Mechanical Engineering Chapter 15 Homework By considering the x and y motion of the marble from A to B

554

15–78.

Using a slingshot, the boy fires the 0.2-lb marble at the

concrete wall, striking it at B. If the coefficient of restitution

between the marble and the wall is , determine the

speed of the marble after it rebounds from the wall.

e=0.5

5ft

A

B

C

100 ft

vA75 ft>s

45

60

SOLUTION

Kinematics: By considering the xand ymotion of the marble from Ato B,Fig. a,

and

and the direction angle of vBis

Conservation of Linear Momentum: Since no impulsive force acts on the marble

along the inclined surface of the concrete wall ( axis) during the impact, the linear

momentum of the marble is conserved along the axis. Referring to Fig. b,

x¿

a+cb

A

vB

B

y=

A

vA

B

y+ayt

Coefficient of Restitution: S

i

nce t

h

e concrete wa

ll

d

oes not move

d

ur

i

ng t

h

e

i

mpact

,

the coefficient of restitution can be written as

Solving Eqs. (1) and (2) yields

15–78. Continued

15–79.

The two disks Aand Bhave a mass of 3 kg and 5 kg,

respectively. If they collide with the initial velocities shown,

determine their velocities just after impact. The coefficient

of restitution is e=0.65.

SOLUTION

Solving,

(yAx)2=-3.80 m>s(yBx)2=2.378 m>s

(yBx)1=-7 cos 60° =-3.5 m>s

A

yBy

B

1=-7 cos 60° =-6.062 m>s

(yAx)=6m>s(yAy)1=0

Line of impact

AB

60

(v

A

)

1

fi6m/s

(v

B

)

1

fi7m/s

557

*15–80.

A ball of negligible size and mass m is given a velocity of v0

on the center of the cart which has a mass M and is originally

at rest. If the coefficient of restitution between the ball and

walls A and B is e, determine the velocity of the ball and the

cart just after the ball strikes A. Also, determine the total

time needed for the ball to strike A, rebound, then strike B,

and rebound and then return to the center of the cart.

Neglect friction.

SOLUTION

After the first collision;

(

+

)

(

d

+

)

e=

v

c

–v

b

v

0

v0(1 +e)=

a

1+

M

m

b

vc

The relative velocity on the cart after the first collision is

Similarly, the relative velocity after the second collision is

Total time is

v0

A B

d

558

15–81.

The girl throws the 0.5-kg ball toward the wall with an

initial velocity . Determine (a) the velocity at

which it strikes the wall at B, (b) the velocity at which it

rebounds from the wall if the coefficient of restitution

, and (c) the distance sfrom the wall to where it

strikes the ground at C.

e=0.5

vA=10 m>s

vA10 m/s

1.5 m

30

3m

s

A

C

B

SOLUTION

Kinematics: By considering the horizontal motion of the ball before the impact,

we have

By considering the vertical motion of the ball before the impact, we have

The vertical position of point Babove the ground is given by

Thus, the magnitude of the velocity and its directional angle are

Conservation of “y” Momentum: When the ball strikes the wall with a speed of

, it rebounds with a speed of .

Coefficient of Restitution (x):

(vb)2

(vb)1=8.807 m>s

559

Solving Eqs.

(

1

)

and

(

2

)

yields

Kinematics:By considering the vertical motion of the ball after the impact,we have

By considering the horizontal motion of the ball after the impact, we have

15–81. Continued

Ans:

15–82.

The 20-lb box slides on the surface for which The

box has a velocity when it is 2ft from the plate.

If it strikes the smooth plate, which has a weight of 10 lb and

is held in position by an unstretched spring of stiffness

determine the maximum compression

imparted to the spring.Take between the box and

the plate. Assume that the plate slides smoothly.

e=0.8

k=400 lb>ft,

v=15 ft>s

mk

=0.3.

SOLUTION

Solving,

T

1+aU

1–2=T

2

v15 ft/s

k

2ft

Ans:

561

15–83.

SOLUTION

Collar

Bafter impact:

System:

Solving:

Collar

A:

A

:

+

B

©m1v1=©m1v2

The 10-lb collar Bis at rest, and when it is in the position

shown the spring is unstretched. If another 1-lb collar A

strikes it so that Bslides 4 ft on the smooth rod before

momentarily stopping, determine the velocity of Ajust after

impact, and the average force exerted between Aand B

during the impact if the impact occurs in 0.002 s. The

coefficient of restitution between Aand Bis e=0.5.

AB

k20 lb/ft 3ft

Ans:

562

*15–84.

SOLUTION

Since , from Eqs. (2) and (3)

mv1cos u–mv2cos f

¢t=m1mv1sin u+mv2sin f)

¢t

F

x=mF

y

(+T)m1vy21+Lt2

F

ydx =m1vy22

Aball is thrown onto arough floor at an angle If it rebounds

at an angle and the coefficient of kinetic friction

is determine the coefficient of restitution e.Neglect the size

of the ball. Hint: Show that during impact, the average

impulses in the xand ydirections are related by

Since the time of impact is the same,or

F

x=mF

y.

F

x¢t=mF

y¢t

Ix=mIy.

m,

f

u

.

y

x

uf

Ans:

563

15–85.

A ball is thrown onto a rough floor at an angle of If

it rebounds at the same angle determine the

coefficient of kinetic friction between the floor and the ball.

The coefficient of restitution is Hint: Show that

during impact, the average impulses in the xand y

directions are related by Since the time of impact

is the same, or F

x=mF

y.F

x¢t=mF

y¢t

Ix=mIy.

e=0.6.

f=45°,

u

=45°.

