534
15–59.
SOLUTION
and .
By referring to Fig. a,
(1)
Coefficient of Restitution: Here
,.
Applying the relative velocity equation,
(2)
Applying the coefficient of restitution equation,
(3)e=(vc)2–(vt)2
8.333 –2.778
A
:
+
B
e=(vc)2–(vt)2
(vt)1–(vc)1
(vc)2–(vt)2=4.167
A
:
+
B
(vc)2=(vt)2+4.167
(vc)2=(vt)2+(vc>t)2
(vc>t)=c15
A
103
B
m
hda 1h
3600 s b=4.167 m>s:
5
A
vt
B
2+2
A
vc
B
2=47.22
5000(8.333) +2000(2.778) =5000
A
vt
B
2+2000
A
vc
B
2
a:
+bmt
A
vt
B
1+mc
A
vc
B
1=mt
A
vt
B
2+mc
A
vc
B
2
(vc)1=c10
A
103
B
m
hda 1h
3600 s b=2.778 m>s
The 5-Mg truck and 2-Mg car are traveling with the free-
rolling velocities shown just before they collide. After the
collision, the car moves with a velocity of to the
right relative to the truck. Determine the coefficient of
restitution between the truck and car and the loss of energy
due to the collision.
15 km>h
30 km/h
10 km/h
535
15–59. Continued
Ans:
e=0.75
∆T=–9.65 kJ
Substituting Eq. (2) into Eq. (3),
Ans.
Solving Eqs. (1) and (2) yields
Kinetic Energy: The kinetic energy of the system just before and just after the
collision are
Thus,
Ans.=9.65 kJ
=9.645
A
103
B
J
¢T=T1–T2=181.33
A
103
B
–171.68
A
103
B
=171.68
A
103
B
J
=1
2(5000)(5.5562)+1
2(2000)(9.7222)
T2=1
2mt(vt)22+1
2mc(vc)22
=181.33
A
103
B
J
=1
2(5000)(8.3332)+1
2(2000)(2.7782)
T1=1
2mt(vt)12+1
2mc(vc)12
(vc)2=9.722 m>s
(vt)2=5.556 m>s
e=4.167
8.333 –2.778 =0.75
*15–60.
Disk Ahas a mass of 2 kg and is sliding forward on the
smooth surface with a velocity when it strikes
the 4-kg disk B, which is sliding towards Aat
with direct central impact. If the coefficient of restitution
between the disks is compute the velocities of A
and Bjust after collision.
e=0.4,
1vB21=2m/s,
1vA21=5m/s
SOLUTION
(1)
Coefficient of Restitution :
(2)
Solving Eqs. (1) and (2) yields
Ans.(vA)2=-1.53 m s=1.53 m s ;(vB)2=1.27 m s :
A
:
+
B
0.4 =(vB)2–(vA)2
5–(–2)
e=(vB)2–(vA)2
(vA)1–(vB)1
A
:
+
B
2(5) +4(–2) =2(vA)2+4(vB)2
(vA)1=5m/s(vB)1=2m/s
AB
SOLUTION
xmax =0.8394 m =0.839 m
15–61.
The 15-kg block A slides on the surface for which
mk=0.3.
The block has a velocity
v=
10 m
>
s when it is s
=
4mfrom
the 10-kg block B. If the unstretched spring hasastiffness
k
=
1000 N
>
m, determine the maximum compression of
the spring due to the collision. Take e
=
0.6.
k 1000 N/m
A
B
10 m/s
15–62.
The four smooth balls each have the same mass m. If A and
B are rolling forward with velocity v and strike C, explain
why after collision C and D each move off with velocity v.
Why doesn’t D move off with velocity 2v? The collision is
elastic,
e=1
. Neglect the size of each ball.
v
AB CD
v
15–63.
The four balls each have the same mass m. If A andB are
rolling forward with velocity v and strike C, determine the
velocity of each ball after the first three collisions. Take
e=0.5
between each ball.
v
AB CD
v
*15–64.
Ball A has a mass of 3 kg and is moving with a velocity of
8m
>
s when it makes a direct collision with ball B, which has
a mass of 2 kg and is moving with a velocity of 4 m
>
s.
If
e=0.7
, determine the velocity of each ball just after the
collision. Neglect the size of the balls.
SOLUTION
AB
8 m
/
s4 m
/
s
15–65.
A 1-lb ball Ais traveling horizontally at when it
strikes a 10-lb block Bthat is at rest. If the coefficient of
restitution between Aand Bis , and the coefficient
of kinetic friction between the plane and the block is
, determine the time for the block Bto stop sliding.mk=0.4
e=0.6
20 ft
>
s
542
15–66.
SOLUTION
(1)
(2)
Solving Eqs. (1) and (2) for (vB)2yields;
Ans.(vB)2=1
32gh(1 +e)
m(vA)+0=m(vA)2+2m(vB)2
(+T)©mv1=©mv2
e22gh =(vB)2–(vA)2
(+T)e=(vB)2–(vA)2
22gh
vA=22gh
0+0=1
2mv2
A–mgh
Block A, having a mass m, is released from rest, falls a
distance hand strikes the plate Bhaving a mass 2m. If the
coefficient of restitution between Aand Bis e, determine
the velocity of the plate just after collision. The spring has
a stiffness k.
