P1-Allens Made Physics Theory {PART-1}

MAXIMA Just before the maximum the slope is positive, at the maximum it dy is zero and just after the maximum it is negative. Thus, dx dy decreases at a maximum and hence the rate of change of dx is negative at a maximum i.e. d  dy  < 0 at maximum. dx    dx  The quantity d  dy  is the rate of change of the slope. It is written as d2y dx   dx2 .  dx  dy d2y Conditions for maxima are: (a) dx = 0 (b) dx2 < 0 MINIMA Similarly, at a minimum the slope changes from negative to positive. Hence with the increases of x. the slope is increasing that means the rate of change of slope with respect to x is positive hence d  dy  > 0. dx    dx  Conditions for minima are: dy d2y (a) dx = 0 (b) dx2 > 0 Quite often it is known from the physical situation whether the d2y quantity is a maximum or a minimum. The test on dx2 may then be omitted. Example 38. Find minimum value of y = 1 + x2 – 2x dy dx = 2x – 2 for minima dy  0 dx 2x – 2 = 0 x=1 d2y  2 dx 2 d2y  0 dx 2 at x = 1 there is minima for minimum value of y yminimum = 1 + 1 – 2 = 0 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 17

 4. INTEGRATION In mathematics, for each mathematical operation, there has been defined an inverse operation. For example- Inverse operation of addition is subtruction, inverse operation of multiplication is division and inverse operation of square is square root. Similarly there is a inverse operation for differentiation which is known as integration 4.1 ANTIDERIVATIVES OR INDEFINITE INTEGRALS Definitions : A function F(x) is an antiderivative of a function f(x) if F´(x) = f(x) for all x in the domain of f. The set of all antiderivatives of f is the indefinite integral of f with respect to x, denoted by The symbol is an integral sign. The function f is the integrand of the integral and x is the variable of integration. For example f(x) = x3 then f(x) = 3x2 So the integral of 3x2 is x3 Similarly if f(x) = x3 + 4 then f(x) = 3x2 So the integral of 3x2 is x3 + 4 there for general integral of 3x2 is x3 + c where c is a constant One antiderivative F of a function f, the other antiderivatives of f differ from F by a constant. We indicate this in integral notation in the following way :  f(x)dx  F(x)  C. .............(i) The constant C is the constant of integration or arbitrary constant, Equation (1) is read, “The indefinite integral of f with respect to x is F(x) + C.” When we find F(x)+ C, we say that we have integrated f and evaluated the integral. Example 39. Evaluate 2x dx. Solution : an antiderivative of 2x  2x dx  x2  C the arbitrary constant The formula x2 + C generates all the antiderivatives of the function 2x. The function x2 + 1, x2 – , and x2 + 2 are all antiderivatives of the function 2x, as you can check by differentiation. Many of the indefinite integrals needed in scientific work are found by reversing derivative formulas. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 18

 Reversed derivative formula 4.2 INTEGRAL FORMULAS Indefinite Integral xndx xn1 d  xn1  n1  1.   C ,n  –1, n rational dx  n 1 = xn    dx = 1dx  x  C (special case) d (x) = 1 dx 2. sin(Ax  B)dx  cos(Ax  B) C d  coskx  A   = sin kx   dx  k  3. coskx dx sin kx C d  sinkx  k   = cos kx   dx  k  Example 40. Examples based on above formulas : (a) x 5 dx  x6 C Formula 1 with n = 5 6 1 dx = x1/ 2dx  2x1/ 2  C  2 x  C  (b) x Formula 1 with n = –1/2 (c) sin2x dx   cos 2x  C Formula 2 with k = 2 2 cos x dx = cos 1 xdx sin(1/ 2)x C = 2 sin x C Formula 3 with k = 1/2 2 2 1/ 2 2  (d)   Example 41. Right :  x cosx dx = x sin x + cos x + C Reason : The derivative of the right-hand side is the integrand: Check : d dx (x sin x + cos x + C) = x cos x + sin x – sin x + 0 = x cos x. Wrong :  x cosx dx = x sin x + C Reason : The derivative of the right-hand side is not the integrand: Check : d dx (x sin x + C) = x cos x + sin x + 0  x cos x.  4.3 RULES FOR INTEGRATION RULE NO. 1 : CONSTANT MULTIPLE RULE A function is an antiderivative of a constant multiple kf of a function f if and only if it is k times an antiderivative of f.  k f(x)dx  k f(x)dx ; where k is a constant RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 19

Example 42. 5 x 2 dx  5x3 C 3 7 dx = 7x 2 dx 7x 1 C = 7 C x2 1 x  Example 43.     t dt t1/ 2dt t3 / 2 C 2 t3 / 2 C t 3/2 3  Example 44. = =    RULE NO. 2 : SUM AND DIFFERENCE RULE A function is an antiderivative of a sum or difference f  g if and only if it is the sum or difference of an antiderivative of f an antiderivative of g.   [f(x)  g(x)]dx  f(x)dx  g(x)dx Example 45. Term–by–term integration Evaluate :  (x2 – 2x + 5) dx. Solution. If we recognize that (x3 /3) – x2 + 5x is an antiderivative of x2 – 2x + 5, we can evaluate the integral as If we do not recognize the antiderivative right away, we can generate it term by term with the sum and difference Rule:    (x2  2x  5)dx = x2dx – x3 2xdx + 5dx = 3 + C1 – x2 + C2 + 5x + C3. This formula is more complicated than it needs to be. If we combine C ,C and C into a single constant 12 3 C = C + C + C , the formula simplifies to 123 x3 3 – x2 + 5x + C and still gives all the antiderivatives there are. For this reason we recommend that you go right to the final form even if you elect to integrate term by term. Write x2dx – x3    (x2  2x  5)dx = 3 2xdx + 5dx = – x2 + 5x + C. Find the simplest antiderivative you can for each part add the constant at the end. Example 46. Find a body velocity from its acceleration and initial velocity. The acceleration of gravity near the surface of the earth is 9.8 m/sec2. This means that the velocity v of a body falling freely in a vacuum dv changes at the rate of dt = 9.8 m/sec2. If the body is dropped from rest, what will its velocity be t seconds after it is released? RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 20

Solution. In mathematical terms, we want to solve the initial value problem that consists of The differential condition : dv dt = 9.8 The initial condition: v = 0 when t = 0 ( abbreviated as v (0) = 0 ) We first solve the differential equation by integrating both sides with respect to t: dv The differential equation dt = 9.8  dv dt = 9.8dt Integrate with respect to t. dt v + C1 = 9.8t + C2 Integrals evaluated v = 9.8t + C. Constants combined as one This last equation tells us that the body’s velocity t seconds into the fall is 9.8t + C m/sec. For value of C : What value? We find out from the initial condition : v = 9.8t + C 0 = 9.8(0) + C v( 0) = 0 C = 0. Conclusion : The body’s velocity t seconds into the fall is v = 9.8t + 0 = 9.8t m/sec. The indefinite integral F(x) + C of the function f(x) gives the general solution y = F(x) + C of the differential equation dy/dx = f(x). The general solution gives all the solutions of the equation ( there are infinitely many, one for each value of C). We solve the differential equation by finding its general solution. We then solve the initial value problem by finding the particular solution that satisfies the initial condition y(xo) = yo ( y has the value y when x = x .). oo 4.4 DEFINITE INTEGRATION OR INTEGRATION WITH LIMITS b  f (x)dx  g( x ) b  g(b)  g(a) a a where g(x) is the antiderivative of f(x) i.e. g´(x) = f(x) Example 47.  4  4dx 3 x 4 = 3[4 – (–1)] = (3) (5) = 15 3dx = 3 1 1 1  /2 sin xdx   cos x / 2 =  cos    cos(0) = –0 + 1 = 1 0 0  2   RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 21

 4.5 APPLICATION OF DEFINITE INTEGRAL : CALCULATION OF AREA OF A CURVE From graph shown in figure if we divide whole area in infinitely small strips of dx width. We take a strip at x position of dx width. Small area of this strip dA = f(x) dx b So, the total area between the curve and x–axis = sum of area of all strips = f(x)dx a Let f(x)  0 be continuous on [a,b]. The area of the region between the graph of f and the x-axis is b A = f (x)dx a Example 48. Find area under the curve of y = x from x = 0 to x = a a x2 a a2 ydx 2 2 Answer : = 00  5. VECTOR In physics we deal with two type of physical quantity one is scalar and other is vector. Each scalar quantities has magnitude. Magnitude of a physical quantity means product of numerical value and unit of that physical quantity. For example mass = 4 kg Magnitude of mass = 4 kg and unit of mass = kg Example of scalar quantities : mass, speed, distance etc. Scalar quantities can be added, subtracted and multiplied by simple laws of algebra. 5.1 DEFINITION OF VECTOR If a physical quantity in addition to magnitude - (a) has a specified direction. (b) It should obey commutative law of additions     A B B A (c) obeys the law of parallelogram of addition, then and then only it is said to be a vector. If any of the above conditions is not satisfied the physical quantity cannot be a vector. If a physical quantity is a vector it has a direction, but the converse may or may not be true, i.e. if a physical quantity has a direction, it may or may not a be vector. e.g. time, pressure, surface tension or current etc. have directions but are not vectors because they do not obey parallelogram law of addition.  The magnitude of a vector ( A ) is the absolute value of a vector and is indicated by  | A | or A. Example of vector quantity : Displacement, velocity, acceleration, force etc. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 22

Representation of vector : Geometrically, the vector is represented by a line with an arrow indicating the direction of vector as Mathematically, vector is represented by  A Sometimes it is represented by bold letter A . IMPORTANT POINTS : If a vector is displaced parallel to itself it does not change (see Figure) C A=B =C Transition of a vector A B parallel to itself If a vector is rotated through an angle other than multiple of 2 (or 360º) it changes (see Figure). If the frame of reference is translated or rotated the vector does not change (though its components may change). (see Figure). Two vectors are called equal if their magnitudes and directions are same, and they represent values of same physical quantity. Angle between two vectors means smaller of the two angles between the vectors when they are placed tail to tail by displacing either of the vectors parallel to itself (i.e. 0    ). 5.2 UNIT VECTOR  Unit vector is a vector which has a unit magnitude and points in a particular direction. Any vector ( A ) can be written as the product of unit vector ( Aˆ ) in that direction and magnitude of the given vector.  or Aˆ = A  = A Aˆ A A A unit vector has no dimensions and unit. Unit vectors along the positive x-, y- and z-axes of a rectangular coordinate system are denoted by ˆi , ˆj and kˆ respectively such that ˆi  ˆj  kˆ = 1. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 23

Example 49. Three vectors  ,  ,  are shown in the figure. Find angle between (i)  and  , (ii)  and  , (iii) A B C A B B C  and  . A C Solution. To find the angle between two vectors we connect the tails of the two vectors. We can shift  such that tails of  ,  and  are B A B C connected as shown in figure. Now we can easily observe that angle between  and  is 60º, A B  and   and  is 75º. B C is 15º and between A C Example 50. A unit vector along East is defined as ˆi . A force of 105 dynes acts west wards. Represent the force in Solution. terms of ˆi .  = – 105 ˆi dynes F  5.3 MULTIPLICATION OF A VECTOR BY A SCALAR Multiplying a vector  with a positive number  gives a vector B (=   ) whose magnitude is changed by A A the factor  but the direction is the same as that of  . Multiplying a vector  by a negative number  gives A A a vector   whose direction is opposite to the direction of  and whose magnitude is  times  B A A. Example 51. A physical quantity (m = 3kg) is multiplied by a vector a such that   ma . Find the magnitude and F Solution. direction of  if  F (i) a = 3m/s2 East wards (ii) a = –4m/s2 North wards (i)   ma = 3 × 3 ms–2 East wards = 9 N East wards F (ii)   ma = 3 × (–4) N North wards F = – 12N North wards = 12 N South wards 5.4 ADDITION OF VECTORS Addition of vectors is done by parallelogram law or the triangle law :  (a) Parallelogram law of addition of vectors : If two vectors A and B are represented by two adjacent sides of a parallelogram both pointing outwards (and their tails coinciding) as shown. Then the diagonal drawn through the intersection of the two vectors represents the resultant (i.e., vector  sum of A and B ). RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 24

