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
vA75 ft>s
45
60
SOLUTION
and
and
Since , the magnitude of vBis
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,
(1)vœ
Bcos f=19.862
0.2
32.2
A
53.59 sin 21.756°
B
=0.2
32.2
A
vœ
Bcos f
B
A
+Q
B
mB
A
vœ
B
B
x¿=mB
A
vœ
B
B
x¿
x¿
x¿
u=tan–1C
A
vB
B
y
A
vB
B
xS=tan–1
¢
7.684
53.03
≤
=8.244°
vB=2
A
vB
B
x2+
A
vB
B
y2=253.032+7.6842=53.59 ft>s
A
vB
B
x=
A
vA
B
x=75 cos 45° =53.03 ft>s
A
vB
B
y=75 sin 45° +(–32.2)(1.886) =-7.684 ft>s=7.684 ft>sT
a+cb
A
vB
B
y=
A
vA
B
y+ayt
=42.76 ft
A
sB
B
y=0+75 sin 45°(1.886) +1
2(–32.2)(1.8862)
a+cb
A
sB
B
y=
A
sA
B
y+
A
vA
B
yt+1
2ayt2
t=1.886 s
100 =0+75 cos 45° t
555
(2
)
Solving Eqs. (1) and (2) yields
Ans
.
vœ
B=
31.8 ft>s
vœ
Bsin f=24.885
0.5 =–vœ
Bsin f
–53.59 cos 21.756°
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
Ans.(uB)2=tan–1a6.062
2.378
b=68.6°
(yBx)2–(yAx)2=6.175
a:
+be=(yBx)2–(yAx)2
(yAx)1–(yBx)1
; 0.65 =(yBx)2–(yAx)2
6–(–3.5)
3(6)–5(3.5) =3(yA)x2+5(yB)x2
a:
+bmA(yAx)1+mB(yBx)1=mA(yAx)2+mB(yBx)2
(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
*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.
v0
A B
d
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
A
C
B
SOLUTION
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
Ans.
Ans.
Conservation of “y” Momentum: When the ball strikes the wall with a speed of
, it rebounds with a speed of .
(1)
Coefficient of Restitution (x):
(2)
A
:
+
B
0.5 =0–
C
–(vb)2cos f
D
10 cos 30° –0
e=(vw)2–
A
vbx
B
2
A
vbx
B
1–(vw)1
(vb)2sin f=1.602
A
;
+
B
mb(1.602) =mb
C
(vb)2sin f
D
mb
A
vby
B
1=mb
A
vby
B
2
(vb)2
(vb)1=8.807 m>s
u=tan–11.602
10 cos 30° =10.48° =10.5°
(vb)1=2(10 cos 30°)2+1.6022=8.807 m>s=8.81 m>s
(sB)y=1.5 +10 sin 30°(0.3464) +1
2(–9.81)
A
0.34642
B
=2.643 m
(+c)sy=(s0)y+(v0)yt+1
2(ac)yt2
=1.602 m>s
=10 sin 30° +(–9.81)(0.3464)
(+c)vy=(v0)y+(ac)yt
3=0+10 cos 30°tt=0.3464 s
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
Ans.s=0+4.617 cos 20.30°(0.9153) =3.96 m
A
;
+
B
sx=(s0)x+vxt
t1=0.9153 s
–2.643 =0+4.617 sin 20.30°t1+1
2(–9.81)t2
1
(+c)sy=(s0)y+(v0)yt+1
2(ac)yt2
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.
Solving,
Ans.s=0.456 ft
1
2a10
32.2 b(16.38)2+0=0+1
2(400)(s)2
T
1+V
1=T
2+V
2
vP=16.38 ft>s,
vA=5.46 ft>s
0.8 =vP–vA
13.65
e=(vB)2–(vA)2
(vA)1–(vB)1
a20
32.2 b(13.65) =a20
32.2 bvA+10
32.2 vB
(:
+)amv1=amv2
v2=13.65 ft>s
1
2a20
32.2 b(15)2–(0.3)(20)(2) =1
2a20
32.2 b(v2)2
v15 ft/s
k
2ft
15–83.
SOLUTION
Collar
Bafter impact:
System:
Solving:
Ans.
Collar
A:
Ans.
F
=2492.2 lb =2.49 kip
a1
32.2 b1117.72–F10.0022=a1
32.2 b1–42.82
A
:
+
B
mv1+©
LFdt=mv
2
(vA)2=-42.8 ft>s=42.8 ft>s;
(vA)1=117.7 ft>s=118 ft/s :
0.51vA21+1vA22=16.05
0.5 =16.05 –1vA22
1vA21–0
1vA21–1vB21
1vA21–1vA22=160.5
1
32.2 (vA)1+0=1
32.2 (vA)2+10
32.2 (16.05)
A
:
+
B
©m1v1=©m1v2
(vB)2=16.05 ft>s
1
2a10
32.2 b1vB22
2+0=0+1
2120215–322
T
2+V
2=T
3+V
3
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
k20 lb/ft 3ft
*15–84.
SOLUTION
(2)
(3)
Since , from Eqs. (2) and (3)
(4)
Substituting Eq. (4) into (1) yields:
Ans.e=sin f
sin uacos u–msin u
msin f+cos fb
v2
v1
=cos u–msin u
msin f+cos f
mv1cos u–mv2cos f
¢t=m1mv1sin u+mv2sin f)
¢t
F
x=mF
y
F
y=mv1sin u+mv2sin f
¢t
mv1sin u–F
y¢t=-mv2sin f
(+T)m1vy21+Lt2
t1
F
ydx =m1vy22
F
x=mv1cos u–mv2cos f
¢t
mv1cos u–F
x¢t=mv2cos f
(:
+)m1vx21+Lt2
t1
F
xdx =m1vx22
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
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°.
