Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 13.4, Problem 13.162P
To determine

(a)

The velocities of each car if A and C hits B at the same time.

Expert Solution
Check Mark

Answer to Problem 13.162P

vA1=1.288m/svB1=0.312m/svC1=1.512m/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.162P , additional homework tip  1

Mass of each bumper car is equal to 200kg.

Riders A, B and C have mass of 40kg,60kg,35kg.

vA=2m/svC=1.5m/se=0.8

The total linear momentum of two particles is conserved. Therefore,

mAvA+mBvB=mAvA1+mBvB1

The co-efficient of restitution is defined as:

vB1vA1=e(vAvB)

Calculation:

The mass of each bumper car with riders:

mA=240kgmB=260kgmC=235kg

Apply conservation of linear momentum for all 3 bumper cars:

mAvA+mBvB+mCvC=mAvA1+mBvB1+mCvC1

Substitute,

240(2)+0+235(1.5)=240vA1+260vB1+235vC1240vA1+260vB1+235vC1=127.5(1)

The co-efficient of restitution for A and B:

vB1vA1=e(vAvB)

Substitute,

vB1vA1=0.8(2)vB1vA1=1.6(2)

The co-efficient of restitution for B and C,

vC1vB1=e(vBvC)

Substitute,

vC1vB1=0.8(0(1.5))vC1vB1=1.2(3)

By solving equations 1,2 and 3,

vA1=1.288m/svB1=0.312m/svC1=1.512m/s

Conclusion:

The velocities of each car are calculated by solving equations 1, 2 and 3.

To determine

(b)

The velocities of each car if A hits B before C does.

Expert Solution
Check Mark

Answer to Problem 13.162P

vA111=0.9563m/svB111=0.0296m/svC11=1.552m/s

Explanation of Solution

Given information:

Vector Mechanics For Engineers, Chapter 13.4, Problem 13.162P , additional homework tip  2

Mass of each bumper car is equal to 200kg

Riders A, B and C have mass of 40kg,60kg,35kg

vA=2m/svC=1.5m/se=0.8

The total linear momentum of two particles is conserved. Therefore,

mAvA+mBvB=mAvA1+mBvB1

The co-efficient of restitution is defined as,

vB1vA1=e(vAvB)

Calculation:

The mass of each bumper car with riders,

mA=240kgmB=260kgmC=235kg

Apply conservation of linear momentum for A and B,

mAvA+mBvB=mAvA1+mBvB1

Substitute,

240(2)+0=240vA1+260vB1240vA1+260vB1=480(1)

The co-efficient of restitution for A and B,

vB1vA1=e(vAvB)

Substitute,

vB1vA1=0.8(2)vB1vA1=1.6(2)

Solve equations 1 and 2,

vA1=0.128m/svB1=1.728m/s

For the second impact,

Apply conservation of linear momentum for A and B,

mBvB1+mCvC=mBvB11+mCvC1

Substitute,

260(1.728m/s)+235(1.5)=260vB11+235vC11260vB11+235vC11=96.78(1)

The co-efficient of restitution for A and B,

vC11vB11=e(vB1vC)

Substitute,

vC11vB11=0.8(1.728(1.5))vC11vB11=2.5824(2)

Solve above equations,

vB11=1.0304m/svC11=1.552m/s

For the third impact,

Apply conservation of linear momentum for A and B,

mAvA1+mBvB11=mAvA111+mBvB111

Substitute,

240(0.128)+260(1.0304)=240vA111+260vB111240vA111+260vB111=237.184(3)

The co-efficient of restitution for A and B,

(vB111vA111)=e(vA1vB11)

Substitute,

vB111vA111=0.8(0.128(1.0304))vB111vA111=0.92672(4)

Solve equations 3 and 4,

vA111=0.9563m/svB111=0.02955m/s

Therefore, the final velocities are,

vA111=0.9563m/svB111=0.0296m/svC11=1.552m/s

Conclusion:

The velocities of all 3 cars after, car A hits B before C does,

vA111=0.9563m/svB111=0.0296m/svC11=1.552m/s

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Chapter 13 Solutions

Vector Mechanics For Engineers

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