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Chapter 5
Impulse and Momentum
Chapter Abstract. As discussed in Chapter 4, we can reduce the complexity of dynamics
problems for which the change of a particle’s speed is known. We accomplish this through
the Principle of Work and Energy by integrating Newton’s Second Law over a given particle’s
trajectory, which relates the spatial integral of an applied force to the particle’s kinetic energy.
In this chapter, we develop the Principle of Impulse and Momentum by integrating
Newton’s Second Law over time. This results in a vector equation relating the temporal
integral of an applied force, which we call impulse, to the attending change in a particle’s
linear-momentum vector.
The Principle of Impulse and Momentum is especially helpful for solving dynamics problems involving an impulse of very short duration. This limiting case of impulsive motion is
termed impact. Examples of impact are a baseball bat hitting a pitched baseball and the collision of two billiard balls. We will analyze several types of impacts including direct central
impact, oblique central impact and constrained motion, i.e., impacts for which motion in
one or more directions is constrained by a surface or other boundary. The chapter concludes
with applications in which we achieve a solution through a mix of the three methods developed
in this and the preceding two chapters, viz., Newton’s Second Law in differential form, the
Principle of Work and Energy and the Principle of Impulse and Momentum.
5.1 Principle of Impulse and Momentum
In order to establish the Principle of Impulse and Momentum, we begin with Newton’s
Second Law for a particle of mass m written as
F=m
dv
d
= (mv)
dt
dt
(5.1)
where F is the sum of the external forces acting on the particle. This equation says the external
force acting on a particle is equal to the rate of change of its momentum. Integrating over
time from t = t1 to t = t2 , we arrive at the following relation between the integrated force
and the particle’s change in momentum.
t2
t2
F dt =
t1
t1
d
(mv) dt = mv2 − mv1
dt
145
(5.2)
146
CHAPTER 5. IMPULSE AND MOMENTUM
The quantities v1 and v2 are the particle’s velocity vectors at times t1 and t2 , respectively.
We define the integral of the force over the time interval t1 to t2 as the linear impulse, which
we denote by Imp1−2 , viz.,
t2
Imp1−2 ≡
F dt
(5.3)
t1
Thus, the Principle of Impulse and Momentum is
(5.4)
Imp1−2 = mv2 − mv1
The fact that Equation (5.4) is a vector equation constitutes one of the most significant differences between the Principle of Impulse and Momentum and the Principle of Work and
Energy.
Impulse of Typical Forces. Another significant difference concerns the impulse of the standard forces we deal with in dynamics. Only constant forces and some time-dependent forces
can be integrated over time in closed form. For a constant force, F, the impulse is
Imp1−2 = F (t2 − t1 )
(F = constant)
(5.5)
(Dashpot)
(5.6)
Also, for a dashpot we have
t2
Imp1−2 =
t1
−c
dx
i dt = −c (x2 − x1 ) i
dt
Impulse of Internal Forces. How internal forces affect the Principle of Impulse and Momentum is of great importance. As an example, consider a man in Boat A pulling Boat B toward
him as illustrated in Figure 5.1. As he pulls on the rope, the tension forces on Boats A and B
are TA = TA et and TB = −TB et , where T A and TB denote tension in the rope acting on
Boats A and B, respectively, and et is a unit vector tangent to the rope. Clearly, the impulse
of the force provided by the man as he pulls for a time interval ∆t is TA ∆t et . The impulse
transmitted to Boat B during the time interval is −TB ∆t et . But, the tension in the rope is
constant so that TA = TB . Consequently, the net impulse acting on the boats taken as a whole,
Imp1−2 = TA + TB , is zero. Therefore, we conclude that
Imp1−2 = 0
(Internal forces)
(5.7)
For zero net impulse, Equation (5.4) gives us the unsurprising result that the sum of the
boats’ momentum is conserved. This is an example of what we refer to as a non-impulsive
force. Most importantly, since internal forces are non-impulsive, Equation (5.7) underscores
the fact that only external forces must be considered when using the Principle of Impulse and
Momentum.
T
A
..................................................
.............................B
......
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Boat A
T
Boat B
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Figure 5.1: A man in a boat pulling another boat toward him.
5.2. IMPULSIVE MOTION
147
5.2 Impulsive Motion
Because most of the forces encountered in dynamics applications cannot be integrated over time
in closed form, either we compute the change in momentum attending a prescribed impulse
or we compute the impulse corresponding to a known change in momentum. If we know the
duration of the impulse, ∆t ≡ t2 − t1 , we can infer the average force acting on a particle, F,
by noting that, in general (see Figure 5.2),
F≡
1
t2 − t1
t2
F dt
(5.8)
t1
We can rewrite the Principle of Impulse and Momentum, Equation (5.4), in terms of the
average impulsive force as follows.
F ∆t = mv2 − mv1
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F ..........
F
t1
t2
(5.9)
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t1 t2
(b) Short duration
t
(a) Long duration
t
Figure 5.2: Relation between impulse and average force.
Figure 5.2(a) depicts a force that varies gradually as t increases from t1 to t2 , while
Figure 5.2(b) illustrates a force that varies abruptly over a very short time interval. In both
cases, the integration operation masks details of the impulse. The Principle of Impulse and
Momentum is equally effective for either case. As we will discover in this chapter, it is
particularly useful for applications involving a force that acts over a very short time interval.
Example 5.1 A ball of mass m moving in a horizontal plane bounces off of a surface as shown with
velocity vectors before and after the collision v1 = v(cos φ i−sin φ k) and v2 = v(cos φ i+sin φ k),
respectively. If the impulse exerted on the ball by the wall is I k, determine the angle φ. Compute
φ for m = 1 kg, I = 10 N·sec and v = 10 m/sec.
•
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z
v
φ
v
φ
•
x
Solution. The Principle of Impulse and Momentum tells us that
I k = m (v2 − v1 ) = 2mv sin φ k
=⇒
φ = sin−1
I
2mv
For the given values, we have
φ = sin−1
10 N · sec
1
= sin−1
= 30o
2(1 kg)(10 m/sec)
2
148
CHAPTER 5. IMPULSE AND MOMENTUM
The magnitude of the average impulsive force can be astonishingly large for some practical
dynamics examples. As observed by Adair (1990) and Nathan (2000), when a Major League
Baseball player hits a baseball, the bat and ball are in contact for about 0.7 millisecond. We
can use the Principle of Impulse and Momentum to estimate the average force the bat exerts
on the baseball if we know the baseball’s velocity before and after the interaction between the
bat and the ball.
z .............
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v
φ
v
x
Figure 5.3: Pittsburgh Pirates superstar Roberto Clemente hitting a baseball.
Consider a baseball that is pitched with a horizontal speed v approaching a batter as shown
in Figure 5.3. After the batter hits the ball, its speed is 98 v and its path is directed upward at
an angle φ to the ground plane. We wish to compute the average impulsive force of the bat,
F, corresponding to a time of contact between the ball and the bat given by t2 − t1 = τ . Note
that, in addition to the force of the bat, gravity acts on the baseball. For the sake of simplicity,
we will ignore the contribution of gravity. After we have determined F, we will show that
gravity is essentially a non-impulsive force in this problem.
As the first step in our solution, we let x and z denote the horizontal and vertical directions,
respectively. The velocity vectors before and after the bat strikes the ball are
v1 = −v i
and
v2 =
9
v(cos φ i + sin φ k)
8
(5.10)
So, from the Principle of Impulse and Momentum in the form stated in Equation (5.9), we
have
9
9
F τ = mv 1 + cos φ i + mv sin φ k
(5.11)
8
8
so that the average impulsive force is
F=
mv
τ
1+
9
9
cos φ i + sin φ k
8
8
(5.12)
A short computation yields the magnitude of F, viz.,
F =
mv
8τ
145 + 144 cos φ
(5.13)
An official Major League baseball weighs mg = 5 18 ounces, where g = 32.174 ft/sec2 .
Hence, a little arithmetic shows that its mass is m = 0.01 slug. A typical Major League
5.2. IMPULSIVE MOTION
149
pitcher can throw a baseball at v = 90 mph = 132 ft/sec. As noted above, according to
Adair (1990) and Nathan (2000), the contact time is τ = 0.0007 sec. Thus, if the angle is
φ = 30o , the magnitude of the bat’s impulsive force is
F =
(0.01 slug) (132 ft/sec) √
145 + 144 cos 30o = 3871 lb
8(0.0007 sec)
(5.14)
Thus, the average force exerted by a bat on a baseball for these conditions is nearly two tons!
Finally, we can determine the average impulsive force corresponding to the ball’s weight,
|Fg |. Because gravity is constant and the ball’s mass does not change, the average force due
to gravity during the interaction between the ball and the bat is independent of time and
Fg = mg = (5.125 oz)
1 lb
16 oz
(5.15)
= 0.32 lb
Thus, the gravitational force is 0.008% of the bat’s average impulsive force, wherefore gravity
is so small that it is effectively a non-impulsive force.
