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path at that instant, which is perpendicular to, the position vector,, , 29. (b): Given, | Ax Bl=J3 AB, or AB sin@ = J3 ABcos 0, or tan@ = V3 or 6=60°, , 4» 2, [A+ Bl= JA? +B? +2ABcos60°, = VA? +B? + AB, 3, , +, 30.(2): As A-B=0 ;s0 x is perpendicular to, 2 32S >, , B and A-C=0, so A _ is perpendicular, , > > 3, , to C . Now (B xC) will be perpendicular, to the plane containing Band c ie,, parallel to A., , 2 gt 2, ul sin2x45° _ 1099 or = 1000, , & ‘ &, , u? sin? 45° u2 [= *], , 28 “gp 2, , 31. (a):, , , , Max. height, H =, , =1000x 42 =250m, , 2ucos®, 8g, 2usin@ 2ucos@_, u?sin20, x—=2, , & & 8, =2R or tyt,~R, , 33. (a): t, eeusnd and 1, =, 1 g 2, , , , , , , , , , “he, , Pradecp's Fundamental Physics (XI) tefay);, , dytorn! ! ‘dx nih Ape,, p =-%=8-10t v, =—=6, 35.(0)1 vy =F oe, , When = 0, then vy = 85, =6, , 2 fo + v2 = vB? +6? =10 mst, , II. Completion Type Questions |: .., , “ UV, , , , Ld, 2 si u? sin? @, u? sin 20 s, 17. HRs DO My ve “ad, u2sin2@ _ u?sin?@ |, AS —_— =, 8 28, so 2 sin @ cos 6 = sin? 6/2, or sin® _ 4 or tan0=4 or 0=tan7! (4), cos®, , we. uw, 18. Max. range, R,, = 7 sin2 x 45°= —, , , , &, 2 sin? 45° 2, : u*sin“ 45° u, Maximum height, Ha. oR ae, Fmax _ 1, Rae 4, , Ill. True/False Type Questions, , 11. Angle discribed at the centre in one rotation, =2 mrad., =2x3-14, = 6-28 rad., , , , ConcePrruaL IPRros_ems, , peers 16 2iy St, ‘I, Vectors and Vectors addition, , 1. Can three vectors not in one plane give a zero, resultant ? Can four vectors do ?, , Sol. Three vectors which are not in one plane can, , not give zero resultant. This is because resultant, of two vectors (in a plane) lies in their plane, It, can not balance the third vector which is in a, different plane., The resultant of four coplanar vectors can be, zero ; if they are represented in magnitude and, direction by four sides of a polygon taken in, the same order. The resultant of four noncoplanar vectors may be zero., , 2. What is the magnitude and direction of, a A ®, (i+ j) 2, , . A A A A, Sol. Magnitude of G+jfy=Hlit+jl, , = (a? +0? =%2, , AA, Let (7+ j') make an angle B with the direction, of ?, then, tanB=W/1=1=tan4s°; p= 45°, , 3. We can order events in time and there is 4, , Sense of time, distinguishing and, J past, presen!, future. Is therefore, time a vector 7, , Scanned with CamScanner
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r, , :, , yoTionin A PLANE lainsmsbur, , , Time ¢ always flows on and on i.e. fro, so! resent and then to future. Therefore, aeivcetiod, , can be assigned to time. Since, the direction of, time is unique, it is not to be specified or stated., It is due to this reason that time can not be a, vector t though it has a direction., , , , 4. 