SOLUTION

15–2.
A 20-lb block slides down a 30° inclined plane with an
initial velocity of 2 ft>s. Determine the velocity of the block
in 3 s if the coefficient of kinetic friction between the block
and the plane is mk = 0.25.
SOLUTION
A +aB
m1vvœ2 + ©
t2
Lt1
Fyœdt = m1vyœ22
0 + N(3) - 20 cos 30°(3) = 0
A +bB
N = 17.32 lb
t2
m1vxœ21 + ©
Lt1
Fxœdt = m1vxœ22
20
20
(2) + 20 sin 30°(3) - 0.25(17.32)(3) =
v
32.2
32.2
v = 29.4 ft>s
Ans.
15–5.
A man hits the 50-g golf ball such that it leaves the tee at an
angle of 40° with the horizontal and strikes the ground at
the same elevation a distance of 20 m away. Determine the
impulse of the club C on the ball. Neglect the impulse
caused by the ball’s weight while the club is striking the ball.
v2
C
SOLUTION
+ )
(:
sx = (s0)x + (v0)x t
20 = 0 + v cos 40°(t)
(+ c )
s = s0 + v0t +
1 2
at
2 c
0 = 0 + v sin 40°(t) -
1
(9.81)t2
2
t = 1.85 s
v = 14.115 m>s
( +Q)
mv1 + ©
0 +
L
L
L
F dt = mv2
F dt = 10.052114.1152
F dt = 0.706 N # s au 40°
Ans.
40⬚
15–9.
Under a constant thrust of T = 40 kN, the 1.5-Mg dragster
reaches its maximum speed of 125 m>s in 8 s starting from
rest. Determine the average drag resistance FD during this
period of time.
FD
SOLUTION
Principle of Impulse and Momentum: The final speed of the dragster is v2 = 125 m>s.
Referring to the free-body diagram of the dragster shown in Fig. a,
t2
+ )
(;
Fx dt = m(v2)x
Lt1
1500(0) + 40(103)(8) - (FD)avg (8) = 1500(125)
m(v1)x + ©
(FD)avg = 16 562.5 N = 16.6 kN
Ans.
T ⫽ 40 kN
15–13.
The 2.5-Mg van is traveling with a speed of 100 km>h when
the brakes are applied and all four wheels lock. If the speed
decreases to 40 km>h in 5 s, determine the coefficient of
kinetic friction between the tires and the road.
SOLUTION
Free-Body Diagram: The free-body diagram of the van is shown in Fig. a. The frictional
force is Ff = mkN since all the wheels of the van are locked and will cause the van
to slide.
Principle of Impulse and Momentum: The initial and final speeds of the van are
m
1h
m
1h
v1 = c100(103) d c
d = 27.78 m>s and v2 = c40(103) d c
d = 11.11 m>s.
h 3600 s
h 3600 s
Referring to Fig. a,
t2
(+ c )
m(v1)y + ©
Fy dt = m(v2)y
Lt1
2500(0) + N(5) - 2500(9.81)(5) = 2500(0)
N = 24 525 N
t2
+ )
(;
Fx dt = m(v2)x
Lt1
2500(27.78) + [-mk(24525)(5)] = 2500(11.1)
m(v1)x + ©
mk = 0.340
Ans.
15–18.
The 40-kg slider block is moving to the right with a speed of
1.5 m>s when it is acted upon by the forces F1 and F2. If
these loadings vary in the manner shown on the graph,
determine the speed of the block at t = 6 s. Neglect friction
and the mass of the pulleys and cords.
F2
F1
F (N)
SOLUTION
40
The impulses acting on the block are equal to the areas under the graph.
30
+ b
a:
20
m(vx)1 + ©
L
Fx dt = m(vx)2
F1
10
40(1.5) + 4[(30)4 + 10(6 - 4)] - [10(2) + 20(4 - 2)
+ 40(6 - 4)] = 40v2
v2 = 12.0 m>s ( : )
F2
0
Ans.
2
4
6
t (s)
15–22.
