Vector Mechanics for Engineers: Statics and Dynamics
Vector Mechanics for Engineers: Statics and Dynamics
12th Edition
ISBN: 9781259638091
Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self
Publisher: McGraw-Hill Education
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Chapter 14, Problem 14.105RP

Three identical cars are being unloaded from an automobile carrier. Cars B and C have just been unloaded and are at rest with their brakes off when car A leaves the unloading ramp with a velocity of 5.76 ft/s and hits car B, which hits car C. Car A then again hits car B. Knowing that the velocity of car B is 5.04 ft/s after the first collision, 0.630 ft/s after the second collision, and 0.709 ft/s after the third collision, determine (a) the final velocities of cars A and C, (b) the coefficient of restitution for each of the collisions.

Fig. P14.105

Chapter 14, Problem 14.105RP, Three identical cars are being unloaded from an automobile carrier. Cars B and C have just been

(a)

Expert Solution
Check Mark
To determine

Find the final velocity of the car A and C.

Answer to Problem 14.105RP

The final velocity of the car A and C are 0.641ft/s_ and 4.41ft/s_.

Explanation of Solution

Given information:

Consider the mass of car A, car B, and C is denoted by m.

The initial velocity of the car is (vA)1=5.76ft/s.

The initial velocity of the car B and C is zero.

The velocity of the car B after the first, second and third collisions are (vB)2=5.04ft/s, (vB)3=0.630ft/s, and (vB)4=0.709ft/s

Calculation:

The Horizontal momentum of the system is conserved as no horizontal force is acting on the system.

Show the Event 12 of car A hitting the car B as shown in Figure 1.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 14, Problem 14.105RP , additional homework tip  1

Refer to Figure 1.

Show the conservation of horizontal momentum as follows:

m(vA)1+m(vB)1=m(vA)2+m(vB)2 (1)

Substitute 5.76ft/s for (vA)1, 0 for (vB)1, and 5.04ft/s for (vB)2 in Equation (1).

m(5.76)+m×0=m(vA)2+m(5.04)5.76m=m(vA)2+5.04m(vA)2=0.72ft/s

Show the Event 23 of car B hitting the car C as shown in Figure 2.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 14, Problem 14.105RP , additional homework tip  2

Refer to Figure 2.

Show the conservation of horizontal momentum as follows:

m(vB)2+m(vC)1=m(vB)3+m(vC)3 (2)

Substitute 0.630ft/s for (vB)3, 0 for (vC)1, and 5.04ft/s for (vB)2 in Equation (2).

m(5.04)+m×0=m(0.630)+m(vC)35.04m=0.63m+m(vC)3(vC)3=4.41ft/s

Thus, the final velocity of the car C is 4.41ft/s_.

Show the Event 34 of car A hitting the car B as shown in Figure 3.

Vector Mechanics for Engineers: Statics and Dynamics, Chapter 14, Problem 14.105RP , additional homework tip  3

Refer to Figure 3.

Show the conservation of horizontal momentum as follows:

m(vA)2+m(vB)3=m(vA)4+m(vB)4 (3)

Substitute 0.630ft/s for (vB)3, 0.72ft/s for (vA)2, and 0.709ft/s for (vB)4 in Equation (3).

m(0.72)+m(0.63)=m(vA)4+0.709m1.35m=m(vA)4+0.709m(vA)4=0.641ft/s

Thus, the final velocity of the car A is 0.641ft/s_.

(b)

Expert Solution
Check Mark
To determine

Find the coefficients of restitution of each collision.

Answer to Problem 14.105RP

The coefficients of restitution of first, second and third collision are 0.75_, 0.75_, and 0.756_.

Explanation of Solution

Given information:

Calculation:

Refer Part (a).

Consider the coefficient of restitution corresponding to collision Event 12, 23, 34 are denoted by e12, e23, and e34.

Calculate the coefficient of restitution corresponding to collision Event 12 using the relation:

e12=|(vA)2(vB)2(vA)1(vB)1| (4)

Substitute 0.72ft/s for (vA)2, 5.04ft/s for (vB)2, 5.76ft/s for (vA)1, and 0 for (vB)1 in Equation (4).

e12=|0.725.045.760|=0.75

Calculate the coefficient of restitution corresponding to collision Event 23 using the relation:

e23=|(vB)3(vC)3(vB)2(vC)2| (5)

Substitute 0.63ft/s for (vB)3, 4.41ft/s for (vC)3, 5.04ft/s for (vB)2, and 0 for (vC)2 in Equation (4).

e23=|0.634.415.040|=0.75

Calculate the coefficient of restitution corresponding to collision Event 34 using the relation:

e34=|(vA)4(vB)4(vA)3(vB)3| (6)

Substitute 0.641ft/s for (vA)4, 0.709ft/s for (vB)4, 0.72ft/s for (vA)3, and 0.63 for (vB)3 in Equation (6).

e34=|0.6410.7090.720.63|=0.756

Thus, the coefficients of restitution of first, second and third collision are 0.75_, 0.75_, and 0.756_.

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Chapter 14 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

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