By Ahmad A. Kamal

ISBN-10: 3642119425

ISBN-13: 9783642119422

This publication primarily caters to the desires of undergraduates and graduates physics scholars within the quarter of classical physics, particularly Classical Mechanics and electrical energy and Electromagnetism. teachers/ Tutors might use it as a source booklet. The contents of the booklet are in keeping with the syllabi at the moment utilized in the undergraduate classes in united states, U.K., and different international locations. The ebook is split into 15 chapters, every one bankruptcy starting with a quick yet enough precis and beneficial formulation and Line diagrams via a number of standard difficulties important for assignments and checks. exact strategies are supplied on the finish of every bankruptcy.

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**Extra resources for 1000 Solved Problems in Classical Physics: An Exercise Book**

**Example text**

28. The N–atom lies vertically above C, along the z-axis. 38 Å Now, the centre of mass of the three H atoms 3m H lies at C. The centre of mass of the NH3 molecule must lie along the line of symmetry joining N and C and is located below N atom at a distance 40 1 Kinematics and Statics Fig. 50 Take the origin at A at the left end of the boat, Fig. 29. Let the boy of mass m be initially at B, the other end of the boat. The boat of mass M and length L has its centre of mass at C. Let the centre of mass of the boat + boy system be located at G, at a distance x from the origin.

27 Take upward direction as positive, Fig. 18. At time t the velocities of the objects will be v 1 = u 1 iˆ − gt jˆ (1) v 2 = −u 2 iˆ − gt jˆ (2) If v 1 and v 2 are to be perpendicular to each other, then v 1 · v 2 = 0, that is u 1 iˆ − gt jˆ · −u 2 iˆ − gt jˆ = 0 ∴ or −u 1 u 2 + g 2 t 2 = 0 1√ t= u1u2 g ˆ The position vectors are r 1 = u 1 t iˆ − 12 gt 2 j, The distance of separation of the objects will be r12 = |r 1 − r 2 | = (u 1 + u 2 )t Fig. 18 (3) ˆ r 2 = −u 2 t iˆ − 12 gt 2 j. 28 1 Kinematics and Statics or r12 = (u 1 + u 2 ) √ u1u2 g (4) where we have used (2).

42 Taking torque about D, the corner of the obstacle, (F)CD = (W )BD (Fig. 43 Let λ be the linear mass density (mass per unit length) of the wire. Consider an infinitesimal line element ds = R dθ on the wire, Fig. 24. The corresponding mass element will be dm = λds = λR dθ . Then Fig. 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.

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