# Impulse and Linear Momentum - Pearson-Global-Schools

## Impulse and Linear Momentum - Pearson-Global-Schools

5 Impulse and Linear Momentum How does jet propulsion work? How can you measure the speed of a bullet? Would a meteorite collision significantly cha...

5

Impulse and Linear Momentum

How does jet propulsion work? How can you measure the speed of a bullet? Would a meteorite collision significantly change Earth’s orbit?

In previous chapters we discovered that the pushing interaction between car tires and the road allows a car to change its velocity. Likewise, a ship’s propellers

Be sure you know how to:

push water backward; in turn, water pushes the ship forward.

But how does a rocket, far above Earth’s atmosphere, change velocity with no object to push against? Less than 100 years ago, rocket flight was considered impossible. When U. S. rocket pioneer Robert Goddard published an article

Construct a force diagram for an object (Section 2.1). ■ Use Newton’s second law in ­component form (Section 3.2). ■ Use kinematics to describe an ­object’s motion (Section 1.7).

151

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152        Chapter 5  Impulse and Linear Momentum in 1920 about rocketry and even suggested a rocket flight to the Moon, he was ridiculed by the press. A New York Times editorial dismissed his idea, ­s aying, “. . . even a schoolboy knows that rockets cannot fly in space ­b ecause a vacuum is devoid of anything to push on.” We know now that Goddard was correct—but why? What does the rocket push on?

Figure 5.1  The mass is the same in the closed flask (an isolated system) (a) before burning the steel wool and (b) after burning the steel wool. (c) However, the mass increases when the steel wool is burned in (a) the open flask (a non-isolated system). (a) The block balances (a) blockthe The steel woolthe balances The block and steelflask. wool balances and flask.the steel wool and flask.

Closed flask Closed flask Steel wool Steel wool Balancing Closed flask block Balancing Steel wool block 0 Balancing 0 block 0

(b) (b) The steel wool is burned in flask. block (b)a closed The steel wool is The burned still balances. in a closed flask. The block The steel wool is burned still balances. in a closed flask. The block 0 still balances. 0 0

(c) (c) When the steel wool is burned in the open flask, the ismass (c) When the steel wool burned in in the the flask open increases. flask, the mass When the steel wool is burned in the flask increases. in the open flask, the mass in the flask increases. 0 0 0

u

u

We can use Newton’s second law 1a =  F >m2 to relate the acceleration of a system object to the forces being exerted on it. However, to use this law effectively we need quantitative information about the forces that objects exert on each other. Unfortunately, if two cars collide, we don’t know the force that one car exerts on the other during the collision. When fireworks explode, we don’t know the forces that are exerted on the  pieces flying apart. In this chapter you will learn a new approach that helps us analyze and predict mechanical phenomena when the forces are not known.

5.1 Mass accounting We begin our investigation by analyzing the physical quantity of mass. Earlier (in Chapter 2), we found that the acceleration of an object depended on its mass—the greater its mass, the less it accelerated due to an unbalanced external force. We ignored the possibility that an object’s mass might change during some process. Is the mass in a system always a constant value? You have probably observed countless physical processes in which mass seems to change. For example, the mass of a log in a campfire decreases as the log burns; the mass of a seedling increases as the plant grows. What happens to the “lost” mass from the log? Where does the seedling’s increased mass come from? A system perspective helps us understand what happens to the burning log. If we choose only the log as the system, the mass of the system decreases as it burns. However, air is needed for burning. What happens to the mass if we choose the surrounding air and the log as the system? Suppose that we place steel wool in a closed flask on one side of a balance scale and a metal block of equal mass on the other side (Figure 5.1a). In one experiment, we burn the steel wool in the closed flask (the flask also contains air), forming an oxide of iron. We find that the total mass of the closed flask containing burned steel wool (iron oxide) is the same as the mass of the balancing metal block (Figure 5.1b). Next, we burn the steel wool in an open flask and observe that the mass of that flask increases (Figure 5.1c). The steel wool in the open flask burns more completely and absorbs some external oxygen from the air as it burns. Eighteenth-century French chemist Antoine Lavoisier actually performed such experiments. He realized that the choice of the system was very important. Lavoisier defined an isolated system as a group of objects that interact with each other but not with external objects outside the system. The mass of an isolated system is the sum of the masses of all objects in the system. He then used the concept of an isolated system to summarize his (and our) ­experiments in the following way: Law of constancy of mass  When a system of objects is isolated (a closed c­ ontainer), its mass equals the sum of the masses of its components and does not change—it remains constant in time.

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5.2  Linear momentum        153

When the system is not isolated (an open container system), the mass might change. However, this change is not random—it is always equal to the amount of mass leaving or entering the system from the environment. Thus, even when the mass of a system is not constant, we can keep track of the changes if we take into account how much is leaving or entering the system: initial mass of new mass entering or final mass of ° system at earlier ¢ + ° leaving system between ¢ = ° system at later ¢ clock reading the two clock readings clock reading

The above equation helps describe the change of mass in any system. The mass is constant if there is no flow of mass in or out of the system, or the mass changes in a predictable way if there is some flow of mass between the system and the environment. Basically, mass cannot appear from nowhere and does not disappear without a trace. Imagine you have a system that has a total mass of mi = 3 kg (a bag of oranges). You add some more oranges to the bag 1m = 1 kg2. The final mass of the system equals exactly the sum of the initial mass and the added mass: m i + m = m f or 3 kg + 1 kg = 4 kg (Figure 5.2a). We can represent this process with a bar chart (Figure 5.2b). The bar on the left represents the initial mass of the system, the central bar represents the mass added or taken away, and the bar on the right represents the mass of the system in the final situation. As a result, the height of the left bar plus the height of the central bar equals the height of the right bar. The bar chart allows us to keep track of the changes in mass of a system even if the system is not isolated. Mass is called a conserved quantity. A conserved quantity is constant in an isolated system. When the system is not isolated, we can account for the changes in the conserved quantity by what is added to or subtracted from the system. Just as with every idea in physics, the law of constancy of mass in an isolated system does not apply in all cases. We will discover later in this book (Chapters 28 and 29) that in situations involving atomic particles, mass is not constant even in an isolated system; instead, what is constant is a new quantity that includes mass as a component.

Figure 5.2  (a) The initial mass of the oranges plus the mass of the oranges that were added (or subtracted) equals the final mass of the oranges. (b) The mass change process is represented by a mass bar chart. (a)

1 kg

4 kg

mi

m

mf

m

5.2 Linear momentum

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(b)

Review Question 5.1  When you burn a log in a fire pit, the mass of wood clearly decreases. How can you define the system so as to have the mass of the objects in that system constant?

We now know that mass is an example of a conserved quantity. Is there a quantity related to motion that is conserved? When you kick a stationary ball, there seems to be a transfer of motion from your foot to the ball. When you knock bowling pins down with a bowling ball, a similar transfer occurs. However, motion is not a physical quantity. What physical quantities describing motion are constant in an isolated system? Can we describe the changes in these quantities using a bar chart? Let’s conduct a few experiments to find out. In Observational Experiment Table 5.1 we observe two carts of different masses that collide on a smooth track. For these experiments, the system will include both carts. A collision is a process that occurs when two (or more) objects come into direct contact with each other. The system is isolated since the forces that the carts exert on each other are internal, and external forces are either balanced (as the vertical forces are) or negligible (the horizontal friction force).

3 kg

0

0

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154        Chapter 5  Impulse and Linear Momentum Observational Experiment Table

5.1 Collisions in a system of two carts (all velocities are with respect      to the track). Observational experiment Experiment 1. Cart A (0.20 kg) ­moving right at 1.0 m/s collides with cart B (0.20 kg), which is stationary. Cart A stops and cart B moves right at 1.0 m/s.

Video 5.1

Analysis vAix  1.0 m/s

vBix  0

vAfx  0

vBfx  1.0 m/s

A 0.20 kg

B 0.20 kg

A

B

x

The direction of motion is indicated with a plus and a minus sign. ■

S peed: The sum of the speeds of the system objects is the same before and after the collision: 1.0 m>s + 0 m>s = 0 m>s + 1.0 m>s. ■  Mass # speed: The sum of the products of mass and speed is the same before and after the collision: 0.20 kg11.0 m>s2 + 0.20 kg10 m>s2 = 0.20 kg10 m>s2 + 0.20 kg11.0 m>s2. ■  Mass # velocity: The sum of the products of mass and the x-component of velocity is the same before and after the collision: 0.20 kg1+ 1.0 m>s2 + 0.20 kg102 = 0.20 kg102 + 0.20 kg1+1.0 m>s2. Experiment 2. Cart A (0.40 kg) ­moving right at 1.0 m/s collides with cart B (0.20 kg), which is stationary. After the collision, both carts move right, cart B at 1.2 m/s, and cart A at 0.4 m/s.

vAix  1.0 m/s vBix  0 A 0.40 kg

vAfx  0.4 m/s v  1.2 m/s Bfx

B 0.20 kg

A B

x

■

S peed: The sum of the speeds of the system objects is not the same before and after the collision: 1.0 m>s + 0 m>s  0.4 m>s + 1.2 m>s. ■  Mass # speed: The sum of the products of mass and speed is the same before and after the collision: 0.40 kg11.0 m>s2 + 0.20 kg10 m>s2 = 0.40 kg10.4 m>s2 + 0.20 kg11.2 m>s2. ■  Mass # velocity: The sum of the products of mass and the x-component of velocity is the same before and after the collision: 0.40 kg1+ 1.0 m>s2 + 0.20 kg102 = 0.40 kg1+0.4 m>s2 + 0.20 kg1+ 1.2 m>s2. Experiment 3. Cart A (0.20 kg) with a piece of clay attached to the front moves right at 1.0 m/s. Cart B (0.20 kg) moves left at 1.0 m/s. The carts collide, stick together, and stop.

vAix  1.0 m/s vBix  1.0 m/s A 0.20 kg

B 0.20 kg

vAfx  0

vBfx  0

A

B

x

■

S peed: The sum of the speeds of the system objects is not the same ­before and ­after the collision: 1.0 m>s + 1.0 m>s  0 m>s + 0 m>s. ■  Mass # speed: The sum of the products of mass and speed is not the same before and after the collision: 0.20 kg11.0 m>s2 + 0.20 kg11.0 m>s2  0.20 kg10 m>s2 + 0.20 kg10 m>s2. ■  Mass # velocity: The sum of the products of mass and the x-component of velocity is the same before and after the collision: 0.20 kg1+ 1.0 m>s2 + 0.20 kg1-1.0 m>s2 = 0.20 kg10 m>s2 + 0.20 kg10 m>s2. Patterns One quantity remains the same before and after the collision in each experiment—the sum of the products of the mass and ­x-velocity component of the system objects.

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5.2  Linear momentum        155

In the three experiments in Observational Experiment Table 5.1, only one quantity—the sum of the products of mass and the x-component of velocity mvx —remained the same before and after the carts collided. Note also that the sum of the products of the mass and the y-component of velocity mvy u did not change—it remained zero. Perhaps mv is the quantity characterizing motion that is constant in an isolated system. But will this pattern persist in other situations? Let’s test this idea by using it to predict the outcome of the experiment in Testing Experiment Table 5.2.