SOLUTION

(1)

Since , from Eqs. (2) and (3)

Substituting Eq. (4) into (1) yields:

F

x=mF

y

(+T)e=0–[–v2sin f]

v1sin u–0e=v2sin f

v1sin u

y

x

uf

m

k=0.25

564

15–86.

Two smooth billiard balls A and B each have a mass of

200 g. If A strikes B with a velocity

(

vA

)1=1.5 m>s

as

shown, determine their final velocities just after collision.

Ball B is originally at rest and the coefficient of restitution is

e=0.85

. Neglect the size of each ball.

SOLUTION

(v

A

x)

1

=–1.5 cos 40°=–1.1491 m>s

Solving,

(

vA

x

)

2=–

0.08618 m

>

s

(

x

)

For B:

Hence.

(

40

x

y

B

(vA)1  1.5 m/s

A

Ans:

565

15–87.

The “stone” A used in the sport of curling slides over the ice

track and strikes another “stone” B as shown. If each “stone”

is smooth and has a weight of 47 lb, and the coefficient of

restitution between the “stones” is

e=0.8

, determine their

speeds just after collision. Initially A has a velocity of

8 ft>s

and B is at rest. Neglect friction.

SOLUTION

Line of impact (x-axis):

Σmv1=Σmv2

Solving:

Plane of impact (y-axis):

Stone A:

Stone B:

30

x

y

B

(vA)1  8 ft/s

3 ft

A

Ans:

566

*15–88.

The “stone” A used in the sport of curling slides over the ice

track and strikes another “stone” B as shown. If each “stone”

is smooth and has a weight of 47 lb, and the coefficient of

restitution between the “stone” is

e=0.8

, determine the

time required just after collision for B to slide off the

runway. This requires the horizontal component of

displacement to be 3 ft.

SOLUTION

See solution to Prob. 15–87.

30

y

B

3 ft

A

Ans:

567

15–89.

Two smooth disks A and B have the initial velocities shown

just before they collide. If they have masses

mA=4 kg

and

mB=2 kg,

determine their speeds just after impact. The

coefficient of restitution is

e=0.8.

SOLUTION

Impact. The line of impact is along the line joining the centers of disks A and B

represented by y axis in Fig. a. Thus

Coefcient of Restitution. Along the line of impact (y axis),

Conservation of ‘y’ Momentum.

Conservation of ‘x’ Momentum. Since no impact occurs along the x axis, the

component of velocity of each disk remain constant before and after the impact.

A

B

4

5

3

vA  15 m/s

vB  8 m/s

568

15–90.

Before a cranberry can make it to your dinner plate, it must

pass a bouncing test which rates its quality. If cranberries

having an are to be accepted, determine the

dimensions dand hfor the barrier so that when a cranberry

falls from rest at Ait strikes the incline at Band bounces

over the barrier at C.

eÚ0.8

SOLUTION

Conservation of Energy: The datum is set at point B.When the cranberry falls from

a height of 3.5 ft above the datum, its initial gravitational potential energy is

. Applying Eq. 14–21, we have

Coefficient of Restitution ():

Solving Eqs. (1) and (2) yields

Kinematics: By considering the vertical motion of the cranberry after the impact,

we have

(+c)vy=(v0)y+act

y¿

W(3.5) =3.5 W

53

4

3.5 ft

h

C

B

A

d

569

By considering the horizontal motion of the cranberry after the impact, we have

15–90. Continued

Ans:

570

15–91.

SOLUTION

At A:

At B:

(vA)y1=2.5(sin 45°) =1.7678 m>s:

The 200-g billiard ball is moving with a speed of

when it strikes the side of the pool table at A. If the

coefficient of restitution between the ball and the side of

the table is determine the speed of the ball just

after striking the table twice, i.e., at A, then at B. Neglect the

size of the ball.

e=0.6,

2.5 m>s

v2.5m/s45fi

A

B

571

*15–92.

SOLUTION

Conservation of “x” momentum:

Thetwo billiard balls Aand Bare originally in contact with

one another when athird ball Cstrikes each of them at the

same time as shown. If ball Cremains at rest after the

collision, determine the coefficient of restitution.All the balls

have the same mass.Neglect the size of each ball.

CB

A

v

Ans:

572

15–93.

Disks Aand Bhave a mass of 15 kg and 10 kg, respectively.

If they are sliding on a smooth horizontal plane with the

velocities shown, determine their speeds just after impact.

The coefficient of restitution between them is .e=0.8

y

x

A

B

10 m/s

Line of

impact

8m/s

4

3

5

SOLUTION

Conservation of Linear Momentum: By referring to the impulse and momentum of

the system of disks shown in Fig. a, notice that the linear momentum of the system is

conserved along the naxis (line of impact). Thus,

Also, we notice that the linear momentum of disks Aand Bare conserved along the

taxis (tangent to? plane of impact). Thus,

and

Coefficient of Restitution:The coefficient of restitution equation written along the n

axis (line of impact) gives

Solving Eqs. (1), (2), (3), and (4), yeilds

573

15–94.

Determine the angular momentum

HO

of the 6-lb particle

about point O.

SOLUTION

Position and Velocity Vector. The coordinates of points A and B are

A(–8, 8, 12) ft and B(0, 18, 0) ft.

Then

Angular Momentum about Point O.

HO=rOB *mVA

5

i j k

Also,

HO=rOA *mVA

5

y

z

8 ft

8 ft 10 ft

12 ft

O

P

4 ft/s

6 lb

B

A