Question

1a.(i). The motion of an insect is displaced on a straight track by joining the base and the apex of two
vertical poles each 2cm high, and 3m apart. Calculate
(a)The straight track covered by the insect along the diagonal
(b)The angle above the horizontal, and ;
(c)Express the information analytically.
(ii) Find the scalar product A= 4î – 8ĵ – 3ǩ and B= 6î + 3ĵ + 5ǩ.Hence determine the angle between them.
b.(i) State’s Newton’s law of Universal Gravitation
(ii) If two cars 5m apart has masses 1000kg and 1200kg respectively ,Calculate
(a) The Gravitational attraction
(b) The Gravitational Potential

Answer 1

APEX 2m 3m BASE (a) The straight track covered by the insect along the diagonal AB = Use Pythagoras theorem, AB2- BC+CA AB2- 32 22 AB square root (13) AB 3.6055 m. (b) The angle above the horizontal e From triangle ABC tane AC/BC = = 2/3 e 33.69 degrees (c) Dot product or scalar product of two vectors A B ax x bx+ay x by + az x bz A= 41-81-3k and B- 6131 +5k A.B 4x6(-8x3) +(-3x5); = -15

AV(16+64 +9) V(89) BI V(36+9+25) V(70) Now for the other formula: A B IAI x IBI x cos(e) -15 v(89) x V(70) x cos(e) cos(e) -0.19 0 100.955 degrees (Angle between the vectors) State's Newton's law of Universal Gravitation: Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The magnitude of the gravitational force is: mim2 F12 G Where G is the gravitational constant given as G 6,67 x 10 m3 kg-1 s-2 If two cars 5m apart has masses 1000kg and 1200kg respectively ,Calculate (a) The Gravitational attraction F12 = 6.67x 10 x 1000 x 1200/5 3.2 x 105 N (b) The Gravitational Potential gravitational potential energy U Gmam2/r 6.67 x 101x1000 x 1200/5 1.6 10 N.