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Testing Experiment table u

5.2  Testing the idea that mv in an isolated system remains constant (all velocities are with respect to the track). Testing experiment

Video 5.2

Prediction

Cart A (0.40 kg) has a piece of modeling clay attached to its front and is moving right at 1.0 m/s. Cart B (0.20 kg) is moving left at 1.0 m/s. The carts ­collide and stick together. Predict the velocity of the carts after the collision.

Outcome

vAix  1.0 m/s

vfx  ?

vBix  1.0 m/s A 0.40 kg

A

B 0.20 kg

B

x

After the collision, the carts move together toward the right at close to the ­predicted speed.

The system consists of the two carts. The direction of velocity is noted with a plus or minus sign of the velocity component: 10.40 kg21+1.0 m>s2 + 10.20 kg21-1.0 m>s2 = 10.40 kg + 0.20 kg2vf x

or

vf x = 1+ 0.20 kg # m>s2>10.60 kg2 = +0.33 m>s

After the collision, the two carts should move right at a speed of about 0.33 m/s. Conclusion Our prediction matched the outcome. This result gives us increased confidence that   u this new quantity mv might be the quantity whose sum is constant in an isolated system.

u

This new quantity is called linear momentum p.

Figure 5.3  Momentum is a vector quantity with components.

Linear Momentum  The linear momentum pu of a single object is the product of its mass m and velocity v : u

u

u

p  mv

(5.1)

Linear momentum is a vector quantity that points in the same direction as the u ­object’s velocity v (Figure 5.3). The SI unit of linear momentum is (kg # m/s). The total linear momentum of a system containing multiple objects is the vector sum of the momenta (plural of momentum) of the individual objects. u

u

u

u

u

pnet  m1v1  m2v2  P  mnvn  mv

y m

vr pr  mvr The components of a skydiver’s momentum: px  0 py  mv x

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156        Chapter 5  Impulse and Linear Momentum Note the following three important points. u

u

1. Unlike mass, which is a scalar quantity, p = mv is a vector quantity. Therefore, it is important to consider the direction in which the colliding objects are moving before and after the collision. For example, because cart B in Table 5.2 was moving left along the x-axis, the x-component of its momentum was negative before the collision. 2. Because momentum depends on the velocity of the object, and the velocity depends on the choice of the reference frame, different observers will measure different momenta for the same object. As a passenger, the momentum of a car with respect to you is zero. However, it is not zero for an observer on the ground watching the car move away from him. 3. We chose an isolated system (the two carts) for our investigation. The u sum of the products of mass and velocity mv of all objects in the isolated system remained constant even though the carts collided with each other. However, if we had chosen the system to be just one of the carts, we would u u see that the linear momentum p = mv of the cart before the collision is different than it is after the collision. Thus, to establish that momentum u p is a conserved quantity, we need to make sure that the momentum of a system changes in a predictable way for systems that are not isolated. We chose a system in Observational Experiment Table 5.1 so that the sum of the external forces was zero, making it an isolated system. Based on the results of Table 5.1 and Table 5.2, it appears that the total momentum of an isolated system is constant. Momentum constancy of an isolated system  The momentum of an isolated system is constant. For an isolated two-object system:

u

u

u

u

m1v1i  m2v2i  m1v1f  m2v2f

(5.2)

Because momentum is a vector quantity and Eq. (5.2) is a vector equation, we will work with its x- and y-component forms:

m1v1i x  m2v2i x  m1v1f x  m2v2f x

(5.3x)

m1v1i y  m2v2i y  m1v1f y  m2v2f y

(5.3y)

For a system with more than two objects, we simply include a term on each side of the equation for each object in the system. Let’s test the idea that the momentum of an isolated system is constant in another situation. Example 5.1  Two rollerbladers Jen (50 kg) and David (75 kg), both on rollerblades, push off each other abruptly. Each person coasts backward at approximately constant speed. During a certain time interval, Jen travels 3.0 m. How far does David travel during that same time interval? Sketch and translate  The process is sketched at the right. All motion is with respect to the floor and is along the x-axis. We choose the two rollerbladers as the system. Initially, the two rollerbladers are at rest. After pushing off, Jen (J) moves to the left and David (D) moves to the right. We can use momentum constancy to calculate David’s velocity component and predict the distance he will travel during that same time interval.

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Later Jen has traveled 3.0 m.

How far has David traveled?

Simplify and diagram  We model each person as a point-like object and assume that the friction force exerted on the rollerbladers does not affect their motion. Thus there are no horizontal external forces exerted on

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5.3  Impulse and momentum        157

the system. In addition, the two vertical forces, an upu ward normal force NF on P that the floor exerts on each person and an equalmagnitude downw ard u gravitational force FE on P that Earth exerts on each person, cancel, as we see in the force diagrams. Since the net external force exerted on the system is zero, the system is isolated. The forces that the rollerbladers exert on each other are internal forces and should not affect the momentum of the system. Represent mathematically  The initial state (i) of the system is before they start pushing on each other, and the final state (f) is when Jen has traveled 3.0 m. mJvJi x + mDvDi x = mJvJf x + mDvDf x We choose the positive direction toward the right. Because the initial velocity of each person is zero, the above equation becomes 0 + 0 = mJvJf x + mDvDf x or mDvDf x = -mJvJf x

Solve and evaluate  The x-component of Jen’s velocity after the push-off is vJf x = -13.0 m2> t, where t is the time interval needed for her to travel 3.0 m. We solve the above equation for David’s final x-velocity component to determine how far David should travel during that same time interval: mJvJf x

vDf x = -

mD

= -

= -

mJ mD

vJf x

150 kg2 1- 3.0 m2 175 kg2

t

=

12.0 m2 t

Since momentum is constant in this isolated system, we predict that David will travel 2.0 m in the positive direction during t. The measured value is very close to the predicted value. Try it yourself: Estimate the magnitude of your momentum when walking and when jogging. Assume your mass is 60 kg. Answer: When walking, you travel at a speed of about 1 to 2 m/s. So the magnitude of your momentum will be p = mv  160 kg211.5 m>s2  90 kg # m>s. When jogging, your speed is about 2 to 5 m/s or a momentum of magnitude p = mv  160 kg213.5 m>s2  200 kg # m>s.

Notice that in Example 5.1 we were able to determine David’s velocity by using the principle of momentum constancy. We did not need any information about the forces involved. This is a very powerful result, since in all likelihood the forces they exerted on each other were not constant. The kinematics equations we have used up to this point have assumed constant acceleration of the system (and thus constant forces). Using the idea of momentum constancy has allowed us to analyze a situation involving nonconstant forces. So far, we have investigated situations involving isolated systems. In the next section, we will investigate momentum in nonisolated systems.

Review Question 5.2  Two identical carts are traveling toward each other at the same speed. One of the carts has a piece of modeling clay on its front. The carts collide, stick together, and stop. The momentum of each cart is now zero. If the system includes both carts, did the momentum of the system disappear? Explain your answer.

5.3 Impulse and momentum So far, we have found that the linear momentum of a system is constant if that system is isolated (the net external force exerted on the system is zero). How do we account for the change in momentum of a system when the net external force exerted on it is not zero? We can use Newton’s laws to derive an expression relating forces and momentum change.

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158        Chapter 5  Impulse and Linear Momentum

Impulse due to a force exerted on a single object When you push a bowling ball, you exert a force on it, causing the ball to acu u u celerate. The average acceleration a is defined as the change in velocity vf - vi divided by the time interval t = tf - ti during which that change occurs: u

u

vf - vi a = tf - ti u

We can also use Newton’s second law to determine an object’s acceleration if we know its mass and the sum of the forces that other objects exert on it: u

F a = m u

We now have two expressions for an object’s acceleration. Setting these two expressions for acceleration equal to each other, we get u

u

u

vf - vi F = m tf - ti

Now multiply both sides by m1tf - ti2 and get the following: u

u

u

u

u

mvf - mvi = pf - pi =  F1tf - ti2

(5.4)

The left side of the above equation is the change in momentum of the object. This change depends on the product of the net external force and the time interval during which the forces are exerted on the object (the right side of the equation). Note these two important points: 1. Equation (5.4) is just Newton’s second law written in a different form— one that involves the physical quantity momentum. 2. Both force and time interval affect momentum—the longer the time interval, the greater the momentum change. A small force exerted for a long time interval can change the momentum of an object by the same amount as a large force exerted for a short time interval.

Figure 5.4  The impulse of a force is the area under the [email protected]@t graph line. F (N)

Fav

The product of the external force exerted on an object during a time interval and the time interval gives us a new quantity, the impulse of the force. When you kick a football or hit a baseball with a bat, your foot or the bat exerts an impulse on the ball. The forces in these situations are not constant but instead vary in time (see the example in Figure 5.4). The shaded area under the  varying force curve represents the impulse of the force. We can estimate the impulse by drawing a horizontal line that is approximately the average force exerted during the time interval of the impulse. The area under the rectangular average force-impulse curve equals the product of the height of the rectangle (the average force) and the width of the rectangle (the time interval over which the average force is exerted). The product Fav1tf - ti2 equals the magnitude of the impulse. u

u

Impulse  The impulse J of a force is the product of the average force Fav exerted

on an object during a time interval 1t f - ti2 and that time interval:

ti

tf

The area of the shaded rectangle is about the same as the area under the curved line and equals the impulse of the force.

t (s)

u

u

J  F av1t f  ti2

(5.5)

Impulse is a vector quantity that points in the direction of the force. The impulse has a plus or minus sign depending on the orientation of the force relative to a coordinate axis. The SI unit for impulse is N # s  1kg # m>s22 # s  kg # m>s, the same unit as momentum.

It is often difficult to measure directly the impulse of the net average force during a time interval. However, we can determine the net force on the right

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5.3  Impulse and momentum        159

side of Eq. (5.4) indirectly by measuring or calculating the momentum change on the left side of the equation. For this reason, the combination of impulse and momentum change provides a powerful tool for analyzing interactions between objects. We can now write Eq. (5.4) as the impulse-momentum equation for a single object. Impulse-momentum equation for a single object  If several external objects exert forces on a single-object system during a time interval 1tf  ti2, the sum of u their impulses  J causes a change in momentum of the system object: u

u

u

u

pf  pi   J   F on System1tf  ti2

(5.6)

p f x  p i x  Fon System x1t f  t i2

(5.7x)

The x- and y-scalar component forms of the impulse-momentum equation are

p f y  p i y  Fon System y1t f  t i2

(5.7y)

A few points are worth emphasizing. First, notice that Eq. (5.6) is a vector equation, as both the momentum and the impulse are vector quantities. Vector equations are not easy to manipulate mathematically. Therefore, we will use the scalar component forms of Eq. (5.6)—Eqs. (5.7x) and (5.7y). Second, the time interval in the impulse-momentum equation is very important. When object 2 exerts a force on object 1, the momentum of object 1 changes by an amount equal to u

u

u

u

u

p1 = p1f - p1i = F2 on 11tf - ti2 = F2 on 1 t

The longer that object 2 exerts the force on object 1, the greater the momentum change of object 1. This explains why a fast-moving object might have less of an effect on a stationary object during a collision than a slow-moving object interacting with the stationary object over a longer time interval. For example, a fast-moving bullet passing through a partially closed wooden door might not open the door (it will just make a hole in the door), whereas your little finger, moving much slower than the bullet, could open the door. Although the bullet moves at high speed and exerts a large force on the door, the time interval during which it interacts with the door is very small (milliseconds). Hence, it ­e xerts a relatively small impulse on the door—too small to significantly change the door’s momentum. A photo of a bullet shot through an apple illustrates the effect of a short impulse time (Figure 5.5). The impulse exerted by the bullet on the apple was too small to knock the apple off its support. Third, if the magnitude of the force changes during the time interval considered in the process, we use the average force. Finally, if the same amount of force is exerted for the same time interval on a large-mass object and on a small-mass object, the objects will have an equal change in momentum (the same impulse was exerted on them). However, the small-mass object would experience a greater change in velocity than the large-mass object.