1s, \A+Bl greater than or less than, } Alsi Bl 2 Explain., Sol. Fe Be- -0 A141 BY?, aii +1 BP +21 ANB coso, , -| 22 -1BR-2 ANB, , Yow, =—2| All BI (i-cos8), , [27, : , 2-21 All BI (2sin? 6/2), , 2 3, =—41 All Bl sin? 0/2, It is a negative quantity for all values of @ and, , is zero if © = 0°. Hence, [AEBISIAl+IBl, , 4 Oo, 5. Is 1|A-B1| greater than or less than, , > 2, 1AI+IBI ? Explain., nd ° >, so. VA- BP -CA I+) By?, > 2 27 2, =I Al+1 BP 21 All Bl cos®, , 2 2 32 2, -|AP-1BP-21ANB!, , aft,, , =-21 All BI (1+cos®), SS, , =—21 All B | (200s? 0/2), , 3 3, =—4| All Bl cos” 6/2, It is a negative quantity for all values of ® and, to ovis it value for @ = 180°. Hence,, , \A- BisiasiBl, , 3, 6. The resultant of t two vectors A andB is, , perpendicular to x and its magnitude is half, , +) that: of B . What is the angle between, > >, o, AandB ?, Sol. Refer to Fig. 4(Q).1,, , y, Here, |A I= OP, 1Bl=0Q=PS,, , AR ls OS = B/2, In AOPS, (op)? + (05) = PSP, , , , , , , , , , A24 (BR? = B2 or A? = <5, os _ BI2, , 2 tan@ =, Let ZOPS 6, then tan oP. A, , dapung with SVE Bo, TA7 axW3inB V3, or 9= 30°, , = tan30°, , Zz, , Therefore angle between A and B, B= (180° - 8)= ale 30°) = 150°, , are are represented, , OP, 0a and, , 22, 7, The three vectors A, B and Cc, in om and direction’ by, , os Fig. (Q)2.1 A+ B= 2C, show that, , Sis the mid point of PQ., FIGURE 4(Q).2, P, , , , , , Sol. In A OPS, by ve vectors addition,, , os = OP + PS oi), Similarly, in A OQS, we have, OS'= 00+ QS a= (ii), , Adding @ (i) and pai i iwey we havent, 208 = OP + 00+ Ps + Os., , "nn >, or 2¢= A+B 4 Ps +98, o> —, 2-4 -B)= PS + 08, or 0= PS HOS Aaks B, +QS 9 (2 2C=A+B), a —>, or 5 =- OS, So, efor point of PQ, eee a, , , , Scanned with CamScanner, , , , cial
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4182, , 8. ABCD isa parallelogram Fig. 4, ig- 4(Q).3, AC and, BD are its diagonals. Show that, , (@) AC + BD =28C, , —, () AC- BD =2AB, , , , FIGURE 4(Q).3, D, c, , , , , , A B, , , , Sol. (a) Refer to Fig. 4(Q).3, using triangle law of, vectors, we have,, ome aero eas > GS o>, , AC + BD =( AB + BC)+(BC + CD), , => oOo o>, = AB+2BC+CD, CoS Ul OOO, = AB+2 BC-— AB, , —? —, , (- CD =— AB), , , , =, , — —7, | or AC + BD =2BC, , aoe Or iar a GCS o>, (b) AC = BD = (AB + BC)-(BC + CD), —3 ss, = AB-CD, somo eam >, , = —, AC — BD = AB-(- AB)=2 AB, , =, 9. The greatest resultant of two vectors p ‘and, , o is n times their least resultant. Given, , > _, | P1>1@Q1. When @ is the angle between the, , two vectors, their resultant is half the sum of, the two vectors. Show that,, , cos 0 == (n? +2 )M(n?- 1), Sol. The greatest resultant of given two vectors, =(P+Q), and the least resultant of given two vectors, =(P-Q), According to questions ;, , -1, @-)p, (n+l), , If angle between the vectors is 8, then the, , resultant is given by, R? = P* +0? +2 PQ cos @ wi), , , , (P+Q)=n(P-Q) or Q=, , , , Pradeop's Fundamental Physics’ Coy nen, , ' Given, P01 ps( |, a nP_ :, “Gah, Putting the values in. (i), we get ;, , , , 2, nPt_ pp, W= WP pr yp p=1), , (n+l)? (n+1)? (n+) ny, On solving we get,, (n? +2), cos 8 = “WD, , 10. ABCDEF is a regular hexagon, Fig. 4(Q).4,, What is the value of, , aS llaoeron asm oT a Ta., (AB + AC + AD+ AE+ AF)?, , FIGURE 4(Q).4, Fens D, , , , , , , , B, , , , ==> o>, , Sol. 4B + AC + AD+ AE+ AF, , =) Poe, ee