity is 3 ms-'. What is a. 4S\" b. 60\" the range of the projectile? c. 30() d. tan' '(3/4) a. 18 m b. 16 m 71. A projectile is thrown in the upward direction making' an angle of 60\" with the horizontal direction with a velocity c. 12m d. 2l.6 m of 147 ms' . Then the time after whieh its inclination with the horizontal is 45°, is· 64. The range R of projectile is same when its maximum a. 15(./3 -I)s h. 15(v'3 + I) s heights arc h, and hz. What is the relation between R, h\" c,7.5(v'3-I)s d. 7.5(./3 + 1)s and hz? 72. A numherofbullets are fired in all possible directions with the same initial velocity u. The maximum area of ground a. R = vh,hz covered by bullets is b. R = v2h,h2 a. 7T (-/8')2 c. R = 2$1li2 c. Jr (,:)2 d. R = 4vh,hz .g 65. i. The friction of the air causes a vertical retardation equal to 10 % of the acceleration due to gravity (take g = 10 ms-z). The maximum height will be decreased by a. 8 % b. 9 % c. 10 % d. II %
5.38 Physics for IIT-JEE: Mechanics I 73. A person takes an aim at a monkey sItting on a tree and 80. Three particles A. B. and C arc projected li'om the same fires a bullet. Seeking the smoke the monkey begins to fall point with t.he samc initial speeds making angles 30°, 4Y, freely; then the bullet will and 60°, respectively, with the horizontal. Which of the following statements is correct? a. always hit the monkey a. A, B, and C have unequal ranges. b. go above the monkey b. Ranges of A, and C are equal and less than that of B. c. go below the monkey c. Ranges of A, and C arc equal and greater than that of d. hit the monkey if the initial velodty of the bullet is' B. more than a certain velocily d. fl, B, and C have equal ranges. 74. A person siUing at the rear end ufthe compartment throws 81. There are two values of time for which a projectile is to a ball towards the front end. The ball follows a parabolic the same height.. The sum of these two times is equal to path. The train is moving with uniform velocity of 20 (1' = lime of flight of the projectile) ms-l. A person standing outside on the ground also ob- serves the ball. How will the maximum heights (h m ) at- a. 31'12 b, 41'13 c. 31'/4 d. l' tained and the ranges (II) seen by the thrower and the outside observer compare each other? 82. A golfer standing on level ground hits a ball with a ve- locity of u = 52 rnls at an angle a above the horizontal. ff a. same hm different R b. same hm and R tan a = 51l2. then the time for which the ball is at leasll5 c. different hili same R m above the ground will be (take g = 10 rn/s2) d. different hill and R a.ls b.2s c.3s d.4s 83. i. The ceiling of a hall is 40 m high. For max imum horizontal distance, the angle at which the hal! may be thrown with a 75. Two stones arc projected with the same speed but mak- speed of 56 ms ..·! without hitting the ceiling of the hall is ing different angles with the horizontal. Their ranges are a. 25\" b. 30\" c. 45\" d. 60\" equal. If the angle of projection of one is 1f13 and its max- ii. In. question 83's part (i), the maximum horizontal distance imum height is hI then the maximum height of the other will be: a. 160V3 111 will be c. 120V3 In h. 140V3 111 d. lOOV3 m a. 3il l b. 2hl c. h l /2 d. h l /3 84. A body is projected up a smooth inclined plane with a velocity u from the point A as shown in the Fig. 5.110. 76. A ball is projected from the ground at angle 0 with the The angle of inclination is 45() and the top is connected to horizontaL After I s it is moving at angle 45() with the a well of diameter 40 m. If the hody just manages to cross horizontal and after 2 s if is moving horizontally. Wh~t is the well, what is the value of u? Length of inclined plane is 20v'2m. the velocity of projection of the ball? a. 10V3 mls b. 20V3 m/s' a. 40 m/s b. 40v'2 m/s d. 20v'2 m/s c. 20 mls d. 20v'2 mls e. 10../5 mls 77. A body is projected horizontally from the top of a tower Bc with an initial velocity of 18 ms'\" I. It hits the ground at an angle of 45\". What is the vertical component of velocity -A when the body strikes the ground? 40m Fig. 5.110 a. 9 mls b. 9v'2 m/s c. 18 m/s d. 18v'2 mls 78. A plane flying horizontally at 100 m/s releases an object 85. A rille shoots a bullet with a muzzlevelocity of 400 m/s which reaches the ground in 10 s. At what angle with at a small target 400 m away. The height above the tar- horizontal it hits the ground? get at which the bullet must be aimed to hit the target is a. 55° b. 45\" c. 60\" d. 75° (g = 10 ms·2 ). 79. A hose lying on the ground shoots a stream of water up- a. 1 m b. 5 m ward at an angle of 6(Y' to the horizontal with the velocity of 16 ms~ I. The height at which the water strikes the wall c. 10 m d. 0.5 m 8 m away is 86. A projectile is fried from the level ground at an angle 0 above the horizontal. The elevation angle () of the highest a. 8.9 m b.l0.9111 point as seen from the launch point is related to () by the relation c. 12.9111 d.6.9m
Motion in Two Dimensions 5.39 a. tani/! = 2tanO a. y = 2x - 5x 2 b. tani/! = tanO b. y = x - 5x2 1 c. 4 y = 2x - 5x2 c. tan,/> = 2tanO d. y = 2x - 25x2 1 92. A particlc is projected from the ground with an initial d. tani/! = ;;(tanO espeed of v at an angle with horizontal. The average 87. A projectile has initially the same horizontal velocity as velocity of the particle between its point of projection and it would acquire if it had moved from rest with uniform acceleration 01'3 ms-~2 for 0.5 min. If the maximum height highest point of trajectory is reached by it is 80 m, then the angle of projection is ¥a. vT+-z cos2 e (g = 10 ms' '2). b. ~ /I;-2-cos2 0 h. tan- I(312) a. tan- 13 d. sin- I(4/9) 2 c. tan- I(4/9) v c. - e3 cos2 2 88. If a stone s to hit at a point which is at a distance d away d. vcosO and at a height h (sec Fig. 5.111) above the point from where the stone starts, then what is the value of initial 93. Two balls A and B are thrown with speeds u and- u/2. speed II if the stonc is launched at an angle O? respectively_ Both the balls