Vector Mechanics for Engineers: Dynamics
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 13.182P

Block A is released from rest and slides down the frictionless surface of B until it hits a bumper on the right end of B. Block A has a mass of 10 kg and object B has a mass of 30 kg and B can roll freely on the ground. Determine the velocities of A and B immediately after impact when (a) e = 0 . (b) e = 0.7 .

Chapter 13.4, Problem 13.182P, Block A is released from rest and slides down the frictionless surface of B until it hits a bumper

Expert Solution
Check Mark
To determine

(a)

The velocities of A and B just after the impact if e=0.

Answer to Problem 13.182P

velocities of A and B just after the impact if : e=0 is (vA1)x=0(vB1)x=0.

Explanation of Solution

Given information:

Vector Mechanics for Engineers: Dynamics, Chapter 13.4, Problem 13.182P , additional homework tip  1

Mass of A is 10kg.

Mass of B is 30kg.

Conclusion:

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)

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.

T1+V1=T2+V2

Calculation:

Let (vA)x,(vA)y,(vB)x,(vB)y velocities before the impact and (vA1)x,(vA1)y,(vB1)x,(vB1)y velocities after the impact.

Apply conservation of linear momentum while block is sliding down.

0+0=mA(vA)x+mB(vB)x

Therefore.

(vB)x=mAmB(vA)x

Apply conservation of linear momentum at impact.

0+0=mA(vA1)x+mB(vB1)x

Therefore.

(vB1)x=mAmB(vA1)x

Assume. mAmB=β

According to conservation of energy.

The initial potential energy of A is equal to:

V0=mAgh

For B its zero.

Initial kinetic energy is zero.

And just before the impact.

T1=12mAvA2+12mBvB2

Therefore, according to:

T0+V0=T1+V1

Substitute.

mAgh=12mAvA2+12mBvB2mAgh=12(mA+mBβ2)vA2mAgh=12mA(1+β)vA2

Therefore.

vA2=(vA)x2=2gh(1+β)vA=2gh(1+β)

Therefore. the velocities just before the impact.

vA=2gh(1+β)vB=β2gh(1+β)

Apply co-efficient of restitution equation.

vB1vA1=e(vAvB)

Substitute.

(vB1)x(vA1)x=e[(vA)x(vB)x]β(vA1)x(vA1)x=e[(vA)xβ(vA)x]

Therefore:

(vA1)x=e(vA)x

But we know that:

e=0

As a result of it:

(vA1)x=0(vB1)x=0

Conclusion:

The velocities of A and B just after the impact. if e=0.

(vA1)x=0(vB1)x=0

Expert Solution
Check Mark
To determine

(b)

The velocities of A and B just after the impact if e=0.7.

Answer to Problem 13.182P

(vA1)x=1.201m/s(vB1)x=0.4002m/s

Explanation of Solution

Given information:

Vector Mechanics for Engineers: Dynamics, Chapter 13.4, Problem 13.182P , additional homework tip  2

Mass of A is 10kg

Mass of B is 30kg

Conclusion:

According to sub part ‘a’ we have found.

vA=2gh(1+β)vB=β2gh(1+β)

Where.

mAmB=β

Calculation:

Find the exact value of vA

For that.

β=mAmB=1030=0.3333

Then.

vA=2gh(1+β)=2(9.81m/s2)(0.2m)(1+0.3333)=1.7155m/s

But according to co-efficient of restitution equation.

(vA1)x=e(vA)x

Therefore.

(vA1)x=e(vA)x=0.7(1.7155m/s)=1.201m/s

Then.

(vB1)x=β(vA1)x=0.3333(1.201m/s)=0.4002m/s

Conclusion:

The velocities of A and B just after the impact. if e=0.7

(vA1)x=1.201m/s(vB1)x=0.4002m/s

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Chapter 13 Solutions

Vector Mechanics for Engineers: Dynamics

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