Center of Mass - General Physics - Solved Past Paper, Exams of Physics

Download Center of Mass - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 3. (30 pts) A 40 kg child stands at one end of a 70 kg canoe that is 4 m long. The end of the canoe at which the child is standing is initially 3 m from the end of a pier. The child notices a turtle on a rock and proceeds to walk to the other end of the canoe (away from the pier). Ignore friction between the water and the canoe. You can also assume that the canoe has been carefully balanced such that its center of mass is at the center of the canoe. a) (10 pts) Taking the end of the pier as the origin, where is the center of mass of the child/canoe system initially? Here’s what the child/canoe look like at the start: 3 m 4 m The child, mass 40 kg, is at x = 3 m. The canoe, mass 70 kg, is at x = 3 + 2 = 5 m. The center of mass is then rcm = (40)(3) + (70)(5) 70 + 40 = 4.27 m b) (10 pts) After the child reaches the other end of the canoe, where is the center of mass of the child/canoe system (again, use the end of the pier as your measurement origin)? Since (a) the child/canoe system start at rest and (b) there is no net force on the child/canoe system, the center of mass will not move as the child walks to the other end of the canoe. It therefore stays at 4.27 m from the pier. Of course, in order for the center of mass to remain stationary as the child moves, the canoe moves the other way. c) (10 pts) How far from the pier is the child now? There are two ways you could do this. 1. By symmetry: The child is now at the other end of the canoe, so the situation at the start of the problem should be reversed. Rather than the child being 1.27 m to the left of the C.M., he/she should now be 1.27 m to the right of the center of mass (which is still at the same point, 4.27 m from the pier). Thus the child is 4.26 + 1.27 = 5.55 m from the pier. 2. By brute calculation: If the new location of the center of mass of the canoe is r1, then the location of the child with respect to the canoe is r1 + 2 (the C.M. of the canoe is 1/2 way along it and the child is at the end of the canoe, 2 m further out). Then just compute the new C.M. of the system and set it equal to 4.27 m. 4.27 = 70r1 + 40(r1 + 2) 70 + 40 and solve for r1. We get r1 = 3.55 m. Then the location of the child is 3.55 + 2 = 5.55 m from the pier. The answer is the same either way.