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 ...... .................... ......

Boat A

T

Boat B

........................................................................................................................................................................................................................................................................................................................................................................................................................................ ................................................... ...................................................

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 .......... ............... .. ......... ....... ... .... ...... . . . . . . . . ... . .... ... .. ......... . . ......... .... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .................... .. .. .. .. .. .. .. .. .. .. .. ...... .. .. .. .. .. .. .. ......... . .. . . . . ... ... . . . . .. ... . .... . . . . ... . ... . .... .. ... ......................... ....... . .. ... ... .... ................................................................................................................................................... ..

F .......... F

t1

t2

(5.9)

.... F ........ ..... .. .. ... .... .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ................ .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. . . F .... .. ...

.... ... ... .. ... ... ... .. .. .. .. ... ... .... ... .. .. .. ... ... ... .. ... .. .. ... ... .... . . . .................................................................................................................................................

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.

•

... ....... ....... ...... ... ........ ............ ... ....... 2 ................. .. ....... 1 ... ....... .... . . . . . ........... . . . .. ....... ............... ..... ..... .... .. .... .... .... .... ... .... ...... . . . . . . . .. .. . ... .... ... .... .. ........................................................................................................................................................................................................................

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 .............

.... ... .. ... ... ... ... ... ... .... .. ... ... ... ... ... 9 ... ... 8 ... .................. ... . . ....... . ... ......... . . . ... . . .. ... ... ....... ... ....... ... .. ....... ... ..... ... .. .. . . .. ... . . ... . . ... . . ... . . . .... .. .. .. .. .. .. .. .. .. .. .............................................. ... ... ... ... ... ... ... ... ... ........................................................................................................................................................................................................................................................................................................

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 .. .. ........ Line . ... ................... ... .......................... . . . .. . . . . ............ ..... .. .... . . . . . . . . . . . . . ... . ... ... ... ... .. .. ... ..... .. .. .. .. .. .. .. .. .. ........ .. .. .. .. .. .. .. .. .. ....... .. .. .. .. .. .. .. .. .. ........ .. .. ....... .. .. .. .. . ... . . . . .. ...................................... ..................................... .... ... .. .... .... ...... ......... A B ............................ ... ................................. .. ...

A

v

B

(a) Direct central impact

v

.. ... ... ..... ........ ..................... ... ........................... . . ... . . . . ........... ..... ............ .... ... . . ... ............... . ... . B ... .. ... . . . . . . ... ... ... .... . .. .. .. .. .. ....... .. .. ........ .. .. .. .. .. .. .. .. .. ....... .. .. .. .. .. .. .. .. .. ........ .. .. .. .. .. .. .. .. . .. . ... .. . ..................................... ..... .. . . ... . .... . . . . . . . ...... . ... A ............................ ... ................................. .. .. .

of impact...........

v

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. ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ... .... .... .... ..... ................ ................. ... ... ... ... ... .......... .......... ... ... ... ... ... ... ...... ......... ...... ......... ....................... .. ....................... ... . . . . . . . . . ... . . . . . . . . . . . . . . . . . ... . . . ... ... ..... .. ... ... ... . . ... .. . . .. . ... ... ... .... .... . . . . . . ... ... ... ... . .. . ... . . . ... B ... A .. A B . . . . . . . . . ... ... . . . . . . . .... ................................ .. ..................................... .. . ... ..................................... ... ... . . .... . . . . ... .... ............................... .... ........................................ ..... . . . . . . ... .. . . . ... . . . . . . . . . . . . . . . . . ... ... . . . . . . . ... . ... ... .. .... .. ... ..... ....... ... ... ... ... .... .... .... ..... ... .... ...... ... ......... ............. ..... ...... ... .................... ... ... ... ... ... ... ... ... ...... .................... .................. .... ...... ...... ..... ... ... ... ... .......... .......... ... .... .... .... .. .. .. ... ........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................ ... .... .... .... .. . .. ... ............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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