Example 5.2  Abrupt stop in a car A 60-kg person is traveling in a car that is moving at 16 m/s with respect to the ground when the car hits a barrier. The person is not wearing a seat belt, but is stopped by an air bag in a time interval of 0.20 s. Determine the

Figure 5.5  The bullet’s time of interaction with the apple is very short, causing a small impulse that does not knock the apple over.

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average force that the air bag exerts on the person while stopping him. Sketch and translate  First we draw an initial-final sketch of the process. We choose the person as the system since we are investigating a force being exerted on him. (continued )

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160        Chapter 5  Impulse and Linear Momentum The person’s initial x-component of velocity vPi x = +16 m>s decreases to the final x-component of velocity vPf x = 0 in a time interval 1tf - ti2 = 0.20 s. Thus the average force exerted by the air bag on the person in the x-direction is FB on P x =

160 kg210 - 16 m>s2

10.20 s - 02 = -4800 N

The negative sign in –4800 N indicates that the average force points in the negative x-direction. The magnitude of this force is about 1000 lb! Try it yourself: Suppose a 60-kg crash test dummy is in a car traveling at 16 m/s. The dummy is not wearing a seat belt and the car has no air bags. During a collision, the dummy flies forward and stops when it hits the dashboard. The stopping time interval for the dummy is 0.02 s. What is the average magnitude of the stopping force that the dashboard exerts on the dummy?

Simplify and diagram  The force diagram shows u the average force FB on P exerted in the negative direction by the bag on the person. The vertical normal force and gravitational forces cancel. Represent mathematically  The x-component form of the impulse-momentum equation is mPvPi x + FB on P x1tf - ti2 = mPvPf x

Solve and evaluate  Solve for the force exerted by the air bag on the person: mP1vPf x - vPi x2

FB on P x =

1tf - ti2

Figure 5.6  Analyzing the collision of two carts in order to develop the momentum constancy idea. Initial r

v1i

r

1

2

v2i x

During collision r

F2 on 1 1

r

F1 on 2 2

x

Final r

v1f 1

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Answer: The average force that the hard surface exerts on the dummy would be about 50,000 N, extremely unsafe for a human. Note that the momentum change of the person in Example 5.2 was the same. However, since the change for the dummy occurs during a shorter time interval (0.02 s instead of 0.20 s), the force exerted on the dummy is much greater. This is why air bags save lives.

r

v2f 2

x

Using Newton’s laws to understand the constancy of momentum in an isolated system of two or more objects Let’s apply the impulse-momentum equation Eq. (5.4) to the scenario we ­described in Observational Experiment Table 5.1 in order to explore momentum constancy in a two-object isolated system. Two carts travel toward each other at different speeds, collide, and rebound backward (Figure 5.6). We first analyze each cart as a separate system and then analyze them together as a single system. Assume that the vertical forces exerted on the carts are balanced and that the friction force exerted by the surface on the carts does not significantly affect their motion. Cart 1: In the initial state, before the collision, cart 1 with mass m1 travu els in the positive direction at velocity v1i . In the final state, after the colliu sion, cart 1 moves with a different velocity v1f in the opposite direction. To determine the effect of the impulse exerted by cart 2 on cart 1, we apply the impulse-momentum equation to cart 1 only: u

u

u

m11v1f - v1i2 = F2 on 11t f - t i2

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5.4  The generalized impulse-momentum principle        161

Cart 2: We repeat this analysis with cart 2 as the system. Its velocity and momentum change because of the impulse exerted on it by cart 1: u

u

u

m21v2f - v2i2 = F1 on 21tf - ti2

Newton’s third law provides a connection between our analyses of the two carts; interacting objects at each instant exert equal-magnitude but oppositely directed forces on each other: u

u

F1 on 2 = - F2 on 1 Substituting the expressions for the forces from above and simplifying, we get u

u

m21v2f - v2i2 t f - ti u

u

u

= -

u

m11v1f - v1i2 t f - ti

u

u

m21v2f - v2i2 = -m11v1f - v1i2

We now move the initial momentum for both objects to the left side and the final momentum for both objects to the right side: u

u

u

u

m1v1i + m2v2i = m1v1f + m2v2f Initial momentum  Final momentum

This is the same equation we arrived at in Section 5.2, where we observed and analyzed collisions to understand the constant momentum of an isolated system. Here we have reached the same conclusions using only our knowledge of Newton’s laws, momentum, and impulse.

Review Question 5.3  An apple is falling from a tree. Why does its momentum change? Specify the external force responsible. Find a system in which the momentum is constant during this process.

5.4 The generalized impulse-momentum principle We can summarize what we have learned about momentum in isolated and nonisolated systems. The change in momentum of a system is equal to the net external impulse exerted on it. If the net impulse is zero, then the momentum of the system is constant. This idea, expressed mathematically as the generalized impulsemomentum principle, accounts for situations in which the system includes one or more objects and may or may not be isolated. The generalized impulse-momentum principle means that we can treat momentum as a conserved quantity. Generalized impulse-momentum principle  For a system containing one or

more objects, the initial momentum of the system plus the sum of the impulses that external objects exert on the system objects during the time interval 1tf - ti2 equals the final momentum of the system: u

u

u

u

u

1m1v1i  m2v2i  P2  F on Sys1t f  ti2  1m1v1f  m2v2f  P2    (5.8) Initial momentum of     the system

Net impulse exerted on the system

Final momentum of the system

The x- and y-component forms of the generalized impulse-momentum principle are  1m1v1i x  m2v2i x  P2  Fon Sys x1t f  t i2  1m1v1f x  m2v2f x  P2 (5.9x)

1m1v1i y  m2v2i y  P2  Fon Sys y1t f  ti2  1m1v1f y  m2v2f y  P2 (5.9y)

Note: If the net impulse exerted in a particular direction is zero, then the component of the momentum of the system in that direction is constant.

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162        Chapter 5  Impulse and Linear Momentum Equations (5.8) and (5.9) are useful in two ways. First, any time we choose to analyze a situation using the ideas of impulse and momentum, we can start from a single principle, regardless of the situation. Second, the equations remind us that we need to consider all the interactions between the environment and the system that might cause a change in the momentum of the system.

Impulse-momentum bar charts We can describe an impulse-momentum process mathematically using Eqs. (5.9x and y). These equations help us see that we can represent the changes of a system’s ­momentum using a bar chart similar to the one used to represent the changes of a system’s mass. The Reasoning Skill box shows the steps for constructing an impulse-momentum bar chart for a simple system of two carts of equal mass traveling toward each other. Reasoning Skill   Constructing a qualitative impulse-momentum bar chart. 1. Sketch the process, choose the initial and final states, and choose a system. Initial vr1i

vr2i

1

2

Final vrf 1

p1ix  p2ix  Jx  p1fx p2fx

0 Slower in negative direction (m1v1ix  m2v2ix)  Jx  (m1v1fx  m2v2fx) or m1v1i  m2(v2i)  0  (m1  m2)vf

2

x 2. Draw initial and final momentum bars for each object in the system. (Note cart directions and bar directions.) 3. Draw an impulse (J) bar if there is an external nonzero impulse. 4. Convert each bar in the chart into a term in the component form of the impulse-momentum equation.

Note that before constructing the bar chart, we represent the process in an initial-final sketch (Step 1 in the Skill box). We then use the sketch to help construct the impulse-momentum bar chart. The lengths of the bars are qualitative indicators of the relative magnitudes of the momenta. In the final state in the example shown, the carts are stuck together and are moving in the positive direction. Since they have the same mass and velocity, they each have the same final momentum. The middle shaded column in the bar chart represents the net external impulse exerted on the system objects during the time interval 1tf - ti2—there is no impulse for the process shown. The shading reminds us that impulse does not reside in the system; it is the influence of the external objects on the momentum of the system. Notice that the sum of the heights of the bars on the left plus the height of the shaded impulse bar should equal the sum of the heights of the bars on the right. This “conservation of bar heights” reflects the conservation of momentum. We can use the bar chart to apply the generalized impulse-momentum equation (Step 4). Each nonzero bar corresponds to a nonzero term in the equation; the sign of the term depends on the orientation of the bar.

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5.4  The generalized impulse-momentum principle        163

Using impulse-momentum to investigate forces Can we use the ideas of impulse and momentum to learn something about the forces that two objects exert on each other during a collision? Consider a collision between two cars (Figure 5.7). To analyze the force that each car exerts on the other, we will define the system to include only one of the cars. Let’s choose car 1 and construct a bar chart for it. Car 2 exerts an impulse on car 1 during the collision that changes the momentum of car 1. If the initial momentum of car 1 is in the positive ­direction, then the impulse exerted by car 2 on car 1 points in the negative direction. Because of this, the impulse bar on the bar chart points downward. Note that the total height of the initial momentum bar on the left side of the chart and the height of the impulse bar add up to the total height of the final momentum bar on the right side. Using the bar chart, we can apply the component form of the impulse-momentum equation:

Figure 5.7  A bar chart analysis of the collision of car 2 with car 1. Initial vr1i 1

Final

r

vr1f

F2 on 1 2

1

2

1

m1

x

2

m1

p1ix  J2 on 1x p1fx

0

m1v1i x + Jx = m1v1f x

The components of the initial and final momentum are positive. As the force is exerted in the negative direction, the x-component of the impulse is negative and equal to -F2 on 1 t. Thus, +m1v1i + 1-F2 on 1 t2 = +m1v1f

If we know the initial and final momentum of the car and the time interval of interaction, we can use this equation to determine the magnitude of the average force that car 2 exerted on car 1 during the collision.

Car 1 has considerable momentum in the positive direction.

The force exerted by 2 on 1 is in the negative direction.

Car 1 has momentum in the positive direction.

Active Learning Guide

Tip   When you draw a bar chart, always specify the reference frame (the object of reference and the coordinate system). The direction of the bars on the bar chart (up for positive and down for negative) should match the direction of the momentum or impulse based on the chosen coordinate system.