cover the same horizontal dis- tance before returning to the plane of projection. .If the ey/2'0-(,;1, ()_- ,- angle of projection of ball B is 15\" with the horizontal, a. g h) then the angle of projection of A is cos (Da. sin- I C)b. IsmI dd 2 \"8 c. ~ sin I ( ~) C)d. ;;I( sm-I \"8 eb. - - - - - - cos 2(dtanO - h) 38 hgdo- 94. A body of mass m is projected horizontally with a velocity v from the top of a tower of height h and it reaches the c. ground at a distance x from the foot of the tower_ If a h cos2 0 second body of mass 2 m is projected horizontally from d. Yjc;d;=dh25 the top of a tower of height 217, it reaches the ground at a distance 2x from the foot of thc t.ower. The horizontal 11 velocity of the second body is a, v b. 2v c. .,fiv . d. v/2 95. A car is moving horizontally along a straight'line with a uniform velocity of 25 ms-- 1 A projectile is to be fired +-d~ from this .car in such a way that it will return t.o it after it has moved 100 m_ The speed of the projection must be Fig. 5.111 89. The speed of a projectile at its maximum height is ../3/2 a. 10 ms- I b. 20 I11S-- 1 tirnes its initial speed. If the range of the,projectile is P times the maximum height attained by it, P is equal to c. 15 ms- I d. 25 ms- I a. 4/3 b. 2../3 e. 4../3 d. 3/4 96. The horizontal range and maximum hcight attained by a projectile are R and If, respectively. If a constant horizon- 90. The trajectory of a projectile in a vertical plane is tal acceleration a = g/4 is imparted to t.he projectile due to wind, then its horizontal range and maximum height will y = ax -- b)._2, where a and b are constants and x and yare be respectively horizontal and vertical distances of the pro- a. (R + H), -H2 jectile from the point of projection_ Thc maximum height attained by the particle and the angle of projection from ~)b. (R+ ,2H the horizontal are a. -b2 tmf-I(b) a2 e. (R+2H),H b. - talC I(2b) 2a! d. (R + H), H Ii a2 . d. -2a 2 tan-I(a) 97. A particle is projected with a certain velocity at an angle Ci above the horizontal from the foot of an inclined plane of c. ~\"- tan-lea) b' inclination 30°. If the particle strikes the plane normally then a is equal to 4b' 2J.91. A projectile is given an initial velocity of 7+ The cartesian equa!ion of its path is (g:::::: 10 m/s2).
5.40 Physics for IIT-JEE: Mechanics I ~)a. a. 8 mls 30\" + CI ( b. 6 mls tan c. ]() mls b. 45' d. not obtainable from the data c. 60' 103. In the above problem, the angle with the horizontal at which the projectile is projected is d. 30' + tan-I (2../3) (Da, tan- I 98. In Fig. 5.112, the time taKen by the projectile to reach G)b. tan- I from A to B is I. Then the distance A B is equal to ul ../3UI a. h b. -Z- v3 d. Zul c, ../3 ut u c. sin- 1 ( 43) . d. not obtainable from the given data. 104, A body is projected with velocity VI from the point A as shown in the Fig. 5.114. At the same time another body is projected vertically upwards from B with velocity \"2. AC The point B lies vertically below the highest point. For Fig. 5,112 , . V2 99. A motor cyclist is trying to jump across a path as shown in both the bodIes to collIde, - should be Fig. 5.113 by driving horizontally off a cliff A at a speed of 5 ms~' I, Ignore air resistance and take g;;;;;; 10 ms-2 . The VI ![a.2 b. e. 0.5 d.1 speed with which he touches the peak B is --\"I-- a. 2.0ms- 1 b, 12ms- 1 v, tv, c.25ms-- 1 d. 15 ms- 1 .. -13.00 B A Fig. 5.114 ,'- \" t~~B 105, A ball is projected from a point A with some velocity at an 60m angle 30\" with the horizontal as shown in the Fig. 5.115. 70111 ~ Consider a target at point B. The ball will hit the target if ~ it is thrown with a velocity Vo equal to a. 5 ms,·,·j b, 6 ms--' Fig. 5.113 c. 7 ms- 1 d, none of these 100. An aeroplane is flying horizontally with a velocity of 600 kmlhr and at a height of 2,000 m. When it is vertically Ak\"O above a point A on the ground, a bomb is released from it. The bomb strikes the ground at point B. The distance AB is (neglecting air resistance) a. 1,200 m b. 0.33 km 51 III c. 3.33 km . d. 33 j<:m Fig, 5.115 101. The range of a projectile which is launched at an angle of 106. A body is moving with a constant speed of 5 mls in a IS' with the horizontal is 1.5 km. What is the range of the projectile if is projected at an angle 45\" to the horizontal? radius of 2 m. The acceleration of the body is a. 1.5 km b. 3 km a. 0 b. 12.5 ms~-2 c. 6 km d. 0.75 km c, 25 ms- 2 d. None of these 102, The height y and the distance x along the horizontal plane 107. The velocity of a body moving in a circle in the radical of a projectile on a certain planet (with no surrounding direction is atmosphere) are given by y = (8t - 5t 2) m and x = 6t a.O m, where t is in seconds. The velocity with which the projectile is projected at t = 0, is b, speed of the body
Motion in Two Dimensions 5.41 c. v2!R Multiple Correct Solutions on page 5.62 d. none of the above Answers type 108. A body is moving in a circle with a speed of I m/s. This 1. A river is flowing towards east with a velocity of 5 ms-- l . speed increases at a constant rate of 2 m/s ever second. Assume that the radius of the circle described is 25 m. The boat velocity is 10 ms\"\" I. The boat crosses the river by The total acceleration of the body after 2 s is shortest path, hence, a. the direction of boat's 'velocity is 30° west of north. a. 2 m/s2 b. 25 m!s2 h. the direction of boat's velocity is north-west. c. v'sm/s2 d. V7m/s2 c. resultant velocity is 5v'3 ms\"'. 109. A body is moving in a circular path with a constant speed. d. resultant velocity of boat is 5,/2 ms-I. It has a. a constant velocity. 2. A stationary person observes that rain is falling vertically down at 30 krn/h. A cyclist is moving up on an inclined plane b. a constant acceleration, makiug an angle 30° with horizontal at 10 km/h. In which di- c. an acceleration of constant magnitude. d. an acceleration which varies with time in magnitude. rection should the cyclist hold his umbrella to prevent himself from the rain? (3~)a. At an angle tan-I with the inclined plane. 