T (lb)
If the force T exerted on the cable by the motor M is indicated by the graph, determine the speed of the 500-lb crate
when t = 4 s, starting from rest. The coefficients of static and
kinetic friction are ms = 0.3 and mk = 0.25, respectively.
60
30
t (s)
2
SOLUTION
Free-Body Diagram: Here, force 3T must overcome the friction Ff before the crate
T - 30
60 - 30
moves. For 0 … t … 2 s,
or T = A 15t + 30 B lb. Considering the
=
t - 0
2 - 0
free-body diagram of the crate shown in Fig. a, where Ff = mk N = 0.3N,
+ c ©Fy = 0;
N - 500 = 0
N = 500 lb
+
: ©Fx = 0;
3(15t + 30) - 0.3(500) = 0
t = 1.333 s
Principle of Impulse and Momentum: Only the impulse of 3T after t = 1.333 s contributes to the motion. The impulse of T is equal to the area under the T vs. t graph.
At t = 1.333 s, T = 50 lb. Thus,
I =
L
1
3Tdt = 3c (50 + 60)(2 - 1.333) + 60(4 - 2) d = 470 lb # s
2
Since the crate moves, Ff = mkN = 0.25(500) = 125 lb. Referring to Fig. a,
t2
+ )
(:
m(v1)x + ©
Lt1
Fx dt = m(v2)x
500
500
(0) + 470 - 125(4 - 1.333) = a
bv
32.2
32.2
v = 8.80 ft>s
Ans.
T
M
15–29.
The train consists of a 30-Mg engine E, and cars A, B, and C,
which have a mass of 15 Mg, 10 Mg, and 8 Mg, respectively.
If the tracks provide a traction force of F = 30 kN on the
engine wheels, determine the speed of the train when
t = 30 s, starting from rest. Also, find the horizontal
coupling force at D between the engine E and car A.
Neglect rolling resistance.
C
Principle of Impulse and Momentum: By referring to the free-body diagram of the
entire train shown in Fig. a, we can write
m A v1 B x + ©
t2
Lt1
Fxdt = m A v2 B x
63 000(0) + 30(103)(30) = 63 000v
v = 14.29 m>s
Ans.
Using this result and referring to the free-body diagram of the train’s car shown in
Fig. b,
+ b
a:
m A v1 B x + ©
t2
Lt1
Fx dt = m A v2 B x
33000(0) + FD(30) = 33 000 A 14.29 B
FD = 15 714.29 N = 15.7 kN
A
E
D
SOLUTION
+ b
a:
B
Ans.
F
30 kN
15–31.
Block A weighs 10 lb and block B weighs 3 lb. If B is
moving downward with a velocity 1vB21 = 3 ft>s at t = 0,
determine the velocity of A when t = 1 s. Assume that the
horizontal plane is smooth. Neglect the mass of the pulleys
and cords.
A
SOLUTION
sA + 2sB = l
(vB)1
vA = -2vB
+ 2
1;
mv1 + ©
-
1+ T2
L
F dt = mv2
10
10
(vA)2
(2)(3) - T(1) =
32.2
32.2
mv1 + ©
L
F dt = mv2
(vA)2
3
3
()
(3) + 3(1) - 2T(1) =
32.2
32.2
2
- 32.2T - 10(vA)2 = 60
- 64.4T + 1.5(vA)2 = -105.6
T = 1.40 lb
(vA22 = -10.5 ft>s = 10.5 ft>s :
Ans.
3 ft/s
B
15–34.
The 50-kg block is hoisted up the incline using the cable and
motor arrangement shown. The coefficient of kinetic
friction between the block and the surface is mk = 0.4. If the
block is initially moving up the plane at v0 = 2 m>s, and at
this instant (t = 0) the motor develops a tension in the cord
of T = (300 + 120 2 t) N, where t is in seconds, determine
the velocity of the block when t = 2 s.
v0
30
SOLUTION
+a©Fx = 0;
(+Q)
NB - 50(9.81)cos 30° = 0
m(vx)1 + ©
L
NB = 424.79 N
Fx dt = m(vx)2
2
A 300 + 120 2t B dt - 0.4(424.79)(2)
L0
- 50(9.81)sin 30°(2) = 50v2
50(2) +
v2 = 1.92 m>s
2 m/s
Ans.
