## College Physics (4th Edition)

Published by McGraw-Hill Education

# Review & Synthesis: Chapters 6-8 - Review Exercises - Page 323: 13

#### Answer

The speed of the ball when it reaches the top of the pinball machine is $1.53~m/s$

#### Work Step by Step

We can find an expression for the total kinetic energy of the solid sphere: $KE = KE_{tr}+KE_{rot}$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}I~\omega^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}MR^2)~(\frac{v}{R})^2$ $KE = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$ $KE = \frac{7}{10}Mv^2$ We can use conservation of energy to find the speed of the ball at the top of the ramp: $KE_f+U_f = KE_0+U_0$ $\frac{7}{10}Mv_f^2+Mgh = \frac{7}{10}Mv_0^2+0$ $\frac{7}{10}Mv_f^2 = \frac{7}{10}Mv_0^2-Mgh$ $v_f^2 = v_0^2-\frac{10gh}{7}$ $v_f = \sqrt{v_0^2-\frac{10gh}{7}}$ $v_f = \sqrt{(2.20~m/s)^2-\frac{(10)(9.80~m/s^2)(2.05~m)~sin~5.00^{\circ}}{7}}$ $v_f = 1.53~m/s$ The speed of the ball when it reaches the top of the pinball machine is $1.53~m/s$.

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