110. A stone of mass In is tied to a string of length land rotated 5b. At an angle tan - I (3v'3) wI\"th the hon.zontal. in a circle with a constant speed v. If the string is releases, the stone flies ~)c. At an angle tan-I ( with the inclined plane. a. radially outward rl. At an angle tan- ,(7~),WIth the verti.cal. b. radially inward c. tangentially outward d. with an acceleration mv2/l 111. A particle is moving along a circular path with uni- 3. Two cities A and B are connected by a regular bus services form speed. Through what angle does its angular velocity change when it completes half of the circular path? with buses plyiIlg in either direction every T seconds. The a. 0\" c. 180\" speed of each bus is uniform and equal to Vb. A cyclist cycles b. 45\" d. 360\" from A to B with a uniform speed of Ve. A bus goes past the· 112. A particle is moving along a circular path. The angular cyclist in TJ second in the direction A to B and every 72 velocity, linear velocity, angular acceleration, and cen- second in the direction B to A. Then tripetal acceleration of the particle at any instant, respec-, a. 7,, = -V-bT -- b. T, = -\"Vb-T-- w, v,tivcly, are Vb-Vc ct, and {ic. Which of the following rela- Vb+Vc , V bT tions is not correct? h. wJ..a c. T, = VbT -\"-- d. 7, = -Vb-+\"-V-c a. wJ..v Vii - Vc 4. Suppose two particles, I and 2, are projected in vertical plane simultaneously. 113. What is the angular velocity in rad!s of a fly wheel making 160 m/s 100 mis 300 r.p.m? a. 600n b. 20n c. JOn d. 30 114. An air craft executes a horizontal loop of radius 1 krri with a steady speed of 900 kmh\" I. Find its centripetal Fig. 5.116 acceleration. e,Their angles of projection are 30° and respectively, with a. 62.5 m/s2 b. 30 m/s2 the horizontal. Let they collide after a time t in air. Then c. 32.5 m/s2 d. 72.5 mis' ea. = sin\"-'(4/5) and they will have same speed just before US. The angular velocity of a particle moving in a circle of the collision. radius 50 cm is increased in 5 min from 100 revolutions per minute to 400 revolutions per minute. Find tangential eb. = sin-I (4/5) and they will have different speed just be- acceleration of the particle. fore the collision. a. 60 m/s2 b. n 130 mis' c. x < I 280v'3 - 960 TIl. c. n 1 J5 m/s2 d. n 160 m/s2 d. It is possible that the particles collide when both of them arc at their highest point. 5. All the particles thrown with the same initial velocity would strike the ground,
5.42 Physics for IlT-JEE: Mechanics I \"\"\"~ \"~~a. 0 t b. 0 x ,. \"'\"~\"'\"\"~c. 0 0\" CD \" t d. () x CD v v 6 CD 8) \" Fig. S.117 10. The coordinates of a particle moving in a,plane are given by: x = a COS(pl) and y = h sin(pl), where a. h « a) and I' arc a. with the same speed positive constants of appropriate dimensions. then b. simultaneously. c. time would he least for the particle thrown with velocity a. The path of the particle is an ellipse. v downward, i.e.• particle (I). b. The velocity and acceleration of the particle arc normal to d. time would be maximum for the particle (2). each other at I = Jf /21'. 6. A particle is moving in xy-plane with y = x /2 and c. The acceleration of the particle is always directed towards y = 4 - 2t. The correct options are a. Initial velocities of x and y directions are negative. a focus. b. Initial velocities of x and y directions are positive. c. Motion is first retarded, then accelerated. d. The distance travelled by the particle in time interval, t = 0 d. Motion is first accelerated, then retarded. to 1= ,,/21' is a. 7. A particle is dropped from a tower in a unifonn gra-vitational field at t = O. The particle is blown over by a horizontal wind t 1. A body is projected with a velocity u at an angle of projection with constant velocity. Slope (m) of trajectory of the particle with horizontal and its kinetic energy vary according to the owith the horizontal. The direction of velocity of the hody curves. makes angle 30\" with the horizontal at. t :::: 2 s and then after 1 s it reaches the maximum height. Then eb. = 60' a. If = 20~m/s, c. 0 = 30' d. If = IO~ Ill/S 12. A particle moves in a circle of radius 20 cm. Jts linear speed h. is given hy v :::: 2t where t is in s and v in m/s. Then a. The radial acceleration at t :::: 2 s is 80 ms-2 . t h. Tangential acceleration at t = 2 s is 2 ms~\"2. c. Net acceleration at t ;::::: 2 s is greater than 80 ms·-2 . m d. Tangential acceleration remains constant in magnitude o-::+-:::----~x Ct 'k. Asseittion-Reasoning Solution's on page, 5.63 KE . dt 'k Tyge . o I~ KE o 11 __ x Tn the following questions, each question contains STATE- MENT I (Assertion) ancI STATEMENT II (Reason). Each 'lues, 8. A car moves 011 a circular road, describing equal angles about tion has 4 choices (a). (b). (c). ancI (d) out of which ONLY ONE the centre in equal intervals of time. Which of the following is correct. statements about the velocity of car are not true? a. Velocity is constanl. (a) Statement l'is True, Statement II is True; Statement n is a h. Magnitude of velocity is constant but the direction changes. correct explanation for Statement 1. c. Both magnitude and direction of velocity change. (b) Statement I is True. Statement II is True; Statement II is NOT d. Velocity is directed towards the centre of circle. a correct explanation for Statement L 9. A heavy particle is projected with a velocity at an angle with (c) Statement 1 is True, Statement It is False. the horizontal into a uniform gravitational field. The slope of (d) Statement I is False, Statement [! is True. the trajectory of the particle varies not according to which of the following curves? 1. Statement I: The projectile has only vertical component of velocity at the highest point of its trajectory. Statement II: At the highest point only one component of velocity is present. 2. Statement I: The time of night of a' body becomes n. trmes of original value, if its speed is made 11- times.