The interaction between a baseball bat and ball is an example of a special type of interaction
that we call an impact. In general, an impact occurs when two objects collide and the
collision is such that they exert large forces on each other for a very short time interval. In the
following sections, we will focus on two common types of impacts. Referring to Figure 5.4,
we distinguish between a direct central impact and an oblique central impact. In both
cases, we define the line of impact as the line perpendicular to the parts of the two objects’
surfaces that are in contact during the impact. That is, as illustrated in the figure, the two
objects deform during an impact and their surfaces might even be temporarily “flattened”
where they make contact.1
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(a) Direct central impact
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of impact...........
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A
v
B
(b) Oblique central impact
Figure 5.4: Direct central and oblique central impacts.
By definition, direct central and oblique central impacts are as follows.
• Direct Central Impact: The velocities of the objects, vA and vB , are parallel to the
line of impact.
• Oblique Central Impact: At least one object moves along a path other than the line
of impact.
Obviously, direct central impacts involve one-dimensional motion during the impact, while
oblique central impacts involve two or three dimensions. The next two sections develop a
straightforward procedure for computing properties of impacts.
1 For a baseball bat and ball impact the ball can be compressed along the line of impact by as much as one half
of its diameter.
150
CHAPTER 5. IMPULSE AND MOMENTUM
5.3 Direct Central Impact
Consider Particles A and B shown in Figure 5.5, with Particle A moving faster than Particle B,
i.e., |vA | > |vB |. We idealize the impact process in terms of the following four phases.
1. Prior to Impact. Particles A and B are in their initial states. Figure 5.5(a) shows the
particles with their undeformed shapes.
2. Period of Deformation. When the particles collide they deform. During the period
of deformation, the particles’ shapes change. Figure 5.5(b) shows the particles at the
time of maximum deformation, which marks the end of the period of deformation. At
this time, the particles are moving with a common velocity, u.
3. Period of Restitution. During the period of restitution the particles move apart and
approach a new equilibrium state.
4. After Impact. As shown in Figure 5.5(c), Particles A and B move with new velocities
vA and vB , respectively, corresponding to the new equilibrium state. Depending on their
constitution, the particles may or may not be deformed after restitution.
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A
v
B
v
A
(a) Prior to Impact
B
u
v
v
B
A
(b) Maximum Deformation
(c) After Impact
Figure 5.5: Phases of the motion of two particles before, during and after an impact.
Since there are no external forces acting, we know that momentum is conserved. Thus, in
component form, we have
(5.16)
mA vA + mB vB = mA vA + mB vB
Assuming mA , vA , mB and vB are known, this is a single equation involving the two unknown
post-impact speeds vA and vB . In order to find another equation to complete the solution, we
will consider each particle separately, taking into account the internal forces. While they are
non-impulsive for the combined motion of Particles A and B, they do affect the motion of
each particle in isolation. We define the following forces (see Figure 5.6).
• Dba : Force exerted by B on A during the period of deformation
• Dab : Force exerted by A on B during the period of deformation
• Rba : Force exerted by B on A during the period of restitution
• Rab : Force exerted by A on B during the period of restitution
Because these are internal forces, letting their magnitudes be D and R for the deformation
and restitution forces, respectively, in vector form we have
Dba = −D i,
Dab = D i,
Rba = −R i,
where i is a unit vector parallel to the line of impact.
Rab = R i
(5.17)
5.3. DIRECT CENTRAL IMPACT
151
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D
A
•
B
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D
A
•
R
(a) Deformation
B
•
R
(b) Restitution
Figure 5.6: Deformation and restitution forces during an impact.
Focusing first on Particle A, the Principle of Impulse and Momentum tells us that
tm
−
t1
tm
D dt = mA u − mA vA
=⇒
R dt = mA vA − mA u
=⇒
t1
t2
−
D dt = mA (vA − u)
(5.18)
R dt = mA (u − vA )
(5.19)
t2
tm
tm
The integration limits t1 , tm and t2 denote the times at the beginning of the period of deformation, at maximum deformation and at the end of the period of restitution, respectively. We
can solve for the ratio of the impulse of the restitution force to the impulse of the deformation
force in terms of velocity differences by combining Equations (5.18) and (5.19), which yields
t2
R dt
tm
tm
e=
=
D dt
u − vA
vA − u
(5.20)
t1
The quantity e is called the coefficient of restitution. Because its value depends upon the
average values of the unknown deformation and restitution forces, it is not known a priori. Its
value depends upon a variety of factors, most significantly the material of which the particles
are made. For now, we will regard it as a dimensionless parameter. We will return to a
detailed discussion of e after we complete the solution for the impact of Particles A and B.
Turning now to Particle B, a similar calculation gives
tm
t1
tm
D dt = mB u − mB vB
=⇒
R dt = mB vB − mB u
=⇒
t2
tm
D dt = mB (u − vB )
(5.21)
R dt = mB (vB − u)
(5.22)
t1
t2
tm
As with Particle A, we can solve for the ratio of the impulse of the restitution force to the
impulse of the deformation force in terms of velocity differences. Combining Equations (5.21)
and (5.22), we find
t2
R dt
e=
tm
tm
=
D dt
vB − u
u − vB
(5.23)
t1
We can rewrite Equations (5.20) and (5.23) as follows.
u − vA = e (vA − u)
and
vB − u = e (u − vB )
(5.24)
152
CHAPTER 5. IMPULSE AND MOMENTUM
Adding left-hand and right-hand sides of these equations to eliminate u and rearranging terms
shows that the relative velocities of the particles before and after the impact are related by
(5.25)
vB − vA = e (vA − vB )
We refer to this equation as the impact relation. At this point, we have derived a second
equation that can be used in conjunction with Equation (5.16) to solve for the unknown
velocities, vA and vB .
Example 5.2 Ball A of mass m is moving to the right with speed 2v. It approaches Ball B of
mass 2m that is moving to the right with speed v. If the coefficient of restitution is e = 12 , what
are the speeds of the balls after the impact?
m
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2m
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B
v
Solution. Momentum conservation tells us that
m(2v) + 2m · v = mvA + 2mvB
=⇒
vA + 2vB = 4v
From the impact relation with e = 12 ,
vB − vA = e (2v − v) = ev
vB − vA =
=⇒
1
v
2
Solving these two simultaneous equations, the speeds of the balls after the impact are
vA = v
and
vB =
3
v
2
It is worthwhile to pause at this point and make some salient observations. We have
arrived at Equations (5.16) and (5.25), which are sufficient to solve for the velocities after the
impact. We accomplished this by analyzing the dynamics of the two particles separately. In the
process, we introduced two additional equations [Equations (5.20) and (5.23)] and unknown
quantities, viz., u and e. Although we have eliminated u, our solution involves the unknown
coefficient of restitution, e. Thus, we actually have two equations for three unknowns, vA , vB
and e, which still leaves us short an equation. This reflects the fact that we have no details
of the deformation and restitution forces. If we did, we could compute the coefficient e and
our algebraic system would consist of two equations for two unknowns.
In the absence of an extremely complex first-principles computation of the coefficient, e,
we can write Equation (5.25) in the following form.
e=
vB − vA
Relative speed after impact
=
vA − vB
Relative speed before impact
(5.26)
While we have assumed Particle B moves in the same direction as Particle A, this equation
also holds when Particle B moves toward Particle A. In this case, we simply observe that vB
is negative.
Equation (5.26) gives us an indirect way of determining values of e for various materials
in laboratory experiments. By simply measuring the velocity differences before and after an
impact, we can compute and tabulate values of e for specific materials. Measurements show
that the coefficient of restitution has a range of values for common materials. It is also affected
by a variety of factors including temperature, internal pressure for a hollow ball, and even
5.3. DIRECT CENTRAL IMPACT
e
153
1.0
◦
◦
◦ Golf ball ◦
....................................................................
...........................
.........................
...................
...
..................
...
..................
..................
...
..................
...
........
...
... ..........................................................................
...........................
............................
...................
.....
.
.
.
...............
......
................
........
....................
...........
.....................
........................
....................................................................................................................................................
0.8
• •
0.6
0.4
Baseball
•
◦
◦
Tennis ball
•
0.2
0
0
20
40
60
80
100 vrel (mph)
Figure 5.7: Variation of the coefficient of restitution with impact speed; the points corresponding to •, ◦ and depict measured values from various sources.
humidity for baseballs. Generally speaking, it is somewhat sensitive to relative impact speed,
e.g., vrel = vA −vB . As shown in Figure 5.7, for example, e decreases with increasing relative
impact speed for common objects such as golf balls, baseballs and tennis balls. Table 5.1
includes values of e for a variety of balls.