15–37.
30 km/h
The 2.5-Mg pickup truck is towing the 1.5-Mg car using a
cable as shown. If the car is initially at rest and the truck is
coasting with a velocity of 30 km>h when the cable is slack,
determine the common velocity of the truck and the car just
after the cable becomes taut. Also, find the loss of energy.
SOLUTION
Free-Body Diagram: The free-body diagram of the truck and car system is shown in
Fig. a. Here, Wt,WC, Nt, and NC are nonimpulsive forces. The pair of impulsive forces
F generated at the instant the cable becomes taut are internal to the system and thus
cancel each other out.
Conservation of Linear Momentum: Since the resultant of the impulsive force is
zero, the linear momentum of the system is conserved along the x axis. The initial
1h
m
speed of the truck is A vt B 1 = c 30(103) d c
d = 8.333 m>s.
h 3600 s
+ )
(;
mt A vt B 1 + mC A vC B 1 = A mt + mC B v2
2500(8.333) + 0 = (2500 + 1500)v2
v2 = 5.208 m>s = 5.21 m>s ;
Ans.
Kinetic Energy: The initial and final kinetic energy of the system is
T1 =
=
1
1
m (v ) 2 + mC(vC)12
2 t t1
2
1
(2500)(8.3332) + 0
2
= 86 805.56 J
and
T2 = (mt + mC)v22
=
1
(2500 + 1500)(5.2082)
2
= 54 253.47
Thus, the loss of energy during the impact is
¢E = T1 - T2 = 86 805.56 - 54 253.47 = 32.55(103) J = 32.6 kJ
Ans.
15–41.
The block has a mass of 50 kg and rests on the surface of the
cart having a mass of 75 kg. If the spring which is attached to
the cart and not the block is compressed 0.2 m and the system
is released from rest, determine the speed of the block
relative to the ground after the spring becomes undeformed.
Neglect the mass of the cart’s wheels and the spring in the
calculation. Also neglect friction. Take k = 300 N>m.
B
C
SOLUTION
T1 + V1 = T2 + V2
(0 + 0) +
1
1
1
(300)(0.2)2 = (50)(vb)2 + (75)(vc)2
2
2
2
12 = 50 v2b + 75 v2c
+ )
(:
©mv1 = ©mv2
0 + 0 = 50 vb - 75 vc
vb = 1.5vc
vc = 0.253 m>s ;
vb = 0.379 m s :
Ans.
15–42.
The block has a mass of 50 kg and rests on the surface of
the cart having a mass of 75 kg. If the spring which is
attached to the cart and not the block is compressed 0.2 m
and the system is released from rest, determine the speed
of the block with respect to the cart after the spring becomes
undeformed. Neglect the mass of the wheels and the spring
in the calculation. Also neglect friction. Take k = 300 N>m.
B
C
SOLUTION
T1 + V1 = T2 + V2
(0 + 0) +
1
1
1
(300)(0.2)2 = (50)(vb)2 + (75)(vc)2
2
2
2
12 = = 50 v2b + 75 v2c
+ )
(:
©mv1 = ©mv2
0 + 0 = 50 vb - 75 vc
vb = 1.5vc
vc = 0.253 m>s ;
vb = 0.379 m>s :
vb = vc + vb>c
+ )
(:
0.379 = - 0.253 + vb>c
vb c = 0.632 m s :
Ans.
15–43.
20 km/h
The three freight cars A, B, and C have masses of 10 Mg, 5 Mg,
and 20 Mg, respectively. They are traveling along the track
with the velocities shown. Car A collides with car B first, followed by car C. If the three cars couple together after collision, determine the common velocity of the cars after the two
collisions have taken place.
A
SOLUTION
Free-Body Diagram: The free-body diagram of the system of cars A and B when they
collide is shown in Fig. a. The pair of impulsive forces F1 generated during the collision cancel each other since they are internal to the system. The free-body diagram
of the coupled system composed of cars A and B and car C when they collide is
shown in Fig. b. Again, the internal pair of impulsive forces F2 generated during the
collision cancel each other.
