By Ahmad A. Kamal
This publication essentially caters to the wishes of undergraduates and graduates physics scholars within the region of classical physics, in particular Classical Mechanics and electrical energy and Electromagnetism. academics/ Tutors may well use it as a source publication. The contents of the publication are according to the syllabi at present utilized in the undergraduate classes in united states, U.K., and different nations. The publication is split into 15 chapters, every one bankruptcy starting with a short yet sufficient precis and helpful formulation and Line diagrams through a number of average difficulties necessary for assignments and tests. particular 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
1 m. 098 2 dx 10 v0 ρ= (1) (2) (3) Using (2) and (3) in (1) we find ρ = 305 m. 37 Let P be the position of the boat at any time, Let AP = r , angle B AP and let v be the magnitude of each velocity, Fig. 5: dr = −v + v sin θ dt r dθ = v cos θ and dt −1 + sin θ 1 dr = ∴ r dθ cos θ dr ∴ = [− sec θ + tan θ ] dθ r π θ + − ln cos θ + ln C (a constant) ∴ ln r = − ln tan 2 4 When θ = 0, r = a, so that C = a a ∴ r= θ tan 2 + π4 cos θ The denominator can be shown to be equal to 1 + sin θ : ∴ r= a 1 + sin θ This is the equation of a parabola with AB as semi-latus rectum.
22. Draw the reference line OC parallel to AB, the ground level. Let the ball hit the wall at a height H above C. Initially at O, 34 1 Kinematics and Statics Fig. 22 u u x = u cos α = u cos 45◦ = √ 2 u ◦ u y = u sin α = u sin 45 = √ 2 When the ball hits the wall, y = x tan α − Using y = H , x = d and α = 45◦ H =d 1− 1 gx 2 2 u 2 cos2 α gd u2 (1) If the collision of the ball with the wall is perfectly elastic then at P, the horizontal component of the velocity (u x ) will be reversed, the magnitude remaining constant, while both the direction and magnitude of the vertical component v y are unaltered.
19. The equation for the projectile’s motion is y = x tan θ − Now y = R sin α x = R cos α Fig. 3 Solutions 29 Using (2) and (3) in (1) and simplifying R= 2u 2 cos θ sin(θ − α) g cos2 α The maximum range is obtained by setting constant. 30 As the outer walls are equal in height (h) they are equally distant (c) from the extremities of the parabolic trajectory whose general form may be written as (Fig. 20) Fig. 20 y = ax − bx 2 (1) y = 0 at x = R = nr , when R is the range This gives a = bnr (2) The range R = c + r + 2r + c = nr , by problem ∴ c = (n − 3) r 2 (3) The trajectory passes through the top of the three walls whose coordinates are (c, h), c + r, 15 7 h , (c + 3r, h), respectively.
1000 Solved Problems in Classical Physics: An Exercise Book by Ahmad A. Kamal