Title Notes_Exercise_02.Physics And Mathematics.pdf ✅

2.1 SOLUTIONS TO CONCEPTS CHAPTER – 2 1. As shown in the figure, The angle between A  and B  = 110° – 20° = 90° | A |  = 3 and |B |  = 4m Resultant R = A  B  2ABcos  2 2 = 5 m Let  be the angle between R  and A   =           3 4cos90 4sin90 tan 1 = tan–1 (4/3) = 53°  Resultant vector makes angle (53° + 20°) = 73° with x-axis. 2. Angle between A  and B  is  = 60° – 30° =30° | A  | and |B  | = 10 unit R = 2 2 10 10  2.10.10.cos30 = 19.3  be the angle between R  and A   = tan–1 10sin30 1 1 tan 10 10cos30 2 3                   = tan–1 (0.26795) = 15°  Resultant makes 15° + 30° = 45° angle with x-axis. 3. x component of A  = 100 cos 45° = 100 / 2 unit x component of B  = 100 cos 135° = 100 / 2 x component of C  = 100 cos 315° = 100 / 2 Resultant x component = 100 / 2 – 100 / 2 + 100 / 2 = 100 / 2 y component of A  = 100 sin 45° = 100 / 2 unit y component of B  = 100 sin 135° = 100 / 2 y component of C  = 100 sin 315° = –100 / 2 Resultant y component = 100 / 2 + 100 / 2 – 100 / 2 = 100 / 2 Resultant = 100 Tan  = x component y component = 1   = tan–1 (1) = 45° The resultant is 100 unit at 45° with x-axis. 4. a i4 j3      , b i3 j4      a) 2 2 |a|  4  3  = 5 b) |b|  9 16  = 5 c) a|  |b  i7|  |j7  7 2     d) a b ˆ ˆ ˆ ˆ     ( 3 4)i    ( 4 3)j  i j   2 2 | a b |   1  ( 1)  2   . x y  R  B  A  20 x y  B  A  30° 60° 315° 45° 135° Page 1 PHYSICS AND MATHEMATICS
Chapter-2 2.2 5. x component of OA = 2cos30° = 3 x component of BC = 1.5 cos 120° = –0.75 x component of DE = 1 cos 270° = 0 y component of OA = 2 sin 30° = 1 y component of BC = 1.5 sin 120° = 1.3 y component of DE = 1 sin 270° = –1 Rx = x component of resultant = 3  .0 75  0 = 0.98 m Ry = resultant y component = 1 + 1.3 – 1 = 1.3 m So, R = Resultant = 1.6 m If it makes and angle  with positive x-axis Tan  = x component y component = 1.32   = tan–1 1.32 6. a| |  = 3m |b|  = 4 a) If R = 1 unit  3  4  .4.3.2 cos 2 2 = 1  = 180° b) 3  4  .4.3.2 cos 2 2 = 5  = 90° c) 3  4  .4.3.2 cos 2 2 = 7  = 0° Angle between them is 0°. 7. Kˆ J 4 ˆ i 5.0 ˆ AD  2   = 6iˆ  0.5jˆ AD = 2 2 AE  DE = 6.02 KM Tan  = DE / AE = 1/12 = tan–1 (1/12) The displacement of the car is 6.02 km along the distance tan–1 (1/12) with positive x-axis. 8. In ABC, tan = x/2 and in DCE, tan = (2 – x)/4 tan  = (x/2) = (2 – x)/4 = 4x  4 – 2x = 4x  6x = 4  x = 2/3 ft a) In ABC, AC = 2 2 AB  BC = 2 10 3 ft b) In CDE, DE = 1 – (2/3) = 4/3 ft CD = 4 ft. So, CE = 2 2 CD  DE = 4 10 3 ft c) In AGE, AE = 2 2 AG  GE = 2 2 ft. 9. Here the displacement vector k ˆ j 3 ˆ i 4 ˆ r  7    a) magnitude of displacement = 74 ft b) the components of the displacement vector are 7 ft, 4 ft and 3 ft. 2m D A E B x O y 1m 1.5m 90° 30° 60° 6m A E D B 0.5 km  C 2m 4m 0.5 km 2–x G A D B BC = 2 ft AF = 2 ft DE = 2x x C E F r z Y Page 2 PHYSICS AND MATHEMATICS