Example 5.3  Happy and sad balls You have two balls of identical mass and size that behave very differently. When you drop the so-called “sad” ball, it thuds on the floor and does not bounce at all. When you drop the so-called “happy” ball from the same height, it bounces back to almost the same height from which it was dropped. The difference in the bouncing ability of the happy ball is due its internal structure; it is made of different material. You hang each ball from Sad ball Happy ball a string of identical length and place a wood board on its end directly below the support for each string. You pull each ball back to an equal height and release the balls one at a time. When each ball hits the board, which has the best chance of knocking the board over: the sad ball or the happy ball? Sketch and translate  Initial and final sketches of the process are shown at the right. The system is just the ball. In the initial state, the ball is just about to hit the board, moving horizontally toward the left (the

M05_VANH5357_01_SE_C05.indd 163

balls are moving equally fast). The final state is just after the collision with the board. The happy ball (H) bounces back, whereas the sad ball (S) does not.

Same initial momentum Same initial for both balls momentum justboth before for balls hitting the just before board hitting the board

Sad ball stops. Happy Sad ballball stops. bounces back. Happy ball bounces back.

(continued )

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164        Chapter 5  Impulse and Linear Momentum

Simplify and diagram  Assume that the collision time interval t for each ball is about the same. We analyze only the horizontal x-component of the process, the component that is relevant to whether or not each of the boards is knocked over. Each board exerts an ­impulse on the ball that causes the momentum of the ball to change. Therefore, each ball, according to Newton’s third law, exerts an impulse on the board that it hits. A larger force exerted on the board means a larger impulse and a better chance to tip the board. A bar chart for each ball-board collision is shown below.

Note that the x-component of the final velocity of the sad ball is vSf x = 0 (it does not bounce) and that the x-component of the final velocity of the happy ball is vHf x = -vi (it bounces). Solve and evaluate  We can now get an expression for the force exerted by each board on each ball: Sad ball:

FB on S x =

Happy ball:

FB on H x =

m10 - vi2 mvi = . t t

m 31-vi2 - vi 4 2mvi = . t t

Because we assumed that the time of collision is the same, the board exerts twice the force on the happy ball as on the sad ball, since the board causes the happy ball’s momentum to change by an amount twice that of the sad ball. According to Newton’s third law, this means that the happy ball will exert twice as large a force on the board as the sad ball. Thus, the happy ball has a greater chance of tipping the board. The impulse of the happy ball is twice as large in magnitude as that of the sad ball and causes twice as large a momentum change.

Try it yourself: Is it less safe for a football player to bounce backward off a goal post or to hit the goal post and stop?

Represent mathematically  The x-component form Answer: Although any collision is dangerous, it is better to of the impulse-momentum Eq. (5.5x) applied to each ball hit the goal post and stop. If the football player bounces is as follows: back off the goal post, his momentum will have changed by a greater amount (like the happy ball in the last exam# Sad ball: mvi + FB on S x t = m 0 ple). This means that the goal post exerts a greater force on Happy ball: mvi + FB on H x t = m1-vi2 him, which means there is a greater chance for injury. The pattern we found in the example above is true for all collisions—when an object bounces back after a collision, we know that a larger magnitude force is exerted on it than if the object had stopped and did not bounce after the collision. For that reason, bulletproof vests for law enforcement agents are ­designed so that the bullet embeds in the vest rather than bouncing off it.

Review Question 5.4  If in solving the problem in Example 5.3 we chose the system to be the ball and the board, how would the mathematical description for each ball-board collision change?

5.5 Skills for analyzing problems using the impulse-momentum equation Initial and final sketches and bar charts are useful tools to help analyze processes using the impulse-momentum principle. Let’s investigate further how these tools work together. A general strategy for analyzing such processes is

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5.5  Skills for analyzing problems using the impulse-momentum equation        165

described on the left side of the table in Example 5.4 and illustrated on the right side for a specific process.

PROBLEM-SOLVING STRATEGY   Applying the impulse-momentum equation Active Learning Guide›

Sketch and translate ■ Sketch the initial and final states and include appropriate coordinate axes. Label the sketches with the known information. Decide on the object of reference. ■ Choose a system based on the quantity you are interested in; for example, a multi-object isolated system to determine the velocity of an object, or a singleobject nonisolated system to determine an impulse or force.

Example 5.4  Bullet hits wood block A 0.020-kg bullet traveling horizontally at 250 m>s embeds in a 1.0-kg block of wood resting on a table. Determine the speed of the bullet and wood block together immediately after the bullet embeds in the block. The left side of the sketch below shows the bullet traveling in the positive x-direction with respect to the ground; it then joins the wood. All motion is along the x-axis; the object of reference is Earth. The system includes the bullet and wood; it is an isolated system since the vertical forces balance. The initial state is immediately before the collision; the final state is immediately after. Initial

Final

mB  0.020 kg, mW  1.0 kg

vB-Wfx  ? vrB-Wf

vBix  250 m/s, vWix  0 vrBi

x

Simplify and diagram ■ Determine if there are any external impulses exerted on the system. Drawing a force diagram could help determine the external forces and their directions. ■ Draw an impulse-momentum bar chart for the system for the chosen direction(s) to help you understand the situation, formulate a mathematical representation of the process, and evaluate your results.

x

Assume that the friction force exerted by the tabletop on the bottom of the wood does not change the momentum of the system during the very short collision time interval. The bar chart represents the process. The bar for the bullet is shorter than that for the block—their velocities are the same after the collision, but the mass of the bullet is much smaller. We do not draw a force diagram here, as the system is isolated. pBix

pWix

Jx

pBfx

pWfx

0

Represent mathematically ■ Use the bar chart to apply the generalized impulse-momentum equation along the chosen axis. Each nonzero bar becomes a nonzero term in the equation. The orientation of the bar determines the sign in front of the corresponding term in the equation. ■ Remember that momentum and impulse are vector quantities, so include the plus or minus signs of the components based on the chosen coordinate system.

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mBvBi x + mW # 0 + 1Jx2 = [email protected] x + [email protected] x Since Jx = 0,   [email protected] x =

mBvBi x 1mB + mW2

(continued )

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166        Chapter 5  Impulse and Linear Momentum

Solve and evaluate Insert the known information to determine the unknown quantity. ■ Check if your answer is reasonable with respect to sign, unit, and magnitude. Also make sure it applies for limiting cases, such as objects of very small or very large mass. ■

[email protected] x =

10.020 kg21250 m>s2 10.020 kg + 1.0 kg2

= +4.9 m>s

The magnitude of the answer seems reasonable given how fast the bullet was initially traveling. The plus sign indicates the direction, which makes sense, too. The units are also correct (m/s). We can test this using a limiting case: if the mass or speed of the bullet is zero, the block remains stationary after the collision. Try it yourself: A 0.020-kg bullet is fired horizontally into a 2.00-kg block of wood resting on a table. Immediately after the bullet joins the block, the block and bullet move in the positive x-direction at 4.0 m/s. What was the initial speed of the bullet? Answer: 400 m/s.

We could have worked Example 5.4 backward to determine the initial speed of the bullet before hitting the block (like the Try It Yourself question). This exercise would be useful since the bullet travels so fast that it is difficult to measure its speed. Variations of this method are used, for example, to decide whether or not golf balls conform to the necessary rules. The balls are hit by the same mechanical launching impulse and the moving balls embed in another object. The balls’ speeds are determined by measuring the speed of the object they embed in.

Determining the stopping time interval from the stopping distance When a system object collides with another object and stops—a car collides with a tree or a wall, a person jumps and lands on a solid surface, or a meteorite collides with Earth—the system object travels what is called its stopping distance. By estimating the stopping distance of the system object, we can estimate the stopping time interval. Suppose that a car runs into a large tree and its front end crumples about 0.5 m. This 0.5 m, the distance that the center of the car traveled from the beginning of the impact to the end, is the car’s stopping distance. Similarly, the depth of the hole left by a meteorite provides a rough estimate of its stopping distance when it collided with Earth. However, to use the impulsemomentum principle, we need the stopping time interval associated with the collision, not the stopping distance. Here’s how we can use a known stopping distance to estimate the stopping time interval. Assume that the acceleration of the object while stopping is constant. In that case, the average velocity of the object while stopping is just the sum of the initial and final velocities divided by 2: vaverage x = 1vf x + vi x2>2. ■ Thus, the stopping displacement 1x f - x i2 and the stopping time interval 1tf - ti2 are related by the kinematics equation ■

x f - x i = vaverage x 1t f - t i2 =

M05_VANH5357_01_SE_C05.indd 166

1vf x + vi x2 2

1t f - t i2

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5.5  Skills for analyzing problems using the impulse-momentum equation        167 ■

Rearrange this equation to determine the stopping time interval:

tf - ti =

21xf - xi2 vf x + vi x

(5.10)

Equation (5.10) provides a method to convert stopping distance x f - x i into stopping time interval tf - ti. Equation (5.10) can be applied to horizontal or vertical stopping.

Active Learning Guide

Example 5.5  Stopping the fall of a movie

stunt diver

The record for the highest movie stunt fall without a parachute is 71 m (230 ft), held by 80-kg A. J. Bakunas. His fall was stopped by a large air cushion, into which he sank about 4.0 m. His speed was about 36 m/s (80 mi/h) when he reached the top of the air cushion. Estimate the average force that the cushion exerted on his body while stopping him. Sketch and translate  We focus only on the part of the fall when Bakunas is sinking into the cushion. The situation is sketched below. We choose Bakunas as the system and the y-axis pointing up. The initial state is just as he touches the cushion at position yi = +4.0 m, and the final state is when the cushion has stopped him, at position yf = 0. All motion is with respect to Earth. The other information about the process is given in the figure. Be sure to pay attention to the signs of the quantities (especially the initial velocity).

Each external force causes an impulse.

Bakunas has zero momentum in the final state.

Represent mathematically  Since all motion and all of the forces are in the vertical direction, we use the bar chart to help construct the vertical y-component form of the impulse-momentum equation [Eq. (5.6y)] to determine the force that the cushion exerts on Bakunas as he sinks into it: mBvi y + 1N C on B y + F E on B y21tf - ti2 = mBvf y

Simplify and diagram  We draw a force diagram top right, modeling Bakunas as a point-like object. Since Bakunas’s downward speed decreases, the cushion must be exerting an upward force on Bakunas of greater magnitude than the downward force that Earth exerts on him. Thus, the net force exerted on him points upward, in the positive y-­direction. Using this information, we can draw a qualitative impulse-momentum bar chart for the process.

M05_VANH5357_01_SE_C05.indd 167

Using the force diagram, we see that the y-components of the forces are NC on B y = +NC on B and FE on B y = -F E on B = -mB g, where NC on B is the magnitude of the average normal force that the cushion exerts on Bakunas, the force we are trying to estimate. Noting that vf y = 0 and substituting the force components into the above equation, we get mBvi y + 31+N C on B2 + 1-mBg24 1tf - ti2 = mB # 0 1 mBvi y + 1N C on B - mBg21tf - ti2 = 0.