Table 5.1: Coefficient of Restitution for Various Balls
Type of Ball
Table-Tennis Ball
Golf Ball (Low Speed)
Ball of Rubber Bands
Billiard Ball
Hand Ball
Tennis Ball (Low Speed)
Hard-Hollow Plastic Ball
Glass Marble
e
0.91
0.83
0.83
0.80
0.75
0.71
0.69
0.66
Type of Ball
Golf Ball (High Speed)
Basketball
Steel Ball Bearing
Hard Wooden Ball
Baseball (Low Speed)
Baseball (High Speed)
Softball
Tennis Ball (High Speed)
e
0.63
0.60
0.60
0.60
0.55
0.45
0.44
0.40
The deformation process, which involves compression and subsequent expansion of the
impacting particles, has an effect on kinetic energy. Clearly, kinetic energy is lost during
the deformation phase and some is regained during the restitution phase. It is not obvious
that the kinetic energy regained is equal to the kinetic energy that is lost and, in general, it
is not. We can determine the change in kinetic energy by revisiting the equations for the
individual particles. Rearranging terms in Equations (5.18), (5.19), (5.21) and (5.22), we have
the following.
tm
t1
t2
tm
tm
t1
t2
tm
D dt = mA (vA − u) =⇒ mA vA2 = vA
tm
D dt + mA uvA
R dt = mA (u − vA ) =⇒ mA vA2 = −vA
D dt = mB (u − vB ) =⇒ mA vB2 = −vB
R dt = mB (vB − u) =⇒ mB vB2 = vB
(5.27)
t1
t2
R dt + mA uvA
(5.28)
D dt + mB uvB
(5.29)
tm
tm
t1
t2
R dt + mB uvB
tm
(5.30)
154
CHAPTER 5. IMPULSE AND MOMENTUM
Combining these equations, we can compute the kinetic energy before the impact, T , and after
the impact, T , viz.,
T
T
=
1
1
mA vA2 + mB vB2 = (vA − vB )
2
2
=
1
1
mA vA2 + mB vB2 = (vB − vA )
2
2
tm
t1
t2
tm
1
D dt + u (mA vA + mB vB ) (5.31)
2
1
R dt + u (mA vA + mB vB ) (5.32)
2
Appealing to momentum conservation [Equation (5.16)], the definition of the coefficient of
restitution [Equation (5.20)], and the impact relation [Equation (5.25)], tells us that
tm
mA vA + mB vB = mA vA + mB vB ,
tm
R dt = e
t1
D dt,
t1
vB − vA = e (vA − vB )
(5.33)
Therefore, the kinetic energy after the impact can be rewritten as
T =
1 2
e (vA − vB )
2
tm
t1
1
D dt + u (mA vA + mB vB )
2
(5.34)
Subtracting Equation (5.34) from Equation (5.31) shows that the loss of kinetic energy is
T −T =
1
1 − e2 (vA − vB )
2
tm
D dt
(5.35)
t1
The impulse of the deformation force, D, is inherently positive. Also, we have chosen vA to
be positive and greater than vB , so that (vA − vB ) is positive. Consequently, the sign of the
kinetic-energy loss depends upon the sign of the factor 1 − e2 . We can use this information
to establish the upper and lower bounds for the numerical value of the coefficient of restitution.
Upper Bound. The upper bound for the coefficient of restitution is obviously e = 1. If it
were greater than one, two colliding particles would gain energy as a result of the impact,
which is physically unrealistic. At best, all of the kinetic energy lost during the deformation
phase will be recovered during the restitution phase. In the limiting case e = 1, we describe
the collision of two particles as a perfectly-elastic impact.
Lower Bound. We can establish a lower bound on e by reasoning as follows. Obviously,
Equation (5.35) tells us that the amount of kinetic energy lost in an impact increases as e
decreases. By definition, e is the ratio of the impulse of the restitution force, R, to the impulse
of the deformation force, D. Thus, as e decreases, the amount of energy recovered during
the restitution phase decreases. We can place a lower bound of e = 0, which corresponds
to none of the kinetic energy lost during the deformation phase being recovered. Inspection
of Equation (5.24) shows that in this case, the particles move as a single larger particle with
the same speed after the impact. In the limiting case e = 0, we describe the collision of two
particles as a perfectly-plastic impact.
Summarizing, we conclude that
0≤e≤1
All of the data in Figure 5.7 and Table 5.1 fall within this range.
(5.36)
5.3. DIRECT CENTRAL IMPACT
155
Example 5.3 When a Major League “slugger” swings a bat, the speed of the bat, va , varies with
the bat’s weight, ma g, according to va = 75 − 0.42ma g, where va is expressed in mph and ma g
in ounces [Bahill and Freitas (1995)]. Determine the speed, vb , of line drive hits by Hank Aaron,
Roberto Clemente and Babe Ruth for an incoming pitch with speed vb = 90 mph.
Roberto Clemente
ma g = 42 oz
Hank Aaron
ma g = 32 oz
Babe Ruth
ma g = 47 oz
Solution. In hitting a line drive, we will assume the batter swings level at the same height above
the ground as the pitched ball and that the ball leaves the bat moving parallel to the ground.
...........
.
.....................
... ..... .......... va..
.. ..................................
ma ................. ~
..
.. .
... ...
.... ............ .....
................
..................
......
.....
b
... .
... ....
..................................................... ..
.. ..
... ..
...
......
...................
v
mb
Momentum Conservation. This is a direct central impact. Noting that the baseball initially moves
to the left so that its momentum is negative, we have
ma va + mb (−vb ) = ma va + mb vb
=⇒
mb vb + ma va = ma va − mb vb
Impact Relation. Again, accounting for the fact that the ball initially moves toward the bat, the
impact relation is
vb − va = e [va − (−vb )]
=⇒
vb − va = e (va + vb )
Multiplying the impact relation through by ma , adding to the momentum equation, and dividing
the resulting equation through by ma + mb , the speed of the ball after the impact is
vb =
(1 + e)ma
ema − mb
va +
vb
ma + mb
ma + mb
An official Major League baseball weighs mb g = 5 18 oz and reference to Figure 5.7 and Table 5.1
shows that its coefficient of restitution for relative impact speeds in excess of 100 mph is e = 0.45.
Also, from the given formula, the initial bat speeds, va , for each of the three Major Leaguers are
va =
⎧
⎪
⎨ 61.56 mph, Hank Aaron
⎪
⎩
57.36 mph,
Roberto Clemente
55.26 mph,
Babe Ruth
Substituting these values into the equation developed above for vb yields the following.
vb =
⎧
⎪
⎨
⎪
⎩
99.5 mph,
100.4 mph,
99.9 mph,
Hank Aaron
Roberto Clemente
Babe Ruth
Remarkably, the ball’s speed rounds to 100 mph for all three players.
156
CHAPTER 5. IMPULSE AND MOMENTUM
5.4 Oblique Central Impact
Because at least one of the particles involved in an oblique central impact moves along a path
other than the line of impact, we must alter the approach developed in the preceding section.
Specifically, we must take account of the fact that the Principle of Impulse and Momentum is
a vector equation. Thus, we consider the components of the equations of motion parallel to
and normal to the line of impact. Referring to Figure 5.8, we select a coordinate system in
which the t axis is parallel to the line of impact and the n axis is normal to the line of impact.
We represent the velocity vectors of Particles A and B by
vA = vAt et + vAn en
vB = vBt et + vBn en
and
(5.37)
where et and en are unit vectors parallel to and normal to the line of impact, respectively.
....
.......
......
.
........
....................... ... .............................
.
.
.
.
.
.
............
.....
............
...
..............
...
...
....
B
...
.
..
.
.
A ..
B
...
.
...
.. .. .. .. .. .. .. .. .. ....... .. .. .. .. .. .. .. .. .. ...... .. .. .. .. .. .. .. .. .. ....... .. .. .. .. .. .. .............
...
..
....
...
..
.
.
.
.
...
.....
...
.....
....................
.....
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.............
.............
.........
...
A............
..
...
..
......
n
m
m
A
B
v
t
en ...............
.
....
...
.................................
et
v
Figure 5.8: Oblique central impact.
As with direct central impacts, we assume that the deformation and restitution forces acting
on the particles are parallel to the line of impact. We ignore any friction forces between the
particles resulting from relative motion normal to the line of impact. We do this because such
forces will be trivially small compared to the deformation and restitution forces and will thus
be effectively non-impulsive.
Thus, the equations we developed for direct central impacts still hold along the line of
impact, without modification. That is, tangential momentum is conserved, and the impact
relation holds for the tangential velocity components. Turning to momentum normal to the
line of impact, the sum of the two particles’ momenta is conserved. However, we can make
an even stronger statement regarding the individual particles. That is, since both particles are
subject to neither an external nor an internal force along the n axis, their individual momenta
will also be conserved. Therefore, we implement the following equations for an oblique central
impact.
Conservation of Tangential Momentum. Because no external forces act on the particles,
mA vAt + mB vBt = mA vAt + mB vBt
(5.38)
where mA and mB denote the masses of Particles A and B, respectively. Also, vAt and vBt
are the tangential components of the velocity vectors after the impact for Particles A and B,
respectively.
