Solucionario Fundamentos de Física 9na edición Capitulo 6

6Momentum and Collisions

CLICKER QUESTIONS

Question D1.01

Description: Introducing impulse.

Question

A 100-g lump of clay hits a wall at 70 cm/s and sticks. A 100-g rubber ball hits the same wall at 60 cm/s and rebounds with a speed of 30 cm/s. Which object experiences the larger impulse delivered by the wall during the collision?

70cm/s

100g

100g

60cm/s

1. The clay 2. The ball 3. Both impulses are the same. 4. Cannot be determined without knowing the duration of the collision 5. Cannot be determined without knowing the forces exerted by the wall 6. Both #4 and #5 are true. 7. Cannot be determined for some other reason

Commentary

Purpose: To explore momentum and impulse, especially its vector nature.

Discussion: Impulse is defi ned as the integral of force over time; one must know the force exerted at each point in time to determine an impulse via this defi nition. Since we don’t know either the force exerted by the wall or for how long it is exerted in this situation, it would seem impossible to determine which

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projectile experiences the larger impulse. In other words, we cannot apply the defi nition of impulse to answer this question, because we do not have enough information to do so.

However, the impulse–momentum theorem states that the net impulse delivered to a body is equal to the change in the body’s momentum. This is not another defi nition of impulse, but rather a consequence of Newton’s second law. Since we have enough information to determine momentum and changes in momentum for the objects in this question, we can answer the question.

Both the clay and the ball are 100 g, so we can focus on velocity to compare momentum. The object with the larger change in velocity has the larger change in momentum, and therefore, it has the larger net impulse delivered to it as well.

The clay has the larger velocity initially, and zero velocity after sticking to the wall. You might think that the ball has a smaller change in velocity, since its speed changes only from 60 cm/s to 30 cm/s.

But velocity and momentum are vectors, and the change in a vector depends on direction. When the ball’s velocity changes from 60 cm/s (right) to 30 cm/s (left), the change in velocity is 90 cm/s (left), which is larger than the change in the clay’s velocity. So, the ball experiences the larger impulse delivered by the wall.

Note that we started with a quantity that was impossible to determine: the impulse delivered by the wall. Then, by shifting our focus multiple times, we transformed the question into one that we could answer: compare changes in velocity. Note also that we did not compute any quantities, that is, we did not compute the impulse delivered by the wall. We simply determined, by the most effi cient method possible, which projectile experiences the larger impulse.

Key Points:

• The impulse–momentum theorem states that the impulse delivered to an object is equal to the change in its momentum.

• Applying the defi nition of impulse requires you to know the details of a force—how long it acted for and what the force was equal to at every moment of that time—but applying the impulse–momentum theorem does not.

• Velocity, momentum, and impulse are vector quantities. Impulse is equal to the change in an object’s momentum, not to the change in the magnitude of its momentum. Signs and directions must be accounted for.

• Effi cient analysis and problem solving often involves changing the question from an unanswerable one to a manageable one.

For Instructors Only

We are taking for granted that the net impulse is the impulse delivered by the wall during the collision.

Students often treat momentum as a scalar quantity, rather than as a vector. Since the magnitude of momentum changes by a smaller amount for the ball, students often conclude that its change in momentum is smaller as well.

Some students might fi nd components more useful. The clay’s horizontal component of velocity changes from +70 cm/s to 0, for a change of –70 cm/s. The ball’s horizontal component of velocity changes from +60 cm/s to –30 cm/s, for a change of –90 cm/s.

It is useful to compare the ball’s situation to a hypothetical one in which the fi nal velocity of the ball is 30 cm/s (right). (For instance, imagine that it is thrown at a piece of paper such that it breaks through the


Momentum and Collisions 247

paper but slows down during the process.) The change in velocity is now only 30 cm/s (left), which is smaller than the change in the clay’s velocity.

