A 45'1-kg girl is standing on a 148-kg plank. Both originally at rest on a frozen lake that constitutes a frictionless, flat surface. The girl begins to walk along the plank at a constant velocity of 1.61? m/s relative to the plank. (a) What is the velocity of the plank relative to the ice surface? (b) What is the girl's velocity relative to the ice surface? Part1 of 4 - Conceptualize The plank slips backward, so the girl moves forward relative to the ice at a bit less than her speed relative to the plank, and the plank moves backward at around half the speed of the girl moving forward. Part 2 of 4 - Categorize Conservation of momentum for the girl-plank system is the key to our calculations. We could use the center of mass not moving as an approach, but this would involve using more algebra. Pan 3 of 4 - Analyze We represent the velocity of the girl relative to the ice with: 39, the velocity of the girl relative to the plank with 39p, and the velocity of the plank relative to the ice with 3,, and the speeds as v9, vgp, and vp, respectively. The girl and the plank exert forces on each other, but the ice isolates them from outside horizontal forces. Therefore, the net momentum is zero for the girl-plank system, as it was before motion began, We have the following equation. _ . . 0 177ng + 177pr The motion is in one dimension so we can write the following. . 7 . . V9 ' Van + Vp In unit-vector notation, we have Vgl = Vgpl + Vpl and therefore vg : vgp + vp. The momentum equation becomes 0 = (El v kg)(1.61 + Vp)i + (El v kg)v Part 2 of 4 - Categorize Conservation of momentum for the girl-plank system is the key to our calculations We could use the center of mass not moving as an approach, but this would involve using more algebra. Part 3 of 4 - Analyze We represent the velocity of the girl relative to the ice with: 39, the velocity of the girl relative to the plank with 39p, and the velocity of the plank relative to the ice with 3p, and the speeds as v9, vgp, and vp, respectively. The girl and the plank exert forces on each other, but the ice isolates them from outside horizontal forces. Therefore, the net momentum is zero for the girl-plank system, as it was before motion began. We have the following equation. _ . 0 mgvg + mpvp The motion is in one dimension so we can write the following. _ 7 v9 vgp + vp In unit-vector notation, we have Vgl = Vgpl + Vpl and therefore V9 = Vgp + Vp. The momentum equation becomes 0 = (|45.1 iv; kg)(1.61 + vp)i +( 143 1.; kg)vpi. (3) Solving for vp, we have _ ' EVX kg \"Liv rn/s)=|le Your response differs from the correct answer by more than 10%, Double check your calculations. kg Your response differs from the correct answer by more than 100%. m/s VD For the velocity of the plank relative to the ice in unit-vector notation, we have 3p = [i m/s. Submit ED. (you cannot come back) Submit