By Ahmad A. Kamal
This e-book primarily caters to the wishes of undergraduate and graduate physics scholars in classical physics, in particular Classical Mechanics and electrical energy and Electromagnetism. Lecturers/Tutors may perhaps use it as a source booklet. The contents of the e-book are in accordance with the syllabi at the moment utilized in the undergraduate classes within the united states, U.K., and different international locations. The booklet includes 15 chapters, each starting with a quick yet sufficient precis and valuable formulation and Line diagrams via a number of standard difficulties helpful for assignments and tests. designated ideas are supplied on the finish of every chapter.
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Extra info for 1000 Solved Problems in Classical Physics: An Exercise Book
Air resistance F actually increases with the velocity of the object (F ∝ v or v 2 or v 3 ). Here for simplicity we assume it to be constant. For upward motion, the equation of motion is ma1 = −(F + mg) 22 1 Kinematics and Statics or F +g m a1 = − (1) For downward motion, the equation of motion is ma2 = mg − F or F m a2 = g − (2) For ascent v1 = 0 = u + a1 t = u − t1 = v1 2 F + g t1 m u F m = 0 = u 2 + 2a1 h (3) g+ 2h u= g+ (4) F m where we have used (1). 3 Solutions 23 It follows that t2 > t1 , that is, time of descent is greater than the time of ascent.
32 Let the rifle be fixed at A and point in the direction AB at an angle α with the horizontal, the monkey sitting on the tree top at B at height h, Fig. 21. The bullet follows the parabolic path and reaches point D, at height H , in time t. Fig. 21 The horizontal and initial vertical components of velocity of bullet are u x = u cos α; u y = u sin α Let the bullet reach the point D, vertically below B in time t, the coordinates of D being (d, H ). As the horizontal component of velocity is constant d = u x t = (u cos α)t = udt s where s = AB: t= s u The vertical component of velocity is reduced due to gravity.
44 Let the x-axis lie along the diameter of the semicircle. The centre of mass must lie on y-axis perpendicular to the flat base of the semicircle and through O, the centre of the base, Fig. 25. 3 Solutions 37 Fig. 25 For continuous mass distribution yCM = 1 M y dm Let σ be the surface density (mass per unit area), so that M= 1 π R2σ 2 In polar coordinates dm = σ dA = σ r dθ dr where dA is the element of area. 45 Let O be the origin, the centre of the base of the hemisphere, the z-axis being perpendicular to the base.