Conservation of Linear Momentum: When A collides with B, and then the coupled
cars A and B collide with car C, the resultant impulsive force along the x axis is zero.
Thus, the linear momentum of the system is conserved along the x axis. The initial
speed of the cars A, B, and C are
A vA B 1 = c20(103)
A vB B 1 = c5(103)
1h
m
da
b = 5.556 m>s
h 3600 s
m
1h
da
b = 1.389 m>s,
h 3600 s
and A vC B 1 = c 25(103)
m
1h
da
b = 6.944 m>s
h 3600 s
For the first case,
+ B
A:
mA(vA)1 + mB(vB)1 = (mA + mB)v2
10000(5.556) + 5000(1.389) = (10000 + 5000)vAB
vAB = 4.167 m>s :
Using the result of vAB and considering the second case,
+ B
A:
(mA + mB)vAB + mC(vC)1 = (mA + mB + mC)vABC
(10000 + 5000)(4.167) + [-20000(6.944)] = (10000 + 5000 + 20000)vABC
vABC = -2.183 m>s = 2.18 m>s ;
Ans.
5 km/h
B
25 km/h
C
15–47.
20 km/h
The 30-Mg freight car A and 15-Mg freight car B are moving
towards each other with the velocities shown. Determine the
maximum compression of the spring mounted on car A.
Neglect rolling resistance.
A
SOLUTION
Conservation of Linear Momentum: Referring to the free-body diagram of the freight
cars A and B shown in Fig. a, notice that the linear momentum of the system is conserved along the x axis. The initial speed of freight cars A and B are
1h
1h
m
m
(vA)1 = c20(103) d a
b = 5.556 m>s and (vB)1 = c10(103) d a
b
h 3600 s
h 3600 s
= 2.778 m>s. At this instant, the spring is compressed to its maximum, and no relative
motion occurs between freight cars A and B and they move with a common speed.
+ )
(:
mA(vA)1 + mB(vB)1 = (mA + mB)v2
30(103)(5.556) + c -15(103)(2.778) d = c30(103) + 15(103) dv2
v2 = 2.778 m>s :
Conservation of Energy: The initial and final elastic potential energy of the spring
1
1
1
is (Ve)1 = ks12 = 0 and (Ve)2 = ks22 = (3)(106)smax2 = 1.5(106)smax2.
2
2
2
©T1 + ©V1 = ©T2 + ©V2
1
1
1
c mA(vA)12 + mB(vB)12 d + (Ve)1 = (mA + mB)v22 + (Ve)2
2
2
2
1
1
(30) A 103 B (5.5562) + (15) A 103 B A 2.7782 B + 0
2
2
=
1
c30 A 103 B + 15 A 103 B d A 2.7782 B + 1.5 A 106 B smax2
2
smax = 0.4811 m = 481 mm
Ans.
10 km/h
k ⫽ 3 MN/m
B
*15–52.
The free-rolling ramp has a mass of 40 kg. A 10-kg crate is
released from rest at A and slides down 3.5 m to point B. If
the surface of the ramp is smooth, determine the ramp’s
speed when the crate reaches B. Also, what is the velocity of
the crate?
3.5 m
A
30
SOLUTION
Conservation of Energy: The datum is set at lowest point B. When the crate is at
point A, it is 3.5 sin 30° = 1.75 m above the datum. Its gravitational potential energy
is 1019.81211.752 = 171.675 N # m. Applying Eq. 14–21, we have
T1 + V1 = T2 + V2
0 + 171.675 =
1
1
1102v2C + 1402v2R
2
2
171.675 = 5 v2C + 20 v2R
(1)
Relative Velocity: The velocity of the crate is given by
vC = vR + vC>R
= - vRi + 1vC>R cos 30°i - vC>R sin 30°j2
= 10.8660 vC>R - vR2i - 0.5 vC>Rj
(2)
The magnitude of vC is
vC = 2(0.8660 vC>R - vR22 + 1- 0.5 vC>R22
= 2v2C>R + v2R - 1.732 vR vC>R
(3)
Conservation of Linear Momentum: If we consider the crate and the ramp as a
system, from the FBD, one realizes that the normal reaction NC (impulsive force) is
internal to the system and will cancel each other. As the result, the linear momentum
is conserved along the x axis.