Motion in Two Dimensions 5.43 Statement II: This due to the rauge of the projectile which 2. The direction of rainfall with vertical is becomes n times. a. 45' b. 30\" d. 90\" 3. Statement I: If the string of an oscillating simple pendulum is cut, when the hob is at the mean position, the bob falls For Problems 3-4 along a p~~rabolic path. A person can swim in still water at the rate of 1 km/hr. He tries Statement II: The bob possesses horizontal velocity, at the to cross a river by swimming perpendicular to the river flowing mean position. at the rate of 2 krn/hr, If the width of the river is 10m, then 4. Statement I: In a plane projectile motion, the angle between 3. The location of the point where he lands on the other side instantaneous velocity vector and acceleration vector can be of the river ahead of the point exactly opposite to his start anything between 0 or n (excluding the limiting case). point is Statement II: In plane to plane projectile motion, accelera- tion vector is always pointing vertical downwards. (Neglect a.20m b. 10m e. 15 m d.25m air friction). 4. The time taken by him to cross the river is 5. Statement I: A hody with constant acceleration always a. 30 s b. 36 s c. 54 s d. 45 s moves along a straight line. Statement II: A -body with constant magnitude of accelera- For Problems 5-6 A river 400 m wide is flowing at a rate of 2.0 m/s. A boat 'is tion may not speed up. .. sailing at a velocity of to m/s with respect to the water, in a direction perpendicufar'to the river. 6. Statement I: In a non-uniform circular motion tangential ac- celeration arises due to change in magnitude of velocity. 5. Find the time taken by the boat to reach the opposite bank. Statement II: In a non-uniform circular motion centripetal acceleration is produced due to change in direction of veloc- a. lO s b. 30 s ity, c. 40 s d. 20 s 7. St.atement I: In a uniform circular motion angle between ve- 6. How far from the point directly opposite to the starting point docs the boat reach the opposite bank? locity vector and acceleration vector is always \"n2' Stat.ement II: For any type of motion, angle between accel- a. 100 m b.80m eration and velocity is always is ::. c. nO m d.40m 2 For Problems 7-10 8. Statement I: Two particles start from the rest simultaneously A man can row a boat 4 km/hl' in still water. If he is crossing a river where the CUlTcnt is 2 kmJhr, then answer the followings and proceed with the same acceleration. The relative velo,,?ily 7. In what direction should his boat be headed if he wants of these particles will be zero throughout the,motion. to reach a point on the other bank, directly opposite to Statement ll: At every moment the two particles will have starting point? the same velocity. 9. Statement I: A river is flowing from east to west at a speed of a. 120° downstream Sm/min. A man on south bank of river, capable of swimming b. 12(P upstream 10m/min in still water, wants to swim across the river in shortest time. He should swim due north. c. 60° downstream Statement II: For the shortest time the man needs to swim perpendicular to the bank. 10. Statement 1: Rain is falling vertically downwards with ve- d. Perpendicular to the flOW of river locity 6 km/h, A man walks witha velocity of8 km/h, Relative velocity of rain \\V.r.t. the man is 10 kmih. 8. If width of (he river is 4 km how long will it take him to Statement II: Relative velocity is the ratio of two velocities. cross the river, with the condition in question 7? , (iO'fllprel1ensi1l6 a. 5)7 hr 2~ 4 b. - - hr 3 [ype, \" Solutions on page 5.64 4 3./2 c. - - hr d. -4hr 2-/3 For Problems 1-2 9. In what direction should he head the boat if he wants to A man is walking due east at the speed 01'3 kmh-- 1, Rain appears cross river in shortest time? to fall down vertically at the rate of' 3 kmh'-!. a. 1200 downstream 1. The actual velocity of the rainfall is b. 120\" upstream a. 3~krn/h b. 2./2km/h c. 60° downstream c. 4~km/h d. 3./2km/h d. Perpendicular to the flow of river
5.44 Physics for IIT-JEE: Mechanics I 10. How long will it takc him to row 2 km up thc stream and 18. The horizontal distance between two bodies, when their then back to his starting point? velocity are perpendicular to each other is a. 3/4 hr b. 3/2 hr a. 1 m b . 0.5 m c. 2/3 hr d. 4/3 hr c.2m d.4m For Problems 11-12 19. The time taken forthe displacement vectors of two bodies A car is moving towards south with a speed of 20 m/s. A mo- torcyclist is moving towards east with a speed of IS mls. At a to be come perpendicular to each other is certain instant, the motorcyclist is due south of the car and is at a.O.ls b.0.2s a distance of 50 m from the car. c. 0.8 s d. 0.ji).8 s 11. The shortest distance between the motorcyclist and the For Problems 20-21 car is A particle starts from origin at t = 0 with a velocity 5.0 1m/s and a. 10m b.20m moves in x-y plane under the action of a force which produces a constant acceleration of 3.01+2.0) m/s2 c.30m d.40m 20. y-coordinatc of the particle at the instant its x-coordinate 12. The time after which they are closest to each other is 84 m a. 84 m a. 113 s b. 8/3 s b. 56 m c. liS s d. 8/5 s c. 48 m d. 36 m 21. The speed of the particle when its y-coordinate is 84 m For Problems 13-14 a. 48 mls b. 36 mls A man can swim at a speed of 3 kmlh in still water. He wants c. 26 mls d. 3,000 mls to cross a 500 m wide river flowing at 2 kmlh. He keeps himself always at an angle to 120D with the river flow while swimming. For Problems 22-23 13. The time taken to cross the river is b; I eA particle is projected at an angle with the horizontal such that a. -3 h -h it just able to clear (horizontally) a vertical wall of height h at a 6 distance h from point of projection as shown in Fig. 5.118. 