.......................................................................................................................................................................................................................................................................................... ... ... ... ...... ...... ...... ...... ... ... ... ..... ....... ....... ....... ..... ....... ....... ........ ... ... .. ... .. ... ... ... .... ... ... .. . ... ... .. ... .. . .... . . ... ... ... . ...... ... ...... ... . ... . . ... .... ... . . . . . . . ab ba ab . . . . . . . . . ... ...........ba . .. ... ........................................................ ... ............................................................ .......................................................... ... .................................................. . ... . . ... . ... . . . ... ... . . . . . ... . . . . . ... ... ... .... .... . . . . . . . . . . . . . . ... . ... . ... . ... .... .... ... ... ... ... ... ... ... ... .. .. ..... ....... ...... ...... .... ...... ...... ...... ... . . ......... ......... ......... ........ ... ... ... ... ... ... ....................................................................................................................................................................................................................................................................................... ... ... ... ... .. ... .......................................................................................................................................................................................................................................................................................

D

A •

B •

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

................... ....... .... .... ... ... ... .... ..... ............................................................................. ... .. ... ... .... . . . . ........ ....... .........

2v

A

2m

................... ....... .... .... ... ... ... .... ..... . ........................................... ... .. ... ... ..... . . . . ........ ......... .......

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, z(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

... .. ... .......... ... ... ... ... ............ ........ .... ... ..... . ... ......................................... ......................................... ... . .. ... .. . . ..... .. ... ... . ... . . . ....... .... .. .. . ....................................................................................................................... .. .. ......................................................................................................................... ..... ....................................................................................................................... .. .. .. .. .. .. .. .. .. .. ............................................... .. .. .. .. .. .. ................ ............................................................................................. ... . ............................................................................................ ............................................................................................ . . . . . .... ..... ..... ............. ............. ............. . . . . .......................................................................................................... .......... .......... ............... ............... ............... ................................................................................................ ....... ...... . . . . . . . . . . . . . . . . . . . . . . . . ......... ......... ......... ............ . . . ........ ............... ............... ............... ................... . . . .............. ............ ............ ............ ......... . . . ............................................................................................... ............................................................................................... ............................................................................................... ............................................................................................... . .. .. .. .. ... ... ... ........................................................................................... ........................................................................................... ........................................................................................... .............................................................................................. ........................................ ... .... ... ... ... ......................................

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. ................... ... ... .. ... ... ... ... ... ... ... . . . ................. ......... ........... . . .. ...... .... .. . .. .. .. . ........ ........ ........ ................... . ................................................................... ..................................................................... . . .................................................................... ..................................................................... ................

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

.... ....... . . ............... ............... .. . .. .......... o ... ... .. .............. ............... ... .... .............................................................................................................................................................................................. ... . ... .... .. .......................................... ........................................ .... .. .. ....... .... o .. . ............................................................. ........................................................... . .... .

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. .. ............................................................. ... .. .. ..... ... ....... ... ...... .... ... ...... ..... ... ..... .. .. .. .. .. .. ................................... ..... ....... ... .. ...... .... . . . . . . . 2 .... ........ . ... . .... .... ... .... ... ...... . 3 ... ... .. ..... ... ............. ........................................................................................................................................................................................................................................ ............. .............. .... .. ... .

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? ......... ....... ......... ... 2 ..... ........................................ . .. ... ... ..... . . . ................

................... ..... ... 1 ... .. .... .............................................. ... .. .... ... . . . ..................

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. ........................................ ... ... ... ... ... ... . ........................ . 2 ..... . 2 . ..........................1 . . . ... . . ............................ .... .............................................. 1 .... ... . . ................................................................................................................................................................................................................ . . .............................................................................................................................................................................................................