Conservation of Normal Momentum. Because the momentum of each particle is conserved,
and since each particle’s mass does not change as a result of the impact, the normal velocity
components for both particles must be unchanged. Therefore, we have
vAn = vAn
and
vBn = vBn
(5.39)
Impact Relation. As with direct central impacts, the impact relation is
vBt − vAt = e (vAt − vBt )
(5.40)
5.4. OBLIQUE CENTRAL IMPACT
157
Equations (5.38) through (5.40) are sufficient to solve oblique central impact problems. However, the question of where the line of impact lies must be established before the solution
can proceed. This is a nontrivial issue because, in general, it will depend on the shapes and
constitution of the impacting objects. Obviously, we can solve problems for which the line of
impact is specified. As shown in the following example, there are some problems for which
the line of impact can be deduced from specified velocities before and after the impact.
Consider the game of billiards depicted in Figure 5.9. The cue ball impacts the eight ball
as shown. Both balls have mass m, and we ignore effects of friction. The cue ball has an
initial speed u, and moves on a path at an angle φ to the horizontal, while the eight ball is
initially at rest. After the impact, the eight ball has speed v and goes into the corner pocket.
The angle between its path and the horizontal is θ. The cue ball moves toward the cushion of
the pool table after the impact. The angle between the paths followed by the balls is ψ. Our
goal is to determine the angle ψ and the speed of the cue ball after the impact.
..
...
...
..
... ...........
..
..
......
...........
......
.
.
.
.
.
.
.
.
.
.
.
.
.
.. .....
....
.. ........................ ...
. ......
..8.....
...
.
.
.
...
...
...
...
.
.
.
...
....
..............
..... .
.
.
.
.....
.....
.....
........
..... ....
.
.
.
... ......................
θ
ψ
t
|
v
u
φ
l
|
Figure 5.9: Motion of balls in a game of billiards.
Kinematics. The first step in the solution to this problem is establishing the line of impact.
We can do this by focusing on the eight ball, noting that its velocity components normal to
the line of impact before and after the impact must be equal. Since the ball is initially at rest,
its normal velocity component must be zero after the impact. Therefore, the line of impact is
parallel to the eight ball’s post-impact velocity vector. Figure 5.10 shows the line of impact
at the instant that the cue ball impacts the eight ball.
Inspection of Figure 5.10 shows that if we express the initial velocity of the cue ball in
terms of the t and n axes, we can do all of our computations in terms of those axes. With et
and en denoting unit vectors along the t and n axes, respectively, the velocity vectors before
and after the impact are as follows.
vC
v8
vC
v8
=
=
=
=
(Cue ball before impact)
(Eight ball before impact)
(Cue ball after impact)
(Eight ball after impact)
u cos(φ − θ)et + u sin(φ − θ)en ,
0,
vCt et + vCn en ,
v et ,
.....
.........
.........
.......
......
..
............. ..
..
...................
.
.
.
.
..
.
.
.
.
.
.
.
..
.......
..............
. ...... ................................
.....
....... 8
.....
...... ...
.....
.....
..
....... .........................
.
..
... ........
....
..
. ..
......
..
......
......
..
.....
......
.
.........
.
.
.
.
. .
..................................
n
Line of
impact
v
v
n ~
~
φ−θ
t
θ
u
φ
Figure 5.10: Line of impact for the cue ball and the eight ball.
(5.41)
(5.42)
(5.43)
(5.44)
158
CHAPTER 5. IMPULSE AND MOMENTUM
Tangential-Momentum Conservation. The t component of the momentum equation is
mvCt + mv8t = mvCt + mv8t
(5.45)
Dividing through by m and making use of Equations (5.41) through (5.44) tells us that
u cos(φ − θ) + 0 = vCt + v
(5.46)
Therefore, the cue ball’s tangential velocity component after the impact is
vCt = u cos(φ − θ) − v
(5.47)
Normal-Momentum Conservation. Because the line of impact is parallel to the eight ball’s
velocity vector after the impact, and since the eight ball was initially at rest, we automatically
satisfy continuity of the eight-ball’s normal velocity components. For the cue ball, we have
vCn = vCn
(5.48)
Making use of Equation (5.41), we conclude that the cue ball’s normal velocity component
after the impact is
(5.49)
vCn = u sin(φ − θ)
Impact Relation. The impact relation is
v8t − vCt = e [vCt − v8t ]
(5.50)
Substituting known quantities from Equations (5.41) through (5.44), we find
v − vCt = e [u cos(φ − θ) − 0]
(5.51)
Rearranging terms, we arrive at a second equation for the cue ball’s tangential velocity component after the impact, viz.,
(5.52)
vCt = v − eu cos(φ − θ)
Final Computations. We now have sufficient information to complete the solution to this
problem. First, substituting for vCt from Equation (5.47) into Equation (5.52) yields
u cos(φ − θ) − v = v − eu cos(φ − θ)
(5.53)
Solving for v in terms of the given quantities, there follows
v=
1+e
u cos(φ − θ)
2
(5.54)
Substituting the solution for v into Equation (5.52) gives the tangential component of the cue
ball after the impact. The solution is
vCt =
1−e
u cos(φ − θ)
2
(5.55)
Therefore, the velocity vector of the cue ball after the impact is
vC =
1−e
u cos(φ − θ)et + u sin(φ − θ)en
2
(5.56)
5.5. CONSTRAINED MOTION
159
v
Ct
...
.... .. .. ..................
.......
........
.
C
...
..
...
...
.. .........................................
..
.
.
...
...............
... ...
.. ............................. π
... ..........
2
Cn .................
. .
... ....
... ............
.............................
.. .. ......
.....
....
... ....
.....
...
.....
......
...
.....
.
....
.
.
.
.
..
.....................................................................................................................................................................................
.
v
v
Cue-ball velocity
after the impact
−ψ
Eight-ball velocity
after the impact
ψ
v8
Figure 5.11: Velocity-vector geometry.
The final quantity we must determine is the angle ψ between the velocity vectors of the
cue ball and the eight ball after the impact. The simplest way to do this is to observe from
the geometry (Figure 5.11) that
tan
π
v
− ψ = Ct
2
vCn
(5.57)
Noting that tan(π/2 − ψ) = cot ψ and substituting for the components of the cue ball’s
velocity after the impact, we obtain
1−e
u cos(φ − θ)
1−e
cot ψ = 2
=
cot(φ − θ)
u sin(φ − θ)
2
(5.58)
The solution for the angle ψ is
ψ = tan−1
2
tan(φ − θ)
1−e
(5.59)
5.5 Constrained Motion
In some applications, one or even both of the impacting objects may be constrained in their
motion. For example, consider a ball moving vertically downward toward a cart as shown
in Figure 5.12. The horizontal surface is immovable and thus exerts a reaction force, N, on
the cart. While the ball is free to move in the plane of the figure, the cart can only move
z ...........
j
z
..
...
...
..
...
...
...
.
...
.
.
..........
.....
.
...
.
.
.
.. b
...
... ....
...
....
.
... ...
....
...
...
... ...
....
.
.
..
...
.
.
.
... ..
..
a
...
... ..
...
.................... Line of impact
.......................................................
..... .......
...
........
....
...
.......
..
.
..
.
.
.
... ....
...
... ......
...
a
...
.
............................................
..
...
....
..
..
.......
...
..........
...
.......
...
.
.
...
...
........ ..b
..
...
.
.
.
.............
.
......
...
...............................................................................
.
...
.
....................................................•
..........................................................................
...........................................•
.
......................................................................................................................................................
..
v
v
v
j
z
N
e
u
e
u
v
x
Figure 5.12: Constrained-motion example.
160
CHAPTER 5. IMPULSE AND MOMENTUM
horizontally. On the one hand, a component of the reaction force acts in the direction of the
line of impact during the impact process. Therefore, momentum is not conserved along the
line of impact. On the other hand, the reaction force will be tiny compared to the deformation
and restitution forces, and will thus be effectively non-impulsive. Hence, the impact relation
will be valid and momentum in the direction normal to the line of impact will be conserved.
Thus, the solution to an oblique central impact problem with constrained motion is very
similar to the free-motion case discussed in the preceding subsection. The only change is
in the momentum-conservation equation. While momentum along the line of impact is not
conserved, there will be a direction in which momentum is conserved. For the example of
Figure 5.12, there are no external forces acting in the x direction so that x momentum is
conserved. Combined with the impact relation and normal-momentum relations, the solution
can be obtained. The following example illustrates how we solve an oblique central impact
with constrained motion.
Example 5.4 If the mass of Block A is λm and the mass of Ball B is m, determine the value of
the constant λ for which the block will be at rest after the impact. Ignore effects of friction.
......
.......
...
.........................
..............................
.
...
...
....................
....
...
.....
... ...
...
......
...
..... ..
.....
..
.
.
.
.
.
.
.........
...
...
...................................................................
.........
...
...
...
...
.......
...
.....
.......
...
...
.......
...
.........
.
.
.
.
.
.
.
...
.
.
.
.
... ..
....
...
...
... ... ................
............. ..
...
...
................................
...........................................
...
...
..
...
.
...
...
....
...
...
...
......
π
...
...
.... ...
.
.. ....
...
.
.
.
.
.
...........................................................................................2
.