We can find the time interval that the cushion takes to stop Bakunas using Eq. (5.10) and noting that vf y = 0: tf - ti =

21yf - yi2 0 + vi y

(continued )

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168        Chapter 5  Impulse and Linear Momentum

Solve and evaluate  The stopping time interval while Bakunas sinks 4.0 m into the cushion is tf - ti =

210 - 4.0 m2 = 0.22 s 0 + 1-36 m>s2

Solving for N C on B, we get N C on B =

-mBvi y 1tf - ti2

+ mBg

-180 kg21-36 m>s2 + 180 kg219.8 N>kg2 10.22 s2 = +13,000 N + 780 N = 14,000 N =

Wow, that is a huge force! To reduce the risk of injury, stunt divers practice landing so that the stopping force that a cushion exerts on them is distributed evenly over the entire body. The cushions must be deep enough so that they provide a long stopping time interval and thus a smaller stopping force. The same strategy is applied to developing air bags and collapsible frames for

automobiles to make them safer for passengers during collisions. Notice four important points. First, we’ve included only two significant digits since that is how many the data had. Second, it is very easy to make sign mistakes. A good way to avoid these is to draw a sketch that ­includes a coordinate system and labels showing the values of known physical quantities, including their signs. Third, the impulse due to Earth’s gravitational force is small in magnitude compared to the impulse exerted by the air cushion. Lastly, the force exerted by the air cushion would be even greater if the stopping distance and consequently the stopping time interval were shorter. Try it yourself: Suppose that the cushion in the last ­e xample stopped Bakunas in 1.0 m instead of 4.0 m. What would be the stopping time interval and the magnitude of the average force of the cushion on Bakunas? Answer: The stopping time interval is 0.056 s, and the ­average stopping force is approximately 50,000 N.

Order-of-magnitude estimate—will bone break?

The strategy that we used in the previous example can be used to analyze skull fracture injuries that might lead to concussions. Laboratory experiments indicate that the human skull can fracture if the compressive force exerted on it per unit area is 1.7 * 108 N>m2. The surface area of the skull is much smaller than 1 m2, so we will use square centimeters, a more reasonable unit of area for this discussion. Since 1 m2 = 1 * 104 cm2, we convert the compressive force per area to 11.7 * 108 N>m22a Example 5.6  Bone fracture estimation1 A bicyclist is watching for traffic from the left while turning toward the right. A street sign hit by an earlier car accident is bent over the side of the road. The cyclist’s head hits the pole holding the sign. Is there a significant chance that his skull will fracture? Sketch and translate  The process is sketched at the right. The initial state is at the instant that the head initially contacts the pole; the final state is when the head

1 m2 b = 1.7 * 104 N>cm2. 1 * 104 cm2

and body have stopped. The person is the system. We have been given little information, so we’ll have to make some reasonable estimates of various quantities in order to make a decision about a possible skull fracture. Simplify and diagram  The bar chart illustrates the ­momentum change of the system and the impulse exerted by the pole that caused the change. The person was initially moving in the horizontal x-­direction with respect to Earth, and not moving after the collision. The pole exerted an impulse in the negative x-­direction on the cyclist. We’ll need to estimate the following quantities: the mass and speed of the cyclist in this situation, the stopping time interval, and the area of contact. Let’s assume that this is a 70-kg cyclist moving at about 3 m/s. The person’s body keeps moving forward for a short distance after the bone makes contact with the pole. The skin indents some during the collision. Because of these two factors, we assume

1

This is a true story—it happened to one of the book’s authors, Alan Van Heuvelen.

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5.6  Jet propulsion        169

a stopping distance of about 10 cm. Finally, we assume an area of contact of about 4 cm2. All of these numbers have large uncertainties and we are not worrying about significant figures, because this is just an estimate.

The pole’s negative impulse on the person causes the momentum to decrease.

Represent mathematically  We now apply the generalized impulse-momentum principle: mPersonvPerson i x + 1FPole on Person x21tf - ti2   = mPersonvPerson f x = 0 1 F Pole on Person x

mPersonvPerson i x = tf - ti

We can use the strategy from the last example to estimate the stopping time interval tf - t i from the stopping distance x f - x i: tf - ti =

21x f - x i2 vf x + vi x

where vix is the initial velocity of the cyclist and vfx = 0 is his final velocity. Solve and evaluate  Substituting the estimated initial velocity and the stopping distance into the above, we get an estimate for the stopping time interval:

tf - t i =

21x f - x i2 vf x + vi x

=

210.1 m - 02 0 + 3 m>s

= 0.067 s

Since this stopping time interval is an intermediate calculated value, we don’t need to worry about its number of significant digits. When we complete our estimate, though, we will keep just one significant digit. We can now insert our estimated values of quantities in the expression for the force exerted by the pole on the person: F Pole on Person x = = -

mPersonvPerson i x 1tf - ti2 170 kg213 m>s2 10.067 s2

= -3000 N

Our estimate of the force per area is Force 3000 N =  800 N>cm2 Area 4 cm2 Is the person likely to fracture his skull? The force per area needed to break a bone is about 1.7 * 104 N>cm2 = 17,000 N>cm2. Our estimate could have been off by at least a factor of 10. The force per area is still too little for a fracture. Try it yourself: What would be the magnitude of the force exerted on the cyclist if he bounced back off the pole instead of stopping, assuming the collision time ­interval remains the same? Answer: 6000 N.

Review Question 5.5  As the bullet enters the block in Example 5.4, the block exerts a force on the bullet, causing the bullet’s speed to decrease to almost zero. Why did we not include the impulse exerted by the block on the bullet in our analysis of this situation?

5.6 Jet propulsion Cars change velocity because of an interaction between the tires and the road. Likewise, a ship’s propellers push water backward; in turn, water pushes the ship forward. Once the ship or car is moving, the external force exerted by the water or the road has to balance the opposing friction force or the vehicle’s velocity will change. What does a rocket push against in empty space to change its velocity? ­Rockets carry fuel that they ignite and then eject at high speed out of the exhaust nozzles (see Figure 5.8). Could this burning fuel ejected from the rocket provide the push to change its velocity? Choose the system to be the rocket and fuel together. If the rocket and fuel are at rest before the rocket fires its engines, then its momentum is zero. If there are no external impulses, then even after the rocket fires its engines, the momentum of the rocket-fuel system should still be zero. However, the burning fuel is ejected backward at high velocity from the exhaust nozzle and has a backward momentum. The rocket must now have a nonzero forward velocity. We test this idea quantitatively in Testing Experiment Table 5.3.

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Figure 5.8  A rocket as it expels fuel. Expelled fuel moves left.

Rocket moves right.

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170        Chapter 5  Impulse and Linear Momentum Testing Experiment table

5.3  Rocket propulsion.

Testing experiment

Prediction

Outcome

You are traveling through space in a rocket and observe another rocket moving with equal velocity next to you. All of a sudden you notice a burst of burning fuel that is ejected from it. Predict what happens to that rocket’s velocity.

Choose the other rocket and its fuel as the system. Your rocket serves as the object   of reference and the +x-direction is in the direction of its motion. The other rocket has zero velocity in the initial state with respect to the object of reference. Its final state is just after it expels fuel backward at high speed; the rocket in turn gains an equal magnitude of momentum in the forward direction. We can represent this process with an initial-final sketch and a momentum bar chart for the rocket-fuel system.

The velocity of the other rocket does increase, and we see it move ahead of our rocket.

vrInitial  0 mFuel mRocket

Initial

x Final

mFuel

mRocket

vrFuel

vrRocket

Fuel and rocket system pRix  pFix  Jx  pRfx  pFfx

0

Assuming that fuel is ejected all at once at constant speed, the velocity of the rocket should be 0 = mRocketvRocket x + mFuelvFuel x 1 vRocket x = -

mFuelvFuel x mRocket

Here mRocket is the mass of the rocket without the fuel. We can also choose the rocket alone as the system. The rocket pushes back on the fuel, expelling it backward at high speed 1vFuel x 6 02; the fuel in turn pushes forward against the rocket, exerting an impulse that causes the rocket’s momentum and the velocity (assuming the mass of the rocket itself does not change) to increase 1vRocket x 7 02.

Rocket alone as system pRix  JF on Rx  pRfx

0

Conclusion The outcome of the experiment is consistent with the prediction, supporting the generalized impulse-momentum principle. We have learned that, independent of the choice of the system, when a rocket expels fuel in one direction, it gains velocity and therefore momentum in the opposite direction. This mechanism of accelerating a rocket or spaceship is called jet propulsion.

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5.7  Meteorites, radioactive decay, and two-dimensional collisions: Putting it all together        171

The force exerted by the fuel on a rocket during jet propulsion is called thrust. Typical large rocket thrusts measure in mega-newtons 1106 N2, and exhaust speeds are more than 10 times the speed of sound. Thrust provides the necessary impulse to change a rocket’s momentum. You can observe the principles of jet propulsion using a long, narrow balloon. Blow up the balloon; then open the valve and release it. The balloon will shoot away rapidly in the opposite direction of the air streaming out of the balloon’s valve. In reality, a rocket burns its fuel gradually rather than in one short burst; thus its mass is not a constant number but changes gradually. However, the same methods we used in Testing Experiment Table 5.3, together with some calculus, can be applied to determine the change in the rocket’s velocity. The main idea behind the jet propulsion method is that when an object ejects some of its mass in one direction, it accelerates in the opposite direction. This means that the same method that is used to speed up a rocket can also be used to slow it down. To do this, the fuel needs to be ejected in the same direction that the rocket is traveling.

T i p   You can become your own jet propulsion machine by standing on rollerblades or a skateboard and throwing a medicine ball or a heavy book forward or backward.

Review Question 5.6  The following equation is a solution for a problem. State a possible problem.

12.0 kg21-8.0 m>s2 + 0 = 12.0 kg + 58 kg2vx .

5.7 Meteorites, radioactive decay, and two-dimensional collisions: Putting it all together In this section we apply impulse-momentum ideas to analyze meteorites colliding with Earth, radioactive decay of radon in the lungs, and two-dimensional car collisions. We start by analyzing a real meteorite collision with Earth that occurred about 50,000 years ago.

Canyon Diablo Crater In this example we use two separate choices of systems to answer different questions about a meteorite collision with Earth.

Example 5.7  Meteorite impact Arizona’s Meteor Crater (also called Canyon Diablo Crater), shown in Figure 5.9 , was produced 50,000 years ago by the impact of a 3 * 108 -kg meteorite traveling at 1.3 * 104 m>s (29,000 mi/h). The crater is about 200 m deep. Estimate (a) the change in Earth’s velocity as a result of the impact and (b) the average force exerted by the meteorite on Earth during the collision. Sketch and translate  A sketch of the process is shown on the next page. To analyze Earth’s motion, we choose a coordinate system at rest with respect to Earth. The origin of the coordinate axis is at the point where the meteorite first hits Earth. We keep track of the dot at the bottom of the meteorite. The axis points in the direction of the meteorite’s motion.

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Figure 5.9  Canyon Diablo Crater, site of a meteorite impact 50,000 years ago.

To answer the first question, we choose Earth and the meteorite as the system and use momentum constancy to determine Earth’s change in velocity due to the (continued )

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172        Chapter 5  Impulse and Linear Momentum

constancy to determine the speeds of Earth and the meteorite after they join together:

Keep track of this point.

