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A 44.3-kg girl is standing on a 154-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 1.491 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? Part 1 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. Part 3 of 4 - Analyze and the velocity of the plank relative to gp' We represent the velocity of the girl relative to the ice with: , the velocity of the girl relative to the plank with the ice with, and the speeds as V₁, Vop, 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. V = Vgp↑ + Vpi 0= m₂v₁ +mp ) The motion is in one dimension so we can write the following. V₁ V₂ + др In unit-vector notation, we have and therefore -(2 (a) Solving for Vg = Vgp + Vp. The momentum equation becomes 0= -44.3 mg Vp² VD=- 8P we have mg mg + mp _kg)(1.49 + vp)i + ( Vgp kg kg m/s) = kg)vpi. m/s For the velocity of the plank relative to the ice in unit-vector notation, we have î m/s.

A 44.3-kg girl is standing on a 154-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 1.491 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? Part 1 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. Part 3 of 4 - Analyze and the velocity of the plank relative to gp' We represent the velocity of the girl relative to the ice with: , the velocity of the girl relative to the plank with the ice with, and the speeds as V₁, Vop, 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. V = Vgp↑ + Vpi 0= m₂v₁ +mp ) The motion is in one dimension so we can write the following. V₁ V₂ + др In unit-vector notation, we have and therefore -(2 (a) Solving for Vg = Vgp + Vp. The momentum equation becomes 0= -44.3 mg Vp² VD=- 8P we have mg mg + mp _kg)(1.49 + vp)i + ( Vgp kg kg m/s) = kg)vpi. m/s For the velocity of the plank relative to the ice in unit-vector notation, we have î m/s.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A 44.3-kg girl is standing on a 154-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 1.491 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? Part 1 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. Part 3 of 4 - Analyze g' We represent the velocity of the girl relative to the ice with: the velocity of the girl relative to the plank with and the velocity of the plank relative to the ice with, and the speeds as V₁₁ Vgp' and v Vp² 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. gp' 0 = m+mp The motion is in one dimension so we can write the following. = ARK m, g gp In unit-vector notation, we have v₁₁=Vgp and therefore 0 = = ( Vp m₁ Р Vg = Vgp p' The momentum equation becomes = = +V Р (a) Solving for v we have p² = ‚Î + vî V₂₁ p + V. -44.3 8P mg mg + mp 18 V kg)(1.49 + vp)i + ( gp kg kg m/s) = kg)vpî. For the velocity of the plank relative to the ice in unit-vector notation, we have Î m/s. m/s
Transcribed Image Text:A 44.3-kg girl is standing on a 154-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 1.491 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? Part 1 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. Part 3 of 4 - Analyze g' We represent the velocity of the girl relative to the ice with: the velocity of the girl relative to the plank with and the velocity of the plank relative to the ice with, and the speeds as V₁₁ Vgp' and v Vp² 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. gp' 0 = m+mp The motion is in one dimension so we can write the following. = ARK m, g gp In unit-vector notation, we have v₁₁=Vgp and therefore 0 = = ( Vp m₁ Р Vg = Vgp p' The momentum equation becomes = = +V Р (a) Solving for v we have p² = ‚Î + vî V₂₁ p + V. -44.3 8P mg mg + mp 18 V kg)(1.49 + vp)i + ( gp kg kg m/s) = kg)vpî. For the velocity of the plank relative to the ice in unit-vector notation, we have Î m/s. m/s
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