Question

Constants A basketball (which can be closely modeled as a hollow spherical shell) rolls down a mountainside into a valley and then up the opposite side, starting from rest at a height Ho above the bottom. In the figure, the rough part of the terrain prevents slipping while the smooth part has no friction. (Figure 1) Part A How high, in terms of Ho, will it go up the other side? O ACOM O Submit Request Answer Figure < 1 of 1 > Part B Why doesn't the ball return to height Ho? Has it lost any of its original potential energy? Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3800 Character(s) remaining Rough Smooth Submit Request Answer

Answer 1

Using law of conservation of energy,

0.5 mv^2 + 0.5 Iw^2 = mgHo

0.5 mv^2 + 0.5 (2/3) mr^2 (v/r)^2 = mg Ho

0.5 mv^2 + mv^2/3 = mg Ho

(5/6) mv^2 = mg Ho

(5/6) (2 mgH) = mgHo

H = (3/5) Ho

B)

It loses some of its potential energy in kinetic energy

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