0 = mC 1vC2x + mR vR
+ )
(:
0 = 1010.8660 vC>R - vR2 + 401-vR2
0 = 8.660 vC>R - 50 vR
(4)
Solving Eqs. (1), (3), and (4) yields
vR = 1.101 m>s = 1.10 m>s vC = 5.43 m>s
Ans.
vC>R = 6.356 m>s
From Eq. (2)
vC = 30.866016.3562 - 1.1014i - 0.516.3562j = 54.403i - 3.178j6 m>s
Thus, the directional angle f of vC is
f = tan - 1
3.178
= 35.8°
4.403
cf
Ans.
B
15–54.
20 ft
The 80-lb boy and 60-lb girl walk towards each other with
constant speed on the 300-lb cart. If their velocities measured relative to the cart are 3 ft>s to the right and 2 ft>s to
the left, respectively, determine the velocity of the cart while
they are walking.
A
SOLUTION
Conservation of Linear Momentum: From the free-body diagram of the body, girl,
and cart shown in Fig. a, the pairs of impulsive forces F1 and F2 generated during the
walk cancel each other since they are internal to the system. Thus, the resultant of the
impulsive forces along the x axis is zero, and the linear momentum of the system is
conserved along the x axis.
+ )
(:
©mv1 = ©mv2
0 + 0 + 0 =
60
300
80
v (v ) v
32.2 b
32.2 g
32.2 c
80vb - 60vg - 300vc = 0
(1)
Kinematics: Applying the relative velocity equation and considering the motion of
the boy,
vb = vc + vb>c
vb = -vc + 3
+ )
(:
(2)
For the girl,
vg = vc + vg>c
+ )
(:
-vg = -vc - 2
vg = vc + 2
(3)
Solving Eqs. (1), (2), and (3), yields
vb = 2.727 ft>s :
vg = 2.273 ft>s ;
vc = 0.2727 ft>s = 0.273 ft>s ;
Ans.
A
15–57.
The 10-kg block is held at rest on the smooth inclined plane
by the stop block at A. If the 10-g bullet is traveling at
300 m>s when it becomes embedded in the 10-kg block,
determine the distance the block will slide up along the
plane before momentarily stopping.
300 m/s
A
30
SOLUTION
Conservation of Linear Momentum: If we consider the block and the bullet as a
system, then from the FBD, the impulsive force F caused by the impact is internal
to the system. Therefore, it will cancel out. Also, the weight of the bullet and the
block are nonimpulsive forces. As the result, linear momentum is conserved along
the x œ axis.
mb(vb)x¿ = (mb + mB) vx œ
0.01(300 cos 30°) = (0.01 + 10) v
v = 0.2595 m>s
Conservation of Energy: The datum is set at the blocks initial position. When the
block and the embedded bullet is at their highest point they are h above the datum.
Their gravitational potential energy is (10 + 0.01)(9.81)h = 98.1981h. Applying
Eq. 14–21, we have
T1 + V1 = T2 + V2
0 +
1
(10 + 0.01) A 0.25952 B = 0 + 98.1981h
2
h = 0.003433 m = 3.43 mm
d = 3.43 > sin 30° = 6.87 mm
Ans.
15–59.
The 5-Mg truck and 2-Mg car are traveling with the freerolling velocities shown just before they collide. After the
collision, the car moves with a velocity of 15 km>h 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.
30 km/h
10 km/h
SOLUTION
Conservation of Linear Momentum: The linear momentum of the system is
conserved along the x axis (line of impact).