1mE # 0 + mMvMi y2 + 3 01t f - ti24 = 1mEvEf y + mMvMf y2 1 mMvMi y = mEvE f y + mMvMf y = 1mE + mM2vf y

To estimate the force that the meteorite exerts on Earth during the collision, we use the y-component form of the impulse-momentum equation with the meteorite alone as the system: mMvMi y + FE on M y1tf - ti2 = mMvMf y

The time interval required for the collision [using Eq. (5.9)] is 21yf - y i2

tf - ti = collision. To estimate the average force that the meteorite exerted on Earth during the collision (and that Earth ­exerted on the meteorite), we choose the meteorite alone as the system and use the impulse-momentum equation to answer that question. Simplify and diagram  Assume that the meteorite hits perpendicular to Earth’s surface in the positive y-direction. The first impulse-momentum bar chart below represents the process for the Earth-meteorite system to answer the first question. The second bar chart represents the meteorite alone as the system during its collision with Earth to answer the second question.

Isolated system: momentum is constant.

vMf y + vMi y

Solve and evaluate  To answer the first question, we solve for the final velocity of Earth and meteorite together: vfy =

mM v mE + mM Mi y 3 * 108 kg

=

24

6 * 10 kg + 3 * 108 kg = 7 * 10-13 m>s

11.3 * 104 m>s2

This is so slow that it would take Earth about 50,000 years to travel just 1 m. Since Earth is so much more massive than the meteorite, the meteorite’s impact has extremely little effect on Earth’s motion. For the second question, the time interval for the impact is about tf - t i =

21yf - yi2

vMf y + vMi y

=

21200 m2

11.3 * 10 m>s + 7 * 10-13 m>s2 = 0.031 s

Earth’s impulse on the meteorite causes the meteorite’s momentum to decrease.

Represent mathematically  The y-component of the meteorite’s initial velocity is vMi y = +1.3 * 104 m>s. Earth’s initial velocity is zero (with respect to the object of reference). The y-component of the meteorite’s ­f inal ­v elocity equals Earth’s since the meteorite embeds in Earth. The meteorite’s mass is about 3 * 108 kg and Earth’s mass is 6 * 1024 kg. We use momentum

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4

Like most impulsive collisions, this one was over quickly! Note that we’ve estimated the displacement of the meteorite to be the depth of the crater. Rearranging the impulse-momentum equation as applied to the collision, we find the average force exerted by Earth on the meteorite: FE on M y = =

mM1vMf y - vMi y2

1tf - ti2 13 * 108 kg217 * 10-13 m>s - 1.3 * 104 m>s2

10.031 s2 = -1.3 * 10 N  -1 * 1014 N 14

The force exerted by Earth on the meteorite is negative— it points opposite the direction of the meteorite’s initial

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5.7  Meteorites, radioactive decay, and two-dimensional collisions: Putting it all together        173

velocity. According to Newton’s third law, the force that the meteorite exerts on Earth is positive and has the same magnitude: FM on E y = +1 * 1014 N This sounds like a very large force, but since the mass of Earth is 6 * 1024 kg, this force will cause an acceleration of a little over 10-11 m>s2, a very small number.

Try it yourself: Estimate the change in Earth’s velocity and acceleration if it were hit by a meteorite traveling at the same speed as in the last example, stopping in the same distance, but having mass of 6 * 1019 kg instead of 3 * 108 kg. Answer: About 0.1 m/s and 4 m>s2.

Tip   Notice how the choice of system in Example 5.7 was motivated by the question being investigated. Always think about your goal when deciding what your system will be.

An object breaks into parts (radioactive decay) We will learn in the chapter on nuclear physics (Chapter 28) that the nuclei of some atoms are unstable and spontaneously break apart. In a process called alpha decay, the nucleus of the atom breaks into a daughter nucleus that is slightly smaller and lighter than the original parent nucleus and an even smaller alpha particle (symbolized by a; actually a helium nucleus). For example, radon decays into a polonium nucleus (the daughter) and an alpha particle. Radon is produced by a series of decay reactions starting with heavy elements in the soil, such as uranium. Radon diffuses out of the soil and can enter a home through cracks in its foundation, where it can be inhaled by people living there. Once in the lungs, the radon undergoes alpha decay, releasing fastmoving alpha particles that may cause mutations that could lead to cancer. In the next example, we will analyze alpha decay by radon by using the idea of momentum constancy.

in lungs

An inhaled radioactive radon nucleus resides more or less at rest in a person’s lungs, where it decays to a polonium nucleus and an alpha particle. With what speed does the alpha particle move if the polonium nucleus moves away at 4.0 * 105 m>s relative to the lung tissue? The mass of the polonium nucleus is 54 times greater than the mass of the alpha particle. Sketch and translate  An initial-final sketch of the situation is shown at the right. We choose the system to be the radon nucleus in the initial state, which converts to the polonium nucleus and the alpha particle in the final state. The coordinate system has a positive x-axis pointing in the direction of motion of the alpha particle, with the object of reference being the lung tissue. The initial velocity of the radon nucleus along the x-axis is 0 and the final velocity component of the polonium daughter nucleus is

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vPo fx = -4.0 * 105 m>s. The final velocity component of the alpha particle vaf x is unknown. If the mass of the alpha particle is m, then the mass of the polonium is 54m.

(continued )

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174        Chapter 5  Impulse and Linear Momentum

Simplify and diagram  Assume that there are no ­external forces exerted on the system, meaning that the system is isolated and thus its momentum is constant. The impulsemomentum bar chart below represents the process.

Isolated system: momentum is constant.

Represent mathematically  Use the bar chart to help apply the impulse-momentum equation for the process: mRn102 + 1021tf - ti2 = mPovPof x + mavaf x 1 0 = mPovPof x + mavaf x

Solve and evaluate  Rearranging, we get an expression for the final velocity of the alpha particle in the x-direction: vaf x = -

mPovPof x ma

The sign indicates that the alpha particle is traveling in the positive x-direction opposite the direction of the polonium. The magnitude of this velocity is huge—about one-tenth the speed of light! The speeding alpha particle passes through lung tissue and collides with atoms and molecules, dislodging electrons and creating ions. Radon exposure causes approximately 15,000 cases of lung cancer each year. Try it yourself: Francium nuclei undergo radioactive decay by emitting either an alpha particle or a beta particle (an electron). The alpha particle is about 8000 times more massive than a beta particle. If the particles are emitted with the same speed, in which case is the recoil speed of the nucleus that is left after an alpha or a beta particle is emitted greatest? Answer: Since the mass of the alpha particle is much greater than the mass of the beta particle, and they are traveling with the same speed, the momentum of the ­alpha particle is much greater than the momentum of the beta particle. Therefore, the nucleus that is left would have a greater recoil speed during alpha decay.

The x-component of the velocity of the alpha particle ­after radon decay is vaf x = -

154ma21-4.0 * 105 m>s2 mPovPof x = ma ma

= +2.2 * 107 m>s

Collisions in two dimensions So far, the collisions we have investigated have occurred along one axis. Often, a motor vehicle accident involves two vehicles traveling along perpendicular paths. For these two-dimensional collisions, we can still apply the ideas of ­impulse and momentum, but we will use one impulse-momentum equation for each coordinate axis.

Example 5.9  A 1600-kg pickup truck traveling east at 20 m/s collides with a 1300-kg car traveling north at 16 m/s. The vehicles remain tangled together after the collision. Determine the velocity (magnitude and direction) of the combined wreck immediately after the collision. Sketch and translate  We sketch the initial and final situations of the vehicles. We use a P subscript for the pickup and a C subscript for the car. The initial state is just before the collision; the final state is just after the vehicles collide and are moving together. We choose the two vehicles as the system. The object of reference is Earth; the positive x-axis points east and the positive y-axis points north.

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Simplify and diagram  Force diagrams represent the side view for each vehicle just before the collision. We

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5.7  Meteorites, radioactive decay, and two-dimensional collisions: Putting it all together        175

assume that the friction force exerted by the road is very small compared to the force that each vehicle exerts on the other. Thus, we ignore the impulse due to surface friction during the short collision time interval of about 0.1 s. We then apply momentum constancy in each direction. Impulse-momentum bar charts for the x-direction and for the y-direction are shown below.

The net external force on the system is zero—an isolated system.

The x and y-components of momentum are independently constant.

y-component equation: 11600 kg210 m>s2 + 11300 kg2116 m>s2   = 12900 [email protected] sin u

Divide the left side of the second equation by the left side of the first equation and the right side of the second equation by the right of the first, and cancel the 2900 kg and vP + Cf on the top and bottom of the right side. We get 11300 kg2116 m>s2 sin u = = tan u = 0.65 11600 kg2120 m>s2 cos u

A 33 angle has a 0.65 tangent. Thus, the vehicles move off at 33 above the +x-axis (the north of east direction). We can now use this angle with either the x-component equation or the y-component equation above to determine the speed of the two vehicles immediately after the collision. Using the x-component equation, we get [email protected] =

11600 kg2120 m>s2 = 13 m>s 12900 kg2cos 33

From the y-component equation, we have [email protected] =

Represent mathematically  Now, convert each momentum bar in the x-component bar chart into a term in the x-component form of the impulse-momentum equation [Eq. (5.7x)] and each bar in the y-component bar chart into a term in the y-component form of the impulsemomentum equation [Eq. (5.7y)]. Notice that the x-component of the final velocity vector is [email protected] x = [email protected] cos u and the y-component is vP + Cf x = [email protected] sin u: [email protected] equation: mPvPi x + m CvCi x = 1m P + m [email protected] cos u

[email protected] equation:

mPvPi y + mCvCi y = 1mP + [email protected] sin u

We have two equations and two unknowns [email protected] and u2. We can solve for both unknowns. Solve and evaluate  x-component equation:

11600 kg2120 m>s2 + 11300 kg210 m>s2   = 12900 [email protected] cos u

11300 kg2116 m>s2 = 13 m>s 12900 kg2sin 33

The two equations give the same result for the final speed, a good consistency check. For collisions in which vehicles lock together like this, police investigators commonly use the lengths of the skid marks along with the direction of the vehicles after the collision to determine their initial speeds. This allows them to decide whether either vehicle was exceeding the speed limit before the collision. Try it yourself: Use a limiting case analysis and the x- and y-component forms of the impulse-momentum equation to predict what would happen during the collision if the pickup had infinite mass. Is the answer reasonable? Answer: If we place  in 11300 kg2116 m>s2 sin u = = tan u 11600 kg2120 m>s2 cos u

in place of the 1600-kg mass of the pickup, the left side of the equation becomes zero. Then tan u = 0. The pickup would move straight ahead when hitting the car. In other words, the collision with the car would not change the direction of travel of the pickup. The result seems reasonable if the mass of the pickup was large compared to the mass of the car.

Review Question 5.7  When a meteorite hits Earth, the meteorite’s

motion apparently disappears completely. How can we claim that ­momentum is conserved?

