Question

A barge of mass M = 300 kg is floating on a lake (assume there are no forces between the barge and the water in the plane of the lake and that the water is not moving). The barge is moving at a velocity vo = 0.4 m/s along the +x direction. As the barge floats by, a man of mass m = 90 kg standing at the edge of the lake takes a running start and jumps in the air, landing on the barge. His velocity in the air is Vm = 1.2 m/s, along the +y direction. 1) At what angle e relative to the +y direction do the barge and the man travel after the jump? = 90 A = 18 = 48 A = 550 - 73 Submit 2) Which of the following would increase the angle e? Increasing the mass M of the barge. Increasing the mass m of the man jumping. Increasing the speed Vm of the jumping man. Submit 3) What is the kinetic energy of the system including the man and the barge, after the man has landed on the barge? KE m+b = 33.4 KE m+b = 61.13 m+b = 88.8 J

Answer 1

M 7 vcoso Assume that the boat and man will travel at an angle o with respect to positive y-axis with a velouty ne. For this problem, the momentin vo must be conserved along a and Ý before and after collision of jump.. . For a- direction, momentum conservation -. . 1 Of My = (M+m) usino ( As man was moving & Ý 1. H o direction, so horizontal component along a For yo direction momentum conservation ai mom to = (M4m) acoso. So, 0 + 0-. tano = 12o 1 MVO mum 300X 0.4 90x12 108 Lemi o = tano (120) [0=480] 02) Dato' Menton) oam and ox imi extreme so to increase A (M) mass of the barge / boat need to o be increased .. . 3) 8² +8, Putting A = 48° in o, we get. - (44 xu & 2 ra mi v Mub . (Mim) singgo 300X0.4 (300+90) sin480 50.414 mis. Kinetic energy of the system Ž (Mtim) we? 2(30090)*(0,414)? E