To connect with and develop students’ physical intuition, it is useful to fi nd an analogous situation in which the student would be providing the impulse. For instance, a friend is rolling toward you at constant speed on a skateboard. In one situation, you apply a force long enough to stop your friend. In another, you apply the same force, but this time you reverse his direction. The force is in the same direction for slowing down and speeding up again, yet the force is exerted over a much longer time period when the direction is reversed, so the impulse is larger.

This is a good opportunity to discuss expert-like problem solving. The original question is unanswerable using the defi nition of impulse. By shifting the focus to net impulse, then to momentum and change in momentum, and fi nally to velocity and change in velocity, the question becomes trivial, provided students understand how to fi nd the change in velocity.

Some students will compute the individual momenta of both projectiles before and after the collision, then compute the net impulses and compare them to answer the question. All of these computation are not needed, as only the comparison is called for. Students should be discouraged from computing quantities when doing a comparison.

Question D1.02a

Description: Distinguishing work from impulse.

Question

A block having mass m travels along a horizontal frictionless surface with speed v. What is the smallest amount of work that must be done on the block to reverse its direction?

1. −mv2

2. − 12

2mv 3. 0 4. 1

22mv

5. mv2 6. None of the above 7. Cannot be determined

Commentary

Purpose: To hone your understanding of work and its connection to energy.

Discussion: You might think that you need to know exactly what is meant by “reverse its direction” to answer this question. Rather, the intent is that you consider all possible motion, and then identify which requires the smallest amount of work.

One interpretation is that the block is stopped and then given a very small velocity in the opposite direction. In this case, the change in kinetic energy is very close to − 1

22mv , so by the work–kinetic energy

theorem, this is also the amount of work done.

Another interpretation is that the block has the same speed v but is moving in the opposite direction. The new kinetic energy is the same as the old, 1

22mv , so the change in kinetic energy is zero, and therefore, the

total work done is zero too. Since zero is the smallest number, this must be the smallest amount of work done on the block to reverse its direction.


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There are many ways to accomplish this. For instance, the block could hit a spring. The spring fi rst does − 1

22mv of work to stop the block, then 1

22mv of work to speed it up again in the opposite direction, for a

total work of zero. The block could also slide up a curved ramp and then slide back down again. You can probably think of lots of other situations in which the block reverses direction without changing its kinetic energy. In each case, the total work done is zero.

Key Points:

• According to the work–kinetic energy theorem, the total work done on an object equals the change in its kinetic energy.

• Energy is a scalar so direction doesn’t matter.

• Work can be positive or negative.

• An object can change direction without having any work done on it.


This question will inspire a lively debate, partially because it is unclear what is meant by “reversing the direction” of the block. However, they do not need to know. In essence, the question is asking them to clari-fy what is meant by it by fi nding the situation with the smallest amount of work being done. It is important, therefore, that you do not clarify the situation at all. When students ask you what is meant by the phrase, tell them that everything they need to answer the question is already there.

Most students will not seriously entertain the “zero” answer. Many will not even believe that it is possible.

All of the other answers will be well represented, and each for good reasons:

• Students who answer (1) (−mv2) will most likely think that the block reverses direction by moving with speed v in the opposite direction. They have misapplied the work–kinetic energy theorem, treating kinetic energy as a vector, whose sign determines the direction.

• Students who answer (2) (− 12

2mv ) will most likely think that the block reverses direction by moving at an infi nitesimally small speed in the opposite direction. They have correctly applied the work–kinetic energy theorem and wisely ignored the negligible amount of kinetic energy the block has, but they have failed to fi nd the “smallest” amount of work, though it is the “most negative.”

• Students who answer (4) ( 12

2mv ) will make many of the same assumptions as those who answer (2). The main difference is that they also think that work is always positive, and they will likely have trouble understanding how work can be negative.

• Students who answer (5) (mv2), like those who answer (1), will likely think that the block reverses direction by moving at the same speed v in the opposite direction. The difference is that they will also think that you must do 1

22mv of work to stop the block and then another 1

22mv of work to get it up to

speed v again.