Se, , = AB+(AD+DC)+ AD+(AD + DE)+ AF, , =_7-l oO o> nS > —_=, , =3 AD +(AB + DE)+(DC + AF) =3 AD,, remy _—, , =3x(2 AO) =6 AO, , — —> —> ~—, [-- AB =— DE, and pc =- AF |, , , , TL. Components of Vectors and Relative Velocity|, , 11. Can a flight of a bird be an example of, composition of vectors ?, , Sol. Yes, the flight of a bird is an example of, composition of vectors as is clear from the, Fig.4(Q).5. As the bird flies, it strikes the aif, downwards with forces W, W with its wings:, along WO. According to Newton’s third Law, of motion, air strikes the wings in opposite, directions with the same force in reaction. Thes?, Teactions are represented in Fig. 4(Q).5 alonB, , Scanned with CamScanner
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16., , Sol., , , , ‘ ba pe:, , Velocity of wind, v,, = OB, , where v,,.= 80 km he!, , The plane will have a resultant velocity v along, , OC. Let B be the angle between D and Up ‘., , v ;, , then tanB = tg = =; =0-2 = tan 11°20’, B = 11° 20’ south of east, , A weight mg is Suspended from the middle, , of a rope whose ends are at the same level., , The rope is no longer horizontal. Find the, , minimum tension required to completely, straighten the rope., , See Fig. 4(Q).8 ; clearly,, , 2Tsin@=mg or Tas, , , , FIGURE 4(Q),, A RE 4(Q).8, , , , Sats yE= =F, T cos 0 T cos 6, , mg, , , , , , , , When the rope is straight, 6 = 0°;, , , , = 00, , mg, Then, T=, 2sin 0°, , , , IIL Scalar product and vector product of vectors, , 17., , Sol., , 18., , 27> FIO, If A.B=A.C , is it correct to conclude, , 72 3S, that B=C ?, , 72> 83, , > >, Given A.B=A.C,, ie., AB cos 8; =AC cos 0, +(8), > >, where 0, is smaller angle between A and B ;, >, and @, is the smaller angle between A and Cc :, From (i), ~B cos 8, = C cos @, «ti), If 0, = @,, then from (ii), B= Cor B=c, =y, , If ; #05, then from (ii), B# Cor B¥C, , >, , : > ~, Three vectors A, B and C satisfy the, , ~~ S ~ SS, relation A.B =0 and A.C =0.To which, , . >, vector, the vector A is parallel ?, , 237 ae 1 2, Sol. As A.B =0;so A is perpendicularto B ., , Pradeep's Fundamental’ Physics (XI) hay, , 19,, , Sol., , 20., , Sol., , ey, Ny 4.C=0350 A is perpendicular to c, =, As:BxC. is perpendicular to both Band C,, >, so BxC. is parallel to A, , If A x B = A x Cc ,is it correct to conclude, > OO, that B=C.?, ny, Let 6, be the smaller angle between A and, = >, B_; and @, be the smaller angle between 4, ~, and C ve, 327m FO FD, Given, AxB=AxXC, 2 A, ABsin®@,m=ACsin®,n, i), where nh and Ny are unit vectors in the., , o> 3 > 3, direction of vectors (A x B) and (A xC), respectively., From (i), Bsin@,m,=Csin®, nj’ (i), A _A ], If @, = 0, and n =n),, > 3, B=C and B=C, A A, If 0; #0, and nj #7),, then, , then, , ee, B#¥C and B4C, A A, If ; #6, and n, =n),, 2 OS, then B+C and B¥C, If 6, = 8, and 1%,, , 27> OS, then B=C and B4¥C, , > 23> > 3, If AxB=C xB , show that © need not, , ae, be equal to 4 ., > 73> 23> > tt, AXB=CxB or AxB-CxBe0, >> OS, or (A-C)x B=0. oi), To satisfy (i), the three Possibilities can be there, > 3, OA-C=00 2-2, , W B=0, 3, , a nad, (i) A-C and B are Parallel to each other, > > >, ie. A-C=nB, where n is a non zero real, number., , Scanned with CamScanner
