Chapter 15
Kinetics of a Particle：
Impulse and Momentum
APPLICATIONS
15-2
15.1 Principle of Linear
Impulse and Momentum
2 nd Law : Σ F = m a = m
dv
dt
⇒ Σ Fdt = md v
t2
v2
⇒
Σ
F
dt
=
m
∫
∫
t1
v1 d v = m v 2 − m v1
Impulse, 衝量
15-3
Principle of linear impulse and momentum
Relation of v, F and time
若以2nd Law 需兩個步驟才可解出
mv2 = mv1 + I
15-4
15-5
15-6
p.231, 15-8.
The 1.5-Mg four-wheel-drive jeep is used to pushtwo identical
crates,each having a mass of 500 kg.If the coefficient of static
friction between the tires and theground is μs = 0.6, determine
the maximum possible speedthe jeep can achieve in 5 s
without causing the tires to slip.The coefficient of kinetic
friction between the crates andthe ground is μk = 0.6.
15-7
15-8
15-9
p. 235, 15-28
The winch delivers a horizontal towing force F to its cable
at A which varies as shown in the graph. Determine the
speed of the 80-kg bucket when t = 24 s. Originally the
bucket is released from rest.
15-10
15-11
15.2 For a system of Particle
Σ m i ( v i )1 + Σ ∫tt12 F i dt = Σ mi ( v i ) 2
if G is mass center
then
m r G = ∑ mi r i
⇒ m v G = ∑ mi v i
⇒ m ( v G )1 + Σ ∫tt12 Fdt = m ( v G ) 2
If external impulse is zero (I = 0)
⇒ ( vG )1 = ( vG ) 2
Σmi ( vi )1 = Σmi ( vi ) 2
Conservation of linear momentum
15-12
15-13
15-14
p.244, 15-34.
The car A has a mass of 2250 kg and is traveling to the
right at 1 m>s. Meanwhile a 1500-kg car B is traveling
at 2 m>s to the left. If the cars crash head-on and
become entangled, determine their common velocity
just after the collision. Assume that the brakes are not
applied during collision.
15-15
p. 246, 15-47, 48
The free-rolling ramp has a weight of 60 kg. If the 40 kg
crate is released from rest at A, determine the speed
and distance the ramp moves when the crate slides 4.5
m down the ramp to the bottom B. Assume the ramp is
smooth and neglect the mass of the wheels.
15-16
15-17
由
及
⇒
⇒
⇒
⇒
4
− 1.5vr = vr − vB / r
5
4
2.5vr = vB / r , integrate both sides
5
4
4
2.5sr = s B / r = (4.5)
5
5
sr = 1.44 m
15-18
15.4 Impact
Def.
two bodies collide with each other
during a very short interval of time,
causing relatively large (impulsive)
forces.
15-19
APPLICATIONS
The quality of a tennis ball is measured by the height of its
bounce. This can be quantified by the coefficient of
restitution of the ball.
If the height from which the ball is dropped and the height of
its resulting bounce are known, how can we determine the
coefficient of restitution of the ball?
15-20
When the direction of motion of the mass centers of
the two colliding particles is along a line passing
through the mass centers of the particles , the
impact is named as central impact.
15-21
When the motion of one or both of the particles is at
an angle with the line of impact, the impact is named
as oblique impact.
15-22
Central Impact (中心碰撞)
15-23
m A ( v A )1 + m B ( v B )1 = m A ( v A ) 2 + m B ( v B ) 2
for particle
m A ( v A )1 m Av -
A
∫ Pdt
∫ Rdt
= m Av
= m A (v A ) 2
Def. e: coefficient of restitution
Rdt v - (vA )2
∫
e=
=
(
v
)
v
Pdt
A
1
∫
15-24
for particle B
Rdt
∫
e=
∫ Pdt
(v B ) 2 - v
=
v - ( v B )1
v - (v A ) 2 (v B ) 2 - v (v B ) 2 - (v A ) 2
=
=
⇒e =
v - ( v B )1
( v A )1 - v
( v A )1 - ( v B )1
( v B / A ) 2 脫離速度
(v B ) 2 - (v A ) 2
==
∴e =
( v A )1 - ( v B )1
( v B / A )1 接近速度
e =1
彈性碰撞 elastic impact
e=0
塑性碰撞 plastic impact
Some typical values of e are:
Steel on steel: 0.5 – 0.8
Wood on wood: 0.4 – 0.6
Lead on lead: 0.12 – 0.18
Glass on glass: 0.93 – 0.95
15-25
Oblique Impact (偏心碰撞)
15-26
15-27
(1) 碰撞線方向，系統動量守恆 Σm(v x )1 = Σm(v x ) 2
(vB x ) 2 - (v Ax ) 2
( 2) 碰撞線方向，相對速度 e =
(v A x )1 - (vB x )1
(3) 質點A動量守恆(切線方向) - y方向速度不變
(4) 質點B動量守恆(切線方向) - y方向速度不變
15-28
p. 257, 15-65
The girl throws the ball with a horizontal velocity of y1 =
2.4 m>s. If the coefficient of restitution between the ball
and the ground is e = 0.8, determine (a) the velocity of
the ball just after it rebounds from the ground and (b)
the maximum height to which the ball rises after the first
bounce.