(2)
(3)
Since , from Eqs. (2) and (3)
(4)
Substituting Eq. (4) into (1) yields:
Ans.0.6 =1–m
1
+
mm=0.25
0.6 =sin 45°
sin 45° acos 45° –msin 45°
msin 45° +cos 45° b
e=sin f
sin uacos u–msin u
msin f+cos fb
v2
v1
=cos u–msin u
msin f+cos f
mv1cos u–mv2cos f
¢t=m(mv1sin u+mv2sin f)
¢t
F
x=mF
y
F
y=mv1sin u+mv2sin f
¢t
mv1sin u–F
y¢t=-mv2sin f
(+c)m
A
vy
B
1+Lt2
t1
Fydx =m
A
yy
B
2
F
x=mv1cos u–mv2cos f
¢t
mv1cos u–F
x¢t=mv2cos f
A
:
+
B
m(vx)1+Lt2
t1
F
xdx =m(vx)2
y
x
uf
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.
0.9642 b
40
x
y
B
(vA)1 1.5 m/s
A
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.
Σmv1=Σmv2
(+ a)
0 +
47
32.2
(8) cos 30°=
47
32.2
(vB)2x+
47
32.2
(vA)2x
(+ a)
e=0.8 =
(
vB
)
2x
–(
vA
)
2x
8 cos 30°–0
Solving:
(vA)2x=0.6928 ft>s
(vB)2x=6.235 ft>s
Plane of impact (y-axis):
Stone A:
mv1=mv2
(Q +)
47
32.2
(8) sin 30°=
47
32.2
(vA)2y
(v
A
)
2y
=4
Stone B:
mv1=mv2
(Q +)
0 =
47
32.2
(vB)2y
(v
B
)
2y
=0
(v
A
)
2=2
(0.6928)2
+
(4)2
=
4.06 ft
>
s Ans.
(v
B
)
2=2
(0)2
+
(6.235)2
=
6.235
=
6.24 ft
>
s Ans.
y
B
3 ft
A
Ans:
(vA)2=4.06 ft>s
(vB)2=6.24 ft>s
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
(vB)2=6.235 ft>s
s=s0+v0t
3=0+(6.235 cos 60°)t
t=0.962 s
Ans.
y
B
3 ft
A
Ans:
t=0.962 s
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.
A
B
4
5
3
vA 15 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 “”Momentum: When the cranberry strikes the plate with a
speed of , it rebounds with a speed of (vc)2.
(1)
Coefficient of Restitution ():
(2)
Solving Eqs. (1) and (2) yields
Kinematics: By considering the vertical motion of the cranberry after the impact,
we have
=0.080864 ft
=0+13.17 sin 9.978° (0.07087) +1
2(–32.2)
A
0.070872
B
(+c)sy=(s0)y+(v0)yt+1
2(ac)yt2
0=13.17 sin 9.978° +(–32.2) tt=0.07087 s
(+c)vy=(v0)y+act
f=46.85° (vc)2=13.17 ft>s
( a+) 0.8 =0–(vc)2sin f
–15.01a4
5b–0
e=(vP)2–
A
vcy¿
B
2
A
vcy¿
B
1–(vP)1
y¿
(vc)2cos f=9.008
(+ b)mc(15.01) a3
5b=mc
C
(vc)2cos f
D
mc
A
vcx¿
B
1=mc
A
vcx¿
B
2
(vc)1=15.01 ft>s
x¿
(vc)1=15.01 ft>s
0+3.5W=1
2aW
32.2 b(vc)2
1+0
T1+V1=T2+V2
53
4
3.5 ft
h
C
B
A
569
Ans.
Thus,
Ans.h=sy+3
5d=0.080864 +3
5(1.149) =0.770 ft
d=1.149 ft =1.15 ft
4
5d=0+13.17 cos 9.978° (0.07087)
Ans:
d=1.15 ft
h=0.770 ft
15–91.
SOLUTION
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
v2.5m/s45fi
A
B
*15–92.
SOLUTION
(1)
Coefficient of restitution:
(2)
Substituting Eq. (1) into Eq. (2) yields:
Ans.e=y
œ
2y
œ
cos230° =2
3
(+Q)e=y
œ
ycos 30°
y=2y
œ
cos 30°
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
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
573
15–94.
Determine the angular momentum
HO
of the 6-lb particle
about point O.
SOLUTION
A(–8, 8, 12) ft and B(0, 18, 0) ft.
Then
VA=vA
a
rAB
rAB
b
=4
•
[0 –(–8)]i+(18 –8)j+(0 –12)k
2
[0 –(–8)]2+(18 –8)2+(0 –12)2
¶
=
•
32
2
308
i+40
2
308
j–48
2
308
k
¶
ft
>
s
Angular Momentum about Point O.
HO=rOB *mVA
=
5
i j k
0 18 0
6
32.2
a
32
2308 b
6
32.2
a
40
2308 b
6
32.2
a
–48
2308 b5
=5–
9.1735i
–
6.1156k
6
slug
#
ft
2>
s
=5–
9.17i
–
6.12k
6
slug
#
ft
2>
s Ans.
Also,
HO=rOA *mVA
=
5
i j k
–8 8 12
6
32.2
a
32
2308 b
6
32.2
a
40
2308 b
6
32.2
a
–48
2308 b5
=5–
9.1735
i–
6.1156
k6
slug
#
ft
2>
s
=5–
9.17
i–
6.12
k6
slug
#
ft
2>
s Ans.
y
z
8 ft
12 ft
P
4 ft/s
6 lb
B
A