A
B
k
h
3
15–67.
SOLUTION
Balls Aand B:
Solving:
Ans.
Balls Band C:
Solving:
Ans.
Ans.1vC23=11.9 ft>s
1vB23=0.964 ft>s
0.85 =1vC23–1vB23
12.86 –0
A
:
+
B
e=1vC23–1vB23
1vB22–1vC22
10.5
32.22112.862+0=10.5
32.221vB23+10.5
32.221vC23
A
:
+
B
©mv2=©mv3
1vB22=12.86 ft>s
1vA22=1.04 ft>s
0.85 =1vB22–1vA22
13.90 –0
A
:
+
B
e=1vB22–1vA222
1vA21–1vB21
10.5
32.22113.902+0=10.5
32.221vA22+10.5
32.221vB22
A
:
+
B
©mv1=©mv2
1vA21=13.90 ft>s
0+10.52132=1
210.5
32.221v
A22
1+0
The three balls each weigh 0.5 lb and have a coefficient of
restitution of If ball Ais released from rest and
strikes ball Band then ball Bstrikes ball C, determine the
velocity of each ball after the second collision has occurred.
The balls slide without friction.
e=0.85.
r3ft
A
BC
*15–68.
A pitching machine throws the 0.5-kg ball toward the wall
with an initial velocity vA
=10 m>s
as shown. Determine
(a) the velocity at which it strikes the wall at B, (b) the
velocity at which it rebounds from the wall if
e=0.5
, and
(c) the distance s from the wall to where it strikes the ground
at C.
SOLUTION
30
1.5 m
vA10 m/s
B
A
15–69.
Conservation of Linear Momentum: Since no impulsive force acts on the football
along the xaxis, the linear momentum of the football is conserved along the xaxis.
Coefficient of Restitution: Since the ground does not move during the impact, the
coefficient of restitution can be written as
Thus, the magnitude of is
Ans.
and the angle of is
Ans.u=tan–1C
A
vœ
B
B
y
A
vœ
B
B
xS=tan–1
¢
5
21.65
≤
=13.0°
vœ
B
vœ
B=2
A
vœ
B
B
x+
A
vœ
B
B
y=221.652+52=22.2 m>s
vœ
B
A
vœ
B
B
y=5m>sc
0.4 =–
A
vœ
B
B
y
–25 sin 30°
A
+c
B
e=0–
A
vœ
B
B
y
A
vB
B
y–0
A
vœ
B
B
x=21.65 m>s;
0.3
A
25 cos 30°
B
=0.3
A
vœ
B
B
x
a;
+bm
A
vB
B
x=m
A
vœ
B
B
x
AB
u30
vA25 m>s
v¿B
A 300-g ball is kicked with a velocity of vA = 25 m>s at point A
as shown. If the coefficient of restitution between the ball and
the field is e = 0.4, determine the magnitude and direction u
of the velocity of the rebounding ball at B.
546
15–70.
A B
v0
Two smooth spheres Aand Beach have a mass m. If Ais
given a velocity of , while sphere Bis at rest, determine
the velocity of Bjust after it strikes the wall.The coefficient
of restitution for any collision is e.
v0
SOLUTION
2
547
15–71.
SOLUTION
Ans.
Ans.
Ans.u=tan–115.384
29.303 =27.7° a
vB2=2(29.303)2+(15.384)2=33.1 ft>s
0.7 =vBy 2
21.977,vBy 2=15.384 ft>sc
e=vBy 2
vBy 1
vB2x=vB1x=29.303 ft>s:
(:
+)mv1=mv2
vBy 1=0+32.2(0.68252) =21.977 ft>s
(+T)v=v0+act
vBx1=29.303 ft>s
vA=29.303 =29.3 ft>s
t=0.682524
7.5 =0+0+1
2(32.2)t2
(+T)s=s0+v0t+1
2act2
20 =0+vAt
(:
+)s=s0+v0t
It was o
b
serve
d
t
h
at a tenn
i
s
b
a
ll
w
h
en serve
d
h
or
i
zonta
ll
y
7.5 ft above the ground strikes the smooth ground at B20 ft
away. Determine the initial velocity of the ball and the
velocity (and ) of the ball just after it strikes the court at
B.Take e=0.7.
uvB
vA
20 ft
v
B
v
A
7.5ft
A
B
u
*15–72.
The tennis ball is struck with a horizontal velocity
strikes the smooth ground at B, and bounces upward at
Determine the initial velocity the final velocity
and the coefficient of restitution between the ball and
the ground.
vB,
vA,u=30°.
v
A
,
SOLUTION
Ans.
Ans.
