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2. The ball is dropped from rest and falls a distance of 1.2 m before striking the smooth plane at A. If e = 0.8, determine the disdance d to where it again strikes the plane at B. (H 15-63, p221) Solution:

2. The ball is dropped from rest and falls a distance of 1.2 m before striking the smooth plane at A. If e = 0.8, determine the disdance d to where it again strikes the plane at B. (H 15-63, p221) Solution:

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Why is et = 4/5 i  - 3/5 j and en = 3/5 i + 4/5 j .

Finding it hard to visualise, are u able to draw it out and derieve it?

I just need to know how to derive et and en

2. The ball is dropped from rest and falls a distance of 1.2 m before striking the smooth plane at A. If e = 0.8, determine the disdance d to where it again strikes the plane at B. (Н 15-63, р221) 1.2 m Solution: (1) From state 1 to state 2, using Principle of W and E U1-2 = T, – T, y mg(1.2) 3 0.5mту' > v= /2.4g = 4.8522 (m/s) 1 (2) Collision at position 2 1.2 m Before collision 2 4 =-v(- 3 v, = -vŷ ê, ê,) in 5 3 = vê, + vê, After collision, assume d 5 v, = v"ê, + v{ ê, In ê, direction the linear momentum is conserved, so 3 vi = vó = 5" v = 0.6v In ê, direction, use the restitution equation (3.82) (v" – 0) = -e(v" – 0) 4 we have v =-e(--v)= 0.64v 5 Different Componenets of V Thus we have i, = v" ê, + v{ ê, obtained as v, = 0.64vê, + 0.6v ê, 3 Using ê, 5 j, ê, + v, can be written as %3D v, = 0.864v î + 0.152vĵ
Transcribed Image Text:2. The ball is dropped from rest and falls a distance of 1.2 m before striking the smooth plane at A. If e = 0.8, determine the disdance d to where it again strikes the plane at B. (Н 15-63, р221) 1.2 m Solution: (1) From state 1 to state 2, using Principle of W and E U1-2 = T, – T, y mg(1.2) 3 0.5mту' > v= /2.4g = 4.8522 (m/s) 1 (2) Collision at position 2 1.2 m Before collision 2 4 =-v(- 3 v, = -vŷ ê, ê,) in 5 3 = vê, + vê, After collision, assume d 5 v, = v"ê, + v{ ê, In ê, direction the linear momentum is conserved, so 3 vi = vó = 5" v = 0.6v In ê, direction, use the restitution equation (3.82) (v" – 0) = -e(v" – 0) 4 we have v =-e(--v)= 0.64v 5 Different Componenets of V Thus we have i, = v" ê, + v{ ê, obtained as v, = 0.64vê, + 0.6v ê, 3 Using ê, 5 j, ê, + v, can be written as %3D v, = 0.864v î + 0.152vĵ
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