R = A2  B2  2ABcos   The direction of resultant vector R from A is given by tan  = MN = MN Bsin  PN PQ  QN = A  B cos    tan 1 A B sin     B cos     (b) Triangle law of addition of vectors : To add two vectors A and B shift any of the two vectors parallel to itself until the tail of  is at the head of   +  is a vector  drawn from B A . The sum A B R    the tail of A to the head of B , i.e., A + B = R . As the figure formed is a triangle, this method is called ‘ triangle method’ of addition of vectors. If the ‘triangle method’ is extended to add any number of vectors in one operation as shown . Then the figure formed is a polygon and hence the name Polygon Law of addition of vectors is given to such type of addition. IMPORTANT POINTS : To a vector only a vector of same type can be added that represents the same physical quantity and the resultant is a vector of the same type. As R = [A2 + B2 + 2AB cos]1/2 so R will be maximum when, cos  = max = 1, i.e.,  = 0º, i.e. vectors are like or parallel and Rmax = A + B. The resultant will be minimum if, cos  = min = -1, i.e.,  = 180º , i.e. vectors are antiparallel and Rmin = A  B. If the vectors A and B are orthogonal, i.e.,  = 90º, R = A2  B2 As previously mentioned that the resultant of two vectors can have any value from (A ~ B) to (A + B) depending on the angle between them and the magnitude of resultant decreases as  increases 0º to 180º Minimum number of unequal coplanar vectors whose sum can be zero is three. The resultant of three non-coplanar vectors can never be zero, or minimum number of non coplanar vectors whose sum can be zero is four. Subtraction of a vector from a vector is the addition of negative vector, i.e.,   A  B = A  ( B ) (a) From figure it is clear that  is equal to addition of  with reverse of  A B A B = RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 25

 | A  B | = [(A)2 + (B2) + 2AB cos (180º  )]1/2  A2  B2  2AB cos  |AB| = (b) Change in a vector physical quantity means subtraction of initial vector from the final vector. Example 52. Find the resultant of two forces each having magnitude F0, and angle between them is . Solution. FR2e sultan t = F02 + F02 + 2 F02 cos  = 2 F02 ( 1 + cos ) = 2 F02 (1 + 2 cos2  – 1) 2 = 2 F02 × 2 cos2  2 Fresultant = 2F0 cos  2 Example 53. Two non zero vectors  and  are such that |  +  | = |  –  |. Find angle between  and  ? Solution. A B A B A B A B |  +  | = |  –  |  A2 + B2 + 2AB cos  = A2 + B2 – 2AB cos  A B A B  4AB cos  = 0  cos  = 0  =  2 Example 54. If the sum of two unit vectors is also a unit vector. Find the magnitude of their difference? Solution. Let Aˆ and Bˆ are the given unit vectors and Rˆ is their resultant then | Rˆ | = | Aˆ + Bˆ | 1 = (Aˆ )2  (Bˆ )2  2 | Aˆ || Bˆ | cos  1 1 = 1 + 1 + 2 cos   cos  = – 2 |  –  | = (Aˆ )2  (Bˆ )2  2 | Aˆ || Bˆ | cos  = 1 1 2  1 1( 1 ) =3 A B 2  5.5 RESOLUTION OF VECTORS RESONACE If a and  be any two nonzero vectors in a plane with different directions and  b A be another vector in the same plane.  can be expressed as a sum of two vectors - one obtained by multiplying a by a real number A  and the other obtained by multiplying b by another real number .  =  a  (where  and  are real numbers) A + b  We say that A has been resolved into two component vectors namely a and  b RESONANCE NEET–MATHEMATICAL TOOLS - 26

 a and   along a and  respectively. Hence one can resolve a given vector into two component vectors b b along a set of two vectors  all the three lie in the same plane. Resolution along rectangular component : It is convenient to resolve a general vector along axes of a rectangular coordinate system using vectors of unit magnitude, which we call as unit vectors. ˆi,ˆj,kˆ are unit vector along x,y and z-axis as shown in figure below: Resolution in two Dimension  Consider a vector A that lies in xy plane as shown in figure,  =  +  A A1 A2  A xˆi ,  = A yˆj   = A xˆi + A yˆj A1 = A2 A  A. The quantities A and A are called x- and y- components of the vector xy Ax is itself not a vector but A xˆi is a vector and so is A yˆj . Ax = A cos  and Ay = A sin  Its clear from above equation that a component of a vector can be positive, negative or zero depending on the  value of . A vector A can be specified in a plane by two ways : (a) its magnitude A and the direction  it makes with the x-axis; or (b) its components Ax and Ay. A= A 2  A 2 ,  = tan1 Ay x y Ax Note : If A = Ax Ay = 0 and if A = Ay  Ax = 0 i.e. components of a vector perpendicular to itself is always zero. The rectangular components of each vector and those of the sum  =  +  are shown in figure. We saw that C A B  =  +  is equivalent to both C A B Cx = Ax + Bx and Cy = Ay + By  Resolution in three dimensions. A vector A in components along x-, y- and z-axis can be written as :  =  +  =  +  +  OP OB BP OC CB BP   =  +  +  =  +  +  A AZ Ax Ay Ax Ay AZ  = Ax i  A y j  Azk RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 27

A= A 2  A 2  A 2 x y z P O B C A = A cos , A = A cos , A = A cos  x y z  where cos , cos  and cos  are termed as Direction Cosines of a given vector A . cos2  + cos2  + cos2  = 1 Example 55. A mass of 2 kg lies on an inclined plane as shown in figure.Resolve its weight along and perpendicular to the plane.(Assumeg=10m/s2) Solution. Component along the plane = 20 sin 30 = 10 N component perpendicular to the plane = 20 cos 30 = 10 3 N Example 56. A vector makes an angle of 30º with the horizontal. If horizontal Solution. component of the vector is 250. Find magnitude of vector and its vertical component? Let vector is  A A3  500 Ax = A cos300 = 250 = 2 A= 3 500 1 250 Ay = A sin300 = 3 × 2 = 3 Example 57.  = ˆi + 2 ˆj – 3 kˆ , when a vector  is added to  , we get a unit vector along x-axis. Find the value Solution. A B A  of  ? Also find its magnitude B  +  = ˆi A B  = ˆi –  = ˆi – ( ˆi + 2 ˆj – 3 kˆ ) = – 2 ˆj + 3 kˆ B A |  | = (2)2  (3)2 = 13 B RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 28

Example 58. In the above question find a unit vector along  ? B   2ˆj  3kˆ B Solution. Bˆ = B = 13 Example 59. Vector  ,  and  have magnitude 5, 5 2 and 5 respectively, direction of   and  are Solution. A B C A, B C towards east, North-East and North respectively. If ˆi and ˆj are unit vectors along East and North respectively. Express the sum  +  +  in terms of ˆi , ˆj . Also Find magnitude and direction of A B C the resultant.  = 5 ˆi  = 5 ˆj North A C C  B B =5 2 cos 45 ˆi + 5 2 sin 45 ˆj = 5 ˆi + 5 ˆj = 10 ˆi + 10 ˆj  +  +  = 5 ˆi + 5 ˆi + 5 ˆj + 5 ˆj East A B C    A A B C | + + | = (10)2  (10)2 = 10 2 10   = 45º from East tan  = 10 = 1 Example 60. You walk 3 Km west and then 4 Km headed 60° north of east. Find your resultant displacement Solution. (a) graphically and (b) using vector components. Picture the Problem : The triangle formed by the three vectors is not a right triangle, so the magnitudes of the vectors are not related by the Pythagoras theorem. We find the resultant graphi- cally by drawing each of the displacements to scale and mea- suring the resultant displacement. (a) If we draw the first displacement vector 3 cm long and the second one 4 cm long, we find the resultant vector to be (b) 1. about 3.5 cm long. Thus the magnitude of the resultant 2. displacement is 3.5 Km. The angle  made between the resultant displacement and the west direction can then be measured with a protractor. It is about 75°. Let  be the first displacement and choose the x-axis to be in the easterly direction. Compute A A x and A , A = – 3 , Ay = 0 y x  Similarly, compute the components of the second displacement B , Bx = 4cos 60° = 2 , By = 4 sin 60° = 2 3 3. The components of the resultant displacement  =  +  are found by addition, C A B  = (–3  2)ˆi + (2 3 )ˆj = – ˆi + 2 3 ˆj C 4. The Pythagorean theorem gives the magnitude of  . C 12  2 2 13 = 3.6  C = = 3 5. The ratio of Cy to Cx gives the tangent of the angle  between  and the x axis. C tan  = 23  = – 74° 1 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 29

Remark : Since the displacement (which is a vector) was asked for, the answer must include either the magnitude and direction, or both components. in (b) we could have stopped at step 3 because the x and y components completely define the displacement vector. We converted to the magnitude and direction to compare with the answer to part (a). Note that in step 5 of (b), a calculator gives the angle as –74°. But the calculator can’t distinguish whether the x or y components is negative. We noted on the figure that the resultant displacement makes an angle of about 75° with the negative x axis and an angle of about 105° with the positive x axis. This agrees with the results in (a) within the accuracy of our measurement.  5.6 MULTIPLICATION OF VECTORS 5.6.1 THE SCALAR PRODUCT The scalar product or dot product of any two vectors  and  denoted as  A B, A.B  (read A dot B ) is defined as the product of their magnitude with cosine of angle  between them. Thus, A . B = AB cos  {here  is the angle between the vectors} PROPERTIES : It is always a scalar which is positive if angle between the vectors is acute (i.e. < 90º) and negative if angle between them is obtuse (i.e. 90º <   180º) It is commutative, i.e.,  .  =  .  A B B A It is distributive, i.e.   +  =  .    A. (B C) A B +A. C    A. B As by definition A . B = AB cos  The angle between the vectors  = cos1  AB     A . B = A (B cos ) = B (A cos ) Geometrically, B cos  is the projection of  onto  and A cos  is the projection of  onto  as shown. B A A B So   is the product of the magnitude of    A .B A and the component of B along A and vice versa.  A.B Component of B along  = B cos= A = Aˆ .  A B  A.B  along  =  Bˆ Component of A B = A cos= B A. Scalar product of two vectors will be maximum when cos  = max = 1, i.e.,  = 0º, i.e., vectors are parallel  (  .  )max = AB A B If the scalar product of two nonzero vectors vanishes then the vectors are perpendicular. The scalar product of a vector by itself is termed as self dot product and is given by    ( A )2 = A . A = AA cos = AAcos0º = AA2  A = A. A RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 30