2 ' i< c. _I_h' d. None L eo 3~ h 14. The drift of the man along the direction of flow, whcn he arrivc at the opposite bank is Fig. 5.118 I b. 6~ e22. The angle of projection is a. -6-~km c. 3~ a. tan\" 2 b. tan-' 3 For Problems 15-16 c. tan-'(2/3) d. tan-'(3/2) To a stationary man, rain appears to be falling at his back at an angle 30° with the vertical. As he starts moving forward with a 23. The velocity of projection l( is b. Viii speed of 0.5 ms\", he finds that the rain is falling vertically, a. cj2gh d. j~gh 15. The speed of rain with respect to stationary man is c. cj5gh a. 0.5 ms\" b. 1.0 ms·· J For Problems 24-25 c. 0.5~ ms\" d. 0.43 ms· J A ball is thrown from a point in level with and at a horizontal distance r from the top of a tower of height h. 16. The speed of rain with respect to the moving man is a. 0.5 ms- 1 b. 1.0 ms~l u c. 0.5 ~ ms-' d. 0.45 ms\" oA For problems 17-19 o - -------~c-------f From a tower of height 40 m, two bodies are simultaneously projected horizontally in opposite directions, with velocities of ..tower h 2 mls and 8 mis, respectively. 8+x+ Fig. 5.119 17. The time taken for the velocity vectors of two bodies to 24. How must the speed and angle of the projection of the ball become perpendicular to each other is be related to r in order that the ball may just go grazing a. 0,1 s b. 0.2 s past the upto edge of the tower? c, 0.4 s d. 0.8 s a. gr = u2 sin 28 b. gr = u2 sinO
Motion in Two Dimensions 5.45 32. hi. 25. At what horizontal distance x froni the foot of the tower a. g/2 b. g does the ball hit the ground? c. 3g/2 d. 2g a. uco.e .in' 0 + 2gh)I/' - . For Problems 33-35 --{(u' U lIn 0) g eA projectile is thrown with velocity v at an angle with the b. -u-s{in(Bu'co')s'O+2gh)1/'-ucosO) horizontal. When the projectile is at a height equ~l to half of the maximum height, then Ii 33. The vertical component of the velocity of projectile is c. _u si_ne {(u 2 co.' 0 +gh)1/2 - ueosO) a. vsinex 3 b. vsine +3 g ed. -uc-o{sO( u 2s.m2 + gh) 1/2 - 11 sinO) v sine d. v sine Ii -/3 c. ../i For Problems 26-29 A 0.098 kg block slides down a frictionless track as shown in 34. The velocity ofthe projectile when it is at a height equal Fig. 5.120. to half of the maximum height is •, a. v cos, 2 0 +si-n22-e b• ../i v cose , A .~\". ----\\.B,- -----1 c. ../i v sin e d. vtane seee 3m!, f35. What is the angle of projectile with the vertical if. the 30° \\C 1m velocity at the highest point is times the velocity when x it is at a height equal to half of the maximum height? Fig. 5.120 a. 15° b.300 d.60° 26. The vertical component of the velocity of block at A is For Problems 36-39 A stone is thrown with a velocity of 20 mls at angle of 30° above a. y'g b. 2y'g horizontal from the top of a building 15 m high. Find c. 3y'g d. 4y'g 36. The time after which the stone strikes the ground. 27. The time taken by the block to move from A to B is a. 1 2 a. 6 s b. 3s y'g b. - c. 1.5 s d. 0.75 s ~g 3 4 37. The distance of the landing point of the stone from the c. d. - building. y'g y'g 28. The time taken by the block to move from A to C is a. 20../i m b. 1O../i m a. fg b. fg c. 20-/3 m d. 30-/3 m 38. The velocity with which the stone hits the ground. I d1.+--/-3 a. 10,(7 mis b. 20,(7mis c. - y'g c. 10-/3 mis d. 20-/3 mis y'g 29. The horizontal distance x traveled, by the block in moving 39. The maximum height attained by the stone above the from A to C is ground. a. (l +-/3)m b. (1- -/3)m a.5m b. 10m c. 20m d.40m c. (-/3+3)m d. gm. For Problems 30-32 For Problems 40-41 A projectile i. thrown with velocity v making an angle 0 with the horizontal. It ju.t crosses the topes of two poles, each of height A block sides off a horizontal table top I m high with a speed h, after I sand 3 s, respectively. of 3m/s. Find 30. The time of flight of the projectile is 40. The horizontal distance from the edge ofthe table at which the block strikes the floor. a. -10m 3 -/3 b. --m a.1s b.3s c.4s d.8s 3 v'1O 31. Maximum height attained by the projectile is c. - m 5 a. g b. 2g c. 3g d.4g ..J5 d. -m -/3
