[...]... velocity and bounce height (the coefficient of restitution is defined as e = -vf/vi, where vf and vi are the vertical velocities just after and before the bounce respectively) ( Wisconsin) Fig 1.5 Solution: Use unit vectors i, j as shown in Fig 1.5 and let the horizontal velocity of the marble be Vh The velocities just before and after a bounce are respectively 12 Problems d Solutions o n Mechanics. .. densities of the earth and the sun from the following approximate data: 6 = angular diameter of the sun seen from the earth = +" 1 = length of 1" of latitude on the earth's surface = 100 km t = one year = 3 x lo7 s g = 10 ms-2 ( UC, Berkeley) Solution: Let r be the distance between the sun and the earth, Me and Ma be the masses and Re and R, be the radii of the earth and the sun respectively, and G be the gravitational... of mechanical energy 1011 Assume all surfaces to be frictionless and the inertia of pulley and cord negligible (Fig 1.6) Find the horizontal force necessary to prevent any relative motion of m l , m2 and M ( Wisconsin ) Newtonian Mechanics 13 Fig 1.6 Solution: The forces f i , F and mg are shown in Fig 1.7 The accelerations of ml, m and M are the same when there is no relative motion among them 2 The... follows Let the radius of the orbit be R and that of the earth be & *For a more accurate calculation, the orbiting period should be taken as 23 hours 56 minutes and 4 seconds 3 Problems €4 Solutions on Mechanics 4 We have mu2 -=- GMm R2 ’ R where v is the speed of the space station, G is the universal constant of gravitation, m and M are the masses of the space station and the earth respectively, giving... tension of the cord ( Wisconsin) Solution: Neglecting the moment of inertia of the pulley, we obtain the equations of motion mlx = mlg - F and m2P = F - m2g Problems d Solutions on Mechanics 6 Hence the tension of the cord and the acceleration are respectively and 2 = (m1 - m 2 ) g rnl+mz = 1.225 m/s2 I* - - 29 16 X mlg m2 9 Fig 1.2 1005 A brick is given an initial speed of 5 ft/s up an inclined... 1006 A person of mass 80 kg jumps from a height of 1 meter and foolishly forgets to buckle his knees as he lands His body decelerates over a distance of only one cm Calculate the total force on his legs during deceleration ( Wisconsin) Problems & Solutions on Mechanics 8 Solution: The person has mechanical energy El = mg(h + s) just before he lands The work done by him during deceleration is E2 = fs,... is the velocity of the earth, m and m, are the maSses of the earth and the sun respectively For the motion of the sun around the center of the galaxy, where R is the distance from the sun to the center of the galaxy, V is the velocity of the sun and M is the mass of the galaxy Hence M=Using V = 2rR/T, v = 2rr/t, where T and t are the periods of revolution of the sun and the earth respectively, we have... . y0 w0 h0" alt="" Pro~lems and Solutions on Mechanics Major American Universities Ph.D. Qualif~~n~ Questions and Solutions Problems and Solutions on Mechanics Co~~il~d by: The. to, such problems. Working out problems is thus an essential and important aspect of the study of physics The series on Problems and Solutions in Physics comprises seven vol- umes and is. to collect and solve these problems and introduce them to physics students and teachers everywhere, even though the work is both tedious and strenuous. About a hundred teachers and graduate