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{MOTION IN A PLANE, , , , 23> 3 >, , ‘\, © A=C+nB, nd, , oD Thus, if AXB=CxB, Cc need not be, , >, equal to A . The given statement is true if B, , . > s > 3, gholoe. is a zero vector or A is equalto C+nB, , >., 21. If three vectors A, B andC are such that, , 27 7D, , A.B=A.C,, > 8S, then prove that B=C., , > 3 349 8 ind, AXB=AxXC, A#0, , 3239 33, , “gol. Given, A.B = A.C or A.B-A.C=0, 77 3, or A.(B-C)=0 (i), 4) > 3, s But A 0 , so either B-C=0 or B=C, ia >. . > 3, or A is perpendicular to (B-C), , 27 3p OP 2 3 8 8, ' Also AXB=AxXC or AXB-AxC=0, , > 3 oe oO ., , ain or Ax(B-C)=0 (ii), 7 3 =e, , But A #0 therefore either B-C = 0. or, , 42 Oo, Be c or a is parallel to (B-C), But at a time a cannot be perpendicular to, , 4 3 > 2 ., (B—C) and parallel to (B— C) . So equations, 32 >, (i) and (ii) will be true at a time if B=C, Inany AABC as shown in Fig. 4(Q).9(a) prove, pee Bs ox anil, , sinA sinB sinC’, , Refer to Fig. 4(Q).9(5),, , 22., , , , , , , , Sol., , ‘ en, , ' vectors @, band ¢ are represented by the three, , '0n Los sides of a triangle taken in one order. Their, resultant is zero. So, , FIGURE 4(Q).9, , , , , , t, iy, or atb=, , , ~, ot ax Pebx eno or —exa+bxe=0, or bxe=cxa (i), Similarly we can get, axe = bx? (ii), From (i) and (ii),, axb=bxe=exa, or laxbl=lbxel=lexal, or ab sin (180° — C) = bc sin (180° — A), = ca sin (180° - B), or ab sin C = be sin A= ca sin B, Dividing it by abc, we get, sinC _ sinA _ sinB, c a b, a b c, or (Proved), , , , sinA sinB_ sinC, , De EPP cro, , 23., , A body slides down a smooth inclined plane, when released from the top, while another, body falls freely from the same point. Which, one will strike the ground earlier ?, , A body falling freely will reach the ground, earlier because its acceleration is g (i.e., acceleration due to gravity) which is greater than, the acceleration of other body = g sin @ ; where, @ is the inclination of the plane with the, horizontal., , . A stone droped from the window of a, stationary bus takes 4 seconds to reach the, ground. In what time the stone will reach the, ground when the bus is moving with, (a) constant velocity of 108 km h-!, (5) constant acceleration of 2 km h., , (a) 4 seconds (b) 4 seconds. In both the cases, the initial vertical downward velocity is zero, and the vertical downward acceleration is, acceleration due to gravity g and the stone is to, cover the same vertical height in each case., When a rifle is fired at a distant target, the, barrel is not lined up exactly on the target., Why ?, , As soon as a bullet is fired from a gun whose, barrel is lined up exactly on the target, it starts, falling downwards on account of acceleration, due to gravity. Due to it, the bullet will hit below, the target. Just to avoid it, the barrel of the gun, , Sol., , Sol., , 25., , Sol., , Scanned with CamScanner, , Wicca oe