The initial speeds of the truck and car are (vt)1 = c 30 A 103 B
and (vc)1 = c10 A 103 B
1h
m
da
b = 8.333 m>s
h 3600 s
1h
m
da
b = 2.778 m>s.
h 3600 s
By referring to Fig. a,
+ b
a:
mt A vt B 1 + mc A vc B 1 = mt A vt B 2 + mc A vc B 2
5000(8.333) + 2000(2.778) = 5000 A vt B 2 + 2000 A vc B 2
5 A vt B 2 + 2 A vc B 2 = 47.22
Coefficient of Restitution: Here, (vc>t) = c15 A 103 B
Applying the relative velocity equation,
(vc)2 = (vt)2 + (vc>t)2
+ B
A:
(vc)2 = (vt)2 + 4.167
(vc)2 - (vt)2 = 4.167
(2)
Applying the coefficient of restitution equation,
+ B
A:
e =
(vc)2 - (vt)2
(vt)1 - (vc)1
e =
(vc)2 - (vt)2
8.333 - 2.778
(3)
(1)
1h
m
da
b = 4.167 m>s : .
h 3600 s
15–59. continued
Substituting Eq. (2) into Eq. (3),
e =
4.167
= 0.75
8.333 - 2.778
Ans.
Solving Eqs. (1) and (2) yields
(vt)2 = 5.556 m>s
(vc)2 = 9.722 m>s
Kinetic Energy: The kinetic energy of the system just before and just after the
collision are
T1 =
=
1
1
mt(vt)1 2 + mc(vc)1 2
2
2
1
1
(5000)(8.3332) + (2000)(2.7782)
2
2
= 181.33 A 103 B J
T2 =
=
1
1
m (v ) 2 + mc(vc)2 2
2 t t2
2
1
1
(5000)(5.5562) + (2000)(9.7222)
2
2
= 171.68 A 103 B J
Thus,
¢E = T1 - T2 = 181.33 A 103 B - 171.68 A 103 B
= 9.645 A 103 B J
= 9.65 kJ
Ans.
15–66.
If the girl throws the ball with a horizontal velocity of 8 ft>s,
determine the distance d so that the ball bounces once on
the smooth surface and then lands in the cup at C. Take
e = 0.8.
A
vA
3 ft
C
B
d
SOLUTION
(+ T )
v2 = v20 + 2ac1s - s02
(v122y = 0 + 2132.22132
1v12y = 13.90 T
(+ T )
s = s0 + v0 t +
3 = 0 + 0 +
1 2
ac t
2
1
132.221tAB22
2
tAB = 0.43167 s
(+ T )
e =
1v22y
1v12y
0.8 =
1v22y
13.90
1v22y = 11.1197 c
(+ T )
v = v0 + ac t
11.1197 = -11.1197 + 32.21tBC2
tBC = 0.6907 s
Total time is tAC = 1.1224 s
Since the x component of momentum is conserved
d = vA1tAC2
d = 811.12242
d = 8.98 ft
Ans.
15–70.
Two smooth spheres A and B each have a mass m. If A is
given a velocity of v0, while sphere B is at rest, determine
the velocity of B just after it strikes the wall. The coefficient
of restitution for any collision is e.
v0
A
SOLUTION
Impact: The first impact occurs when sphere A strikes sphere B. When this occurs, the
linear momentum of the system is conserved along the x axis (line of impact).
Referring to Fig. a,
+ )
(:
mAvA + mBvB = mA(vA)1 + mB(vB)1
mv0 + 0 = m(vA)1 + m(vB)1
(vA)1 + (vB)1 = v0
+ )
(:
e =
(vB)1 - (vA)1
vA - vB
e =
(vB)1 - (vA)1
v0 - 0
(1)
(vB)1 - (vA)1 = ev0
(2)
Solving Eqs. (1) and (2) yields
(vB)1 = a
1 + e
bv0 :
2
(vA)1 = a
1 - e
bv0 :
2
The second impact occurs when sphere B strikes the wall, Fig. b. Since the wall does
not move during the impact, the coefficient of restitution can be written as
+ )
(:
e =
e =
0 - C -(vB)2 D
(vB)1 - 0
0 + (vB)2
c
1 + e
dv0 - 0
2
(vB)2 =
e(1 + e)
v0
2
Ans.
B
15–74.