In case of unit vector nˆ ,  nˆ . nˆ = ˆi . ˆi = ˆj . ˆj = kˆ . kˆ = 1 nˆ . nˆ = 1 x 1 x cos 0º = 1 In case of orthogonal unit vectors ˆi , ˆj and kˆ ; ˆi . ˆj = ˆj . kˆ = kˆ . ˆi = 0  .  = ( ˆi Ax + ˆj Ay + kˆ Az) . ( ˆi Bx + ˆj By + kˆ Bz) = [AxBx + AyBy + AzBz] A B Example 61. If the Vectors  = a ˆi + a ˆj + 3 kˆ  = a ˆi – 2 ˆj – kˆ are perpendicular to each other. P and Q Solution. Find the value of a?   If vectors  and  are perpendicular P Q   ·  =0  (a ˆi + a ˆj + 3 kˆ ) . (a ˆi – 2 ˆj – kˆ ) = 0 P Q  a2 – 3a + a – 3 = 0 a2 – 2a – 3 = 0  a = –1, 3 a(a – 3) + 1(a – 3) = 0 Examplr 62. Find the component of 3 ˆi + 4 ˆj along ˆi + ˆj ? Solution.  A B Componant of  along  is given by B hence required component A B (3ˆi  4ˆj)  (ˆi  ˆj) 7 = 2 =2 Example 63. Find angle between  = 3 ˆi + 4 ˆj and  = 12 ˆi + 5 ˆj ? A B Solution. We have cos  =  = (3ˆi  4ˆj)  (12ˆi  5ˆj) A B 32  42 122  52 AB cos  = 36  20 = 56 56 5 13 65  = cos–1 65  5.6.2 VECTOR PRODUCT  The vector product or cross product of any two vectors A and B , denoted as    (read      = AB sin  nˆ A B A cross B ) is defined as : A B Here  is the angle between the vectors and the direction nˆ is given by the right-hand-thumb rule. Right-Hand-Thumb Rule: V=A×B  To find the direction of nˆ , draw the two vectors A and B with both the tails coincid-  n ing . Now place your stretched right palm perpendicular to the plane of A and B in  B  such a way that the fingers are along the vector A and when the fingers are closed they go towards  . The direction of the thumb gives the direction of nˆ . B A RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 31

PROPERTIES : Vector product of two vectors is always a vector perpendicular to the plane containing the two vectors i.e.   orthogonal to both the vectors A and B , though the vectors A and B may or may not be orthogonal. Vector product of two vectors is not commutative i.e.      . AB B A   But A B = B A = AB sin  The vector product is distributive when the order of the vectors is strictly maintained i.e.  =  +    . A(B  C) AB A C The magnitude of vector product of two vectors will be maximum when sin = max = 1, i.e,,  = 90º  | A  B |max  AB i.e., magnitude of vector product is maximum if the vectors are orthogonal. The magnitude of vector product of two non–zero vectors will be minimum when |sin| = minimum = 0,i.e.,  = 0º or 180º and |  |min  0 i.e., if the vector product of two non–zero vectors vanishes, the vectors are AB collinear. Note : When  = 0º then vectors may be called as like vector or parallel vectors and when  = 180º then vectors may be called as unlike vectors or antiparallel vectors. The self cross product i.e. product of a vector by itself vanishes i.e. is a null vector. Note : Null vector or zero vector : A vector of zero magnitude is called zero vector. The direction of a zero vector is in determinate (unspecified).  = AA sin 0º nˆ =  AA 0.  In case of unit vector nˆ , nˆ  nˆ = 0  ˆi  ˆi  ˆj ˆj  kˆ  kˆ = 0 In case of orthogonal unit vectors ˆi , ˆj and kˆ in accordance with right-hand-thumb-rule, ˆi  ˆj  kˆ ˆj  kˆ  ˆi kˆ  ˆi  ˆj ˆi ˆj kˆ In terms of components,  Ax Ay Az = ˆi A y Az  ˆj Ax Az  kˆ Ax Ay AB = By Bz By Bz Bx Bz Bx By Bx    = ˆi (Ay Bz  Az By)  ˆj (Az Bx  Ax Bz )  kˆ (Ax By  Ay Bx ) A B The magnitude of area of the parallelogram formed by the adjacent sides of vectors  and  equal to A B  | AB| RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 32

Example 64.  is Eastwards and  is downwards. Find the direction of  ×  ? Solution. A B A B Applying right hand thumb rule we find that  ×  is along North. A B Example 65. If  ·  = |  ×  | , find angle between  and  A B A B A B Solution.  ·  = |  ×  | AB cos  = AB sin  tan  = 1   = 45º A B A B Example 66. Two vectors  and  are inclined to each other at an angle . Find a unit vector which is perpen- Solution. A B  dicular to both  and  A B  ×  = AB sin  nˆ A B  AB nˆ = AB sin  here nˆ is perpendicular to both  and  . A B Example 67. Find   if  = ˆi – 2 ˆj + 4 kˆ and  = 2 ˆi – ˆj + 2 kˆ . A ×B A B ˆi ˆj kˆ Solution.   =1 2 4 = ˆi (– 4 – (–4)) – ˆj (2 – 12) + kˆ (–1–(–6)) = 10 ˆj + 5 kˆ A ×B 3 1 2 Problem 1. Find the value of (a) sin ( ) (b) cos ( ) (c) tan ( ) (d) cos (   ) (e) sin (  + ) (f) cos (  + ) 2 2 2 (g) sin (  ) (h) cos (  ) (i) sin ( 3  ) 2 (j) cos ( 3  ) (k) sin ( 3 + ) (l) cos ( 3 + ) 2 2 2 (m) tan (  ) (n) cot (  ) 2 2 Answers : (a) – sin  (b) cos  (c) – tan  (d) sin  (e) cos  (f) – sin  Problem 2. (j) – sin  (k) – cos  (l) sin  Answers : (g) sin  (h) – cos  (i) – cos  (m) cot  (n) tan  (i) For what value of m the vector  = 2 ˆi + 3 ˆj – 6 kˆ is perpendicular to  = 3 ˆi – m ˆj + 6 kˆ A B (ii) Find the components of vector  = 2 ˆi + 3 ˆj along the direction of ˆi + ˆj ? A (i) m = –10 5 (ii) 2 . Problem 3. (i)  is North–East and  is down wards, find the direction of  ×  . Answers : (ii) A B A B (i) Find  ×  if  = 3 ˆi – 2 ˆj + 6 kˆ and  = ˆi – ˆj + kˆ . B A A B North - West. (ii) –4 ˆi – 3 ˆj + kˆ RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 33

PART - I : FUNCTION & DIFFERENTIATION SECTION - (A) : FUNCTION A-2. f(x) = x3 Find f(–3) A-1. y = f(x) = x2 Find f(2) A-3. If S = r2 Find S(2) SECTION - (B) : DIFFERENTIATION OF ELEMENTRY FUNCTIONS Find the derivative of given functions w.r.t. corresponding independent variable. B-1. y = x3 1 B-2. y = x 2 1 B-4. y = 2tan x B-3. S = t Find the first derivative & second derivative of given functions w.r.t. corresponding independent variable. B-5. y = sin x B-6. r = 22 B-7. y = nx SECTION - (C) : DIFFERENTIATION BY PRODUCT RULE Find derivative of given functions w.r.t. the independent variable x. C-1. ex. sinx C-2. x sin x C-3. y = ex nx SECTION - (D) : DIFFERENTIATION BY QUOTIENT RULE Find derivative of given functions w.r.t. the independent variable. sin x D-2. y = nx D-1. y = cos x x SECTION - (E) : DIFFERENTIATION BY CHAIN RULE dy E-2. y = 2 sin (x + ) where  and  constants Find dx as a function of x E-4. y = (4 – 3x)9 E-1. y = sin 5 x E-3. y = (2x + 1)5 SECTION - (G) : DIFFERENTIATION AS A RATE MEASUREMENT G-1. Suppose that the radius r and area A = r2 of a circle are differentiable functions of t.Write an equation that relates dA / dt to dr / dt. G-2. Suppose that the radius r and surface area S = 4r2 of a sphere are differentiable functions of t. Write an ds dr equation that relates dt to dt . SECTION - (H) : MAXIMA & MINIMA H-1. If function is given y = 1 – x2 then find out maximum value of this function. H-2. If function is given y = (x – 2)2 then find out minima of this function. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 34

PART - II : INTEGRATION SECTION - (A) : INTEGRATION OF ELEMENTRY FUNCTIONS Find integrals of given functions A-1. (a) 2x (b) x2 (c) x2 – 2x + 1 1 5 5 A-2. (a) x 2 (b) x 2 (c) 2 – x2 3 3 1 A-3. (a) (b) 2 x (c) x + x 2 x 4 1 1 A-4. (a) 3 3 x (b) 33 x (c) 3 x + 3 x A-5. (1 x2  3x5 ) A-6. 3 sin x A-7. 4 x3  7  x A-8. x8 + 9 9 x2 A-9. x–7 1 A-10. 3x SECTION - (B) : DEFINITE INTEGRATION 1 1 B-1. 5 dx B-2.  d 2 2 4 4 x 2 2 B-3.   3  dx B-4 sin d   0 2   1 B-5. ex dx 0 SECTION - (C) : CALCULATION OF AREA Use a definite integral to find the area of the region between the given curve and the x–axis on the interval [0,b] C-1. y = 2x x C-2. y = +1 2 Use a definite integral to find the area of the region between the given curve and the x–axis on the interval [0, ] C-3. y = sin x PART - III : VECTOR SECTION - (A) : DEFINITION OF VECTOR & ANGLE BETWEEN VECTORS A-1. Vectors  ,  and  are shown in figure. Find angle between A B C (i)  and  , A B (ii)  and  , A C (iii)  and  . B C RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 35

A-2. The forces, each numerically equal to 5 N, are acting as shown in the Figure. Find the angle between forces? A-3. Rain is falling vertically downwards with a speed 5 m/s. If unit vector along upward is defined as ˆj , represent velocity of rain in vector form. A-4. The vector joining the points A (1, 1, –1) and B (2, –3, 4) & pointing from A to B is - (1) – ˆi + 4 ˆj – 5 kˆ (2) ˆi + 4 ˆj + 5 kˆ (3) ˆi – 4 ˆj + 5 kˆ (4) – ˆi – 4 ˆj – 5 kˆ . SECTION - (B) : ADDITION OF VECTORS B-1. A man walks 40 m North, then 30 m East and then 40 m South. Find the displacement from the starting point? B-2. Two force  and  are acting at right angles to each other, find their resultant ? F1 F2 B-3. A vector of magnitude 30 and direction eastwards is added with another vector of magnitude 40 and direction Northwards. Find the magnitude and direction of resultant with the east.   B-4. Two force of F1 = 500 N due east and F2 = 250 N due north , Find F2 – F1 ? B-5. Two vectors a and  inclined at an angle  w.r.t. each other have a resultant c which makes an angle  with b a . If the directions of a and  are interchanged, then the resultant will have the same b (1) magnitude (2) direction (3) magnitude as well as direction (4) neither magnitude nor direction. B-6. Two vectors  and  lie in a plane. Another vector  lies outside this plane. The resultant  of A B C A BC these three vectors (1) can be zero (2) cannot be zero (3) lies in the plane of  &  (4) lies in the plane of  &  +  A B A A B B-7. The vector sum of the forces of 10 N and 6 N can be (1) 2 N (2) 8 N (3) 18 N (4) 20 N. B-8. A set of vectors taken in a given order gives a closed polygon. Then the resultant of these vectors is a (1) scalar quantity (2) pseudo vector (3) unit vector (4) null vector. B-9. The vector sum of two force P and Q is minimum when the angle  between their positive directions, is (1)  (2)  (3)  (4) . 4 3 2 B-10. The vector sum of two vectors  and  is maximum, then the angle  between two vectors is - A B (1) 0º (2) 30° (3) 45° (4) 60° B-11. Given :  =  +  . Also, the magnitude of  ,  and  are 12, 5 and 13 units respectively. The angle C A B A B C between  and  is A B (1) 0º (2)  (3)  (4) . 4 2 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 36

