Explanation Given information: α = 30 ° v 0 = 0.6 m / s l = 0.9 m θ = 20 ° e = 0.8 k = 1500 N / m Mass of sphere is 1 k g . Mass of wedge is 2 k g . 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: At initial stage: h 0 = l ( 1 − c o s α ) = ( 0.9 ) ( 1 − c o s 30 ° ) = 0.12057 m Find the Kinetic and potential energies. V 0 = m A g h 0 T 0 = 1 2 m A v 0 2 Just before the impact. T 1 = 1 2 m A v A 2 V 1 = 0 Therefore: m A g h 0 + 1 2 m A v 0 2 = 1 2 m A v A 2 v A 2 = 2 g h 0 + v 0 2 Substitute and solve. v A 2 = 2 ( 9.81 m / s 2 ) ( 0.12057 m ) + ( 0.6 ) 2 v A = 1 To determine (b) The maximum deflection of the spring.

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 13.4, Problem 13.189P

To determine

(a)

The velocities of A and B just after the impact.

To determine

(b)

The maximum deflection of the spring.

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Chapter 13.4, Problem 13.188P

Chapter 13, Problem 13.190RP

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(a) The velocities of A and B just after the impact.

Solution Summary: The author analyzes the velocities of A and B just after the impact.