...............................................................................................................................
z ............
n
v
Ball B
n
m~
Block A
2θ
v
t
θ
λm
−θ
x
Solution. On the one hand, since the block is constrained to move in the x direction, one of the
relevant equations we will introduce is for horizontal-momentum conservation. On the other hand,
the impact relation holds along the line of impact, which we assume is perpendicular to the block’s
slanted face. This means we must work in both the xz and nt coordinate systems.
Kinematics. To simplify our analysis, we will first develop relations between unit vectors in the
xz and nt coordinate systems. From inspection of the geometry, clearly
et = cos θ i + sin θ k, en = − sin θ i + cos θ k, i = cos θ et − sin θ en , k = sin θ et + cos θ en
The velocities prior to and after the impact are
vA = v i, vB = −v cos 2θ i − v sin 2θ k, vA = vA i, vB = vBx i + vBy k
Substituting for i and k, a simple computation shows that
vA = v cos θ et − v sin θ en , vB = −v cos θ et − v sin θ en
vA = vA cos θ et − vA sin θ en , vB = vBt et + vBn en
Horizontal-Momentum Conservation. Because no external forces act in the x direction, horizontal
momentum is conserved, viz.,
mA vAx + mB vBx = mA vAx + mB vBx
Noting that mA = λm and mB = m, substituting for the given velocities before the impact, and
dividing through by m, we find
λv − v cos 2θ = λvA + vBx
5.6. MIXING SOLUTION METHODS
161
Example 5.4 (Continued)
Normal-Momentum Conservation. The ball’s momentum normal to the line of impact is conserved, wherefore
vBn = −v sin θ
Impact Relation. The impact relation holds along the line of impact, which tells us that
vBt − vAt = e (vAt − vBt )
=⇒
vBt − vAt = e [v cos θ − (−v cos θ)]
Noting from the kinematics relations that vAt = vA cos θ, this simplifies to
vBt = vA cos θ + 2ev cos θ
Final Computations. At this point, we have a sufficient number of equations to complete the
solution. Since both vBx and vBt appear in these equations, we need to determine one as a function
of the other. We can compute vBx from the unit vectors, viz.,
vBx = i · vBt et + vBn en = vBt cos θ − vBn sin θ
We have shown from the impact relation and normal-momentum conservation that
vBt = vA cos θ + 2ev cos θ
Consequently, vBx is
and
vBn = −v sin θ
vBx = vA cos2 θ + v 2e cos2 θ + sin2 θ
Substituting into the horizontal-momentum conservation equation yields
λv − v cos 2θ = λvA + vA cos2 θ + v 2e cos2 θ + sin2 θ
After a little rearrangement of terms, we conclude that
vA =
λ − cos 2θ − sin2 θ − 2e cos2 θ
v
λ + cos2 θ
Finally, using the trigonometric identity cos 2θ = cos2 θ − sin2 θ, the solution for the block’s speed
in the x direction is
λ − (1 + 2e) cos2 θ
v
vA =
λ + cos2 θ
We are seeking the value of the constant λ for which the block is at rest after the impact, i.e., the
value that makes vA = 0. Therefore, λ is
λ = (1 + 2e) cos2 θ
5.6 Mixing Solution Methods
We have now completed formulation of three methods for solving typical dynamics problems.
All are based on Newton’s Second Law, which we focused on in Chapter 3. In Chapter 4, we
developed and applied the Principle of Work and Energy. Finally, in this chapter we explored
the Principle of Impulse and Momentum. In summary the three methods are as follows.
• Newton’s Law in Differential Form. We solve for the motion of a particle by solving
the differential equation md2 r/dt2 = F. In principle, direct solution permits solving for
complete details of a particle’s motion.
162
CHAPTER 5. IMPULSE AND MOMENTUM
• Principle of Work and Energy. By integrating Newton’s Second Law over distance
we relate the work done by a force to the change in a particle’s kinetic energy. When the
forces acting are conservative, the principle simplifies to conservation of total energy.
This method involves a scalar algebraic equation and can be used to relate properties at
two positions.
• Principle of Impulse and Momentum. By integrating Newton’s Second Law over
time we relate the impulse of a force to the change in a particle’s momentum. This
method involves a vector algebraic equation and can be used to relate properties at two
times.
In general, we can use any or all of these solution methods to solve parts of a given problem.
That is, we can tackle dynamics problems by mixing solution methods.
As an example, consider the problem for which Block A of mass 3m drops from a height
h onto a pan of mass m as depicted in Figure 5.13. The pan is supported by a spring of
constant k. The object of the problem is to determine the maximum deflection of the pan, H,
for a perfectly-plastic impact. To solve this problem, we implement a three-step procedure.
In the first step, we use Newton’s Second Law in differential form. Next, we treat the
impact part of the problem using the Principle of Impulse and Momentum. Finally, we use
mechanical-energy conservation to complete the solution.
From Initial State to Impact. During the block’s movement from its initial position, the only
external force acting is gravity. We will use Newton’s Second Law to compute the block’s
speed at the moment of impact with the pan, vA . So, we must solve the following initial-value
problem.
d2 z
m 2 = −mg,
z(0) = h, ż(0) = 0
(5.60)
dt
Integrating over time and using the initial conditions, the block’s position and velocity are
1
z(t) = h − gt2
2
and
v(t) = −gt k
The block impacts the pan when z = 0, which occurs at t =
downward speed at the moment of impact with the pan is
vA = g
2h
=
g
(5.61)
2h/g. Therefore, the block’s
(5.62)
2gh
...
..
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...
...
...
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..............................................................................................
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...
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......................................
A
3m
h
vA
A
•
m
•
k
z
k
•
Initial state
(at rest)
A
•
u
k
•
Beginning of
the impact
k
•
End of the
deformation phase
A
•
•
Maximum
deflection
Figure 5.13: Block impacting a pan supported by a spring.
H
5.6. MIXING SOLUTION METHODS
163
Impact. This is a straightforward direct central impact problem. Because gravity is effectively
non-impulsive, momentum is conserved in the vertical direction from the beginning of the
deformation phase to the end of the restitution phase. Thus, we have
3mvA + m · 0 = 3mvA + mvB
(5.63)
where vB is the pan’s speed after the impact. The impact relation tells us that
(5.64)
vA − vB = e (0 − vA )
But, the impact is perfectly plastic so that e = 0. So, we conclude immediately that vA = vB .
Calling the common value u, the momentum-conservation equation simplifies to
3mvA = 4mu
=⇒
u=
3
vA
4
(5.65)
Substituting for vA from Equation (5.62), the common speed of the block and pan at the end
of the deformation phase is
3
2gh
(5.66)
u=
4
From Impact to Maximum Deflection. As the block and pan move downward, there are
two external forces acting, viz., gravity and the spring force. Since both of these forces are
conservative, total energy is conserved during this final part of the motion. The total energy
is E = T + Vg + Vs , where Vg and Vs are potential energy of the gravity and spring forces,
respectively. Selecting z = 0 as the initial position of the pan, the potential energy of the
spring is
1
(5.67)
Vs = k (z − zo )2
2
where zo is the value of z for the undeformed spring. We will determine the value of zo after
we develop an equation based on energy conservation. First, we compute the total energy at
the beginning of this phase of the motion.
E1 =
1
1
(4m)u2 + 4mg · 0 + kzo2
2
2
(5.68)
Using Equation (5.66), the total energy is
E1 =
9
1
mgh + kzo2
4
2
(5.69)
At maximum deformation, the pan and block are at rest and the location of the pan is z = −H.
Hence, the total energy is
E2 =
1
1
1
2
2
(4m) · 02 + 4mg(−H) + k (−H − zo ) = −4mgH + k (H + zo )
2
2
2
(5.70)
Since E1 = E2 , we can combine Equations (5.69) and (5.70) and rearrange terms to obtain
1
9
2
k (H + zo ) − zo2 − 4mgH − mgh = 0
2
4
(5.71)
Using the fact that (H + zo )2 − zo2 = H (H + 2zo ), conservation of energy for the final phase
of the motion simplifies to
H 2 + 2zo −
8mg
k
H−
9mg
h=0
2k
(5.72)
164
CHAPTER 5. IMPULSE AND MOMENTUM
Finally, we must determine the initial displacement of the spring, zo . Before the block impacts
the pan, the spring force balances the pan’s weight. Thus,
mg = −k (0 − zo )
=⇒
zo =
mg
k
(5.73)
Using this result in Equation (5.72), a bit of straightforward algebra leads to
H2 −
6mg
9mg
H−
h=0
k
2k
(5.74)
Finally, solving this quadratic and rejecting the negative root, which yields a negative value
for H, the maximum-deflection distance is
H =3
mg
1+
k
1+
kh
2mg
(5.75)
As a final comment, note that we could have used energy conservation in the initial phase
of the solution. This shows that, depending upon what information is needed, the various
solution methods are often interchangeable.
Chapter Summary
Key topics discussed in this chapter...
• Impulse. The impulse a force exerts on a particle is equal to the integral over time of
the force from the beginning to the end of the motion.
