a frictionles, vertical wire has two metal beads upon it. the beads are held 30

Question

a frictionles, vertical wire has two metal beads upon it. the beads are held 30 cm apart by horizontal magnets. if the top magnet is removed, the first bead falls under the force of gravity and strikes the second. the first bead bounces back up to a height of 10 cm , and the second is knocked free of the magnet and falls downward. if the two beads each have a mass of 49.7 g, what is the kinetic energy of the second immediately after impact?

The book says the answer is 97.4 kJ. Anyone get this answer too?

Explanation / Answer

Here

Total Energy remains Constant

Therefore

0.5*m*(2gh) = mgh' + E

Therefore

E = (49.7*10^-3*9.8*0.30) - 9.8*0.10*49.7*10^-3

= 0.0974 J

= 97.4 mJ

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