5.46 Physics for IIT-JEE: Mechanics I 41. The speed of the block when it reaches the fioor. 48.. At the point where the particle's velocity makcs an angle 012 with the horizontal. a. ffi mls h. .J24 mls c. v'29 mls d. v08 mls eu2 cos2 sec3 ~ eu2 cos2 sec3 0 a. - g For Problems 42-43 h. A Projectile shot at an angle of 45° above the horizontal strikes a e2u 2 cos2 sec3 ~ 2g building 30m away at a point 15m above the point of projection. Find c. e ~ul cosl sec3 2 42. The speed of projection g d. J]g b. 6J1Omls a. 8,/5mls 49. At the point where the particle is at a height of half the c. SKm/s d. 10/6mls maximum height H attained by it. 43. The magnitude and direction of velocity of projectile 2u 2(1 + cos' 0)·1/2 h. l/2(l + cos2 0)3/2 when it strikes the building. a. \".\"--.-\"~----- g2j2cosO a. IOJ] mis, horizontal g 2,/2 cosO h. IS mis, horizontal c. 8 mis, horizontal ,,2(1 - sin' 0)3/2 [/2( I _ tan2 0)'/' d. 10 mis, vertical d. c. g 2,/2 cos 0 g ,/2 cos I) For Prohlems 44-46 MatcHing , Solutions on-pdge,5,69 (lo[umn Tllpe A particle is projected horizontally with a speed 11 = 5 mls from ethe top of a plane inclined at an angle = 37° to the horizontal as shown in the Fig. 5.121. Fig. 5.121 1. If VIII1/! = velocity of a man relative to water1 VlI! = velocity of water. VIII = velocity of man relative to ground, match the 44. How far from the point of projection will the particle strike the plane? following a. 75 m 65 CohunnI 75 h. -c m ,i., Mihinium distance for c. -m 16 16 85 d. -111 9 45. Find the time taken by the particle to hit the plane. iii.- Minimum distance for ~.\"vmw'J-vw a. 3/4 s b. 3 s iV.-,'Minitrilim' time for v;nm'-< d.e =. sin'\" - - c. 4 s 4 1J,nw d. - s 3 46. What is the velocity of the particle just before it hits the plane? e vmwwhere = angle between and the width of the river, 5ma. mls 2m5 2. A ball is projected from the ground with velocity v such that its range is maximum. h. mls c. 10m mls d. 5J2iim/s --cOIiihiliT---- Column II v For Prohlems 47-49 '~--,-'----c-' A particle is projected with a speed\" at an angle 0 to the hori- 1l. '-- i; Velocity at half of themilximum height zontaL Find the radius of curvature h.-~ ... . . 2 47. At the highest point of its trajectory, ii.Veloci ty at the rnaXilTlUIn height eu2 cos2 J] ,,2 cos2 0 iii. Change in its velocity whenitretuflls to c. v,/2 a. h. lhegt-,,-urtd< .~~_ _ _ _ _~~! 2g 2g iv. Average velocity when it reaches the d,'!.~ eu2 cos2 J] [/, cos2 0 ____2..Y._2__ d. ----_.- '---_,c'm=a=ximc=um=,h:.e:i.g.h..t.:._____........_.. ____........ c. - - - - Ii g
Motion in Two Dimensions 5.47 Column!, i., Fbfapartic]elnoving in a circle' ',' a(rheacceleration 'l11ayheperpendicul!li' _'__'____'_ _-'-~,-'-~'--\"'___\"_~-'-~~-'-f\"-cl\\Q::c.\"-itc:s,velQcity , '' ii.For apartiC!emoyi9g in~straightIiil\" ' h.Theaeceietat1Pill!)aYbeinthedirecti()n pfvel&city, ' ..' ' alo ac:Jheac'1c.(cll.e;r;att)l\"qn,i.m\\1;y·(p?e~Va~tI$o(c)iJt1r\\.\"a\"llgle iv. 'Por aparticleis mOl'illgin spa,,\" d.TheacG~Ierati(jiimayb() opposIte t;' ii\" velocity , , ANSWERS AND SOLUTIONS Subjective Type 192 min =>S. tshort-path - tshort-!ime = 4 1. For D, time taken where v = 17 mis, U = 8 mls 1.5 1.5 solve to get, d = 510 In 11 = - - - + - - - = 3,2 h = 4+3,5 4-3,5 6. vm = 57, Vrjm = -vi, Vr = Vrjm + Vm = Sf - v) •• _3.5kmlh v, = ~52 + v2 = 10 =} v = 5.)3 AB So Vi: = 57 - 5.)3] Fig. 5.122 s For C; time taken. 12 = 41.5' + 41.5' = 43 II = 45 min 8 ,, sfi ----: 2. Letvelocity or boat is v, then Fig. 5.124 v= 8+8 = -- 8 km/h 2 Time taken when the\" water is flowing; 88, e5 = 30' t = --+ - - =2h40mm tanO = ;;; =} 8+4 8 --4 3. a. v, = 18m/s, v, = 13m/s 5,,3 b. v-'le = v, - lie = 18 - 13 = 5 m/s 7. V, = 27+ 1.25] + 0,25 k c. v\" = - 3 mis, V\"/' = v,, - v, = -3 - 13 = -16m/s ~22Magnitude: Iv,,1 = ~JI m/s °-Vs = 0, Vsjc = Vs -'- Vc = 13 = '-13 m/s + 1.252 + 0.252 - Negative sign indicates left direction. 8. Both should reach simultaneously at A, 4. VI =20eos3oo7+20sin30] = 10.)37 + 10] TS = 40 km =? T A = 40 km V2=V], VI/2=VI-V2= 1OJ37+(I0-v)] N Time taken by bomber to reach A: ~W E 40 2 , s t = -]00 = -5 h = 24 mm Fig. 5.123 So time taken by interceptor plane to reach A: For Vl/2 to be towards east, its .7 comp should be zero, =} 10.- v = 0 =? v = 10 km/h 40v'2 400v'2 24 - ]6 = ]8 mm, v = (18/60) = - 3 - kmlh
5.48 Physics for IIT-JEE: Mechanics I 9. x = ( U-VSinB)d- =} dx = 0 =} sm. B = -v cosB v u - dB cosO:' = ~ = a = coS-I (~) IIU Fig. 5.128 ' 1°. -dr= nu sin e -.: u de ld r 0 nu cosB . Fig. 5.126 dsece =}r=-----~ I (u-fH (vu2 - v')d [tan [~+ ~]F _ _. 16 4 gwe can get x = = 13. VAIB = 12i + 16j, tan a = 2\" = 3\" v 2 VAIB = ,,1(12)2 + (16)2 = 20 kmlh . 1-- u2 10. vm = vm/w + Vw vm = v cos(B + 45')1 + v sin(B + 45')J + ucos45'1 - u sin 45'J o vm u= [v cos(B + 45' + cos 45°Jl + [v sin(B + 45°) - u sin4SOJJ JTo reach the point B, component should be zero. 16 kmlh A \"----LL_ __ x Fig. 5.129 B DB=EBsin\" =(OB - OE)sina y OA . A = (AB cos 45° - --)8ma Fig. 5.127 tanO! v sin 45° = [~ - AB cos45° x ~] x ~ v sin(B + 45°) = u sin 45° =} v = = ~[~Jx~=..fikm sin(B + 45°) 14. UI = -u,j, ih = uzcose J+uzsinel eFor Vmin, + 45° = 90° =} Ii = 45° \"2 -=} U, = U2 sinel+ (U2cosB +u,)J vrn\" = ..ufi' 45° wi.th upstream. 11. From A to B: AB = (vcosjJ +ucosa)tI from B to A: AB = (UC08jJ - u cosa)t, (i) a = 60°, v sin,B ~ u sin a- now I = II + 12 A 5x 60 = 1200 + - -1-20-0 - - a vcosjJ+ucosa (vcosjJ-ucosa) e =} put U = y'I7, a = 60°, we get (ii) Fig. 5.130 =} V2COS2 jJ - 8vcosjJ - 417 = 0 =} vcosjJ = 217\" tan\" = -- U2 sin e n;From equations (i) and (ii) tan jJ = =-- ,+- -- Uz cosB UI JUT cosBeAB = 'In=} jJ = tan-I (3 , -;==XOU~2 ~sin~=== u~+ + 2UIUZ and v = V85 mls Xo sm\" = 12. dr = (-u + nu sin e)dl, de = (nu cos e)dt 15. Let \"v, = '\"VO', Vhl' = -2v\" j - v'.T)3',