The 1 lb ball is dropped from rest and falls a distance of 4 ft
before striking the smooth plane at A. If it rebounds and in
t = 0.5 s again strikes the plane at B, determine the
coefficient of restitution e between the ball and the plane.
Also, what is the distance d?
4 ft
A
d
SOLUTION
T1 + V1 = T2 + V2
B
3
5
4
0 + 0 =
1
1m21vA221 - m132.22142
2
1vA21 = 22132.22142 = 16.05 ft>s
3
116.052 = 9.63 ft>s
5
+R
1vA22x¿ =
Q+
4
1vA22y¿ = ea b 116.052 = 12.84e ft>s
5
+ )
(:
s = s0 + v0 t
4
1d2 = 0 + vA2x10.52
5
(+ T )
s = s0 + v0 t +
1 2
a t
2 c
3
1
1d2 = 0 - vA2y 10.52 + 132.2210.522
5
2
+ )
(:
3
4
4
0.5c 9.63a b + 12.84e a b d = d
5
5
5
(+ c)
4
3
3
0.5c -9.63a b + 12.84e a b d = 4.025 - d
5
5
5
Solving,
e = 0.502
Ans.
d = 7.23 ft
Ans.
15–77.
The cue ball A is given an initial velocity (vA)1 = 5 m>s. If
it makes a direct collision with ball B (e = 0.8), determine
the velocity of B and the angle u just after it rebounds from
the cushion at C (e¿ = 0.6). Each ball has a mass of 0.4 kg.
Neglect the size of each ball.
(vA)1 ⫽ 5 m/s
A
B
30⬚
C
u
SOLUTION
Conservation of Momentum: When ball A strikes ball B, we have
mA(vA)1 + mB(vB)1 = mA(vA)2 + mB(vB)2
0.4(5) + 0 = 0.4(vA)2 + 0.4(vB)2
(1)
Coefficient of Restitution:
e =
+)
(;
(vB)2 - (vA)2
(vA)1 - (vB)1
0.8 =
(vB)2 - (vA)2
5 - 0
(2)
Solving Eqs. (1) and (2) yields
(vA)2 = 0.500 m>s
(vB)2 = 4.50 m>s
Conservation of “y” Momentum: When ball B strikes the cushion at C, we have
mB(vBy)2 = mB(vBy)3
(+ T)
0.4(4.50 sin 30°) = 0.4(vB)3 sin u
(vB)3 sin u = 2.25
(3)
Coefficient of Restitution (x):
e =
+ )
(;
(vC)2 - (vBx)3
(vBx)2 - (vC)1
0.6 =
0 - [-(vB)3 cos u]
4.50 cos 30° - 0
(4)
Solving Eqs. (1) and (2) yields
(vB)3 = 3.24 m>s
u = 43.9°
Ans.
15–89.
Two smooth disks A and B have the initial velocities shown
just before they collide at O. If they have masses mA = 8 kg
and mB = 6 kg, determine their speeds just after impact.
The coefficient of restitution is e = 0.5.
y
13 12
5
A
vA
O
SOLUTION
vB
+b©mv1 = ©mv2
- 6(3 cos 67.38°) + 8(7 cos 67.38°) = 6(vB)x¿ + 8(vA)xœ
e =
(+b)
0.5 =
(vB)2 - (vA)2
(vA)1 - (vB )1
(vB)xœ - (vA)xœ
7 cos 67.38° + 3 cos 67.38°
Solving,
(vB)xœ = 2.14 m>s
(vA)xœ = 0.220 m>s
(vB)yœ = 3 sin 67.38° = 2.769 m>s
(vA)y¿ = -7 sin 67.38° = -6.462 m>s
vB = 2(2.14)2 + (2.769)2 = 3.50 m>s
Ans.
vA = 2(0.220)2 + (6.462)2 = 6.47 m>s
Ans.
3 m/s
7 m/s
x
B
15–93.
