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 pool of oil. The platform has a mass of 110 kg. is 5.00 m long, and floats 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. The platform is initially tethered 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? The tether is now untied so that the platform floats freely on the oil, starling 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? If there's a difference in your answers between Parts a and b. explain why that is so. Where did the energy go? 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? Sandy is now given a 10.0 kg lead weight to throw. On land, she can throw it 4.50 m. If she again stands at the left end of the untethered platform and throws the weight to the right, exactly w here docs it land relative to the right end of the platform? Sandy the astronaut trains by walking on a platform floating frictionlessly on a pool of oil.

Explanation / Answer

a) the kinetic energy of the system is only the KE of sandy as the platform is at rest.

KE = (1/2) mV^2 = (1/2)(65)(4)^2 = 520J

b) According to conservation of momentum

m1 V1 = m2 V2 here m1 = 65kg and m2 = 110kg and V1 =4m/s so

(65)(4) = 110 V2 so V2 = 2.36m/s

the platform moves in teh opposite direction with 2.36m/s

The Kinetic energy of the system is K = (1/2) m1 V1^2 - (1/2) m2 V2^2 = 520J - (1/2)(110)(2.36)^2 = 213.6J

Energy is lost in moving the platform in the backward direction.

c) If L=5m where sandy walks on the platform then platform moves back with a distance of

x = m1 L / (m1 + m2) = 65*5 / (65+110) = 1.86m

The platform is already at 1m away from the right end of the pool so total distance is 1.86+1m = 2.86m

d) On land if she throw a lead ball , its horizontal projecctile. Let height of the throw be 2m so time of fall from hand to ground is

t = sqrt [ 2h/g] = 0.639s (kinematics equation)

X = V. t so 5 = V (0.639) speed of throw on ground V = 7.83m

Now using momentum conservation

m1 V = m2 U , here U is the spped of platform after throw so

U = 65(7.83)/110 = 4.624m/s

The relative speed of the throw is 7.83-4.624 = 3.2m/s

the distance of throw on the platform is

X1 = (3.2)0.639 = 2.05m or approx 2m

Hope this helps

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