Title Med-RM_Phy_SP-1_Ch-4_Motion in a Plane.pdf ✅

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph. 011-47623456 Chapter Contents SCALARS AND VECTORS Scalars These are physical quantities which need only a magnitude. For example mass, temperature, etc. Vectors A vector has magnitude and direction as well, and it follows rules of vector addition (Commutative Law i.e.  ABBA ). Some vectors are displacement, velocity, acceleration, force etc. TYPES OF VECTORS (A) Unit Vector : A vector whose magnitude is unity is called a unit vector. If a be a vector whose magnitude is a, then the corresponding unit vector is represented by ˆ | | a a a a a   a a a aa   | | ˆ ˆ Y j ( ) X i ( ) Unit vector has no unit. Z k () The unit vector along axes X, Y and Z are represented by ˆˆ ˆ ij k , and respectively. (B) Like Vectors : Two vectors are said to be like or parallel vectors if their directions are same. Also parallel vectors can be represented as multiples of each other. i.e., if a and b are two parallel vectors then b can be represented in multiples of a as b = m a, where m is a positive constant. (C) Equal Vectors : Two vectors having same magnitude and same direction, are said to be equal vectors. e.g., if a b and are equal vectors then a b  . Chapter 4 Scalars and vectors Types of vectors Multiplication of Vectors by Real Numbers Addition and Subtraction of Vectors Resolution of a Vector Motion in Plane with Constant Acceleration Relative Motion in Two Dimensions Crossing a River Projectile Motion Circular Motion Motion in a Plane

NEET Motion in a Plane 75 Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 Case - IV : | | | | | | 120 CAB      A 120° Case - V :  = 90° A2 + B2 = C 2  A B  i.e.  = 90°, tan B A   A  90° | | (| | | |) and | | (| | ~ | |)    C AB C AB Subtraction of Vectors C ABA B     ( ) 2 2 | | C A B AB     2 cos sin tan cos B A B      A  C  180°  (–B) B If | || | AB AB      2 2 ,| | | | ABAB AB A B If | | | | then | | 2 sin /2 A B AB A    RESOLUTION OF A VECTOR Rectangular Component The vector a shown in figure has magnitude a and is making an angle with positive direction of X-axis. Using triangle law: OA OM MA   ˆ ˆ a ai a j   x y Y A a X O M ay ax  Here, ax is the x-component of a and ay is the y-component of a Also cosθ and sinθ OM MA ax ay OA a OA a    i.e., ax = a cos ay = a sin            2 2 1 and tan y x y x a a aa a From the above relations, if a and  are known, then ax and ay can be found and vice-versa. The rectangular component of vector on that axis from which angle () is given will be of cos and other rectangular component will be of sin. The component of a vector may be greater than vector. Rectangular component of a vector can never be greater than vector, may be equal to vector. If a vector A makes angle ,  and  from x, y and z axis respectively then Ax = A cos Ay = A cos Az = A cos cos, cos and cos called direction cosines.    x y z A
76 Motion in a Plane NEET Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456 If vector  ˆˆˆ A Ai A j Ak xy z   222 | | A AAA xyz Angle of vector from x-axis, 222 cos | | x x xyz A A A A A A     Angle of vector from y-axis,     222 cos y xyz A A A A Angle of vector from z-axis,     222 cos z xyz A A A A cos2 + cos2 + cos2 = 1 sin2 + sin2 + sin2 = 2 Example 1 : Calculate the angle of  345 ˆˆˆ Ai jk from z-axis. Solution : | | 9 16 25 5 2    A    5 1 cos | | 52 2 Az A   = 45° Example 2 : Find the resultant of two vectors P i j and Q i j   3 2 2 3. ˆˆ ˆˆ Solution : Let R PQ     (3 2 ) (2 3 ) ij ij - - Rij   5 5 This gives the resultant vector. Magnitude of R   25 25 5 2 Orientation of R , –1 5 tan 5         = 45° with x-axis Example 3 : If x and y components of a vector P have numerical values 5 and 6 respectively and that of P Q have magnitudes 10 and 9, find the magnitude of Q. Solution : According to the question Pij   5 6 and PQ i j   10 9  Q ijP   (10 9 ) ( ) = (10 9 ) ( ) ij P  