Motion in Two Dimensions 5.49 2' } + 2V\" f!' = - VI, ( Vo ~ Vj -/3)\"I f. V = Ju 2 + g2[2 - 2ugt sin Ii = v'402 + 102 X l' 2 x 40 x 10 x 1 x sin 30\" = 36 mls 40sin30' - 10 x 1 I tan,B = ---:-:--::::-- = - - ~ ,B = 16\" 40 cos 30\" 2v3 19. u = 30 mis, Ii = 60' V = v''''3''02''-+-;lcoO''2-x-;2''2>--;;2:-x-;;-3''O-x\"'-;-;lO~x-;2;-s7il-16'-;0;::u Fig. 5.131 = 16.14 mls -/3 2110 30 sin 60' - 10 x 2 3-/3 - 4 -/3 vI 0~ tan,B = 30cos60\" = 3 = 0.4 2Now: Vo - = VI = '7)also: (~I)' + (VO - VI ~ ,B = tan-I (0.4) = 21.8\" 2 = (5-/3)' . u2 102 20. a. Rmo> = - = - = 10m 411' _0 = 75 x 4 g 10 ~ liT = 75 x 4 ~ 3 b. 100 = u' ~ u = 1O.jjQ 10 110 = 15 mls 16. a. True, the speed of the projectile is minimum atits highest u' (1O.jjQ)' position, because as the projectile goes up its speed first Hm\" = -2g = -'-2::-x--;1-0;c- = 50 m decreases and then increases while coming down, becom- ing minimum at the highest point u2 c. 160 = 10 ~ u = 40 m!s b. True, at highest point acceleration is perpendicular to ve- locity. lu sin e u2 sin2 30° u2 u2 sin2 60° 3u2 8g c. False. T = , so time of flight depends upon angle 21. HI = = - . H2 = 2g 8g 2g g Clearly H2 = 3HI of projection. u2 ~ u2 = 2gH Sum of angles = 60 + 30 = 90\", the horizontal range is same d. False, H = - 2g _ 202 sin(2 x R_ 45\") _ -_ -2 x-2-0 s.in-45-\" __ 2vh2 22. -40m, T s u' 10 10 Rmax = - = 2H l?-'~ g e. True, a body feels weightlessness during projectile motion because it is a frcc fall motion. f. True, direction of velocity is always along tangent to the /<af---25 In ,III x--+l path of a particle. >1 g. False, the instantaneous magnitude of velocity is not equal to the slope of the tangent drawn at the trajectory of the par- I~ R ticle at that instant, but it is equal to the slope of position- time graph. Fig. 5.132 y=300tan45'- -10(300)' =255m x = 40 - 25 = 15 m, h 15 Required speed = h = 5.3 m!s 2v2 e23. Given u = 16 m/s. = 30\" 2 (I OOv 2)' cos' 45' 18. Given Ii = 30\". u = 40 mls u\"\"16m1s a. T= 2u sin Ii 2 x 40 sin 30' =4s g 10 u' sin 20 (40)' sin (600) b. R = = = 80-/3 m h g 10 u2 sin 2 Ii 402 sin' 30\" ~R_ c.H= = =20m 2g 2g Fig. 5.133 16-/3) d. time of ascent = 'T2 = 42' = 2s R = uxt = 16cos30° x 4 = -2-'- x e. 110 = U cos Ii = 40 cos 30\" = ~o-/3 mls it is in horizontal direction. -h = 16sin 30\" x 4 - ~(lO)(4)' ~ h = 48 m
5.50 Physics for IIT·JEE: Mechanics I 24. -30 = -10 x sin30\"t - ~ lOt2 =} 5/2 + 5t - 30 = 0 =} t2 + t - 6 = 0 Fig. 5.134 e ,I 1<I1000c---- x +=} (t 3)(t - 2) = 0 =} I = 2 s Fig. 5.136 R ;X40 i--------------u= 720kmIh= 200mls 25. T= --9=.82.86s,R=. uT=20x2.86=57.2m 26. a. T = )2 x 120000 =20. s, AB = uT = 540 x 158 x 20 = 3000m 1.5 km uo= 400 mls 2 x 245 . b. T = 100 x 10 = 0.7 s, Distance = uT = 54 x 158 x 0.7 = 10.5 m 27. h = 2110(3)2 = 45 m Fig. 5.137 rF\"=---I>u g(50)' 30. 13 = 50tanO - 2(1O~)2cos2e h 17 3 25 tan2 e - 100 tan e + 51 = 0 =} tan 0 = -5'5- time taken will be less for tan 0 = 3/5. 31. X2 = u cos 8 t I X2 - =Xl V - t Xl + v t = u cos 8t ,3D -~------ v v Fig. 5.135 u h 3 hl2 vcos30° = 10 x 3 =} v = 20y') mls e u = v sin 30° = 20y') x ~ = lOy') mls I4--XI~ 28. -H = tan 0 =} H = xtane (see Fig. 5.136) x o ~1.----X2----.0>I1 Let the particle reaches at P at time I, then Fig. 5.138 hi = . - I 2 .usmOt _2gt gx 2 3h u2 sin2 0 h=xtan8- 2u2co2s 0' 2= 2g I Simplify to get x 2 6h cotO x + 6h2cot2e = 0 2Height descended by apple in time I: h2 = gl2 6hcote ± Mhcote X =} x = '-'--~--- hi + h2 = usinOt = usine-- =xtanO = H 2 . u cose = 3hcote ± y') h cote It means particle will hit the apple. X2-XI 2y')hcote 29. uo sin 0 = u =} 400sinO = 200 (see Fig. 5.137) v= --- = . = -I =} 0=30\" II smB 2 X2 (3+y'))hcote Verify that maximum height attained by the shell is u cos 0 = - = -'-----'--- greater than or equal to 1.5 km. It _v_ = 2y') ~: 2y')(3 - y')) = y') _ 1 ucosO 3+y') .A. 9-3
Motion in Two Dimensions 5.51 32. x = 0.2 m, nx = 21g12,nx > ut =} n2x 2 > u2t2 u =}n2x2 >u2-2n-x 2(1.8)' T 6m T g n>--- 15m 18.75 m 2u' 2 x 0.2 =} n>- =} T gx =} n > 3.24 Fig. 5.140 =} 15 =usinex1 --1-lO(1-)2 4 22 2 =} usine = 10 -15 = lOT - ~ lOT2 =} T2 - 2T - 3 = 0 =} T = 3 s 2 36. a. 0 = u sin (a - fJ)1 - ~gCOS fJt2 t = 2u sin(a - fJ)1g cos fJ (i) Fig. 5.139 0= u cos(a - fJ) - g sin fJI =} I = u cos(a - fJ) (ii) 1 (n -1)x = 2gtt =} utI> (n -1)x g sin fJ u2tt > (n - 1)'x' x u u22(n - l)x > (n- 1)2x\" g 2u' Fig. 5.141 =} - > n - 1 =} n - 1 < 3.24 =} n < 4.24 From equations (i) and (ii) gx gR' 2 sin(a - fJ) cos(a - fJ) 33. -h = Rtane - 2u'cos'e = cos fJ sinfJ gR2 sec2 e (i) =} 2 tan(a - fJ) = cot fJ = } h = - R t a n e +2u-2- (i) =} 2(tan a - tan fJ) = cot fJ eDifferentiating w.r.t. l+tanatanfJ dR =} 2tana - 2tanfJ = cotfJ +tana =} 0 = -Rsec'e +tane de +2gu-R2 2 sec2e + g sec'e 2 RdR- =} tana=cotfJ+2tanfJ 2u' de h. 1= 2u sin (a - fJ)/gcosfJ dR u2 Put - = 0, we get R = - - , put R in de· g tane e+ - -u2gsec2 u4 =} h = - - eg 2u2 -,;---oc-c Fig. 5.142 g2 tan' =} sine = 2(u' + gh) AlsO,t = -u sin-a =} 2 sin(a - fJ) sina =e=} 43.78\" g cos fJ =} 2 sin a cos fJ - 2cos a sin fJ = sin a cos fJ 34. Superimpose an upward acceleration a on the system. The =} sin a cos fJ = 2 cos a sin fJ box becomes stationary. The particle has an upward acceler- ation a and a downward acceleration g. If a = g, the particle =} tan fJ = 2 tan fJ has no acceleration and will hit C. If a > g, the particle has 37. A to B: 0' = u2 sin' g - 2(gcosa)h a net upward acceleration, and if a < g, the particle has a net downward acceleration. A to C: -h = us.me1l - 2gc2osal 35. ucose = 12,6 = ucoset =} t = 0.5 s (Fig. 5.140) ucose 3.75 = uSinel - ~gl' And 0 = u sine - g sina I =} t = - - gsina u' e-h = u2 sinecose 1 gcosa cos' g sina 2 g2 sin2 0:
