Sandy is training to spend time on the International Space station by learning t

Question

Sandy is training to spend time on the International Space station by learning to walk on a platform floating in a spool of oil. The platform bas a mass of 110. kg, is 5.00 m long, and Boats on the oil without any friction. Sandy, with all her gear on, has a total mass of 65.0 kg. We consider "the system" to consist of Sandy and the platform. a) The platform is initially tether 1.00 m away from the right end of the pool. Sandy is initially at the left end of the platform, and walks with a constant velocity of 4.00 m/s towards the right. What is the kinetic energy of the system when Sandy reaches the right end of the platform? b) The tether is now untied so that the platform floats freely on the oil, starting 1.00 m away from the right end of the pool. Sandy again starts at the left end of the platform and walks right with a constant velocity of 4.00 m/s relative to the platform. Now what is the kinetic energy of the system when Sandy reaches the right end of the platform? c) If there's a difference in your answer between Parts a and b, explain why that is so. Where did the energy go? d) When Sandy reaches the right end of the platform in Part b, how far away is the platform from the right end of the pool? e) Sandy is now given a 10.0 kg lead weight to throw. On land, she can throw it 4.50 m. She again stands at the left end of the untethered platform, which is again initially 1.00 m away from the right end of the pool. She now throws the weight to the right. Exactly where does it land relative to the right end of the platform? f) Where does it land relative to the right end of the pool?

Explanation / Answer

a)Since the platform is not moving

total KE = KE of sandy=1/2*65*4^2=520J

b)Let the velocity of platform be v tot the right then absolute velocity of sandy=4+v

Now using conservation of momentum as there is no external force in horiznal direction

110*(x)-65*(4+x) = 0

45x=260

x=260/45=5.77m/s

Total KE = 1/2*65*(5.77+4)^2+1/2*110*5.77^2=4933J

c)In the first part the energy goes to the other end of rope connected to the platform. The other end absorbed the energy that could have gone to the platform.

d)The center of mass remains conserved. Considerring x=0 at the right end of rope.

initial center of mass = (110*(-3.5)+65*(-6))/(110+65)

let the platform shift by x to the left then

final center of mass = (110*(-3.5-x)+65*(-1-x))/(110+65)

equating these we get

(110*(-3.5-x)+65*(-1-x))/(110+65)=(110*(-3.5)+65*(-6))/(110+65)

this gives x = 1.857m

so the platform is now at (x+1)=2.857m from the right end

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