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 ............................................... ... . .... . .....................

m

v

.... ........ ........... ... ... 2 .. . ................................................. 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? ................................................................................................................................. . ............................................................................................................................................ .... ... ... ...... .. ...... .... .... ... ... ....... ....... .... .... .... ... ....... . . . . . . . . ... . .. .. ..... . . . . . . . . ............. ......... ... ... .... ...... .... . .... .... . 1 ... .... .... ... ... ... 2.. ... ... ..... ... . . ............. ..............

• • •

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. .............................................................................................. . . ..................................................................................................... .... ...... . ... .... .... ... ... .... ... ... .. ... .. ... . .. ... .. .. ... .. ... .. .. .. ... . .. ... ... .. .. ..... ..... .. ..... ... ... .... ... .. ... ..... ....... .. .. .......... ...................... . .. .. ...... ... ... ...... ... ... .......................

•

m v ..... ................................ ....• .

.........

M

... ... ... ... ... .. ........ ..... .

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. ........................................ . . ... ............................................ ... ... ... .. ... ... ... ... .... ... ... ... ... .. ... ... ......... ... ... ...... ... ... . ... .... ... .. ... ... ... ... ... ....... .. .... . ... ... .. .......... ... .... ...... .. .... . ... .. .. .. ..... ..................................... . . . . . . . . . . . . . .. .................................. ... .. . ... .. ....... . ....... ...... . . . . . . . ..................................................................................................... .............................................................................................

•

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. . ................ ........................................ .... . . . ... ......................................................... ... ........................................... ... ................................................... .... .... ... ................................................. .. ... ... . ... ................................................ . ... ... .................................................. ... . ... ... ................................................ . . . ... ................................................ . . . . . . . . . . . . . .......... .......................................... ... ... . .................................................... . ... . . ................................................... . . ... . ................................................. . . ... ............................................... ... ... .... ................................................. ... ............................................ ... . . . .. .................................................. ....... .. ... .... . ... ....... ................................................................... ... ... ............................................................. ... .......... ... ................................................................. ... .... ... ... ............................................................................................ .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................................................................................................................................................. .... ... .............. ..... ....................................................................................................... .... .. .................................................................................. ............. ............... .................................................................. ......... .... .... .................... ...... ......... .

............. .... .... .. ... ... ... ..

{

•

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. .... ...... .. ......................... .... .............................. . . . . . .... .. ..... .... .. ... .. .. ... .... ....... ... .. . ....... ... A ... ......................................... .............................................. .... . . ... ..... .. . ....... .. ... . . . . .... . .. ... ...... . . . . . . . ...... ............................ ... ................................................. . ..... .... .. .. ... ....... ... ............................ B

y

v

A

B

x

θ

v

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 . ... .. ... ......... ... .. ... ... ......... ... ........ ... ..... i ........... ... ..... . . ....... ...... . . . . ....... .. . f ...... ....... ..... ......... .... .. . ..... ..... ... ... .. ... ... ... ................. ................ . . . . ... .. ... ... ... ... ... .... ... ... .. ... ... i ... . . f .. ... .. ........................... ..... .. .. ..... ... .. .. ...... .... ...................................................................................................................................................................................................................

z

v

v

~

~

θ

θ

x

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 . ... ... ... .... . ... ........... L .. ..... . ... . . . . . . . . . . . ....... .......... .... ... ... ... ..... L ... ... .. ... . .. ...... . . . . . ........................................................................................................................................................ .... .. ... ... . .... S .... ... . .... . . . .................... .. ... ........... .... ....... ..... ... ..... . . . . . ..... .... ... S ...

y ...........

v

m

m

x

v

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. ... .

... .. .... .. .. . .. .. .. .. .. . .. . .. .. . .. .. .. . .. .. . .. .. . . . .. . . . . 8 . . . . . . ..................................... ... .. .. .. .. .. .. .. .. ..... . .. ........ .. .. ........ .. ........... .. .... .. ... .. ... .. ... .. ... .. .. ... .. ... .. ... .. ... .. .. ... .. ... .. ... .. ... .. .. ... . .

