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A spherical ball of mass 20 kg starts from rest at the top of a hill 100 m high. It moves down a frictionless track to the ground, then goes up a second hill of height 30 m, and finally descends to a horizontal platform that is 20 m above the ground. The speed of the ball at that point is

1. 20 m/s
2. 40 m/s
3. 10√(30) m/s
4. 10 m/s

Correct answer: 40 m/s

Solution

Frictionless track conserves energy; speed depends only on net descent to the 20 m platform: v = sqrt(2g(100-20)) = sqrt(2*10*80) = sqrt(1600) = 40 m/s.

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