Tiêu đề 03_VIII_OLY._PHY._VOL-1_WS-3-4 & KEY(VECTORS)_25 to 55.pdf ✅

26 Narayana Group of Schools VIII - PHYSICS (Vol-1) Olympiad Text Book 5. Triangle law of vectors: If two vectors are represented in magnitude and direc- tion by the two sides of a triangle taken in order, the third side of the triangle taken in reverse order represents their resultant in magnitude and direction.(or) If three vectors simultaneously acting at a point have zero resultant then these three vectors can be represented both in magnitude and direction by the sides of a triangle taken in an order. A C B In the above figure A B C O        6. PARALLELOGRAM LAW OF VECTORS : Two vector quantities can be added using parallelogram law ( velocity vector can be added to velocity vector only). This law is useful to find both magnitude and direction of resultant. Statement: If two vectors are represented in magnitude and direction by the adjacent sides of a parallelogram drawn from a point, the diagonal passing through that point represents their resultant both in magnitude and direction. A  B D Q  P  A  B D Q  P  A  B D Q  P  C E R   figure (d) Explanation: P  and Q  are two vectors represented by AB  and AD  . Both vectors act at the common point A and mutually inclined at angle ‘  ’ as shown in fig (d). If the parallelogram ABCD is completed taking AB and AD as adjacent sides, then the diagonal AC  represents their resultant R  both in magnitude and direction. Magnitude of the resultant: The line of action of P  is extended. The perpendicular drawn from ‘C’ meets the extension of AB at E. From the figure, it is obvious that BC AD Q      and    CBE  Length of AB = magnitude of P  = P ; Length of BC = magnitude of Q  = Q Length of AC = magnitude of R  = R From triangle CBE, cos  = BE BC , BE = BC cos   BE = Q cos  ........... [1] From the triangle AEC, EC sin BC   and EC = Q sin ........... [2] (AC)2 = (AE)2 + (EC)2  (AC)2 = (AB + BE)2 + (EC)2  (AC)2 = (AB)2 + (BE)2 + 2AB . BE + (EC)2  R2 = P2 + Q2 cos2  + 2PQ cos  + Q2 sin2  R2 = P2 + Q2 + 2PQ cos   R 2 2      P Q 2PQcos (3)
VIII - PHYSICS (Vol-1) Olympiad Text Book Narayana Group of Schools 27 Direction of the resultant : The resultant makes angle ‘ ’ with A  (say) From triangle CAE : EC EC tan AE AB BE     Qsin tan P Qcos       1 Qsin tan P Qcos               (4) The expression (3) and (4)gives the magnitude and direction of the resultant of P  and Q  . SPECIAL CASES:1) If P  and Q  are in same direction, then    0 , and cos  = 1  From equation (3) and (4), R P Q      and  = 0 Hence the magnitude of resultant is sum of the magnitude of individual vectors. The direction of resultant is same as that of individual vectors. 2)If P  and Q  are opposite, then  = 180° and cos  = –1  R P Q      i.e., R P Q      or Q P   and    0 or 180°. Thus the magnitude of resultant is equal to difference of magnitudes of individual vectors and the direction of resultant is same as that of the vector of larger magnitude. 3) If P  and Q  are perpendicular, then    90 & cos  = 0  R 2 2   P Q and  = Tan–1 (Q/P) 4) If P Q    , then R = 2P cos  /2 and   /2  If the vectors have equal magnitude, then the resultant will bisect the angle between them. Example for a parallelogram law of vectors: A current in a wire is represented by the direction but it is not vector quantity because it does not obey the laws of vector addition. In figure the current flowing in wire OC = current in wire AO + current in wire BO i i1 i2 A B C O  i i1 i2 A B C O  i = i1 +i2 If the current is a vector then i will be i = 2 2 1 2 1 2 i i 2i i cos    according to parallelogram law of vector addition.
28 Narayana Group of Schools VIII - PHYSICS (Vol-1) Olympiad Text Book VECTORS WORKSHEET-3 :1. If A+B = A + B then the angle between the A and B . 1) 0 0 2) 900 3) 1800 4) 600 2. p  and q  are two adjacent sides of a parallelogram. if p q    is one diagonal, then the other diagonal is 1) p q    2) q p    3) q p    4) 1 & 2 3. a-b = a + b then the angle between a and b is 1) 900 2) 1800 3) 450 4) 00 4. If P + Q = R and P - Q = S then 2 2 R + S is equal to 1) 2 2 P + Q 2) 2 2 2(P - Q ) 3) 2 2 2(P + Q ) 4) 4PQ 5. If the magnitudes of _ A and _ B are a and b respectively, the magnitude of the resultant vector is always 1) equal to a b   2) less than a b   3) greater than a b   4)not greater than a b   6. _ _ _ _ A B A B    then the angle between _ _ A B, is 1) 1200 2)00 3)900 4)1800 7. Out of the following the resultant of which cannot be 4N. 1) 2N and 2N 2) 3N and 8N 3) 2N and 6N 4) 4N and 8N 8. The maximum value of magnitude of A B   is 1) A B 2) A 3) A B  4) 2 2 A B  9. Two vectors A and B have precisely equal magitudes. If magnitude of A B  to be larger than the magnitude of A B by a factor n, the angle between them is 1)   1 2 tan 1 n  2)   1 tan 1 n  3)   1 tan 1 2n  4)   1 2 tan 1 2n  10. If the Resultant of P  and Q  makes an angle 1 with P  and 2 with Q  . Then 1. 1 2     if P Q 2. 1 2     if P Q 3. 1 2     if P Q 4. 1 2     if P Q 11. It is found that A B A   . This necessarily implies. 1. B  0 2. A, B are antiparallel 3. A, B are perpendicular 4. A B. 0 