5.52 Physics for IIT-JEE: Mechanics I tlt = I, - 12 = 2 - I = I s \"'c b. The coordinates of P at which the two shots collide are x = X; + v;12 = 0+ (5vS) (1) = 5vS m andy = y;;- ZI gl,2 = 10 - I2 Z(lO)(I) = 5 m Fig. 5.143 39. The motion of the sphere takes place in a plane; the x- and y-components of its acceleration arc Ax = g sin 30°, a y = 0 The x- and y- component of sphere's velocity at time I =2sare Vx = VOx - ax! = 3 sin 60° - g sin 30° x 2 e_u2 sin2 u2 sineeose eu2 cos a cos2 = -7.40 mls 2g sina = gsina 2g sin2 a e- sin2 e sin e cose cosa cos2 2cosa sina 2 sin2 a e ., ' 2e'osa Multiplymg both side by eos2 we get e e -- tan2 = 2 cot a tan cot2 a = 0 Fig. 5.144 tan2 e + 2cota tane - cot2 a =0 Vy = V cos 600 = constant = 1.5 mls So the magnitude of sphere's velocity is -2eota + v'4cot2 a +4eot2 a Jv; v;[vi = tane = 2 + = ,1(-7.40)2 + (1.5)' = 7.55 m!s tane = (Vl - I) cota 40. Horizontal Range = -2u'xu-y g 38. Let gun I and gun 2 be fired at an interval I'll, such that, Let the initial horizontal and vertical components of the 'I = Iz + b.t (i) velocity of the shot be u and v, respectively, where I, and '2 are the respective times taken by the two shots 2(u+vp)v , to reach point p, RJ = platfonn movIng forward g For gun 1: x - Xi = Vi COS 60°11 2(u-vp)v R2 = platform moving backward Y - Yi = Vi sm. 600 t\\ - 12 g - gIl 2 4uv 4vpv R, + R2 = - and R, - R2 = -- I .J3 I2 or, x = + lVif ! Y = Yi + - zgtl g g TVitI Xi 2 16v~v2 For gun 2: x - Xi = Vi COS OOtz Now (R, - R2) = - -gM, - Y-Yi = Vism' 0O tz - 21g122 and (R, - Rz)' 4v~ v =-x- or x = Xi + Vi 12 Y =y; - -Ig122 R, + R2 gu 2 V g (R, - RzJ' a. Now we can equate x- and y-coordinates of shots, i.e., +0 [ - = - U 4v~ (R, R2) lIUilj = Vi!Z or tl = 212 (ii) Elevation of gun and T.JV3 ;I, - Z1gl,2= 1- Z2g12 [g_, (V) _, (R, - R2)'] '\" = tan - = tan 4--2--vxp(R! + R) u 2 or, TvSV;', + ZI g(lz2 - I2,) = 0 (\"1'1') 41. True, tangential acceleration is along or opposite to the di- rection of velocity so it changes the magnitude of velocity. On substituting I, from eqn, (ii) into eqn, (iii), we get, Centripetal acceleration is perpendicular to the velocity so it changes only direction of the velocity. TvS + I , II; ( 212) Zg(-312 ) = 0 )2 X 342. a, = w'R = ( 2n , 24 x 3600 x 6400 10 = O,0338m!s2 or, 12 ( vSv; - ~g(2) = 0 ' or, 12 = 0 and 12 = .J23 IVi; = v2S x (51v0S) = I s , ,5 15 Therefore, I, = 212 = 2(l) = 2 s; 43. v = 27 x 18 = \"2 mls 2n 44. tv = 1200 rpm = 1200 x 60 = 40n radls
Motion in Two Dimensions 5.53 a, = w2r = 30 480rrz m/sz Refer to Fig. 5.147(b), 100 = (40rr)2 x 1 & s.m&= -Vw= - =Vw- o r = 3 0 0 45. a = 2 radls2 , WI = 0, / = 4 s Vm 2vw 2 + +a, Wz = WI a/ =} Wz = 0 2 x 4 = 8 radls So, it is 60e upstream. Iz &= 0 x 4 + I z' 4. a. 102 = vZ + 8Z (Fig. 5.148) x 2 x 4 = 16 rad 2at e 2b.= + WIt =} c, I revolution = 2rr rad =} 1 rad = _I_rev =} 16 rad v 2rr 810 16 8 8 x 7 28 = -rev = -rev = - - = - rev. Fig. 5.148 2rr rr 22 11 or v2 = 100 - 64 = 36 or v = 6 kmh- I d. after 4 s: v = Wzr = 8 x 10/100 = 0,8 mis, 5. d. v, = J102 + 32 = Jf69 km h- I (Fig. 5.149) a= v2 (0,8)2 6A mls z -r = ' c -10-/10-0 = 10 = 0.2 mls2 at = ar = 2 x - 100 Now magnitude of net acceleration: a = Ja~ + af = J6Az + 0.22 = J4f mlsz, eDl·fectl,On 0f acceIera'tlOn: tan = -at = -0.2 = -1 . a, 6.4 32 Fig. 5.149 6. b. The bird keeps on flying with a constant speed till the time ~a, of crash. So, let us first find the time of crash. If the two trains V crash each other after 1 hrs, then the total distance travelled Fig. 5.145 by the two trains in the same time of t hrs should be 60 km. Objective Iype 1. C,v'lg = v, + (-v8)' .. 401 + 60t = 60 =? 60 1 = - = 0.6 hrs JV\"8 = v~ + v~ = JI6 + 9 km h- I = 5 km h- I 100 2. a. For B always to be north of A, the velocity components Now, the distance travelled by the bird in 0.6 h is of both along east should be same (see Fig, 5.146). 0.6 x 30 = 18 km 7. a. Here, VR = 25 ms- I, Vw= 10 ms- I :. Velocity of rain w.r,t. woman: VR/W = VR - Vw eLet vRIW makes an angle with vertical then Vw 10 tane=-=-=O.4 VR 2.5 NB N -----,I-.::V.t.v..-+s ,, Fig. 5.150 l v]=5kmlh eShe should hold her umbrella at an angle of = tan-I (0.4) E with vertical towards south. A 8. d. When the body is dropped from the balloon, it also ac- Fig. 5.146 quires the upward velocity of balloon. So w.r.t. a person on V2 cos 60e = VI =} V2 = 10 kmlh the ground, the ball appears t9 be going up. But a person in the balloon is also going up. SO W.r.t. him the velocity of the 3. d. Refer to Fig. 5. 147(a), body will be zero and he will then see the body to be coming down. 50 1 100 = Of Vm = 2 Vw e '2=Vw tan = Vm 50 2~ 100 e (b) (a) Fig. 5.147
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