B-12. If  +  =  –  and  is the angle between  and  , then P Q P Q P Q (1)  = 0º (2)  = 90º (3) P = 0 (4) Q = 0 B-13. The sum and difference of two perpendicular vectors of equal lengths are (1) of equal lengths and have an acute angle between them (2) of equal length and have an obtuse angle between them (3) also perpendicular to each other and are of different lengths (4) also perpendicular to each other and are of equal lengths. SECTION (C) : RESOLUTION OF VECTORS C-1. Find the magnitude of 3 ˆi + 2 ˆj + kˆ ? C-2. If  = 3 ˆi + 4 ˆj then find Aˆ A C-3. What are the x and the y components of a 25 m displacement at an angle of 210º with the x-axis (anti clockwise)? C-4. One of the rectangular components of a velocity of 60 km h–1 is 30 km h–1. Find other rectangular component? C-5. If 0.5 ˆi + 0.8 ˆj + C kˆ is a unit vector. Find the value of C C-6. The rectangular components of a vector are (2, 2). The corresponding rectangular components of another vector are (1, 3 ). Find the angle between the two vectors C-7. The x and y components of a force are 2 N and – 3 N. The force is (1) 2 ˆi – 3 ˆj (2) 2 ˆi + 3 ˆj (3) –2 ˆi – 3 ˆj (4) 3 ˆi + 2 ˆj SECTION - (D) : PRODUCTS OF VECTORS D-1. If  = ˆi + ˆj + kˆ and  = 2 ˆi + ˆj find A B (a)  .  (b)   A B A× B D-2. If |  | = 4, |  | = 3 and  = 60° in the figure , Find A B B  A (a)  .  (b) |  ×  | A B A B      D-3. Three non zero vectors A, B & C satisfy the relation A . B  0 & A . C  0 . Then A can be parallel to :     (1) B (2) C (3) B . C (4) B x C RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 37

PART - I : FUNCTION & DIFFERENTIATION SECTION - (A) : FUNCTION A-1. f(x) = cos x + sin x Find f(/2) A-2. If f(x) = 4x + 3 Find f(f(2)) A-3. A = 4r2 then A(3) = SECTION (B) : DIFFERENTIATION OF ELEMENTRY FUNCTIONS Find the derivative of given functions w.r.t. corresponding independent variable. B-1. y = x2 + x + 8 B-2. s = 5t3 – 3t5 B-3. y = 5 sin x B-4. y = x2 + sin x B-5. y = tan x + cot x Find the first derivative & second derivative of given functions w.r.t. corresponding independent variable. B-6. y = 6x2 – 10x – 5x- 2 12 4 1 B-8. y = sin x + cos x B-7. r = – 3 + 4   B-9. y = nx + ex SECTION (C) : DIFFERENTIATION BY PRODUCT RULE Find derivative of given functions w.r.t. the corresponding independent variable. C-1. y = ex tan x C-2. y = (x2 + 3x + 2) . (2x4 – 5) C-3. y = sin x cos x C-4 s = (t2 + 1) (t2 – t) SECTION (D) : DIFFERENTIATION BY QUOTIENT RULE Find derivative of given functions w.r.t. the independent variable. D-1. y = x2  1 sin x x D-2. x2 y2 cos x D-3. x = 2y  1 D-4. y = x SECTION (E) : MAXIMA & MINIMA E-1. Particle's position as a function of time is given by x = – t2 + 4t + 4 find the maximum value of position co- ordinate of particle. E-2. Find the values of function y = 2x3 – 15 x2 + 36 x + 11 at the points of maximum and minimum RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 38

PART - II : INTEGRATION SECTION (A): INTEGRATION OF ELEMENTRY FUNCTIONS Find integrals of given functions. A-1. 1 2  2x  dx A-2. x3(x  1) dx     5 x3  A-3.  y 2  1  y2  3  dy A-4. (sint  cos t  t3  3t2  4)dt  2y   A-5.  sin x  2  5x4  e2x  3  dx  x3    SECTION (B) : DEFINITE INTEGRATION 2 37 B-1.  d B-2. x2 dx 0   B-3. cos x dx 0 SECTION (D) : CALCULATION OF AREA Use a definite integral to find the area of the region between the given curve and the x–axis on the interval [0,b], D-1. y = 3x2 PART - III : VECTOR OBJECTIVE QUESTIONS 1. A hall has the dimensions 10 m × 12 m × 14 m. A fly starting at one corner ends up at a diametrically opposite corner. The magnitude of its displacement is nearly (1) 16 m (2) 17 m (3) 18 m (4) 21 m. 2. A vector is not changed if (2) it is rotated through an arbitrary angle (1) it is displaced parallel to itself (4) it is multiplied by an arbitrary scalar. (3) it is cross-multiplied by a unit vector 3. If the angle between two forces increases, the magnitude of their resultant (1) decreases (2) increases (3) remains unchanged (4) first decreases and then increases 4. A car is moving on a straight road due north with a uniform speed of 50 km h–1 when it turns left through 90º. If the speed remains unchanged after turning, the change in the velocity of the car in the turning process is (1) zero (2) 50 2 km h–1 S-W direction (3) 50 2 km h–1 N-W direction (4) 50 km h–1 due west. 50 km h–1 6. When two vector a and  are added, the magnitude of the resultant vector is always b (1) greater than (a + b) (2) less than or equal to (a + b) (3) less than (a + b) (4) equal to (a + b) RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 39

7. Given :  = 2 ˆi + 3 ˆj and  = 5 ˆi – 6 ˆj . The magnitude of  is A B A B (1) 4 units (2) 10 units (3) 58 units (4) 61 units 8. Given :  = 2 ˆi – ˆj + 2 kˆ and  = – ˆi – ˆj + kˆ . The unit vector of  –  is A B A B 3ˆi  kˆ 3ˆi kˆ  3ˆi  kˆ (1) (2) (3) (4) 10 10 10 10 9. If |  +  | = |  |= |  |, then the angle between  and  is A B A B A B (1) 0º (2) 60º (3) 90º (4) 120º. 11. Vector  is of length 2 cm and is 60º above the x-axis in the first quadrant. Vector  is of length 2 cm and A B 60º below the x-axis in the fourth quadrant. The sum  +  is a vector of magnitude - A B (1) 2 along + y-axis (2) 2 along + x-axis (3) 1 along – x axis (4) 2 along – x axis 12. Six forces, 9.81 N each, acting at a point are coplanar. If the angles between neighboring forces are equal, then the resultant is (1) 0 N (2) 9.81 N (3) 2 (9.81) N (4) 3 (9.81) N.   13. A vector A points vertically downward & B points towards east, then the vector product A  B is (1) along west (2) along east (3) zero (4) along south SUBJECTIVE QUESTIONS  14. Vector A points N – E and its magnitude is 3 kg ms–1it is multiplied by the scalar  such that  = –4 second. Find the direction and magnitude of the new vector quantity. Does it represent the same physical quantity or not ? 15. The resultant of two vectors of magnitudes 2A and 2 A acting at an angle  is 10 A. Find the value of  ? 16. A force of 30 N is inclined at an angle  to the horizontal . If its vertical component is 18 N, find the horizontal component & the value of  . 17. Two vectors acting in the opposite directions have a resultant of 10 units . If they act at right angles to each other, then the resultant is 50 units . Calculate the magnitude of two vectors . 18. The angle  between directions of forces  and  is 90º where A = 8 dyne and B = 6 dyne. If the resultant A B  makes an angle  with  then find the value of ‘’ ? R A 19. Find the resultant of the three vectors OA , OB and OC each of magnitude r as shown in figure? 20. If  = 3 ˆi + 4 ˆj and  ˆi + ˆj + 2 kˆ then find out unit vector along  A B= A B  21. The x and y components of vector A are 4m and 6m respectively. The x,y components of vector   A  B are 10m and 9m respectively. Find the length of B and angle that B makes with the x axis. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 40

AIIMS CORNER ASSERTION / REASON 1. Statement-1 : A vector is a quantity that has both magnitude and direction and obeys the triangle law of addition. Statement-2 : The magnitude of the resultant vector of two given vectors can never be less than the magnitude of any of the given vector. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is False 2. Statement-1 : If the rectangular components of a force are 8 N and 6N, then the magnitude of the force is 10N. Statement-2 : If  then   |  | 2  1 . | A || B | 1 | A B |2 A.B (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is False 3. Statement-1 : The minimum number of vectors of unequal magnitude required to produce zero resultant is three. Statement-2 : Three vectors of unequal magnitude which can be represented by the three sides of a triangle taken in order, produce zero resultant. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is False    4.  Statement-1 : The angle between the two vectors ˆi  ˆj and kˆ is 2 radian.   A.B     Statement-2 : Angle between two vectors is given by  = cos1   ˆi  ˆj and kˆ  AB  .  (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True 5. Statement-1 : Distance is a scalar quantity. Statement-2 : Distance is the length of path transversed. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True TRUE / FALSE 6. State True or False    (i) If A & B are two force vectors then A . B = B . A     (ii) If A & B are two force vectors then A × B = B × A (iii) If the vector product of two non-zero vectors vanishes, the vectors are collinear. RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 41

1. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces : [AIPMT Screening 2003] (1) are not equal to each other in magnitude (2) cannot be predicted (3) are equal to each other (4) are equal to eah other in magnitude 2. If |  | = 3  .  , then the value of  is : [AIPMT Screening 2004] A B A B | A B| AB 1/ 2 3 (1) (A2 + B2 + AB)1/2 (2)  A 2  B2       (3) A + B (4) (A2 + B2 + 3 AB)1/2 3. If a verctor 2 ˆi +3 ˆj + 8 kˆ is perpendicular to the vector 4 ˆj – 4 ˆi +  kˆ , then the value of  is : [AIPMT Screening 2005] (1) –1 1 1 (4) 1 (2) (3) – 2 2 4. If the angle betwen the vectors  and  is , the value of the product (  ×  ).  is equal to : A B B A A [AIPMT Screening 2005] (1) BA2 cos  (2) BA2 sin  (3) BA2 sin  cos  (4) zero 5. The vectors  and  are such that : A B |    | = |   | A B A –B The angle between the two vectors is : [AIPMT Screening 2006] (4) 45° (1) 90° (2) 60° (3) 75° 6. A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are ( 3 , 3). The path of the particle makes with the x-axis an angle of : [AIPMT screeing 2007] (1) 30º (2) 45º (3) 60º (4) 0º 7. A and B are two vectors and  is the angle between them, if A  B = 3 (A  B ) the value of  is : (1) 60º (2) 45º (3) 30º [AIPMT screeing 2007] (4) 90º 8. Six vectors, a through  have the mangitudes and directions indicated in the figure.Which of the following f statements is true ? [AIPMT Screening 2010] ` (1)  + c =  (2)  + c =  (3)  + e =  (4)  + e =  b f d f d f b f RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 42

9. A car travels 6 km towards north at an angle of 45° to the east and then travels distance of 4 km towards north at an angle 135° to east. How far is the point from the starting point ? What angle does the straight line joining its initial and final position makes with the east ? [AIIMS 2008] (1) 50 km and tan–1(5) (2) 10 km and tan–1( 5 ) (3) 52 km and tan–1(5) (4) 52 km and tan–1( 5 ) 10. The magnitudes of sum and difference of two vectors are same, then the angle between them is [RPMT 2003] (1) 90º (2) 40º (3) 45º (4) 60º 11. The projection of a vector 3ˆi  4kˆ on y-axis is : [RPMT 2004] (1) 5 (2) 4 (3) 3 (4) zero 12. Two forces of 12N and 8N act upon a body. The resultant force on the body has a maximum value of : [RPMT 2005] (1) 4N (2) 0N (3) 20 N (4) 8N 13. A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are ( 3,3) . The path of the particle makes with the x-axis an angle of [RPMT 2008] (1) 30º (2) 45º (3) 60º (4) 0º 14. A truck travelling due north at 20 ms–1 turns west and travels with same speed. What is the change in velocity ? [RPMT Entrance Exam 2005] (1) 20 2 ms–1 south-west (2) 40 ms–1 south west (3) 20 2 ms–1 north west (4) 40 ms–1 north west RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 43