• Principle of Impulse and Momentum. This basic momentum principle says that the
impulse a force exerts on a particle is equal to its change in linear momentum.
• Impact. An impact occurs when two objects collide and the collision is such that they
exert large forces on each other for a very short time interval. If both objects move
parallel to the line of impact, it is a direct central impact. If at least one object moves
along a path other than the line of impact, it is an oblique central impact.
• Period of Deformation. In an impact, particles deform during the period of deformation. At the time of maximum deformation, which marks the end of the period of
deformation, the particles are moving with a common velocity.
• Period of Restitution. In an impact, particles move apart during the period of restitution and approach a new equilibrium state.
• Coefficient of Restitution. This parameter, denoted by e, is the ratio of the impulse of
the restitution force to the impulse of the deformation force. Its value lies between 0
and 1.
• Perfectly-Plastic Impact. There is no restitution period and the impacting particles
move together after the impact. A perfectly-plastic impact corresponds to e = 0.
• Perfectly-Elastic Impact. All kinetic energy lost in the deformation phase is regained
in the restitution phase of an impact between two particles. A perfectly-elastic impact
corresponds to e = 1.
CHAPTER SUMMARY
165
Important equations introduced in this chapter...
• Impulse: Equation (5.3)
t2
Imp1−2 ≡
F dt
t1
• Principle of Impulse and Momentum: Equation (5.4)
Imp1−2 = mv2 − mv1
• Impact Relation: Equation (5.25)
vB − vA = e (vA − vB )
• Coefficient of Restitution: Equation (5.26)
e=
Relative speed after impact
vB − vA
=
vA − vB
Relative speed before impact
• Oblique-Impact Equations: Equations (5.38), (5.39) and (5.40)
mA vAt + mB vBt = mA vAt + mB vBt
vAn = vAn
and
vBn = vBn
vBt − vAt = e (vAt − vBt )
166
CHAPTER 5. IMPULSE AND MOMENTUM
Problems
5.1 A pile driver of mass m drops from a height h. The time required to stop the pile driver after it
impacts the pile is ∆t. Determine the magnitude of the average pile-driver force, F , as a function of
1
m, ∆t, h, and gravitational acceleration, g. Compute F for m = 500 kg, h = 8 m and ∆t = 20
sec.
...................
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...................................................................
.....................................................................
.
.
....................................................................
.....................................................................
................
m
g = −g k
h
Problem 5.1
5.2 For a baseball-bat collision, Cross (1999) has found from measurements that the force during the
impact is F (t) = Fmax sin2 (πt/τ ), where τ is the total time of contact and Fmax is the maximum
force during the impact. If τ = 0.0007 sec and the average force is F = 4560 lb, what is Fmax ?
5.3 A block of mass m starts from rest on a frictionless horizontal plane under the action of a force
F = Fo 3t/τ − 5(t/τ )2 , where t is time, τ is a constant time scale and Fo is a constant force scale.
Determine the maximum speed, vmax , that the block achieves as a function of Fo , m and τ .
5.4 A glass cube of mass m is resting on a horizontal glass surface. At time t, a horizontal force,
F = 4mgt/τ , acts on the cube, where g is gravitational acceleration and τ is a constant time scale.
Determine the cube’s speed, v, at time t = 2τ as a function of g and τ . HINT: Take account of both
static and kinetic friction.
5.5 A block of mass m is resting on a frictionless horizontal plane. At time t, a constant horizontal force
of magnitude Fo to the right acts on the block for a time 2τ , where τ is a constant time scale. At time
t = 300τ , a constant horizontal force of magnitude Fo to the left acts on the block for a time τ . What
is the block’s speed, v, at time t = 400τ ? Express your answer in terms of Fo , m and τ . Compute v
for Fo = 50 N, τ = 10 msec and m = 1 kg.
F
....
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...........................................................
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2τ
F
τ
0
m
d
t
300τ
−F
t
400τ
Problem 5.5
5.6 A player hits a tennis ball of mass m with initial horizontal speed v at a height h. The ball bounces
at Point A and rises to a maximum height 23 h where the ball’s horizontal speed is 59 v. The duration of
the ball’s impact at Point A is τ .
(a) Determine the velocity at Point A just before, v, and just after, v , the impact. Express your
answers in terms of v, h and g, the acceleration of gravity.
(b) Using the Principle of Impulse and Momentum, determine the average impulsive force vector, F,
exerted on the ball at Point A.
(c) Compute v, v and F for mg = 2 oz, v = 54 ft/sec, h = 4.5 ft and τ = 4 msec.
.. .............................................................
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..............
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.
h
f
v
v
g
f
v
h
A
Problem 5.6
5
v
9
PROBLEMS
167
5.7 Two identical balls of mass m and coefficient of restitution e approach each other with velocities
v1 = v i and v2 = −u i. Ignore effects of friction.
(a) Determine the balls’ velocities after the impact, v1 and v2 , in terms of u, v and e.
(b) If v = 54 u and v1 = − 11
16 v2 , what is e?
.........
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2 .....
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m
v
v
m
Problems 5.7, 5.8
5.8 Two identical balls of mass m and coefficient of restitution e approach each other with velocities
v1 = v i and v2 = −u i. Determine the balls’ velocities after the impact, v1 and v2 , in terms of u, v
and e. Ignore effects of friction. Compute v1 and v2 for v = 6 m/sec, u = 8 m/sec and e = 0.8.
5.9 A series of n identical balls of mass m lies on a frictionless surface. Ball 1 has initial speed v1 = v
and all of the other balls are initially at rest. Ball 1 collides with Ball 2, which in turn collides with
Ball 3, etc. The coefficient of restitution for all of the balls is e. Determine vn , the velocity of Ball n,
in terms of v, e and n. Assuming e = 0.9, compute vn /v for n = 5, 10 and 20.
vv
f
..............................
n
...............1
........................................2
....................................3
....................................4
................................................................................
.
..
............................................................................................................................................................................................................
f
v
f
v
···
f
v
f
v
Problem 5.9
5.10 A block of mass m1 = 4m is moving to the right with speed v1 = v and a block of mass m2 = 5m
is moving to the left with speed v2 = 32 v. The coefficient of restitution is e. Ignoring effects of friction,
determine the velocity vectors for the blocks immediately after impact, v1 and v2 , in terms of v and e.
Compute the velocity vectors for e = 0.6 and v = 9 ft/sec.
........................................
...
...
...
...
...
...
.
........................
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2 .....
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2
.
..........................1
.
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1 ....
...
.
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................................................................................................................................................................................................................
.
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.............................................................................................................................................................................................................
m
v
v
m
Problems 5.10, 5.11
5.11 A block of mass m1 = m is moving to the right with speed v1 = v and a block of mass m2 = 2m
is moving to the left with speed v2 = 12 v. The coefficient of restitution is e. Ignoring effects of friction,
determine the velocity vectors for the blocks immediately after impact, v1 and v2 , in terms of v and e.
Compute the velocity vectors for e = 0.4 and v = 6 m/sec.
5.12 Two balls of mass m1 = m and m2 = 9m and coefficient of restitution e approach each other
with velocities v1 = v i and v2 = −v i. Determine their velocities after the impact, v1 and v2 , in terms
of v and e. If |v1 | = 3|v2 |, what is e?
..................
.....
...
...
1 ....
...
.....
1 ...............................................
...
.
....
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.....................
m
v
....
........ ...........
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2
..
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.................................................
2 .....
...
.
....
.
.
....................
v
m
Problems 5.12, 5.13, 5.14
5.13 Two balls of mass m1 = m and m2 = 2m and coefficient of restitution e approach each other
with velocities v1 = v i and v2 = − 32 v i. Determine their velocities after the impact, v1 and v2 , in terms
of v and e. If v2 = 0, what are e and v1 ?
5.14 Two balls of mass m1 = 4m and m2 = 3m and coefficient of restitution e approach each other
with velocities v1 = v i and v2 = − 43 v i. Determine their velocities after the impact, v1 and v2 , in
terms of v and e. Compute the velocities for a perfectly-plastic impact, a perfectly-elastic impact, and
an impact with e = 12 .
168
CHAPTER 5. IMPULSE AND MOMENTUM
5.15 A ball initially at rest falls from a height H above a flat surface. It rebounds to a height h.
Determine the coefficient of restitution between the ball and the surface, e, as a function of h and H.
What is e if H = 25 ft and h = 16 ft?
5.16 A ball initially at rest falls from a height H above a flat surface. If the coefficient of restitution
between the ball and the surface is e, to what height, h, does it rebound on the first, second and nth
bounces? Compute h/H on the tenth bounce for e = 0.9.
5.17 Three identical spheres of mass m and coefficient of restitution e are initially arranged as shown.
All three spheres are attached to the upper wall with cords of the same length. The distance between
Spheres B and C is negligibly small compared to sphere diameter. Sphere A is released and we wish
to determine the value of e based on C , the maximum height reached by Sphere C.
