12.2 points SerPSET9 9.P.006.MI.SA This question has several parts that must be

Question

12.2 points SerPSET9 9.P.006.MI.SA This question has several parts that must be completed sequentialy. If you skip a part of the question, you will not to the skipped part. Tutorial Exercise A 46.1-kg girl is standing on a 152-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 1.57i m/s relative to the plank. (a) What is the velocity of the plank relative to the ice surface? (b) What is the girt'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. &P; 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: V the velocity of the girl relative to the plank with Vgp and the velocity of the plank relative to the ice with Vp, and the speeds as vo 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. r Yape and vo The motion is in one dimension so we can write the following In unit-vector notation, we have and therefore

Explanation / Answer

here,

mass of the girl = Mg = 46.1 kg
mass of plank = Mp = 152 kg

Vg = 1.57 i m/s

as initially the girl and plank are rest so no extrnal force is acting ont he system therefore

Net momentum = 0

now

we know that the girls speed relative to the ice surface is her speed relative to the plank minus the speed of the plank,

From Conservation of Momentum we have :

Mb*Vg = - Mp*Vp

Vp = - ( 46.1 / 152 ) * 1.57

Vp = ( -0.47 i ) m/s

the velocity of the plank relative to the surface of ice is -0.47 m/s

so:
vgs = 1.46 + 0.47
Vgs = ( 1.93 i ) m/s

Girl velocity relative to the ice surface is 1.93 m/s

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