Exercise # 1 x3 x6 A-6. – 3 cos x + c A-5. x – 3 – 2 + C PART - I SECTION (A) : A-7. 4. x4  7  x2 C A-8. x9  9x C A-1. 4 A-2. –27 A-3. 4 94 x 2 9 SECTION (B) : B-1. 3x2 B-2. 2 B-3. 1 t 3 / 2 A-9. x 6 C A-10. 1 nx + c x3 2 6 3 B-4. 2 sec2 x B-5. cos x , – sin x dr d2r dy 1 d2y 1 SECTION - (B) : d d2 B-6.  4 ,  4 B-7. dx  x , dx 2   x2 1 SECTION (C) : B-1. 5 dx  5[ x]1 2  5[1– (–2)] = 5 × 3 = 15 – –2 C-1. dy  ex. sinx + ex cosx B-2. 3 dx 2 B-3. 21 C-2. sin x + x cos x C-3. ex nx + ex B-4 0 B-5. e – 1 SECTION (D) : x SECTION - (C) : D-1. sec2 x D-2. 1 nx b x2 – x2 2 cos(x + ) C-1. Area = 2x dx = b2 units 0 SECTION (E) : E-1. 5 cos 5 x E-2. C-2. b2 + b = b(4  b) units 44 E-3. With u = (2x + 1) , C-3. 2 units dy dy du y = u5 : dx = du dx = 5u4 . 2= 10 (2x + 1)4 PART - III dy SECTION - (A) : E-4. dx = – 27(4 – 3x)8 A-1. (i) 105º , (ii) 150º , (iii) 105º SECTION - (G) : A-2. 120º A-3.  VR  5ˆj dA dr ds dr A-4. (3) G-1. dt = 2r dt . G-2. dt = 8r dt H-2. 0 SECTION - (B) : SECTION - (H) : H-1. 1 B-1. 30 m East B-2. F12  F22 B-3. 50, 53º with East PART - II B-4. 250 5 N, tan–1 (2) W of N SECTION - (A) : A-1. (a) x2 + c x3 x3 B-5. (1) B-6. (2) B-7. (2) B-8. (4) (b) + c (c) 3 – x2 + x + c B-9. (4) B-10. (1) B-11. (3) B-12. (4) 3 B-13. (4) 15 5 SECTION (C) : A-2. (a) – x + c (b) – x + c (c) 2x + x + c 3ˆi  4ˆj 2 x3 C-1. 14 C-2. 5 A-3. (a) x3 + c (b) 3 x + c (c) 3 + 2 x + c C-3. – 25 cos 30º and –25 sin 30º A-4. (a) x 4 / 3 + c (b) x2 / 3 +c C-4. 30 3 km h–1. 11 C-6. 15º. C-5. ± 10 2 C-7. (1) 3x4 / 3 3x2 / 3 (c) + + c 42 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 44

SECTION (D) : PART - II D-1. (a) 3 (b) – ˆi + 2 ˆj – kˆ SECTION (A): D-2. (a) 6 (b) 6 3 D-3. (4) x1 A-1. 5 + x 2 + x2 + C Exercise # 2 11 A-2. – x – 2x2 + C PART - I SECTION (A) : y3 y4 3 4 A-1. 1 A-2. 47 A-3.  1 logey   3y C 2 A-3. A = 4(3)2 ; A = 36 SECTION (B) : t4 A-4. –cost – sint + 4 + t3 + 4t + C dy ds B-1. dx = 2x + 1 B-2. dt = 15 t2 – 15 t4 e 2 x 2 dy dy A-5.  cos x  1  x5   3x  C B-3. dx = 5 cos x dx = 2x + cos x x2 B-4. SECTION (B) : B-5. sec2 x – cosec2 x B-1. 32 7 2 B-2. 3 dy d2y B-3. 0 B-6. dx = 12x – 10 + 10x - 3 , dx2 = 12 – 30 x - 4 SECTION (D) : B-7. dr d = –12–2 + 12–4 – 4–5 , b d2r D-1. Area = 3x2 dx = b3 d2 = 24–3 – 48–5 + 20–6 0 PART - III dy d2y 1. (4) 2. (1) 3. (1) 4. (2) B-8. dx = cos x – sin x , dx2 = – sin x – cos x 6. (2) 7. (3) 8. (1) 9. (4) 11. (2) 12. (1) 13. (D) dy 1 d2y 1 14. 12 S-W, No it does not represent the same physi- B-9. dx = x + ex , dx2 = – x 2 + ex cal quantity. 15. 45º 16. 24 N ; 370 approx SECTION (C) : 17. P = 40 ; Q = 30 18. 37º . C-1. ex (tan x + sec2x) 19. r(1 + 2 ) dy 20. 4ˆi  5ˆj  2kˆ 21. 3 5 , tan 1 1 C-2. dx = (2x + 3) (2x4 – 5) + (x2 + 3x – 2) (8x3) 45 2 C-3. cos2 x – sin2 x ds Exercise # 3 C-4. dt = (t2 + 1) (2t) + (t2 – 1)2t = 4t3 SECTION (D) : 1. (1) 2. (2) 3. (1) 4. (1) 5. (1) x(2x)  (x2  1) x2 1 6. (i) T (ii) F (iii) T x2 x2 D-1. dy = = = 1 1 Exercise # 4 dx x2 2. (1) 3. (3) D-2. x2(cos x)  sin x(2x) 1. (4) 6. (3) 7. (1) 4. (4) x4 5. (1) 10. (1) 11. (4) 8. (3) 9. (3) 14. (1) 12. (3) 13. (3) dx (2y  1)(2y)  y2(2) dx 2y2  2y D-3. dy  , dy  (2y  1)2 (2y  1)2 D-4. dy = x( sin x)  cos x dx x2 SECTION (E) : E-1. 8 E-2. 39, 38 RESONACE RESONANCE NEET–MATHEMATICAL TOOLS - 45



JEE-Physics Newton’s  Laws  of  Motion Motion Motion  of  a  body  is  its  movement  and  is  identified  by  change  in  either  its  location  or  orientation  or  both,  relative to  other  objects. Location Location  of  a  rigid  body  tells  us  where  it  is  placed  and  can  be  measured  by  position  coordinates  of  any  particle of  the  body  or  its  mass  center.  It  is  also  known  as  position. Orientation Orientation  of  a  body  tells  us  how  it  is  placed  with  respect  to  the  coordinate  axes.  Angles  made  with  the coordinate  axes  by  any  linear  dimension  of  the  body  or  a  straight  line  drawn  on  it,  provide  suitable  measure  of orientation. Translation  and  Rotation  Motion If  a  body  changes  its  location  without  change  in  orientation,  it  is  in  pure  translation  motion  and  if  it  changes orientation  without  change  in  location,  it  is  in  pure  rotation  motion. Translation  Motion Let  us  consider  the  motion  of  a  plate,  which  involves  only  change  in  position  without  change  in  orientation.  It  is in  pure  translation  motion.  The  plate  is  shown  at  two  different  instants  t  and  t+t.  The  coordinate  axes  shown are  in  the  plane  of  the  plate  and  represent  the  reference  frame.  A  careful  observation  makes  the  following points  obvious.               • None  of the  linear  dimension or  any  line drawn  on  the body  changes  its  angles  with  the coordinate.  Therefore, there  is  no  rotation  motion. • All  the  particles  of  the  body  including  its  mass  center  move  on  identical  parallel  trajectories.  Here  trajectories of  corner  A  and  center  C  are  shown  by  dashed  lines. • All  the  particles  and  mass  center  of  the  body  cover  identical  segments  of  their  trajectories  in  a  given  time interval.  Therefore,  at  any  instant  of  time  all  of  them  have  identical  velocities  and  accelerations. Pure  translation  motion  of  a  body  can  be  represented  by  motion  of  any  of  its  particle.  This  is  why,  we  usually consider  a  body  in  pure  translation  motion  as  a  particle. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Momentum:  Amount  of  Motion Amount  of  motion  in  a  body  depends  on  its  velocity  and  mass. Linear  momentum  of  a  body  is  defined  as  product  of  its  mass  and  velocity.  It  provides  measure  of  amount  of motion. Linear  momentum     of  a  body  of  mass  m,  moving  with  velocity  by     is  expressed  by  the  following  equation. p v  p  mv SI  unit  of  momentum  is  kg-m/s. Dimensions  of  momentum  are  MLT–1 E1

JEE-Physics Force The  concept  of  force  is  used  to  explain  mutual  interaction  between  two  material  bodies  as  the  action  of  one body  on  another  in  form  of  push  or  pull,  which  brings  out  or  tries  to  bring  out  a  change  in  the  state  of  motion of  the  two  bodies.    A  mutual  interaction  between  two  bodies,  which  creates  force  on  one  body,  also  creates force  on  the  other  body.  Force  on  body  under  study  is  known  as  action  and  the  force  applied  by  this  body  on  the other  is  known  as  reaction. Contact  and  Field  Forces: When  a  body  applies  force  on  other  by  direct  contact,  the  force  is  known  as  contact  force.  When  two  bodies apply  force  on  each  other  without  any  contact  between  them,  the  force  is  known  as  field  force. When  you  lift  something,  you  first  hold  it  to  establish  contact  between  your  hand  and  that  thing,  and  then  you apply  the  necessary  force  to  lift.  When  you  pull  bucket  of  water  out  of  a  well,  the  necessary  force  you  apply  on the  rope  by  direct  contact  between  your  hand  and  the  rope  and  the  rope  exerts  the  necessary  force  on  the bucket  through  a  direct  contact.  When  you  deform  a  spring,  you  have  to  hold  the  spring  and  establish  contact between  your  hand  and  the  spring  and  then  you  apply  the  necessary  force.  In  this  way,  you  can  find  countless examples  of  contact  forces. Things  left  free,  fall  on  the  ground,  planets  orbit  around  the  sun,  satellites  orbit  around  a  planet  due  to gravitational  force,  which  can  act  without  any  contact  between  the  concerned  bodies.  A  plastic  comb  when rubbed  with  dry  hair,  becomes  electrically  charged.  A  charged  plastic  comb  attracts  small  paper  pieces  without any  physical  contact  due  to  electrostatic  force.  A  bar  magnet  attracts  iron  nails  without  any  physical  contact between  them.  This  force  is  known  as  magnetic  force.  The  gravitation,  electrostatic  and  magnetic  forces  are examples  of  field  forces. Basic  Characteristics  of  a  Force: Force  is  a  vector  quantity  therefore  has  magnitude  as  well  as  direction.  To  predict  how  a  force  affects  motion of  a  body  we  must  know  its  magnitude,  direction  and  point  on  the  body  where  the  force  is  applied.  This  point is  known  a  point  of  application  of  the  force.  The  direction  and  the  point  of  application  of  a  force  both  decide line  of  action  of  the  force.  Magnitude  and  direction  decide  effect  on  translation  motion  and  magnitude  and  line of  action  decides  effects  on  rotation  motion. Newton’s  Laws  of  Motion Newton has published three laws, which describe how forces affect motion of a body on which they act. These laws are fundamental in nature in the sense that the first law gives concept of force, inertia and the inertial frames; the second law defines force and the third law action and reaction as two aspects of mutual interaction between two bodies. The  First  Law Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Every  material  body  has  tendency  to  preserve  its  state  of  rest,  or  of  uniform  motion  in  a  straight  line,  unless  it is  compelled  to  change  that  state  by  external  forces  impressed  on  it. • Inertia The  tendency  of  a  material  body  to  preserve  its  present  state  of  uniform  motion  or  of  rest  is  known  as inertia  of  the  body.  It  was  first  conceived  by  Galileo. Inertia  is  a  physical  quantity  and  mass  of  a  material  body  is  measure  of  its  inertia. • Inertial  Frame  of  Reference The  first  law  requires  a  frame  of  reference  in  which  only  the  forces  acting  on  a  body  can  be  responsible  for any  acceleration  produced  in  the  body  and  not  the  acceleration  of  the  frame  of  reference.  These  frames  of reference  are  known  as  inertial  frames. The  Second  Law The  rate  of  change  in  momentum  of  a  body  is  equal  to,  and  occurs  in  the  direction  of  the  net  applied  force. A  body  of  mass  m  in  translational  motion  with  velocity   ,  if  acted  upon  with  a  net  external  force   ,  the  second v F law  suggests: 2 E