(a) Determine the horizontal speed of Sphere A, vA , at the instant when it strikes Sphere B. Express
your answer in terms of length and gravitational acceleration, g.
(b) Compute the speeds of Sphere A and Sphere B, vA and vB , respectively, immediately after the
impact between Spheres A and B. Express your answers in terms of , g and e.
(c) Compute the speeds of Sphere B and Sphere C, vB and vC , respectively, immediately after the
impact between Spheres B and C. Express your answers in terms of , g and e.
8
(d) If the maximum height reached by Sphere C is C = 81
, what is the value of e?
.................................................................................................................................
.
............................................................................................................................................
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• • •
g
y
Ai
yCi
y
Bi
Problem 5.17
5.18 A ballistic pendulum is used to measure the striking power of a bullet. As shown in the figure,
the pendulum’s length is and the mass of the bob is M . The mass of the bullet is m and its speed
just before it strikes the bob is v. The bullet is made of rubber and its coefficient of restitution is e.
After the impact, the bob rises to a maximum height h. Determine h as a function of m, M , e, v and
gravitational acceleration, g.
..............................................................................................
.
.
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•
m
v
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....•
.
.........
M
...
...
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...
..
........
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.
g
..
....
..............
h
.............
.....
.
Problems 5.18, 5.19
5.19 A ballistic pendulum is used to measure the striking power of a bullet. As shown in the figure, the
pendulum’s length is and the mass of the bob is M . The mass of the bullet is m and its speed just
before it strikes the bob is vb . The bullet embeds itself in the bob and the bullet/bob combination rises
to a maximum height h.
(a) The aerodynamic-drag force on the bullet due to friction is mv 2 /λ, where v is its speed at a
given instant and λ is a constant length scale. If the bullet is fired from a gun with speed vo
a distance L from the pendulum, what is vb ? If λ = 1000 ft, how important do you think the
effect of aerodynamic drag is in a laboratory measurement? HINT: The bullet’s acceleration is
a = vdv/dx.
(b) Determine the height h as a function of M , m, vb , and gravitational acceleration, g.
PROBLEMS
169
5.20 A block of mass 2m is moving with velocity v i when it impacts a pendulum bob of mass m
that is initially motionless. The coefficient of restitution between the block and the bob is e and the
kinetic-friction coefficient between the block and the surface is μk . You can assume that the block and
pendulum move without a secondary impact. Express your answers in terms of v, e, μk and gravitational
acceleration g.
(a) Determine the maximum height, h, to which the pendulum bob will rise.
(b) Determine the distance, d, the block will slide before coming to rest.
........................................
.
.
...
............................................
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.
.
.
.
.
.
..................................................................................................... .............................................................................................
•
g = −g k
{
v
h
{
.. .
........................
.
z ............
...
...
..
...
...
............................................
x
..
d ...........................
Problem 5.20
5.21 A ball of mass 2m drops into a frictionless tube as shown. When the ball emerges from the tube,
it impacts an initially motionless pendulum of length H with a bob of mass m. The coefficient of
restitution between the ball and the bob is e. Determine the speed of the ball and the bob immediately
after the impact. Express your answers in terms of H, e and gravitational acceleration g.
.
................
........................................
....
.
.
.
...
.........................................................
...
...........................................
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...................................................
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.. ...
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...
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.
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.
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.
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................................................................................................................................................. .... ... ..............
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..
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....
.................... ...... .........
.
.............
....
....
..
...
...
...
..
{
•
H
{
{{
h
g = −g k
z ...........
....
...
..
...
..
..........................................
x
Problems 5.21, 5.22, 5.23
5.22 A ball of mass m drops into a tube with rough inner walls as shown. When the ball emerges from
the tube, it impacts an initially motionless pendulum of length H with a bob of mass m. The coefficient
of restitution between the ball and the bob is e = 12 . The maximum height to which the pendulum bob
will rise is h = 14 H. Determine the work done by friction in the tube, Uf . Express your answer in
terms of H, m and gravitational acceleration g.
5.23 A ball of mass m drops into a frictionless tube as shown. When the ball emerges from the tube,
it impacts an initially motionless pendulum of length H with a bob of mass m. The coefficient of
restitution between the ball and the bob is e. The maximum height to which the pendulum bob will rise
is h, which is not given. Determine the speed of the bob when it has risen a distance 12 h. Express your
answer in terms of H, e and gravitational acceleration g.
5.24 Automobile A of mass mA = 4m is traveling on a horizontal road with speed vA = 2v. It
approaches Automobile B from behind. Automobile B has mass mB = 3m and speed vB = v. When
the automobiles collide, they lock bumpers and continue moving as a single object. How much kinetic
energy, ∆T , is lost as a function of m and v?
5.25 Automobile A of mass mA = 5m is traveling on a horizontal road with speed vA = 4v. It
approaches Automobile B from behind. Automobile B has mass mB = 4m and speed vB = 3v. The
coefficient of restitution is e. After the automobiles collide, Automobile B still moves with speed 3v.
What are e and the speed of Automobile A after the impact?
170
CHAPTER 5. IMPULSE AND MOMENTUM
5.26 Two identical hockey pucks moving with initial speeds vA = 65 v and vB = v collide as shown.
The coefficient of restitution of the pucks is e. Ignore effects of friction and assume the line of impact
is parallel to the x axis.
(a) Determine the velocities of the pucks after impact.
(b) Experimentation shows that Puck A is at rest after the impact when θ = 80.13o , which corresponds
to cos θ = 6/35. Using your result from Part (a), determine the value of e.
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Problems 5.26, 5.27
5.27 Two disks moving with initial speeds vA = vB = v collide as shown. The mass of Disk A is 2m
and that of Disk B is m. The coefficient of restitution of the disks is e = 12 . Ignore effects of friction
and assume the line of impact is parallel to the x axis. Determine the velocities of the disks after impact.
5.28 A ball of mass m impacts a wall as shown. The ball’s initial and final speeds are vi and vf ,
respectively. Determine the ball’s coefficient of restitution, e, as a function of angles θi and θf .
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Problem 5.28
5.29 A large sphere of mass mL = M collides with a small sphere of mass mS = 35 M as shown.
Just before the impact, the large sphere’s velocity is vL = −v j and the small sphere’s velocity is
vS = u(i + j).
(a) Determine the velocities of the spheres after the impact, vL and vS , as functions of u, v and e.
(b) Assuming e = 45 , u = 6 m/sec and v = 10 m/sec, compute vL and vS .
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Problems 5.29, 5.30
5.30 A large sphere of mass mL = M collides with a small sphere of mass mS = 45 M as shown.
Just before the impact, the large sphere’s velocity is vL = −v j and the small sphere’s velocity is
vS = v(i + j).
(a) Determine the velocities of the spheres after the impact, vL and vS , as functions of v and e.
(b) If the large sphere is at rest after the impact, what is e?
PROBLEMS
171
5.31 In a game of billiards, a player hits the cue ball at speed v parallel to the x axis as shown. After
the impact, the eight ball moves at an angle α to the x axis and, after impacting the cushion, goes into
the lower-right corner pocket. Both balls have mass m and the coefficient of restitution for the billiard
balls is e. Assume the surface is frictionless and the balls move without rolling. Determine the velocity
of the cue ball after the impact, v . Assuming α = 50o and that the angle φ at which the cue ball moves
after the impact is 35o , determine the coefficient of restitution, e.
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Problem 5.31
5.32 In a game of billiards, a player has decided to do a “combination shot” in order to “sink” the three
ball in the corner pocket. He aims the cue ball at an angle α to the side of the table and imparts a
speed v1 to it. The cue ball impacts the two ball, which moves parallel to the side of the table at a
speed v2 after the impact. The two ball then impacts the three ball, which moves at a speed v3 after
the impact, making the same angle, α, with the side of the table as shown. All balls on the table have
equal mass, m, and coefficient of restitution, e. Assume the surface is frictionless and the balls move
without rolling. Determine v3 in terms of v1 , α and e. If α = 60o and v3 = 15 v1 , what is e?
v
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Problem 5.32
5.33 In a game of billiards, a player hits the cue ball at speed v with an angle β to the x axis as shown.
After the impact, the eight ball moves at an angle α to the x axis and goes directly into the corner
pocket. Both balls have mass m and the coefficient of restitution for the billiard balls is e. Assume the
surface is frictionless and the balls move without rolling. Determine the velocity of the cue ball after
the impact, v . Assuming β = 12 α and that α
1 (in radians), determine the angle φ at which the cue
ball moves after the impact.
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NOTE:
For α
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sin α ≈ α, cos α ≈ 1
tan φ ≈ φ
A + Bαn ≈ A for any
constants A and B and
any positive integer n
Problem 5.33
172
CHAPTER 5. IMPULSE AND MOMENTUM
5.34 In a game of billiards, a player has decided to do a “bank shot” in order to “sink” the eight ball in
the corner pocket. He imparts a velocity v = v i to the cue ball, which impacts the eight ball. After the
impact, the eight ball moves at an angle α to the y axis. The eight ball then impacts the upper cushion.