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Summary Words

Mathematical representation

Pictorial and physical representations

Isolated system An isolated system is one in which the objects interact only with each other and not with the environment, or the sum of external forces exerted on it is zero. (Sections 5.1–5.2)

Isolated system

Nonisolated system

Conservation of mass If the system is isolated, its mass is constant. If the system is not isolated, the change in the system’s mass equals the mass delivered to the system or taken away from it.   (Section 5.1)

3 kg

1 kg

4 kg

m

mi + m = mf

0

0

mi

m

u

Linear momentum p is a vector quantity that is the product of an object’s mass u m and velocity v. The total momentum of the system is the sum of the momenta of all objects in the system. (Section 5.2)

mf

pr m

vr

u

u

p = mv    Eq. (5.1) psystem = p1 + p2 + g u

u

u

u

Impulse J Impulse is the product ofu the average external force Fav exerted on an object during a time interval t and that time interval. (Section 5.3)

u

u

J = Fav1tf - ti2  Eq. (5.5)

176

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Questions        177 Generalized impulse-momentum principle If the system is isolated, its momentum is constant. If the system is not isolated, the change in the system’s momentum equals the sum of the impulses exerted on the system during the time interval t = 1tf - ti2. (Section 5.4)

Initial

r 1f

v

v

1

r

v  v2f

2

1

1m1v1i + m2v2i + g2 u

Final r 2i

r 1i

2

u

u

+  Fon Sys t

= 1m1v1f + m2v2f + g2 Eq. (5.8) u

u

x- and y-component forms:

p1ix  p2ix  Jx  p1fx  p2fx

m1v1i x + m2v2i x + Fon Sys x t

= m1v1f x + m2v2f x Eq. (5.9x) 0

m1v1i y + m2v2i y + Fon Sys y t

vr1i

= m1v1f y + m2v2f y Eq. (5.9y)

vr1f

1

2

1

2

p1ix  J2 on 1x  p1fx

0

For instructor-assigned homework, go to MasteringPhysics. 

Questions Multiple Choice Questions 1. The gravitational force that Earth exerts on an object causes an impulse of + 10 N # s in one experiment and +1 N # s on the same object in another experiment. How can this be? (a) The mass of the obje ct changed. (b) The time intervals during which the force was exerted are different. (c) The magnitudes of the force were different. 2. A bullet fired at a door makes a hole in the door but does not open it. Your finger does not make a hole in the door but does open it. Why? (a) The bullet is too small. (b) The force exerted by the bullet is not enough to open the door. (c) A finger exerts a smaller force but the time interval is much longer. (d) The bullet goes through the door and does not exert a force at all. 3. How would you convince somebody that the momentum of an isolated system is constant? (a) It is a law; thus, you do not need to convince anybody. (b) Use an example from a textbook to show that the sum of the initial and final velocities of the objects involved in a collision are the same. (c) Derive it from Newton’s second and third laws.

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(d) Use it to make predictions about a new experiment, and then compare the outcome to the prediction. (e) Both (c) and (d) will work. 4. A wagon full of medicine balls is rolling along a street. Suddenly one medicine ball (3 kg) falls off the wagon. What happens to the speed of the wagon? (a) The wagon slows down. (b) The speed of the wagon does not change. (c) The wagon speeds up. (d) Additional information about the ball’s motion is needed to answer. 5. When can you apply the idea that momentum is constant to solve a problem? (a) When the system is isolated (b) When the system is not isolated but the time interval when the external forces are exerted is very small (c) When the external forces are much smaller than the internal forces 6. Choose an example in which the momentum of a system is not constant. (a) A bullet shot from a rifle, with the rifle and the bullet as the system (b) A freely falling metal ball, with the ball as the system (c) A freely falling metal ball, with the ball and Earth as the system (d) It is not possible to give an example since the momentum of a system is always constant.

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178        Chapter 5  Impulse and Linear Momentum 7. Why do cannons roll back after each shot? (a) A cannon pushes on a shell and the shell pushes back on the cannon. (b) The momentum of the cannon-shell system is constant. (c) Both a and b are correct. 8. Which is a safer car bumper in a collision: one that is flexible and retracts or one that is rigid? Why? (a) The retractable bumper, because softer things withstand collisions better (b) The retractable bumper, because it extends the time interval of the collision, thus reducing the force exerted on the car (c) The rigid bumper, because it does not change shape so easily 9. Why does an inflated balloon shoot across a room when air is released from it? (a) Because the outside air pushes on the balloon (b) Because the momentum of the balloon-air system is constant (c) Because the air inside the balloon pushes on the balloon, exerting the same force that the balloon exerts on the air (d) Both b and c are correct. 10. In which situation does the momentum of a tennis ball change more? (a) It hits the racket and stops. (b) It hits the racket and flies off in the opposite direction. (c) It misses the racket and continues moving. 11. A toy car with very low friction wheels and axles rests on a level track. In which situation will its speed increase more? (a) It is hit from the rear by a wad of clay that sticks to the car. (b) It is hit by a rubber ball with the same mass and velocity of the clay that rebounds in the opposite direction after hitting the car. 12. A meteorite strikes Earth and forms a crater, decreasing the meteorite’s momentum to zero. Does this phenomenon contradict the conservation of momentum? Choose as many answers as you think are correct. (a) No, because the meteorite system is not isolated (b) No, because in the meteorite-Earth system, Earth acquires momentum lost by the meteorite (c) No, because the meteorite brings momentum from space (d) Yes, because the meteorite is not moving relative to a medium before the collision 13. A 1000-kg car traveling east at 24 m/s collides with a 2000-kg car traveling west at 21 m/s. The cars lock together. What is their velocity immediately after the collision? (a) 3 m/s east (b)  3 m/s west (c) 6 m/s east (d)  6 m/s west (e) 15 m/s east

Conceptual Questions 14. According to a report on traumatic brain injury, woodpeckers smack their heads against trees at a force equivalent to 1200 g’s without suffering brain damage. This statement contains one or more mistakes. Identify the mistakes in this statement. 15. Jim says that momentum is not a conserved quantity because objects can gain and lose momentum. Do you agree or disagree? If you disagree, what can you do to convince Jim of your opinion? 16. Say five important things about momentum (for example, momentum is a vector quantity). How does each statement apply to real life? 17. Three people are observing the same car. One person claims that the car’s momentum is positive, another person claims that it is negative, and the third person says that it is zero. Can they all be right at the same time? Explain. 18. When would a ball hitting a wall have a greater change in ­momentum: when it hits the wall and bounces back at the same speed or when it hits and sticks to the wall? Explain your answer. 19. In the previous question, in which case does the wall exert a greater force on the ball? Explain. 20. Explain the difference between the concepts of constancy and conservation. Provide an example of a conserved quantity and a nonconserved quantity. 21. Why do you believe that momentum is a conserved quantity? 22. A heavy bar falls straight down onto the bed of a rolling truck. What happens to the momentum of the truck at the instant the bar lands on it? Explain. How many correct answers do you think are possible? Make sure you think of what “falls straight down” means. 23. Construct impulse-momentum bar charts to represent a falling ball in (a) a system whose momentum is not constant and (b) a system whose momentum is constant. In the initial state, the ball is at rest; in the final state, the ball is moving. 24. A person moving on rollerblades throws a medicine ball in the direction opposite to her motion. Construct an ­impulse-momentum bar chart for this process. The person is the system. 25. A person moving on rollerblades drops a medicine ball straight down relative to himself. Construct an impulsemomentum bar chart for the system consisting of the ball and Earth for this process. The rollerblader is the object of reference, and the final state is just before the ball hits the ground.

Problems Below, BIO indicates a problem with a biological or medical focus. Problems labeled EST ask you to estimate the answer to a quantitative problem rather than derive a specific answer. Problems marked with require you to make a drawing or graph as part of your solution. Asterisks indicate the level of difficulty of the problem. Problems with no * are considered to be the least difficult. A single * marks intermediate difficult problems. Two ** indicate more difficult problems.

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5.2 Linear momentum 1. You and a friend are playing tennis. (a) What is the magnitude of the momentum of the 0.057-kg tennis ball when it travels at a speed of 30 m/s? (b) At what speed must your 0.32-kg tennis racket move to have the same magnitude momentum as the ball? (c) If you run toward the ball at a speed of 5.0 m/s, and the ball is flying directly at you at a speed of 30 m/s, what

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Problems        179

is the magnitude of the total momentum of the system (you and the ball)? Assume your mass is 60 kg. In ­every case specify the object of reference. 2. You are hitting a tennis ball against a wall. The 0.057-kg tennis ball traveling at 25 m/s strikes the wall and rebounds at the same speed. (a) Determine the ball’s original momentum (magnitude and direction). (b) Determine the ball’s change in momentum (magnitude and direction). What is your object of reference? 3. A ball of mass m and speed v travels horizontally, hits a wall, and rebounds. Another ball of the same mass and traveling at the same speed hits the wall and sticks to it. Which ball has a greater change in momentum as a result of the collision? ­Explain your answer. 4. (a) A 145-g baseball travels at 35 m/s toward a baseball player’s bat (the bat is the object of reference) and rebounds in the opposite direction at 40 m/s. Determine the ball’s momentum change (magnitude and direction). (b) A golfer hits a 0.046-kg golf ball that launches from the grass at a speed of 50 m/s. Determine the ball’s change in momentum. 5. * A 1300-kg car is traveling at a speed of 10 m/s with respect to the ground when the driver accelerates to make a green light. The momentum of the car increases by 12,800 kg # m>s. List all the quantities you can determine using this information and determine three of those quantities. 6. * The rules of tennis specify that the 0.057-kg ball must bounce to a height of between 53 and 58 inches when dropped from a height of 100 inches onto a concrete slab. What is the change in the momentum of the ball during the collision with the concrete? You will have to use some free-fall kinematics to help answer this question. 7. A cart of mass m moving right at speed v with respect to the track collides with a cart of mass 0.7m moving left. What is the initial speed of the second cart if after the collision the carts stick together and stop? 8. A cart of mass m moving right collides with an identical cart moving right at half the speed. The carts stick together. What is their speed after the collision? 9. EST Estimate your momentum when you are walking at your normal pace.

5.3 Impulse and momentum 10. A 100-g apple is falling from a tree. What is the impulse that Earth exerts on it during the first 0.50 s of its fall? The next 0.50 s? 11. * The same 100-g apple is falling from the tree. What is the impulse that Earth exerts on it during the first 0.50 m of its fall? The next 0.50 m? 12. Why does Earth exert the same impulse during the two time intervals in Problem 10 but different impulses during the same distances traveled in Problem 11? 13. * Van hits concrete support In a crash test, a van collides with a concrete support. The stopping time interval for the collision is 0.10 s, and the impulse exerted by the support on the van is 7.5 * 103 N # s. (a) Determine everything you can about the collision using this information. (b) If the van is constructed to collapse more during the collision so that the time interval during which the impulse is exerted is tripled, what is the average force exerted by the concrete support on the van? 14. BIO Force exerted by heart on blood About 80 g of blood is pumped from a person’s heart into the aorta during each

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180        Chapter 5  Impulse and Linear Momentum

Air bag Force (N)

Force (N)

25. * A boxer delivers a Figure P5.25  punch to his oppo5000 nent’s head, which has a mass of 7.0 kg. Use the graph in Fig3000 ure P5.25 to estimate (a) the impulse of the force exerted by 1000 the boxer and (b) the 0 0 0.005 0.010 0.015 speed of the head after Time (s) the punch is delivered. What assumptions did you make? 26. * Air bag force on Figure P5.26  h e a d The graph in Figure P5.26 shows 2000 the time variation of the force that an automobile’s air bag exerts 1000 on a person’s head during a collision. The mass of the head is 0 8.0 kg. Determine (a) 0 50 100 the total impulse of the Time (ms) force exerted by the air bag on the person’s head and (b) the person’s speed just before the collision occurred. 27. * Equation Jeopardy 1 Invent a problem for which the solution is 127 kg21- 3.0 m>s2 + 130 kg21+4.0 m>s2 = 127 kg + 30 kg2v.