JEE-Physics d  F  (mv) dt  If  mass  of  the  body  is  constant,  the  above  equation  relates  the  acceleration     of  the  body  with  the  net  force  F a acting  on  it.   d  F (mv)  ma dt The  first  law  provides  concept  of  force  and  the  second  law  provides  the  quantitative  definition  of  force,  therefore the  second  law  is  also  valid  only  in  inertial  frames. SI  unit  of  force  is  newton.  It  is  abbreviated  as  N.  One  newton  equals  to  one  kilogram-meter  per  second  square. 1  N  =  1  kg-m/s2 Dimensions  of  force  are  MLT–2 The  Third  Law Force  is  always  a  two-body  interaction.  The  first  law  describes  qualitatively  and  the  second  law  describes quantitatively    what  happens  to  a  body  if  a  force  acts  on  it,  but  do  not  reveal  anything  about  what  happens  to the  other  body  participating  in  the  interaction  responsible  for  the  force. The  third  law  accounts  for  this  aspect  of  the  force  and  states  that  every  action  on  a  body  has  equal  and  opposite reaction  on  the  other  body  participating  in  the  interaction. Concept  of  Free  Body  Diagram  (FBD) A  force  on  a  body  can  only  exists  when  there  is  another  body  to  create  it,  therefore  in  every  physical  situation of  concern  there  must  be  two  or  more  bodies  applying  forces  on  each  other.  On  the  other  hand  the  three  laws of  Newton,  describe  motion  of  a  single  body  under  action  of  several  forces,  therefore,  to  analyze  a  given problem,  we  have  to  consider  each  of  the  bodies  separately  one  by  one.  This  idea  provides  us  with  the  concept of  free  body  diagram. A  free  body  diagram  is  a  pictorial  representation  in  which    the  body  under  study  is  assumed  free  from  rest  of  the system  i.e.  assumed  separated  from  rest  of  the  interacting  bodies  and  is  drawn  in  its  actual  shape  and  orientation and  all  the  forces  acting  on  the  body  are  shown. How  to  draw  a  Free  Body  Diagram  (FBD) • Separate  the  body  under  consideration  from  the  rest  of  the  system  and  draw  it  separately  in  actual  shape and  orientation. • Show  all  the  forces  whether  known  or  unknown  acting  on  the  body  at  their  respective  points  of  application. For  the  purpose  count  every  contact  where  we  separate  the  body  under  study  from  other  bodies.  At  every such  point,  there  may  be  a  contact  force.  After  showing,  all  the  contact  forces  show  all  the  field  forces. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Various  Field  Forces Field  forces  include  the  gravitational  force  (weight)  electrostatic  forces  and  magnetic  forces,  which  can  easily  be identified.  At  present,  we  consider  only  gravitational  pull  from  the  earth  i.e.  weight  of  the  body. Weight:  The  net  gravitational  pull  of  the  Ear th The gravitational pull from the earth acts on every particle of the body hence it is a distributed force. The net gravitational pull of the Earth on a body may be considered as weight of the body. It is assumed to act on the center of gravity of the body. For terrestrial bodies or celestial bodies of small size, this force can be assumed uniform throughout its volume. Under such circumstances, center of gravity and center of mass coincide and the weight i s  as s u med  to ac t on t hem. Fu r t her more, c enter  of mas s of u ni for m b o d i e s  li e s  at t hei r  g eometr i c al c enter. At present, we discuss only uniform bodies and assume their weight to act on their geometrical center. In the figure weight of a uniform block is shown acting on its geometrical centre that coincides with the center of mass and the centre of gravity of the body. E3

JEE-Physics Various  Contact  Forces At  every  point  where  a  body  under  consideration  is  supposed  to  be  separated  from  other  bodies  to  draw  its free-body  diagram,  there  may  be  a  contact  force.  Most  common  contact  forces,  which  we  usually  encounter, are  tension  force  of  a  string,  normal  reaction  on  a  surface  in  contact,  friction,  spring  force  etc. Tension  Force  of  Stri ngs A  string  or  similar  flexible  connecting  links  as  a  thread  or  a  chain etc.  we  use  to  transmit  a  force.  Due  to  flexibility,  a  string  can  be used  only  to  pull  a  body  connected  to  it  by  applying  a  force always  along  the  string.  According  to  the  third  law,  the  connected body  must  also  apply  an  equal  and  opposite  force  on  the  string, which  makes  the  string  taut.  Therefore,  this  force  is  known  as tension  force  T  of  the  string.  In  the  given  figure  is  shown  a  block pulled  by  a  string,  which  is  being  pulled  by  a  person. The  tension  force  applied  by  string  on  the  block  and  the  force  applied  by  the  block  on  the  string  shown  in  the figure  constitute  a  third  law  action-reaction  pair.  Similarly,  tension  force  applied  by  the  string  on  hand  and  force applied  by  the  hand  on  string  is  another  third  law  action-reaction  pair. While  studying  motion  of  the  block,  the  force  applied  by  the string  on  it,  weight  of  the  block  and  a  reaction  from  the  floor  has to  be  considered.  In  the  figure  only  weight  and  tension  of  string are  shown. To  study  motion  of  the  string,  the  force  applied  by  the  block  on  the  string  and  the  force  applied  by  the  hand  on the  string  must  be  considered.  These  forces  are  shown  in  the  FBD  of  string. To  study  conditions  of  motion  of  the  person,  the  force  applied  by  the  string  on  the  hand  has  to  be  considered  as shown  in  the  figure. String  passing  over  a  pulley Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 A  pulley  is  a  device  consisting  of  a  wheel,  which  can  rotate  freely  on  its  axel.  A  single  pulley changes  direction  of  tension  force.  At  present  for  simplicity,    we  discuss  only  ideal  pulley,  which E is  massless  i.e.  has  negligible  mass  and  rotates  on  its  axel  without  any  friction.  An  ideal  pulley offers  no  resistance  to  its  rotation,  therefore  tension  force  in  the  string  on  both  sides  of  it  are equal  in  magnitude.  Such  a  pulley  is  known  as  ideal  pulley. 4

Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 JEE-Physics Normal  Reaction Two  bodies  in  contact,  when  press  each  other,  must  apply  equal  and  opposite  forces  on  each  other.  These forces  constitute  a  third  law  action-reaction  pair.  If  surfaces  of  the  bodies  in  contact  are  frictionless,  this  force acts  along  normal  to  the  surface  at  the  point  of  contact.  Therefore,  it  is  known  as  normal  reaction. Consider  a  block  of  weight  W  placed  on  a  frictionless  floor.  Because  of  its  weight  it  presses  the  floor  at  every point  in  contact  and  the  floor  also  applies  equal  and  opposite  reaction  forces  on  every  point  of  contact.    We show  all  of  them  by  a  single  resultant  N  obtained  by  their  vector  addition.         To  apply  Newton's  laws  of  motion  (NLM)  on  the  block,  its  weight  W  and  normal  reaction  N  applied  by  the  floor on  the  block  must  be  considered  as  shown  in  the  following  figure.  It  is  the  FBD  of  the  block. Consider  a  spherical  ball  of  weight  W  placed  on  a  floor.  The  normal  reaction  from  the  floor  on  the  ball  and  from the  ball  on  the  floor  makes  third  law  action-reaction  pair.  These  forces  are  shown  in  the  left  figure.        To  apply  Newton's  laws  of  motion  (NLM)  on  the  ball,  its  weight  W  and  normal  reaction  N  applied  by  the  floor on  the  ball  must  be  considered  as  shown  in  the  above  right  figure.  It  is  the  FBD  of  the  ball. When  two  surfaces  make  contact,  the  normal  reaction  acts  along  the  common  normal  and  when  a  surface  and a  sharp  corner  make  a  contact  the  normal  reaction  acts  along  the  normal  to  the  surface.  Consider  a  block placed  in  a  rectangular  trough  as  shown  in  the  figure. E5

JEE-Physics Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 To  apply  Newton's  laws  of  motion  (NLM)  on  the  block,  its  free  body  diagram  (FBD)  is  shown  in  the  above  right figure. Spring  Force When  no  force  acts  on  a  spring,  it  is  in  relaxed  condition  i.e.  neither  compressed  nor  elongated.  Consider  a spring  attached  to  a  fixed  support  at  one  of  its  end  and  the  other  end  is  free.  If  we  neglect  gravity,  it  remains in  relaxed  state.  When  it  is  pushed  by  a  force  F,  it  is  compressed  and  displacement  x  of  its  free  end  is  called compression.  When  the  spring  is  pulled  by  a  force  F,  it  is  elongated  and  displacement  x  of  its  free  end  is  called elongation.  Various  forces  developed  in  these  situations  are  shown  in  the  following  figure. The  force  applied  by  the  spring  on  the  wall  and  the  force  applied  by  the  wall  on  the  spring  make  a  third  law action-reaction  pair.  Similarly,  force  by  hand  on  the  spring  and  the  force  by  spring  on  the  hand  make  another third  law  action-reaction  pair. Hooke's  Law: How  spring  force  varies  with  deformation  in  length  x  of  the  spring  is  also  shown  in  the  following  figure. 6E

JEE-Physics The  force  F  varies  linearly  with  x  and  acts  in  a  direction  opposite  to  x.  Therefore,  it  is  expressed  by  the following  equation F  =  –  kx Here,  the  minus  (–)  sign  represents  the  fact  that  force  F  is  always  opposite  to  x. The  constant  of  proportionality  k  is  known  as  force  constant  of  the  spring  or  simply  as  spring  constant.  The slope  modulus  of  the  graph  equals  to  the  spring  constant. SI  unit  of  spring  constant  is  newton  per  meter  or  (N/m). Dimensions  of  spring  constant  are  MT–2. Translational  Equilibrium A  body  in  state  of  rest  or  moving  with  constant  velocity  is  said  to  be  in  translational  equilibrium.  Thus  if  a  body is  in  translational  equilibrium  in  a  particular  inertial  frame  of  reference,  it  must  have  no  linear  acceleration. When  it is  at rest,  it  is  in  static  equilibrium,  whereas if  it  is  moving  at  constant  velocity  it  is  in  dynamic  equilibrium. Conditions  for  translational  equilibrium For  a  body  to  be  in  translational  equilibrium,  no  net  force  must  act  on  it  i.e.  vector  sum  of  all  the  forces  acting on  it  must  be  zero.    If  several  external  forces    F1 ,  F2   ..... Fi .....  and    Fn   act  simultaneously  on  a body  and  the  body  is  in  translational  equilibrium,  the  resultant  of  these  forces must  be  zero.   Fi  0    If  the  forces  F1 ,  F2   ..... Fi .....and  Fn   are  expressed  in  Cartesian  components, we  have  :   Fix  0             Fiy  0             Fiz  0 If  a  body  is  acted  upon  by  a  single  external  force,  it  cannot  be  in  equilibrium. If  a  body  is  in  equilibrium  under  the  action  of  only  two  external  forces,  the  forces  must  be  equal  and  opposite. If  a  body  is  in  equilibrium  under  action  of  three  forces,  their  resultant  must  be  zero;  therefore,  according  to  the triangle  law  of  vector  addition  they  must  be  coplanar  and  make  a  closed  triangle.                          F1  F2  F3  0     The  situation  can  be  analyzed  by  either  graphical  method  or  analytical  method. • Graphical  method  makes  use  of  sine  rule  or  Lami's  theorem. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Sine  rule  :  F1  F2  F3 Lami's  theorem  :  F1  F2  F3 sin  sin  sin  sin A sin B sin C E7

