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PROBLEM 14.1
A 30-g bullet is fired with a horizontal velocity of 450 m/s and
becomes embedded in block B which has a mass of 3 kg. After the
impact, block B slides on 30-kg carrier C until it impacts the end of
the carrier. Knowing the impact between B and C is perfectly plastic
and the coefficient of kinetic friction between B and C is 0.2,
determine (a) the velocity of the bullet and B after the first impact,
(b) the final velocity of the carrier.
SOLUTION
For convenience, label the bullet as particle A of the system of three particles A, B, and C.
(a)
Impact between A and B: Use conservation of linear momentum of A and B. Assume that the time
period is so short that any impulse due to the friction force between B and C may be neglected.
Σ mv1 + Σ Imp1 2 = Σ mv 2
Components
:
m A v0 + 0 = (m A + mB )v′
v′ =
m Av0
(30 × 10−3 kg)(450 m/s)
=
= 4.4554 m/s
m A + mB
(30 × 10−3 kg + 3 kg
v′ = 4.46 m/s
(b)

Final velocity of the carrier: Particles A, B, and C have the same velocity v′′ to the left. Use
conservation of linear momentum of all three particles. The friction forces between B and C are internal
forces. Neglect friction at the wheels of the carrier.
Σ mv 2 + Σ Imp 23 = Σ mv3
Components
:
(m A + mB )v′ + 0 = ( mA + mB + mC )v
v′′ =
=
mA v0
(m A + mB )v′
=
mA + mB + mC m A + mB + mC
(30 × 10−3 kg)(450 m/s)
= 0.4087 m/s
30 × 10−3 kg + 3 kg + 30 kg
v′′ = 0.409 m/s

PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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843
PROBLEM 14.12
For the system of particles of Problem 14.11, determine (a) the
components vx and vz of the velocity of particle B for which the angular
momentum HO of the system about O is parallel to the z axis, (b) the value
of HO.
PROBLEM 14.11 A system consists of three particles A, B, and C. We
know that WA = 5lb, WB = 4 lb, and WC = 3 lb and that the velocities of
the particles expressed in ft/s are, respectively, vA = 2i + 3j − 2k ,
vB = vx i + v y j + vz k , and vC = −3i − 2 j + k. Determine (a) the
components vx and vz of the velocity of particle B for which the angular
momentum HO of the system about O is parallel to the x axis, (b) the value
of HO.
SOLUTION
i
H O = Σri × mvi = Σmi xi
(vi ) x
i j k
i
5
4
= 0 5 4 + 4
g
g
2 3 −2
vx
=
HO =
(a)
(b)
j
yi
(vi ) y
k
zi
(vi ) z
j k
i
j k
3
4 3 +
8 6 0
g
2 vz
−3 −2 1
1
[5(−10 − 12) + 4(4vz − 6) + 3(6 − 0)]i
g
+
1
[5(8 − 0) + 4(3vx − 4vz ) + 3(0 − 8)]j
g
+
1
[5(0 − 10) + 4(8 − 4vx ) + 3(−16 + 18)]k
g
1
[(16vz − 116)i + (12vz − 16vz + 16) j + (−16vx − 12)k ]
g
(1)
For H O to be parallel to the z axis, we must have H x = H y = 0:
H x = 0: 16vz − 116 = 0
vz = 7.25 ft/s 
H y = 0: 12vx − 16(7.25) + 16 = 0
vx = 8.33 ft/s 
Substituting into Eq. (1):
HO =
1
[−16(8.33) − 12]k
32.2
H O = −(4.51 ft ⋅ lb ⋅ s)k 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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859
PROBLEM 14.32
In Problem 14.4, determine the energy lost as the bullet (a) passes
through block A, (b) becomes embedded in block B.
SOLUTION
The masses are m for the bullet and m A and mB for the blocks.
The bullet passes through block A and embeds in block B. Momentum is conserved.
Initial momentum:
mv0 + mA (0) + mB (0) = mv0
Final momentum:
mvB + mA vA + mB vB
Equating,
mv0 = mvB + mA vA + mB vB
m=
mA vA + mB vB (6)(5) + (4.95)(9)
=
= 0.0500 lb
1500 − 9
v0 − vB
The bullet passes through block A. Momentum is conserved.
Initial momentum:
mv0 + m A (0) = mv0
Final momentum:
mv1 + mA vA
Equating,
mv0 = mv1 + mA vA
mv0 − mA vA (0.0500)(1500) − (6)(5)
=
= 900 ft/s
0.0500
m
0.05
m=
= 1.5528 × 10−3 lb ⋅ s 2 /ft
The masses are:
32.2
6
mA =
= 0.18633 lb ⋅ s 2/ft
32.2
4.95
mB =
= 0.153727 lb ⋅ s2 /ft
32.2
(a) Bullet passes through block A. Kinetic energies:
1
1
Before:
T0 = mv02 = (1.5528 × 10−3 )(1500) 2 = 1746.9 ft ⋅ lb
2
2
1 2 1
1
1
After: T1 = mv1 + mA v 2A = (1.5528 × 10−3 )(900) 2 + (0.18633)(5)2 = 631.2 ft ⋅ lb
2
2
2
2
T0 − T1 = 1746.9 − 631.2 = 1115.7 ft ⋅ lb
energy lost = 1116 ft ⋅ lb 
Lost:
v1 =
(b)
Bullet becomes embedded in block B. Kinetic energies:
1
1
Before:
T2 = mv12 = (1.5528 × 10−3 )(900) 2 = 628.9 ft ⋅ lb
2
2
1
1
T3 = ( m + mB )vB2 = (0.15528)(9)2 = 6.29 ft ⋅ lb
After:
2
2
T2 − T3 = 628.9 − 6.29 = 622.6 ft ⋅ lb
energy lost = 623 ft ⋅ lb 
Lost:
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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886
PROBLEM 14.38
Two hemispheres are held together by a cord which maintains a spring under
compression (the spring is not attached to the hemispheres). The potential
energy of the compressed spring is 120 J and the assembly has an initial
velocity v 0 of magnitude v0 = 8 m/s. Knowing that the cord is severed when
θ = 30°, causing the hemispheres to fly apart, determine the resulting
velocity of each hemisphere.
SOLUTION
Use a frame of reference moving with the mass center.
Conservation of momentum:
0 = −m Av′A + mB vB′
v′A =
mB
vB′
mA
V=
1
1
m A (v′A ) 2 + mB (vB′ ) 2
2
2
Conservation of energy:
2
m