Ans.e=vBy 2
vBy 1
=16.918
21.9773 =0.770
vBy 2=29.303 tan 30° =16.918 ft>s
vB2=29.303>cos 30° =33.8 ft>s
vBx2=vBx1=vA=29.303
(:
+)mv1=mv2
vA=29.303 =29.3 ft>s
20 =0+vA(0.68252)
(+T)s=s0+v0t
t=0.68252 s
21.9773 =0+32.2 t
(+T)v=v0+act
vBy1=21.9773 m>s
(vBy)2
1=0+2(32.2)(7.5 –0)
(+T)v2=v2
0+2ac(s–s0)
20 ft
v
B
v
A
7.5 ft
A
B
u
549
15–73.
SOLUTION
Ans.
Ans.
Ans.u=tan–13.20
4.95 =32.9 b
yB=2(4.59)2+(3.20)2=5.89 m>s
yA=1.35 m>s:
(yB)2y =3.20 m>sc
0.5(4
5)(4) =0.5(yB)2y
(+c)my1=my2
(yB)2x=4.95 m>s;
(yA)2x=1.35 m>s:
0.75 =(yA)2x–(yB)2x
4
A
3
5
B
–(–6)
(:
+)e=(yA)2–(yB)2
(yB)1–(yA)1
0.5(4)(3
5)–0.5(6) =0.5(yB)2x+0.5(yA)2x
(:
+)©my1=©my2
Two smooth disks Aand Beach have a mass of 0.
5
kg.If
both
disks are moving with the velocities shown when they
collide
, determine their final velocities just after collision.
T
he coefficient of restitution is e=0.75.
y
54
3
x
A
B
(v
A
)
1
6m/s
Ans:
vA=1.35 m>sS
vB=5.89 m>s
u=32.9°
b
15–74.
Two smooth disks Aand Beach have a mass of 0.
5
kg.If
both disks are moving with the velocities shown when they
collide, determine the coefficient of restitution between the
disks if after collision Btravels along a line, 30°
counterclockwise from the yaxis.
SOLUTION
5)–(–6) =0.0113
(:
+)e=(yA)2–(yB)2
(yB)1–(yA)1
(yA)2x=-1.752 m>s=1.752 m>s;
(yB)2x=3.20 tan 30° =1.8475 m>s;
(yB)2y=3.20 m>sc
(+c) 0.5(4)(4
5)=0.5(yB)2y
–3.60 =-(vB)2x+(yA)2x
(:
+) 0.5(4)(3
5)–0.5(6) =-0.5(yB)2x+0.5(yA)2x
©my1=©my2
y
54
3
x
A
B
(v
A
)
1
6m/s
(v
B
)
1
4m/s
15–75.
The 0.5-kg ball is fired from the tube at A with a velocity of
v
=6 m>s
. If the coefficient of restitution between the ball
and the surface is
e=0.8
, determine the height h after it
bounces off the surface.
SOLUTION
2 m
h
30
B
C
A
v 6 m/s
*15–76.
A ball of mass mis dropped vertically from a height
above the ground. If it rebounds to a height of , determine
the coefficient of restitution between the ball and the
ground.
h1
h0
SOLUTION
Subsequently, the ball’s return from position Bto position Cwill be considered.
Coefficient of Restitution: Since the ground does not move,
Ans.e=-22gh1
–
22
gh0
=Ah1
h0
e=-
(vB)2
(vB)1
(+c)
(vB)2=22gh1 c
1
2 m(vB)2
2+0=0+mgh1
1
2 mvB
2+(Vg)B=1
2 mvC
2+(Vg)C
TB+VB=TC+VC
0+mg(h0)=1
2m(vB)1
2+0
1
2 mvA
2+(Vg)A=1
2 mvB
2+(Vg)B
h1
h0
15–77.
The cue ball Ais given an initial velocity If
it makes a direct collision with ball determine
the velocity of Band the angle just after it rebounds from
the cushion at Each ball has a mass of 0.4 kg.C (e¿=0.6).
u
B (e=0.8),
(vA)1=5 m>s.
SOLUTION
(1)
Coefficient of Restitution:
(2)
Solving Eqs. (1) and (2) yields
Conservation of “y” Momentum: When ball Bstrikes the cushion at C, we have
(3)
Coefficient of Restitution (x):
(4)
Solving Eqs. (1) and (2) yields
Ans.(v
B
)
3=
3.24 m
>
s u
=
43.9°
0.6 =0–[–(vB)3 cos u]
4.50 cos 30° –0
(;
+)
e=(vC)2–(vBx)3
(vBx)2–(vC)1
(vB)3 sin u=2.25
0.4(4.50 sin 30°) =0.4(vB)3 sin u(+T)
mB(vBy)2=mB(vBy)3
(vA)2=0.500 m>s (vB)2=4.50 m>s
0.8 =(vB)2–(vA)2
5–0
(;
+)
e=(vB)2–(vA)2
(vA)1–(vB)1
0.4(5) +0=0.4(vA)2+0.4(vB)2
u
(vA)1 5 m/s
30
C
A
B
Neglect their size.