Disks A and B have 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
5
4
Line of
impact
3
A
10 m/s
x
SOLUTION
8 m/s
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 n axis (line of impact). Thus,
¿
+ Q mA A vA B n + mB A vB B n = mA A vA
B n + mB A vB¿ B n
3
3
¿
15(10)a b - 10(8)a b = 15vA
cos fA + 10vB¿ cos fB
5
5
¿
15vA
cos fA + 10vB¿ cos fB = 42
(1)
Also, we notice that the linear momentum of disks A and B are conserved along the
t axis (tangent to? plane of impact). Thus,
¿
+ a mA A vA B t = mA A vA
Bt
4
¿
15(10)a b = 15vA
sin fA
5
¿
vA
sin fA = 8
(2)
and
+a mB A vB B t = mB A vB¿ B t
4
10(8)a b = 10 vB¿ sin fB
5
vB¿ sin fB = 6.4
(3)
Coefficient of Restitution: The coefficient of restitution equation written along the n
axis (line of impact) gives
+Q e =
¿
(vB¿ )n - (vA
)n
(vA)n - (vB)n
0.8 =
¿
cos fA
vB¿ cos fB - vA
3
3
10a b - c -8a b d
5
5
¿
vB¿ cos fB - vA
cos fA = 8.64
(4)
Solving Eqs. (1), (2), (3), and (4), yeilds
¿
= 8.19 m>s
vA
Ans.
fA = 102.52°
vB¿ = 9.38 m>s
fB = 42.99°
Ans.
B
15–97.
z
Determine the total angular momentum H O for the system
of three particles about point O. All the particles are
moving in the x–y plane.
3 kg
6 m/s
C
O
900 mm
SOLUTION
4 m/s
A 1.5 kg
HO = ©r * mv
i
= 3 0.9
0
j
0
- 1.5(4)
= {12.5k} kg # m2>s
x
i
k
0 3 + 3 0.6
-2.5(2)
0
j
0.7
0
i
k
0 3 + 3 -0.8
0
0
j
-0.2
3(- 6)
k
03
0
Ans.
700 mm
200 mm
800 mm
y
B
2 m/s 600 mm
2.5 kg
*15–100.
The small cylinder C has a mass of 10 kg and is attached to
the end of a rod whose mass may be neglected. If the frame
is subjected to a couple M = 18t2 + 52 N # m, where t is in
seconds, and the cylinder is subjected to a force of 60 N,
which is always directed as shown, determine the speed of
the cylinder when t = 2 s. The cylinder has a speed
v0 = 2 m>s when t = 0.
z
60 N
5 4
0.75 m
(Hz)1 + ©
v
x
SOLUTION
C
M
L
Mz dt = (Hz)2
2
3
(8t 2 + 5)dt = 10v(0.75)
(10)(2)(0.75) + 60(2)( )(0.75) +
5
L0
8
69 + [ t3 + 5t]20 = 7.5v
3
v = 13.4 m>s
Ans.
3
(8t2
y
5) N m
15–102.
The 10-lb block is originally at rest on the smooth surface. It
is acted upon by a radial force of 2 lb and a horizontal force
of 7 lb, always directed at 30° from the tangent to the path
as shown. Determine the time required to break the cord,
which requires a tension T = 30 lb. What is the speed of the
block when this occurs? Neglect the size of the block for the
calculation.
A
4 ft
B
7 lb
SOLUTION
©Fn = man ;
30 - 7 sin 30° - 2 =
10 v2
( )
32.2 4
v = 17.764 ft>s
(HA)1 + ©
L
MA dt = (HA)2
0 + (7 cos 30°)(4)(t) =
t = 0.910 s
10
(17.764)(4)
32.2
Ans.
30
2 lb
15–105.
The four 5-lb spheres are rigidly attached to the crossbar frame having
a negligible weight. If a couple moment M = (0.5t + 0.8) lb # ft, where t
is in seconds, is applied asshown, determine the speed of each of the
spheres in 4 seconds starting from rest. Neglect the size of the spheres.
M
SOLUTION
(Hz)1 + ©
L
Mz dt = (Hz)2
4
0 +
L0
0.6 ft
(0.5 t + 0.8) dt = 4 B a
5
b(0.6 v 2) R
32.2
7.2 = 0.37267 v2
v2 = 19.3 ft>s
Ans.
(0.5t
0.8) lb · ft