Explanation Given information: Mass of sphere A is 5 k g . Mass of section B is 12 k g . Sphere is dropped from a height of 3 m . The sphere bounces back upward a height of 44 m m . The maximum deflection of section B is 33 m m . Calculation: The total linear momentum of two particles is conserved. Therefore m A v A + m B v B = m A v A 1 + m B v B 1 The co-efficient of restitution is defined as v B 1 − v A 1 = e ( v A − v B ) The principle of conservation of energy is defined as, “When a particle moves under the action of conservation of forces, the sum of kinetic energy and potential energy of that particle remains constant.” T 1 + V 1 = T 2 + V 2 Calculation: Apply conservation of energy for sphere A. Assume m A as the mass of sphere A and v A as the velocity of sphere A. m A g y = 1 2 m A v A 2 Therefore v A = 2 g y = 2 ( 9.81 m / s 2 ) ( 3 m ) = 7.672 m / s ↓ Apply conservation of momentum Assume m B as the mass of floor section B and v A 1 , v B 1 as the velocities just after the impact. m A v A + m B v B = m A v A 1 + m B v B 1 Substitute 5 ( 7.672 ) + 0 = 5 v A 1 + 12 v B 1 5 v A 1 + 12 v B 1 = 38.36 → ( 1 ) Apply co-efficient of restitution equation: ↑ v B 1 − v A 1 = e ( v A − v B ) Substitute v B 1 − v A 1 = 7 To determine (b) The spring constant k of floor section.

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 13.4, Problem 13.180P

To determine

(a)

Co-efficient of restitution between sphere and floor.

To determine

(b)

The spring constant k of floor section.

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Chapter 13.4, Problem 13.179P

Chapter 13.4, Problem 13.181P

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(a) Co-efficient of restitution between sphere and floor.

Solution Summary: The author calculates the co-efficient of restitution between sphere and floor section.