α ............

y ..........

.. ..... ... ................................ ..

x

h x

v

t y v

φ

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

................ 2.... .... .. . .. ........... ..... 2 ...................................... .......... ...... . . ... 3 .. .. ................ . ... .. . . . . . . ............................. ....... . ............. ........................ ....... .. . . . . . . ........ . ... 9 .. ... 3 ....... ..................... .. ........ .. ... .. . . 1............... .. .. .. .. .. .. .. .. .. .. .. ..

f v

f v ~

v

f v

v

α

α

n ~

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. ... .

. ...

α ............

y ...........

. .... ... ............................... ... .

x

.... .... .... . . ... ...... .. .... .. .. . 8.......... . ... .. . .. .. .... ................ .... ....... .. . ...... ................... .. .. .. .. . . . . . . . . . . . . . . . .. . .... ................ .... .... .

β

h x

t y v

v

φ

NOTE: For α

1 and φ

1,

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? .......................................... ...

y .............

.... ... .. ..................................

x

............. .... ... ... ... ... .. .. .. ... .. ...

. ............................................ . ............. ..... .... .. ....

.. .... .. .. .. .... .. .. . .. . .. .. .. .. .. .. .. . .. ............ .. .. .. . .. . . .. .. .. . .. . . .. . ................................................. .. .. .. .. .. .. .8 .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .

α

2h .

h x

... ... ... ... ... ... ... ... ... .. ........ ..........

v

h.

... ... ... . ....... ..........

t y

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? ............. ....... ............. .... ....

h

... ..... ..............

h

. ................. ..

. .. ... ... ... .... ...... ... ... ... ... ... 2

α

.. . ... ... ... .. .... .... .. ..... .. ........ . . .. .. ...... .. .. ... ... ... ... .. . .. .... ...... ............. ......... .. ... ..

z

{ e u j z v

h

Problem 5.36

8

u {

y ............

... ... .. .. .........................................

x

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 |. ... ... ....... ... ... .. .... .... ........ .... ......... ......... ... .... .. ......... .... .. ....... ...... ............ . ... .............. ....... . ............ ... ..... . ............. ........ . ........ ....... . ....... ...

.. ....... ..... ...... ... ....... ........ ... ....... ....... ... ..........i ... . .............. .. ... .... .... ... .... .... .... .... .. ...... o . .............. ... .. ........... ... ... ....... .. .. .... ... . .... ... ... .... . . ... . . .. .... .. ... .... .... .... . ... . .... ... . . . ......................................................................................................................................................................................................

v

k

v

v

ξ

v

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? .. .. .. .. .. ... .. ... ..... .. .. ....... .. ..... ... ... .. .... .... .. ... . . . .... .. . . . . ... . .. . ... .......... .. . .. ... . . ... ... .. .... ... . .. .. .. ... . . . . . .. . .. ..... ... . ... .. ................. .................... .. .... ... ....... ......... .. ....... .. ... .... ....... ... . . . . . . . . ...... .. .. ... .. ... . . . ... ...... ................. .............. .... .. ... ... ... .

e u

g

H

h

z

β

x

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

.. .. .................................... ... .. ........ ..... ............. ..................................................... .... .... ... ... ... .... ... .... ... ... ... ... ... ... ... ... ... .... ... ... .... . ... . ..... ... .... ..... .... ... ..... ... ... ... ...... ... ............ ..... .... .... . .................. ... ... .. ... ... .. ... .... ............................................................................................................................................... ..............

v

B

v

h

A

................................................. ... .

v

..

L ..................................................

Problem 5.40

θ

... .. .... . .......... ..... .

g

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

... ... ... ... ... .. .......... ..... .