15-29
15-30
15-31
p. 258, 15-70
Two identical balls A and B of mass m are suspended from cords of
length L/2 and L, respectively. Ball A is released from rest when φ =
90° and swings down to φ = 0° , where it strikes B. Determine the
speed of each ball just after impact and the maximum angle through
which B will swing. The coefficient of restitution between the balls is e.
15-32
15-33
15-34
15-35
15-36
p.259, 15-74
The three balls each
have the same mass m.
If A is released from rest
at θ, determine the angle
φ to which C rises after
collision. The coefficient
of restitution between
each ball is e.
1. Conservation of momentum.
2. Conservation of energy.
3. Relative speed, e.
15-37
VA
15-38
15-39
p. 260, 15-80
The 2-kg ball is thrown so that it travels horizontally at 10 m/s when it
strikes the 6-kg block as it travels down the smooth inclined plane at 1
m/s. If the coefficient of restitution between the ball and the block is e =
0.6, and the impact occurs in 0.006 s, determine the average impulsive
force between the ball and block.
15-40
15-41
15-42
p. 261, 15-84
Two disks A and B have a mass of 1 kg and 2.5 kg, respectively. If they
are sliding on the smooth horizontal plane with the velocities shown,
determine their velocities just after impact. The coefficient of restitution
between the disks is e = 0.6.
15-43
15-44
15-45
15.5 Angular Momentum
Def: moment of linear momentum
Scalar Formulation
Vector Formulation
i
H O = r × m v = rx
mv x
j
k
ry
mv y
rz
mv z
( H o ) z = (d )(mv)
P在xy平面運動
P在空間中運動
15-47
15.6
Relation between Moment of
a Force and Angular
Momentum
若 m 為 constant ⇒ Σ F = mv&
⇒
Σ M O = r × Σ F = r × mv&
d
又 H& O = ( r × mv) = r& × mv + r × mv&
dt
而 r& × mv = m( r& × r&) = 0
∴ Σ M O = H& O = r × mv&
角動量的一次導數為力 矩的總和 Resultant Moment
又 Σ F = L& = mv&
( L = mv)
線動量的一次導數為所 有外力的總和 Resultant Force
15-49
System of Particles
( H& i )O = (r i × F i ) + (r i × f i )
內力
0
⇒ Σ( H& i )O = Σ(r i × F i ) + Σ(r i × f i )
∴
Σ M O = H& O
15-50
15.7 Angular Impulse and Momentum
Principles
APPLICATIONS
Planets and most satellites move in elliptical orbits. This
motion is caused by gravitational attraction forces. Since
these forces act in pairs, the sum of the moments of the forces
acting on the system will be zero. This means that angular
momentum is conserved.
If the angular momentum is constant, does it mean the linear
momentum is also constant? Why or why not?
15-52
APPLICATIONS (continued)
The passengers on the amusement-park
ride experience conservation of angular
momentum about the axis of rotation
(the z-axis). As shown on the free body
diagram, the line of action of the normal
force, N, passes through the z-axis and
the weight’s line of action is parallel to
it. Therefore, the sum of moments of
these two forces about the z-axis is zero.
If the passenger moves away from the zaxis, will his speed increase or decrease?
Why?
15-53
Compared to :
t2
L1 + Σ ∫ F dt = L 2
t1
15-54
Conservation of Angular Momentum
當t2→t1的角衝量為0時
→ (HO)1 = (HO)2
Central force
稱為角動量守恆 (與線動量守恆不一定同時存在)
上圖，F 為一中心力，因此 ∫ M O = 0
但 ∫ F dt ≠ 0 (故線動量不守恆)
15-55
p. 274, 15-96
The ball B has a mass of 10 kg and is attached to the end of a rod
whose mass can be neglected. If the shaft is subjected to a torque M
= (2t2 + 4) N⋅m, where t is in seconds, determine the speed of the ball
when t = 2 s. The ball has a speed v = 2 m/s when t = 0.
15-56
15-57