Question

My answer for 1,2, and 4 is right. Only 3 is wrong so please give me the solution and answer of 3.

Playing Catch (w/ Work) X=0 A person with mass m1 = 52 kg stands at the left end of a uniform beam with mass m2 = 105 kg and a length L = 3.3 m. Another person with mass m3 = 59 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 9 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor. 1) What is the location of the center of mass of the system? 1.767333 m Submit Help 2) The medicine ball is throw to the left end of the beam (and caught). What is the location of the center of mass now? 1.767333 m Submit Help 3) What is the new X-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?) 0.2013607778 m Submit Help 4) To return the medicine ball to the other person, both people walk to the center of the beam. At what x-position do they end up? 1.767333 m Submit Help

Answer 1

If, after the ball being caught by the left person, the beam moves a distance x to the right, Then position of each of the four masses including the centre of mass of the beam will be displaced to the right by x. We use this, considering the fact that the centre of mass of the system remains at the same place, to find the value of x which will also be the new position of person at the left because he was at origin initially

3) After person at left catches the ball, let beam mores distance x to the right Thus, positions of each of the 4 masses are displaced to right by a from their initial position. so, com of system is, Xcm = mqx+ m₂ ( ½ + x) + max+ m₂ (L+>() - Matuzt mat ma CM position is fixed. So, Xcm = 1.76733m So, 176733 m = 52% + 105( 3.3 + 2) + 9x + 59(3.3+2) 5279+ 105 + 59 = x= 10-132]m 14