1
1
= m A  B vB′  + mB (vB′ ) 2
2
2
 mA 
m (m + mB )
(vB′ ) 2
= B A
2m A
vB′ =
Data:
2m AV
mB ( mA + mB )
m A = 2.5 kg mB = 1.5 kg
V = 120 J
vB′ =
v′A =
(2)(2.5)(120)
= 10
(1.5)(4.0)
1.5
(10) = 6
2.5
vB′ = 10 m/s
v′A = 6 m/s
30°
30°
Velocities of A and B.
vA = [8 m/s
] + [6 m/s
vB = [8 m/s
] + [10 m/s
30°]
30°]
vA = 4.11 m/s
vB = 17.39 m/s
46.9° 
16.7° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
893
PROBLEM 15.1
The brake drum is attached to a larger flywheel that is not shown. The
motion of the brake drum is defined by the relation θ = 36t − 1.6t 2 ,
where θ is expressed in radians and t in seconds. Determine (a) the
angular velocity at t = 2 s, (b) the number of revolutions executed by the
brake drum before coming to rest.
SOLUTION
Given:
θ = 36t − 1.6t 2
radians
Differentiate to obtain the angular velocity.
ω=
dθ
= 36 − 3.2t
dt
(a)
At t = 2 s,
ω = 36 − (3.2)(2)
(b)
When the rotor stops,
ω = 0.
0 = 36 − 3.2t
rad/s
ω = 29.6 rad/s 
t = 11.25 s
θ = (36)(11.25) − (1.6)(11.25) 2 = 202.5 radians
In revolutions,
θ=
202.5
2π
θ = 32.2 rev 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
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1009
PROBLEM 15.11
In Problem 15.10, determine the velocity and acceleration
of corner B, assuming that the angular velocity is 9 rad/s
and increases at the rate of 45 rad/s 2 .
PROBLEM 15.10 The bent rod ABCDE rotates about a
line joining Points A and E with a constant angular velocity
of 9 rad/s. Knowing that the rotation is clockwise as viewed
from E, determine the velocity and acceleration of corner C.
SOLUTION
EA2 = 0.42 + 0.42 + 0.22
EA = 0.6 m
rB/A = −(0.25 m) j

EA = −(0.4 m)i + (0.4 m) j + (0.2 m)k

EA  −0.4i + 0.4 j + 0.2k  1
λ EA =
=
 = ( −2i + 2 j + k )
EA 
0.6
 3
1
ω = ω AE λ EA = (9 rad/s) (−2 i + 2 j + k )
3
ω = −(6 rad/s)i + (6 rad/s) j + (3 rad/s)k
v B = ω × rB/A = (−6i + 6 j + 3k ) × (−0.25) j = 1.5k + 0.75i
v B = (0.75 m/s)i + (1.5 m/s)k 
1
α = α AE λ EA = (45 rad/s 2 ) (−2i + 2 j + k )
3
2
α = −(30 rad/s )i + (30 rad/s 2 ) j + (15 rad/s 2 )k
a B = α × rB/A + ω × (ω × rB/A ) = α × rB/A + ω × v B
i
j
k
i
j k
30 15 + −6 6 3
a B = −30
0 −0.25 0
0.75 0 1.5
= 3.75i + 7.5k + 9i + (2.25 + 9) j − 4.5k
a B = (12.75 m/s 2 )i + (11.25 m/s 2 ) j + (3 m/s 2 )k 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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you are using it without permission.
1019
PROBLEM 15.18
A series of small machine components being moved by a
conveyor belt pass over a 120 mm radius idler pulley. At the
instant shown, the velocity of Point A is 300 mm/s to the left
and its acceleration is 180 mm/s2 to the right. Determine (a) the
angular velocity and angular acceleration of the idler pulley,
(b) the total acceleration of the machine component at B.
SOLUTION
vB = v A = 300 mm/s
rB = 120 mm
(aB )t = a A = 180 mm/s
(a)
(b)
vB = ω rB ,
ω=
vB 300
=
= 2.5 rad/s
rB 120
(aB )t = α rB ,
α=
(aB )t 180
=
= 1.5 rad/s
rB
120
ω = 2.50 rad/s

α = 1.500 rad/s 2

(aB ) n = rBω 2 = (120)(2.5) 2 = 750 mm/s 2
aB = (aB )t2 + (aB ) n2 = (180) 2 + (750) 2 = 771 mm/s 2
tan β =
750
,
180
β = 76.5°
a B = 771 mm/s 2
76.5° 
PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited
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1026