JEE-Physics • Analytical  method  makes  use  of  Cartesian  components.  Since the  forces  involved  make  a  closed  triangle,  they  lie  in  a  plane and  a  two-dimensional  Cartesian  frame  can  be  used  to  resolve the  forces.  As  far  as  possible  orientation  of  the  x-y  frame  is selected  in  such  a  manner  that  angles  made  by  forces  with  axes should  have  convenient  values.  Fx  0  F1x  F2x  F3x  0  Fy  0  F1y  F2y  F3y  0 Problems  involving  more  than  three  forces  should  be  analyzed  by  analytical  method.  However,  in  some  situations, there  may  be  some  parallel  or  anti-parallel  forces  and  they  should  be  combined  first  to  minimize  the  number  of forces.  This  may  sometimes  lead  a  problem  involving  more  than  three  forces  to  a  three-force  system. Example Consider  a  box  of  mass  10  kg  resting  on  a  horizontal  table  and  acceleration  due  to  gravity  to  be  10  m/s2. (a) Draw  the  free  body  diagram  of  the  box. (b) Find  value  of  the  force  exerted  by  the  table  on  the  box. (c) Find  value  of  the  force  exerted  by  the  box  on  the  table. (d) Are  force  exerted  by  table  on  the  box  and  weight  of  the  box  third  law  action-reaction  pair? Solution (a) N  :  Force  exerted  by  table  on  the  box.     (b) The  block  is  in  equilibrium.  F  0  W  N  0  N  100 N (c) N  =  100  N  :  Because  force  by  table  on  the  box  and  force  by  box  on  table  make  Newton's  third  law  pair. (d) No Example Consider  a  spring  attached  at  one  of  its  ends  to  a  fixed  support  and  at  other  end  to  a  box,  which  rests  on  a smooth  floor  as  shown  in  the  figure.  Denote  mass  of  the  box  by  m,  force  constant  of  the  spring  by  k  and acceleration  due  to  gravity  by  g. The  box  is  pushed  horizontally  displacing  it  by  distance  x  towards  the  fixed  support  and  held  at  rest. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 (a) Draw  free  body  diagram  of  the  box. (b) Find  force  exerted  by  hand  on  the  box. E (c) Write  all  the  third  law  action-reaction  pairs. Solution (a) F  is  push  by  hand.       (b) Since the block is in equilibrium Fx =  0    F    =    kx (c) (i) Force  by  hand  on  box  and  force  by  box  on  hand. (ii) Force  by  spring  on  box  and  force  by  box  on  spring. (iii) Normal  reaction  by  box  on  floor  and  normal  reaction  by  floor  on  box. (iv) Weight  of  the  box  and  the  gravitational  force  by  which  box  pulls  the  earth. (v) Force  by  spring  on  support  and  force  by  support  on  spring. 8

JEE-Physics Example (a) A  box  of  weight  103  N  is  held  in  equilibrium  with  the  help  of  two  strings  OA  and  OB  as  shown  in  figure-I.  The string  OA  is  horizontal.  Find  the  tensions  in  both  the  strings. Fig. I                                          Fig. II      (b) If  you  can  change  location  of  the  point  A  on  the  wall  and  hence  the  orientation  of  the  string  OA  without  altering the  orientation  of  the  string  OB  as  shown  in  figure-II.  What  angle  should  the  string  OA  make  with  the  wall  so  that a  minimum  tension  is  developed  in  it? Solution (a) Free  body  diagram  of  the  box          Graphical  Method  :  Use  triangle  law   T sin60°  =  103    T =20N 2 2 T ta n60 °   =  1 03    T =1 0 N 1 1 Analytical  Method:  Use  Cartesian  components ...(i) ...(ii)  Fx  0          T2 cos 60  T  Fy  0          T2 sin 60  10 3 From  equation  (i)  &  (ii)  we  have  T =10N  and  T =20N 12 (b) Free  body  diagram  of  the  box Graphical  Method  :  Use  triangle  law For  T   to  be  minimum,  it  must  be  perpendicular  to  T . 12 From  figure   =  60° Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Analytical  Method  :  Use    Cartesian  components ...(i)  Fx  0         T2 cos 60  T1 sin   Fy  0          T1 cos   T2 sin 60  10 3 ...(ii) From  equation  (i)  and  (ii),  we  have 10 3 T   =  1 3 sin   cos  If  T  is  minimum,  3sin  +  cos  must  be  maximum.  Maximum  value  of  3  sin  +  cos  is  2. 1 3sin  +  cos  =2 Solving  the  above  equation  we  get    =  60° E9

JEE-Physics Example Two  boxes  A  and  B  of  masses  m  and  M  are  suspended  by  a  system  of pulleys  are  in  equilibrium  as  shown.  Express  M  in  terms  of  m. Solution Since  tension  on  both  sides  of  a  pulley  are  equal  and  string  is  massless therefore  tension  everywhere  on  the  string  must  have  same  magnitude. FBD  of  block  A FBD  of  pulley FBD  of  Block  B For  equilibrium  of  block  A  ...(i) ...(ii)  F  0  T  =  mg ...(iii)  For  equilibrium  of  pulley  attached  to  block  B  F  0  F  =  2T For  equilibrium  of  block  B   F  0  F  =  Mg From  equation  (i),  (ii)  and  (iii),  we  have  M  =  2m Example A  box  of  mass  m  rests  on  a  smooth  slope  with  help  of  a  thread  as  shown  in  the figure.  The  thread  is  parallel  to  the  incline  plane. (a) Draw  free  body  diagram  of  the  box. (b) Find  tension  in  the  thread. (c) If  the  thread  is  replaced  by  a  spring  of  force  constant  k,  find  extension in  the  spring. Solution (a) Free  body  diagram  of  the  block (b) The  block  is  in  equilibrium,  therefore Fx  0         T  mg sin  ...(i) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 ...(ii) (c) If  the  thread  is  replaced  by  a  spring,  spring  force  must  be  equal  to  T, therefore T  =  kx From  equation  (i)  and  (ii),  we  have x  mg sin  k 10 E

JEE-Physics Example Block  A  of  mass  m  placed  on  a  smooth  slope  is  connected  by  a  string  with  another  block  B  of  mass  M  as  shown in  the  figure.  If  the  system  is  in  equilibrium,  express  M  in  terms  of  m. Solution For  equilibrium  of  the  block  A,  net  force  on  it  must  be  zero. N:  Normal  reaction  from  slope  Fx  0         T  mg sin  ...(i)         Fy  0         N  mg cos  ...(ii) For  equilibrium  of  block  B,  the  net  force  on  it  must  be  zero.  Fy  0         T  mg ...  (ii) From  equations  (i)  and  (ii),  we  have        M  =  msin Example A  70  kg  man  standing  on  a  weighing  machine  in  a  50  kg  lift  pulls  on  the  rope,  which  supports  the lift  as  shown  in  the  figure.  Find  the  force  with  which  the  man  should  pull  on  the  rope  to  keep  the lift  stationary  and  the  weight  of  the  man  as  shown  by  the  weighing  machine. Solution Tension  magnitude  everywhere  in  the  string  is  same.  For  equilibrium  of  the  lift.  Fy  0        500  N  2T ...(i) To  analyse  equilibrium  of  the  man  let  us  assume  him  as  a  block  Fy  0         N  T  700Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 ...(ii) From  equations  (i)  &  (ii),  we  have  T  =  400  N  and  N  =  300  N Here,  T  is  the  pull  of  mass  and  N  is  reading  of  the  weighing  machine. E 11

JEE-Physics Example A  block  of  mass  m  placed  on  a  smooth  floor  is  connected  to a  fixed  support  with  the  help  of  a  spring  of  force  constant  k. It  is  pulled  by  a  rope  as  shown  in  the  figure.  Tension  force  T of  the  rope  is  increased  gradually  without  changing  its direction,  until  the  block  losses  contact  from  the  floor.  The increase  in  rope  tension  T  is  so  gradual  that  acceleration  in the  block  can  be  neglected. (a) Well  before  the  block  losses  contact  from  the  floor,  draw  its  free  body  diagram. (b) What  is  the  necessary  tension  in  the  rope  so  that  the  block  looses  contact  from  the  floor? (c) What  is  the  extension  in  the  spring,  when  the  block  looses  contact  with  the  floor? Solution (a) Free  body  diagram  of  the  block,  well  before  it  looses  contact  with  the  floor.        (b) When  the  block  is  about  to  leave  the  floor,  it  is  not  pressing  the  floor. ...(i) Therefore  N  =  0  and  the  block  is  in  equilibrium. ...(ii)  Fx  0          T cos   kx  Fy  0          T sin   mg From  equations  (ii),  we  have  T   =  mgco sec (c) From  equation  (i)  and  (ii),  we  have  x  mg cot  k Dynamics  of  Par ticles:    Translat ion  motion  of  accelerated  bodies Newton's  laws  are  valid  in  inertial  frames,  which  are  un-accelerated  frames.  At  present,  we  are  interested  in motion  of  terrestrial  bodies  and  for  this  purpose;  ground  can  be  assumed  a  satisfactory  inertial  frame. In  particle  dynamics,  according  to  Newton's  second  law,  forces  acting  on  the  body  are  considered  as  cause  and rate  of  change  in  momentum  as  effect.  For  a  rigid  body  of  constant  mass,  the  rate  of  change  in  momentum equals  to  product  of  mass  and  acceleration  vector.  Therefore,  forces  acting  on  it  are  the  cause  and  product  of mass  and  acceleration  vector  is  the  effect. To  write  the  equation  of  motion  it  is  recommended  to  draw  the  free  body diagram,  put  a  sign  of  equality  and  in  front  of  it  draw  the  body  attached with  a  vector  equal  to  mass  times  acceleration  produced.  In  the  figure Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 is  shown  a  body  of  mass  m  on  which  a  single  force    F   acts  and  an observer  in  an  inertial  frame  of  reference  observes  the  body  moving  with  acceleration  a . Acceleration  imparted  to  a  body  by  a  force  is  independent  of  other  forces,   therefore  when  several  forces    F1,  F2   and  Fn act  simultaneously  on  a body,  the  acceleration  imparted  to  the  body  is  the  same  as  a  single force  equal  to  the  vector  sum  of  these  forces  could  produce.  The  vector sum  of  these  forces  is  known  as  the  net  resultant  of  these  forces. 12 E

JEE-Physics    Fnet  F1  F2  ......................  Fn  ma    ma F In  Cartesian  coordinate  system  the  vector  quantities  in  the  above  equation  is  resolved  into  their  components along  x,  y,  and  z  axes  as  follows:  Fx  max  Fy  may  Fz  maz Example Two  forces  F   and  F     of  magnitudes  50  N  and  60  N  act  on  a  free  body  of  mass  m  =  5  kg  in  directions  shown 12 in  the  figure.  What  is  acceleration  of  object  with  respect  to  the  free  space? Solution In  an  inertial  frame  of  reference  with  its  x-axis  along  the  force  F ,  the  forces  are  expressed  in  Cartesian 2 components.    F1  30ˆi  40ˆj  N and  F2  60ˆi   N  Fx  ma x  a x  6   m/s2  Fy  may  a y  8   m/s2   6ˆi  8ˆj   m/s2 a Example Boxes  A  and  B  of  mass  mA  =  1  kg  and  mB  =  2  kg  are  placed  on  a  smooth  horizontal  plane.  A  man  pushes horizontally  the  1  kg  box  with  a  force  F  =  6  N.  Find  the  acceleration  and  the  reaction  force  between  the  boxes. Solution Since  both  the  blocks  move  in  contact  it  is  obvious  that  both  of  them  have  same  acceleration.  Say  it  is  'a'. Applying  NLM  to  block  A N:  Normal  reaction  from  B N :  Normal  reaction  from  floor 1  Fx  ma x  6  –  N  =  a ...(i) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65  Fy = 0  N   =  10  N ...(ii) 1 ...(iii) Applying  NLM  to  block  B N  :  Normal  reaction  from  A N :  Normal  reaction  from  ground 2  Fx = ma x  N    =    2a  Fy = 0  N     =    20  N ...(iv) 2 From  equations  (i)  &  (iii),  we  have  a  =  2  m/s2  and  N  =  4  N E 13