28. * Equation Jeopardy 2 Invent a problem for which the solution is

10.020 kg21300 m>s2 - 110 N210.40 s2 = 10.020 kg21100 m>s2.

29. * Write a general impulse-momentum equation that describes the following process: a person skating on rollerblades releases a backpack that falls toward the ground (the process ends before the backpack hits the ground). What is the system, and what are the physical quantities you will use to describe the process?

5.4  Generalized impulse-momentum principle 30. * Two carts (100 g and 150 g) on an air track are separated by a compressed spring. The spring is released. Represent the process with a momentum bar chart (a) with one cart as the system and (b) with both carts as the system. (c) Write expressions for all of the physical quantities you can from this information. Identify your object of reference. 31. * A tennis ball of mass m hits a wall at speed v and rebounds at about the same speed. Represent the process with an impulse-momentum bar chart for the ball as the system. Using the bar chart, develop an expression for the change in the ball’s momentum. What is the object of reference? 32. * A tennis ball traveling at a speed of v stops after hitting a net. Represent the process with an impulse-momentum bar chart for the ball as the system. Develop an expression for the ball’s change in momentum. What is the object of reference? 33. * You drop a happy ball and a sad ball of the same mass from height h (see Figure 5.10). One ball hits the ground and rebounds almost to the original height. The other ball does not

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bounce. Represent each process with a bar chart, starting just before the balls hit the ground to just after the first ball rebounds and when the other ball stops. Choose the ball as the system. 34. * You experiment again with the balls from Problem 33. You drop them from the same height onto a ruler that is placed on the edge of a table (Figure P5.34). One ball knocks the ruler off; the other does not. Represent each process with an impulse-momentum bar chart with (a) the ball as a system and (b) the ball and the ruler as the system. The process starts just before the balls hit the ruler and ends immediately after they hit Figure P5.34  the ruler. Use the bar charts to help explain the difference in the results of the experiment. vri 35. ** You demonstrate hitting a board in a karate class. The speed of your hand as it hits the thick board is 14 m/s with respect to the board, and the mass of your hand is about 0.80 kg. How deep does your hand go into the board before stopping if the collision lasts for 2.0 * 10-3 s? What assumptions did you make? What other quantities can you determine using this information? 36. * You hold a beach ball with your arms extended above your head and then throw it upward. Represent the motion of the ball with an impulse-momentum bar chart for (a) the ball as the system and Figure P5.38  (b) the ball and Earth as the system. (a) 37. * A basketball player drops a 0.60-kg basketpOix JF on Ox pOfx ball vertically so that it is  traveling at 6.0 m/s when it reaches the floor. The ball rebounds upward 0 at a speed of 4.2 m/s. (a) Determine the magnitude and direction of the ball’s change in momen(b) tum. (b) Determine the average net force that the p1ix  p2ix  Jx  p1fx  p2fx floor exerts on the ball if  the collision lasts 0.12 s. 38. * Bar chart Jeopardy Invent a problem for 0 each of the bar charts shown in Figure P5.38.

5.5  Skills for solving impulse-momentum problems 39. * A baseball bat contacts a 0.145-kg baseball for 1.3 * 10-3 s. The average force exerted by the bat on the ball is 8900 N. If the ball has an initial velocity of 36 m/s toward the bat and the force of the bat causes the ball’s motion to reverse direction, what is the ball’s speed as it leaves the bat?

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Problems        181 40. * A tennis ball traveling horizontally at a speed of 40.0 m/s hits a wall and rebounds in the opposite direction. The time interval for the collision is about 0.013 s, and the mass of the ball is 0.059 kg. Make a list of quantities you can determine using this information and determine four of them. Assume that the ball rebounds at the same speed. 41. A cannon mounted on the back of a ship fires a 50-kg cannonball in the horizontal direction at a speed of 150 m/s. If the cannon and ship have a combined mass of 40000 kg and are initially at rest, what is the speed of the ship just after shooting the cannon? What assumptions did you make? 42. * A team in Quebec is playing ice baseball. A 72-kg player who is initially at rest catches a 145-g ball traveling at 18 m/s. If the player’s skates are frictionless, how much time is required for him to glide 5.0 m after catching the ball? 43. A 10-kg sled carrying a 30-kg child glides on a horizontal, frictionless surface at a speed of 6.0 m/s toward the east. The child jumps off the back of the sled, propelling it forward at 20 m/s. What was the child’s velocity in the horizontal direction relative to the ground at the instant she left the sled? 44. A 10,000-kg coal car on the Great Northern Railroad coasts under a coal storage bin at a speed of 2.0 m/s. As it goes under the bin, 1000 kg of coal is dropped into the car. What is the final speed of the loaded car? 45. * Avoiding chest injury A person in a car during a sudden stop can experience potentially serious chest injuries if the combined force exerted by the seat belt and shoulder strap exceeds 16,000 N. Describe what it would take to avoid injury by estimating (a) the minimum stopping time interval and (b) the corresponding stopping distance, assuming an initial speed of 16 m/s. Indicate any other assumptions you made. 46. * Bruising apples An apple bruises if a force greater than 8.0 N is exerted on it. Would a 0.10-kg apple be likely to bruise if it falls 2.0 m and stops after sinking 0.060 m into the grass? Explain. 47. * Fast tennis serve The fastest server in women’s tennis is Venus Williams, who recorded a serve of 204 km/h at the French Open in 2007. Suppose that the mass of her racket was 328 g and the mass of the ball was 57 g. If her racket was moving at 200 km/h when it hit the ball, approximately what was the racket’s speed after hitting the ball? Indicate any assumptions you made. 48. * You are in an elevator whose cable has just broken. The elevator is falling at 20 m/s when it starts to hit a shock-­ absorbing device at the bottom of the elevator shaft. If you are to avoid injury, the upward force that the floor of the elevator exerts on your upright body while stopping should be no more than 8000 N. Determine the minimum stopping distance needed to avoid injury (do not forget to include your mass in the calculations). What assumptions did you make? Do these assumptions make the stopping distance smaller or larger than the real-world value? 49. * You jump from the window of a burning hotel and land in a safety net that stops your fall in 1.0 m. Estimate the average force that the net exerts on you if you enter the net at a speed of 24 m/s. What assumptions did you make? If you did not make these assumptions, would the stopping distance be smaller or larger? 50. * Skid marks A car skids to a stop. The length of the skid marks is 50 m. What information do you need in order to decide whether the car was speeding before the driver hit the brakes?

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51. * BIO Leg injuries during car collisions During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N. Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car’s dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass. (a) What minimum stopping time interval in needed to avoid injury to the knee if the person is initially traveling at 15 m/s (34 mi/h)? (b) What is the minimum stopping distance? 52. * BIO Bone fracture The zygomatic bone in the upper part of the cheek can be fractured by a 900-N force lasting 6.0 ms or longer. A hockey puck can easily exert such a force when hitting an unprotected face. (a) What change in velocity of a 0.17-kg hockey puck is needed to provide that impulsive force? What assumptions did you make? (b) A padded facemask doubles the stopping time. By how much does it change the force on the face? Explain. 53. * An impulse of 150 N # s stops your head during a car collision. (a) A crash test dummy’s head stops in 0.020 s, when the cheekbone hits the steering wheel. What is the average force that the wheel exerts on the dummy’s cheekbone? (b) Would this crash fracture a human cheekbone (see Problem 52)? (c)  What is the shortest impact time that a person could sustain without breaking the bone? 54. A cart is moving on a horizontal track when a heavy bag falls vertically onto it. What happens to the speed of the cart? Represent the process with an impulse-momentum bar chart. 55. * A cart is moving on a horizontal track. A heavy bag falls off the cart and moves straight down relative to the cart. Describe what happens to the speed of the cart. Represent your answer with the impulse-momentum bar chart. [Hint: What reference frame will you use when you draw the bar chart?]

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182        Chapter 5  Impulse and Linear Momentum 60. * Car collision A 1180-kg car traveling south at 24 m/s with respect to the ground collides with and attaches to a 2470-kg delivery truck traveling east at 16 m/s. Determine the velocity (magnitude and direction) of the two vehicles when locked together just after the collision. 61. * Ice skaters collide While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respect to the ice. The mass of the other person is 80 kg, and he is traveling north at 9.0 m/s with respect to the ice. You hang on to each other after the collision. In what direction and at what speed are you traveling just after the collision? 62. Drifting space mechanic An astronaut with a mass of 90 kg (including spacesuit and equipment) is drifting away from his spaceship at a speed of 0.20 m/s with respect to the spaceship. The astronaut is equipped only with a 0.50-kg wrench to help him get back to the ship. With what speed and in what direction relative to the spaceship must he throw the wrench for his body to acquire a speed of 0.10 m/s and direct him back toward the spaceship? Explain. 63. * Astronaut flings oxygen tank While the astronaut in Problem 62 is trying to get back to the spaceship, his comrade, a 60-kg astronaut, is floating at rest a distance of 10 m from the spaceship when she runs out of oxygen and fuel to power her back to the spaceship. She removes her oxygen tank (3.0 kg) and flings it away from the ship at a speed of 15 m/s relative to the ship. (a) At what speed relative to the ship does she recoil toward the spaceship? (b) How long must she hold her breath before reaching the ship? 64. Rocket stages A 5000-kg rocket ejects a 10,000-kg package of fuel. Before ejection, the rocket and the fuel travel together at a speed of 200 m/s with respect to distant stars. If after the ejection, the fuel package travels at 50 m/s opposite the direction of its initial motion, what is the velocity of the rocket? 65. * A rocket has just ejected fuel. With the fuel and the rocket as the system, construct an impulse-momentum bar chart for (a) the rocket’s increase in speed and (b) the process of a rocket slowing down due to fuel ejection. (c) Finally, draw bar charts for both situations using the rocket without the fuel as the system. 66. ** You have two carts, a force probe connected to a computer, a motion detector, and an assortment of objects of different masses. Design three experiments to test whether momentum is a conserved quantity. Describe carefully what data you will collect and how you will analyze the data.

General Problems 67. ** EST Estimate the recoil speed of Earth if all of the inhabitants of Canada and the United States simultaneously jumped straight upward from Earth’s surface (reaching heights from several centimeters to a meter or more). Indicate any assumptions that you made in your estimate. 68. * A cart of mass m traveling in the negative x-direction at speed v collides head-on with a cart that has triple the mass and is moving at 60% of the speed of the first cart. The carts stick together after the collision. In which direction and at what speed will they move?

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