By Ahmad A. Kamal
This booklet primarily caters to the wishes of undergraduates and graduates physics scholars within the quarter of classical physics, in particular Classical Mechanics and electrical energy and Electromagnetism. academics/ Tutors could use it as a source ebook. The contents of the e-book are in keeping with the syllabi at the moment utilized in the undergraduate classes in united states, U.K., and different nations. The publication is split into 15 chapters, each one bankruptcy starting with a short yet sufficient precis and precious formulation and Line diagrams by way of numerous usual difficulties important for assignments and tests. targeted options are supplied on the finish of every bankruptcy.
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Additional info for 1000 Solved Problems in Classical Physics: An Exercise Book
1 Basic Concepts and Formulae 49 A two-body problem is reduced to a one-body problem through the introduction of the reduced mass μ. Motion of a Body of a Variable Mass It is well known that in relativistic mechanics the mass of a particle increases with increasing velocity. However, in Newtonian mechanics too one can give meaning to variable mass as in the following example. Consider an open wagon moving on rails on a horizontal plane under steady heavy shower. As rain is collected the mass of the wagon increases at constant rate.
Let the mass of the wire be M. Consider mass element dm at angles θ and θ + dθ Fig. 32 dm = M dθ M R dθ = 2α R 2α (1) From symmetry the CM of the wire must be on the y-axis. The y-coordinate of dm is y = R sin θ 90+α yCM = 1 M ydm = 90−α R sin α R sin θ dθ = 2α α Note that the results of prob. 43) follow for α = 12 π . 57 The diagram is the same as for prob. 44) y = r sin θ dA = r dθ dr dm = r dθ drρ = r dθ dr cr 2 = cr 3 dr dθ π R Total mass M= dm = c r dr 0 yCM dθ = 3 π c R4 4 (1) 0 1 1 = y dm = (r sin θ )cr 3 dr dθ M M R π C = r 4 dr sin θ dθ M 0 0 C 2 5 8a = R = M5 5π (2) where we have used (1).
8 (3) g 1 . 2 u 2 cos2 α 32 1 Kinematics and Statics (ii) At the highest point of the trajectory, the velocity of the particle is entirely horizontal, being equal to u cos α. The momentum of this particle at the highest point is p = mu cos α, when m is its mass. After the explosion, one fragment starts falling vertically and so does not carry any momentum initially. It would fall at half of the range, that is R 1 u 2 sin 2α (350)2 sin(2 × 55◦ ) = = = 5873 m, from the firing point. 8 The second part of mass 12 m proceeds horizontally from the highest point with initial momentum p in order to conserve momentum.