JEE-Physics Example Two  blocks  A  and  B  of  masses  m   and  m   connected  by  light  strings  are  placed  on  a  smooth  floor  as  shown  in 12 the  figure.    If  the  block  A  is  pulled  by  a  constant  force  F,  find  accelerations  of  both  the  blocks  and  tension  in  the string  connecting  them. Solution String  connecting  the  blocks  remain  taut  keeping  separation  between  them  constant.  Therefore  it  is  obvious that  both  of  them  move  with  the  same  acceleration.  Say  it  is  'a'. Applying  NLM  to  block  A T:  Tension  of  string N   :  Normal  reaction  from  ground 1  Fx  ma           F  T= m1a ...(i) Applying  NLM  to  block  B.  Fy  =  0            N1  m1g ...(ii) ..(iii) T  :  Tension  of  string N   :  Normal  reaction  from  ground. 2  Fx  ma x         T  m2a .  Fy  0            N2  m2 g ...(iv) From  equation  (i)  and  (iii),  we  have  a  m1 F   and  T  m2F  m2 m1  m2 Example Three  identical  blocks  A,  B  and  C,  each  of  mass  2.0  kg  are  connected  by  light  strings  as  shown  in  the  figure.    If the  block  A  is  pulled  by  an  unknown  force  F,  the  tension  in  the  string  connecting  blocks  A  and  B  is  measured  to be  8.0  N.  Calculate  magnitude  of  the  force  F,  tension  in  the  string  connecting  blocks  B  and  C,  and  accelerations of  the  blocks. Solution It  is  obvious  that  all  the  three  blocks  move  with  the  same  acceleration.  Say  it  is  'a'. Applying  NLM  to  the  block  A. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 T :  Tension  of  string  connecting  blocks  A  and  B. 1 N   :  Normal  reaction  from  floor. 1  Fx  ma x           F  T1  2a ...(i)  Fy  0             N1  20N ...(ii) Applying  NLM  to  the  block  B. 14 E

JEE-Physics T :  Tension  of  string  connecting  blocks  A  and  B. 1 T   :  Tension  of  string  connecting  B  &  C. 2 N   :  Normal  reaction  from  floor. 1  Fx  ma x           T1  T2  2a ...(iii) Applying  NLM  to  the  block  C.  Fy  0             N2  20N ...(iv) ...(v) T   :  Tension  of  string  connecting  B  &  C. 3 N3  :  Normal  reaction  from  floor.  Fx  ma x         T2  2a  Fy  0            N3  20N ...(vi) ...(vii) From  equations  (i),  (iii)  and  (v),  we  have F  =  6a Now  using  the  fact  that  T =  8N  with  equation  (i),    We  have a  =  2  m/s2 1  Now  from  equation  (i)  we  have  F  =12  N            From  equation  (iii),  we  have  T =  4  N 2  Example Two  blocks  A  and  B  of  masses  m   and  m   connected  by  uniform  string  of  mass  m  and  length    are  placed 12 on  smooth  floor  as  shown  in  the  figure.  The  string  also  lies  on  the  floor.  The  block  A  is  pulled  by  a  constant force  F. (a)  Find  accelerations  a  of  both  the  blocks  and  tensions  T   and  T   at  the  ends  of  the  string. AB (b)  Find  an  expression  for  tension  T  in  the  string  at  a  distance  x  from  the  rear  block  in  terms  of  T ,  T ,  m,    and  x. AB Solution It  is  obvious  that  both  the  blocks  and  the  whole  string  move  with  the  same  acceleration  say  it  is  'a'.  Since  string has  mass  it  may  have  different  tensions  at  different  points. (a) Applying  NLM  to  block  A. T   :  Tension  of  the  string  at  end  connected  to  block  A. A N   :  Normal  reaction  of  floor 1  Fx  ma x          F  TA  m1a ...(i)  Fy  0             N1  m1g ...(ii) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Applying  NLM  to  the  rope T   :  Tension  of  string  at  end  connected  to  block  B. B N  :  Normal  reaction  of  floor E 15

JEE-Physics  Fx  ma x          TA  TB  ma ...(iii) Applying  NLM  to  the  block  B  Fy  0            N  mg ...(iv) ...(v) T   :  Tension  of  string B N   :  Normal  reaction  from  floor 2  Fx  ma x         TB  m2a  Fy  0            N2  m2 g ...(vi) From  equations  (i),  (iii)  and  (v),  we  have  a  F  m2 ...(vii)      TA  m  m2 F ...(viii)      TB  m2F ...(ix) m  m1 m  m1  m2 m  m1  m2 (b) To  find  tension  at  a  point  x  distance  away  from  block  B,  we  can  consider  string  of  length  x  or  –x. Let  as  consider  length  of  string  x  and  apply  NLM. mx ...(x)  :  mass  of  length  x. T   =  Tension  at  distance  x x N   =  Normal  reaction  of  floor x  mx Fx  ma x           Tx  TB   a From  equation  (vii),  (viii),  (ix)  and  (x),  we  have  Tx  m x  m 2  F     m1 m  m2 Example The  system  shown  in  the  figure  is  released  from  rest.  Assuming  mass  m   more 2 than  the  mass  m ,  find  the  accelerations  of  the  blocks  and  the  tension  in  the 1 string. Solution It  obvious  that  both  blocks  move  with  same  acceleration  magnitudes.  Say  it  is  'a'. Since  m   is  heavier,  it  moves  downwards  and  m   moves  upwards. 21 Tension  at  both  the  ends  of  the  string  has  same  magnitude.  Say  it  is  'T'. Apply  NLM  to  block  A  of  mass  m 1  Fy  ma y           T  m1g  m1a ...(i) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Apply  NLM  to  block  of  mass  m 2 16 E

JEE-Physics  Fy  ma y           m2 g  T  m2a ...(ii) From  equations  (i)  &  (ii),  we  have  a   m2  m1  g ,  T   2m1m2  g  m1  m2   m1  m2  Example Block  A  of  mass  m  placed  on  a  smooth  slope  is  connected  by  a  string  with  another  block  B  of  mass  M  (>  msin) as  shown  in  the  figure.  Initially  the  block  A  is  held  at  rest  and  then  let  free.  Find  acceleration  of  the  blocks  and tension  in  the  string. Solution Both  the  blocks  must  move  with  the  same  magnitude  of  acceleration. Since  M  >  m  sin,  block  B  move  downward    pulling  block  A  up  the  plane.  Let  acceleration  magnitude  is  'a'. Tension  at  both  the  ends  of  the  string  is  same.  Say  it  is  'T'. Apply  NLM  to  block  A N:  Normal  reaction  from  slope T  :  Tension  of  string  Fx  ma x           T  mg sin   ma ...(i)  Fy  0              N  mg cos  ...(ii) Apply  NLM  to  block  B Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 T:  Tension  of  string ...(iii)  Fy  may           Mg  T  Ma From  equation  (i)  &  (iii),  we  have  a  M  m sin  g ,  T  1  sin  mMg Mm mM E 17

JEE-Physics System  of  Interconnected  bodies In  system  of  interconnected  bodies,  several  bodies  are  interconnected  in  various  manners  through  some  sort  of physical  links.  Sometimes  these  physical  links  includes  ropes  and  pulleys  and  sometimes  the  bodies  under investigation  are  pushing  each  other  through  direct  contact.  In  systems  consisting  of  bodies  interconnected through  ropes  and  pulleys,  relation  between  their  accelerations  depends  on  the  arrangement  of  the  ropes  and pulleys.  In  addition,  in  system  where  bodies  push  each  other,  they  affect  relation  between  their  accelerations due  to  their  shapes. In  kinematics  while  dealing  with  dependant  motion  or  constrained  motion,  we  have  already  learnt  how  to  find relations  between  velocities  and  accelerations  of  interconnected  bodies. Analysis  of  physical  situations  involving  interconnected  bodies  often  demands  relation  between  accelerations  of these  bodies  in  addition  to  the  equations  obtained  by  application  of  Newton’s  laws  of  motion.  Therefore,  while analyzing  problems  of  interconnected  bodies,  it  is  recommended  to  explore  first  the  relations  between  accelerations and  then  apply  Newton’s  laws  of  motion. In  following  few  examples,  we  learn  how  to  deal  with  problems  of  interconnected  bodies. Example Two  boxes  A  and  B  of  masses  m  and  M  interconnected  by  an  ideal  rope  and  ideal  pulleys,  are  held  at  rest  as shown.  When  it  is  released,  box  B  accelerates  downwards.  Find  accelerations  of  the  blocks. AB Solution. We  first  show  tension  forces  applied  by  the  string  on  the  box  A  and  the  pulley  connected  to  box  B.  Since  the string,  as  well  as  the  pulleys,  both  are  ideal;  the  string  applies  tension  force  of  equal  magnitude  everywhere. Denoting  the  tension  force  by  T,  we  show  it  in  the  adjacent  figure. TT T aA A B a B We  first  explore  relation  between  accelerations  a   and  a   of  the  boxes  A  and  B,  which  can  be  written  either  by AB using  constrained  relation  or  method  of  virtual  work  or  by  inspection. y ma T A x A mg a   =  2a ...(i) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 AB ...(i) Applying  Newton’s  Laws  of  motion  to  box  A  Fy  may  T  mg  maA TT y x 0×a B F 18 E

JEE-Physics Applying  Newton’s  Laws  of  motion  to  the  pulley  Fy  may  2T  F  0  aB y Ma 2T B F  =  2T          ...(iii) Applying  Newton’s  Laws  of  motion  to  box  B x A  Fy  may  Mg  2T  MaB ...(iv) Mg From  equations  (i),  (ii),  (iii)  and  (iv),  we  have aA  2  M  2m  g  and a B   M  2m  g  M  4m   M  4m  Example In  the  system  shown  in  figure,  block  m   slides  down  a  friction  less  inclined  plane.  The  pulleys  and  strings  are 1 ideal.  Find  the  accelerations  of  the  blocks. m m1 2  Solution Tension  forces  applied  by  the  strings  are  shown  in  the  adjacent  figure. 2T 2T T m2 a2 m1 a1 Let  the  block  m   is  moving  down  the  plane  with  an  acceleration  a   and  m   is  moving  upwards  with  accelerations 1 12 a .  Relation  between  accelerations  a   and  a   of  the  blocks  can  be  obtained  easily  by  method  of  virtual  work. 2 12 a   =  2a ...(i) m1gsin 12 ...(ii) mg Applying  Newton’s  laws  to  analyze  motion  of  block  m 1 1  Fx  ma x  m1g sin   T  m1a1 m gcos  1 T N m1a1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\NLM\\English\\Theory.p65 Applying  Newton’s  laws  to  analyze  motion  of  block  m 2  Fy  ma y  2T  m2 g  m2a2 ...(iii) From  equation  (i),  (ii)  and  (iii),  we  have 2T y ma x 2 2 (2 m1 sin   m ) 2m1g sin   m2g m 4m1  m2 4m1  m2 2 mg 2 a1  2 g   a  2 E 19