HKIPhO Mechanics Assignment IV Solution1. Two identical small beads with mass m are attached on a smooth circular ring that stands vertically on the ground. Initially the beads are resting on the top of the ring. They then start to slide down. Derive the relationship between the mass of the ring and the mass of the beads such that the ring can jump up from the ground. Also, find the location of the beads when the ring jumps up.Sol: Let the mass of the ring be M and radius be R. At time t, the angle m made with the vertical is q.Since there is no friction between beads and ring, therefore the force acting on the beads are always pointing along N, and Newtons Law for circular motion implies.Conservation of energy implies.Putting back in the first equation we obtain.Notice that N becomes negative when is large enough. In this case the reaction force N of the beads on the ring will try to lift the ring up. This can happen when . Since N = - N, we have the condition for ring to jump up:, or. For this to have a solution, we must have or .2. This problem models the launching of a detector from the surface of the earth to Mars. We assume that the earth and Mars are moving around the sun in circles resting on the same plane. The radius of the orbit of Mars is Rm, which is 1.5 times larger than that of the earth. An economical and simple way to lounge the detector consists of 2 steps. First, a rocket is used to accelerate the detector on the earths surface such that it acquires enough kinetic energy to overcome the gravitational force of the earth and becomes a satellite moving around the earth. Second, at a suitable time, an engine that connects to the detector ignites for a short instant and accelerates the detector along its direction of motion. After a short (negligible) time, the speed of the detector increases to a suitable value such that the detector moves along an elliptical orbit that connects the earth and Mars, with the two planets located at the end points of the ellipse (see Diagram A)(1) In Step I, what is the minimum speed needed for the detector to become an artificial satellite that moves along the earths orbit?(2) After the detector becomes a satellite moving around the earth, on 1 March 00.00am of a certain year, the angular distance between the detector and Mars is measured to be 60o (diagram B). What is the date that the engine of the detector should be fired so that the detector can fall on the surface of Mars (Correct to day)? Given: radius of the earth: 6.4x106 m, acceleration due to gravity = 9.8m/s2 .(1) For circular orbit, .(2) Again we use , and , where . Therefore for the detector orbit.Next we use Keplers third law which implies . Therefore the time taken for the detector to go to Mars is year. During this period, the angular distance Mars travel is , where . Therefore . Therefore the angular distance between the detector and Mars should be 180-137=43o at the moment when the detector fires. The angular velocity of earth and Mars are and , respectively. The angle that earth and Mars travel in time t (in days) is and , respectively. Therefore, the detector should be fired t days after March 1st, when . Solving the equation we obtain days, corresponding to April 7th of the same year.3. The perigee and apogee of the first artificial satellite of our mother country is 8754 km and 6809 km respectively. Find the speed of the satellite when it passes through the perigee and apogee. What is the period of the satellite?Sol: Use , where a + c = 8734km and a c = 6809km, and Keplers third law . We obtain and T = 114 mins.4. Consider the attractive electrostatic force between an electron and a proton inside a nucleus , where k 0 is a constant. a) Assuming that the 2 particles circle around the center of mass in a circular orbit, determine the relation between the radius of the orbit and velocities of the 2 moving particles.b) Discuss how the Keplers Three Laws will be modified in this situation?a) Using Newtons second law,Considering the distances from the center of mass, Hence by eliminating re and rp, .Similarly,.b) Keplers three laws are modified as follows:First law: The orbits of both the electron and proton are elliptical with the center of mass being one focus.Second law: The vector from the center of mass to the electron sweeps out area at a constant rate. The same for the proton.Third law: The square of the period is proportional to the cube of the semimajor axis of the elliptical orbits. The proportionality constant in the center-of-mass frame is modified:.Compared with the case that the proton is assumed stationary, the mass of the electron is replaced by the reduced mass memp/(me + mp).5. A solid ball with uniform density and radius R (originally resting on a smooth desk) is hit by a hori