. ........ ... .... ...... .. ... ... .. .. .. . .. .. . . . . .. . .. .. ... .. .. .. ... .. .. ... .. .. .. .. . . . .. .. .. .. .. .. .. ... .. . ... .. .. . . . . . . .. .. ... .. .. .. .. .. .. .... ............ .. ... ..... .. .. ... ........................................ .. .. .. .. .. .. .. .. .. .. .. .. ..... .. . ... ... ... ... . ... ... .. ....... ..... .. ... ............................................................................................................. ... . ..............................................

c s

c s

v

h

θ

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 . .. ............ .......... ....... . .......... ............................... .......... ................... . . . . . . . ... . ... . ..... ..... . ... .......... .......... ... A .... ................... ................... ... ... . . .. .......... .......... ... ........................... ................... .. .. ... . . . . . . . . . . . . . . . . . . . .......... ...... ...... . . . . . . . . . . . . . . . . . . .... ...... ...... . . . . . . . . . . . . . . . . . . ......... ..................... . .......... ... .......... ................... . . .......... ......................................................................................................................................................................................................................

mB

.........• ................ o .... ... ............... ............................................

θ

v

m

θ

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. .................................. .. Line of impact ..... ... ... ..... ... ... ..... (normal to wedge) ... ..... . . ... . ..... ... ... ..... ..... ..... ... ....... . ... . . ........................................... ..... ..... ... ..... ... ..... ... ..... ... ....... ... ... .... .. ...... . ............................................................................................................................................................................................

v

λm

n m~

θ

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. z ............

j z

.. ... ... .. ... . ... .......... ... ... . ...... .. .... ... . ... .. ... .... b ... .... . ... .. .... .. ... .... ... .. .. . . . . . ... . . . .. a ............................ .... .................... Line of impact ... .............................................. ..... ....... ... ......... ... ... (normal to cart) ......... .. ... . . ... ... .. ... ...... .. .. a .. ... . .... ....................................................... ... . ... . . ........ ... ....... ... ... ... ...........b .. ... ... ................ ............................................................................ ... ............................................• • ............................................................................................................................ .. ...................................................................................................................................................

v

v

2θ

v

j z

e u

e u

v

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. .......................................................... . .. ............................................................. .. .. ... .... . . .. ... .... ... .... .. . . ... ... ... ... ... . . . ... .. . . ... . . . ... . . . ... .. . .. . . ... . . . . .. . .. ..... .................. . ....... ......... . . .... ......... .. . . . . . . . .. .. .. .. .. ....... .... ... .... .... .. ... ..................... ... .............. .. .. .. .. ..... ..... ............................ . . .. ............ ..... ............. ... .... .... ............... ..

•

θ

B~

h

v

B~

A

Problem 5.47

... ... ... ... ... .. ........ ..... .

v

g

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. z ............

... ............................................................ ... ... ..... ... .......... ... ... ..... ... ........ ........ ... ... ... .. ... ....... ........ ... ... ... .. ..... .... . .. . .. ..... ..... . .. . ... . ... ..... ..... ......... .. . . ... . . . . ...... .... ..... .. ..... ..... .... ... . ... . . . .. ....... ...... ... ...... .. ... . . . . ... .... ........................................... ..... ........... .. ... ... . .............................................................................................................................................................................................................

m

g = −g k

M

h

θ

x

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. z ............

............. ... ... ... ... . ... ... .... ... .. . . .... .. ... ... . ... . ... ... . . . . . . ... ... ... ... ... . . . ... . ... .. ... ... . ... . . . ... ... .. ... .... . . . . ... ..... .... . . . . . . . ... .... . a .... . ........ ............ ..... ..... ..... ..... ..... ..... ........ . . ................................. . . . . . . . . . . ... .... .... .... .... .... .... .... ... ... .. ..... . . . ..... .... .... .... ..... . . . . .... ... . .... .... .... . . . ... ... . .. .. .... ...................................................................................................................................................................................... .................................... .. ... ....................................................................................................................................................................................... ..

Line of impact ...... .. (normal to block) .......

n m~

k

v

M •

•

θ

x

Problem 5.50