Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.4 - Impact - Problems - Page 276: 76



Work Step by Step

We can determine the required coefficient of restitution as follows: First, we apply the conservation of total energy from position A to position B $\frac{1}{2}mv_A^2+mgh_{\circ}=\frac{1}{2}mv_{B_1}^2+mgh_B$ $0+mgh_{\circ}=\frac{1}{2}mv_{B1}^2+0$ This simplifies to: $v_{B1}=\sqrt{2gh_{\circ}}\uparrow$ The conservation of energy when the ball returns from position B to C is: $\frac{1}{2}mv_B^2+0=\frac{1}{2}mv_C^2+mgh_1$ This simplifies to: $v_{B2}=\sqrt{2gh_1}\uparrow$ We know that: $e=\frac{v_{B2}}{v_{B1}}$ We plug in the known values to obtain: $e=\frac{\sqrt{2gh_1}}{\sqrt{2gh_{\circ}}}$ This simplifies to: $e=\sqrt{\frac{h_1}{h_{\circ}}}$
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