After impacting the cushion, the eight ball follows the path shown and goes into the corner pocket. The
cue ball and the eight ball have equal mass, m, and coefficient of restitution, e. The impact between
the eight ball and the cushion is perfectly elastic. Assume the surface is frictionless and the balls move
without rolling. Determine the eight ball’s speed, v8 , after reflecting from the cushion in terms of v, α
and e. If the length = 32 h, what is α?
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Problems 5.34, 5.35
5.35 In a game of billiards, a player has decided to do a “bank shot” in order to “sink” the eight ball in
the corner pocket. He imparts a velocity v = v i to the cue ball, which impacts the eight ball. After the
impact, the eight ball moves at an angle α to the y axis. The eight ball then impacts the upper cushion.
After impacting the cushion, the eight ball follows the path shown and goes into the corner pocket. The
cue ball and the eight ball have equal mass, m, and coefficient of restitution, e. Assume the surface is
frictionless and the balls move without rolling. Determine the cue ball’s velocity, v , after impacting the
eight ball in terms of v, α and e. If α = 45o and e = 45 , for what length ratio, /h, will the cue ball
fall into the other corner pocket?
5.36 In a game of billiards, a player has decided to “play position” in order to “sink” the two ball in the
corner pocket in such a way that he will have an “easy shot” on the eight ball. He imparts a velocity to
the cue ball such that its speed is v just before it impacts the two ball. After the impact, the two ball
moves at an angle α = 45o to the x axis and approaches the upper corner pocket. The cue ball then
moves toward the lower cushion. If the cue ball travels a maximum distance of h before it stops rolling,
it will be in an excellent position for an easy shot on the eight ball. All balls on the table have equal
mass, m, and coefficient of restitution, e. Assume the surface has a rolling-friction coefficient of μr ,
and that rolling has no effect on the impact.
(a) Determine v in terms of e, μr , h and gravitational acceleration, g.
(b) Compute v for μr = 0.002, e =
4
5
and h = 1 ft.
(c) Based on your result of Part (b), will the two ball have sufficient speed to reach the corner pocket?
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Problem 5.36
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PROBLEMS
173
5.37 A ball of mass m rolls into a horizontal corner with initial velocity vi . After it reflects from the
two walls forming the corner, it encounters a spring. The coefficient of restitution of the ball is e and
the spring constant is k = m/τ 2 , where τ is a characteristic time.
(a) Verify that the velocity of the ball when it emerges from the corner, vo , is parallel to vi .
(b) Determine the maximum deflection of the spring, ξ, as a function of e, τ and v = |vi |.
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Problem 5.37
5.38 A ball of mass m and coefficient of restitution e is dropped from a height H = 4h above a fixed
incline of angle β = 30o to the horizontal as shown. The height of the point of impact relative to the
ground is h.
(a) Determine the ball’s vertical speed, vz , immediately after bouncing from the incline. Express
your answer in terms of e, h and gravitational acceleration, g.
(b) If the maximum height to which the ball rises after the impact is zmax = 54 h, what is e?
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Problems 5.38, 5.39
5.39 A ball of mass m and coefficient of restitution e is dropped from a height H above a fixed incline
of angle β to the horizontal as shown. The height of the point of impact relative to the ground is h.
(a) Determine the ball’s velocity, v = vx i + vz k, immediately after bouncing from the incline.
Express your answer in terms of e, H, β and gravitational acceleration, g.
5
(b) Now, assume the impact is perfectly elastic, β = 30o and h = 16
H. Compute the time at which
the ball first strikes the ground. Express your answer in terms of g and H.
5.40 A ball impacts the ground at Point A with speed v at an angle θ as shown. After the impact, the
ball reaches Point B, where its velocity vector is exactly horizontal. Its speed, u, at Point B is not
given. If the ball’s coefficient of restitution is e, determine the distances h and L. Express your answers
in terms of e, v, θ, and gravitational acceleration, g.
u
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Problem 5.40
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174
CHAPTER 5. IMPULSE AND MOMENTUM
5.41 A girl throws a ball at an inclined wall from height h, with velocity v. After bouncing off the wall,
the ball hits her on her head. Note that, because she has just bent down to tie a loose shoelace, her head
is at height h when the ball hits her. For simplicity, assume the ball travels horizontally as it approaches
the wall.
(a) If the coefficient of restitution of the ball is e, determine the velocity of the ball, v , just after it
bounces from the wall as a function of v, θ and e.
(b) Ignoring effects of friction as the ball moves, determine the time at which the ball lands on the
girl’s head as a function of v, θ, e and gravitational acceleration, g. HINT: Make use of the fact
that sin θ cos θ = 12 sin 2θ to simplify your answer.
(c) The solution developed in Part (b) is possible only if θ exceeds a special angle θmin . Explain
why and determine θmin for e = 0.4.
g
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Problem 5.41
5.42 A woman is driving her classic 1962 Jaguar Mark II on a highway at constant speed v. Suddenly,
a large rubber rock of mass m and coefficient of restitution e falls vertically and strikes the Jaguar’s
grill at speed v. The line of impact points upward at an angle of 45o to the road. If the Jaguar’s mass
is M , what is its speed, v , immediately after the impact as a function of M , m, e and v? Compute v
for M = 1510 kg, m = 100 kg, e = 0.7 and v = 40 km/hr.
Problem 5.42
5.43 A bullet of mass mB is fired into a wooden block of mass mA and becomes embedded in it. The
block and bullet then move up the frictionless incline of angle θ for a time τ before they come to a stop.
(a) Determine the magnitude of the bullet’s initial speed, vo , as a function of mA , mB , τ , θ and
gravitational acceleration, g.
(b) Using the Principle of Impulse and Momentum, compute the magnitude of the impulse of the
force exerted by the bullet on the block, |Imp|, as a function of mB , g, τ , θ, and vo .
(c) Compute vo and |Imp| for mB g = 1 oz, mA g = 8 lb, τ = 1.2 sec and θ = 15o .
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Problem 5.43
g
PROBLEMS
175
5.44 A sphere is moving with speed v as shown, and it impacts the inclined face of a wedge. The
sphere’s mass is m and the wedge’s mass is λm, where λ is a constant. The wedge is initially at rest
and is free to move horizontally without friction. After the impact, the sphere moves vertically upward.
(a) Determine the wedge angle, θ, as a function of λ and e, the sphere’s coefficient of restitution.
(b) Compute the kinetic energy change, ∆T . NOTE: Your answer should depend only upon e, m
and v.
..................................
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Problems 5.44, 5.45
5.45 A sphere is moving with speed v as shown, and it impacts the inclined face of a wedge. The
sphere’s mass is m and the wedge’s mass is λm, where λ is a constant. The wedge is initially at rest
and is free to move horizontally without friction. After the impact, the sphere moves vertically upward.
If λ = 5, what is the smallest wedge angle, θ, for which this motion is possible?
5.46 A ball of mass m is moving downward with velocity vb = −vb k toward a cart of mass m that is
moving to the left with velocity va = −va i. The coefficient of restitution between the ball and the cart
is e. Ignoring effects of rolling friction, determine the velocity of the cart, va , after the ball strikes the
cart. Express your answer in terms of va , vb , e and the angle θ. HINT: After setting up the equations
that govern the motion, solve for vbx as a function of vbt , vb and θ. Then, complete the solution.
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u
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u
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x
Problem 5.46
5.47 Ball A of mass m strikes pendulum Bob B of mass 2m with a velocity v as shown. Assuming
the collision is perfectly elastic and that all effects of friction can be ignored, determine the height, h,
reached by Bob B. Express your answer as a function of v = |v|, θ and gravitational acceleration, g.
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Problem 5.47
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176
CHAPTER 5. IMPULSE AND MOMENTUM
5.48 A block of mass M rests on a horizontal surface. The block has a frictionless slot inclined to
the horizontal at an angle θ as shown. A small block of mass m starts from rest and slides down the
slot through a vertical distance h. The coefficient of restitution of the two blocks is e. Determine the
velocity of the large block immediately after the impact. Express your answer in terms of M , m, e, h,
θ and gravitational acceleration, g.
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Problems 5.48, 5.49
5.49 A block of mass M rests on a horizontal surface. The block has a slot inclined to the horizontal
at an angle θ as shown. A small block of mass m starts from rest and slides down the slot through a
vertical distance h. The coefficient of kinetic friction in the slanted slot is μ. When the small block
reaches the bottom of the slot, both blocks move in unison. Determine the velocity of the blocks after
the impact. Express your answer in terms of M , m, μ, h, θ and gravitational acceleration, g.
5.50 A ball of mass m is moving with velocity va = v i. It impacts a wedge-shaped block of mass M .
The block then compresses a spring of constant k through a distance δ. The surface is frictionless, and
the ball’s coefficient of restitution is e. Determine the velocities of the ball and the block, va and vb ,
respectively, immediately after the impact. Also, find δ. Express your answers in terms of m, M , e, v,
k, and the wedge angle, θ.
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Problem 5.50