• Students who answer (6) (none of the above) will likely make the same assumption about the motion as those who answer (2) and (4), but since the block is moving, however slightly, the amount of work will be slightly different from − 1

22mv or 1

22mv .

• Students who answer (7) (impossible to determine) most likely do not appreciate the connection between work and change in kinetic energy. They might be trying to apply the defi nition of work. Since there is no information about forces or displacements, then it is impossible to determine. Another reason to choose (7) is that it has not been specifi ed exactly what is meant by “reverse its direction.” (This is largely irrelevant, because students should consider all possible interpretations, and then choose the one requiring the smallest amount of work.)



This is an excellent problem for engaging students in a discussion of work and energy. A mass traveling in the opposite direction with the same speed would have the same kinetic energy. The work–kinetic energy theorem then states that no total work need be done on the mass. The work–kinetic energy theorem also resolves any ambiguity in the sign of the work if the mass is just brought to rest.

It is easy to demonstrate several situations for which an object reverses its direction and no total work is done. All it requires is a conservative force. For example, let a ball roll up an incline and then back down. Or, allow a mass to encounter a spring. Or, have a marble roll around a semicircular track. This last case is interesting because the force acting on the mass (Normal) does no work; the kinetic energy of the marble is constant, even though it has reversed direction.

Discussion Questions:

• Draw a diagram indicating the direction of motion and the direction of the force acting on the mass. What is the direction of the displacement?

• If the surface had friction and the mass just slid until it stopped, how much work would the friction force do?

Question D1.02b

Description: Distinguishing work from impulse.

Question

A block having mass m travels along a horizontal frictionless surface with velocity vx. What impulse must be delivered to the block to reverse its direction?

1. –2mvx

2. –mvx

3. 0 4. mv

x

5. 2mvx

6. None of the above 7. Cannot be determined

Commentary

Purpose: To hone the concept of impulse and its connection to changes in momentum, and (in conjunction with the previous question in the set) distinguish it from work.

Discussion: There are many defensible answers to this question, but all involve the same principle. The impulse–momentum theorem states that the net impulse delivered to an object is exactly equal to its change in momentum.

One way to reverse direction is to move at a very small velocity in the opposite direction. The initial momentum is mvx, and the fi nal momentum is very close to zero, so the change in momentum is very close to –mvx. Thus, the impulse delivered to the block is also –mvx.

Another way to reverse direction is to move in the opposite direction with the same speed. Mathematically, the fi nal velocity is –vx, so the fi nal momentum is –mvx, and the change in momentum is –2mvx. Note that the change in momentum is not zero, even though the speed has not changed. This is because momentum, unlike energy, is a vector quantity, so direction matters. Momentum is mass velocity× , not mass speed× .


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Note that neither of the “positive” answers is appropriate. An impulse of mvx or 2mvx would cause in increase in speed without changing the direction of the block.

If you are unsure what to assume for “reversing direction,” then “impossible to determine” is a valid choice.

Key Points:

• According to the impulse–momentum theorem, the impulse delivered to an object is equal to the change in its momentum.

• Impulse and momentum are vector quantities, so direction matters.

• The best answer to a question can depend on how you interpret the question.


(1), (2), and (6) are all defensible answers depending upon how students interpret the question. This is a good opportunity for stressing to students that it is their assumptions and their reasoning that are most important, not whether an answer is correct. It is simply too easy to get the right answer for the wrong reason.

Impulse is a vector. Using the impulse–momentum theorem, the change in momentum must be at least –mvx to reverse direction.

Students might answer (3) if they think that the momentum does not change since the speed is the same as before.

Students might answer (4) or (5) if they think that impulse must be a positive number.

When presented back to back, questions 163a and 163b highlight the similarities and differences between work and impulse.

Question D1.03

Description: Introducing momentum conservation for an inelastic collision.

Question

A 500-g lump of clay is dropped onto a 1-kg cart moving at 60 cm/s. The clay is moving at 30 cm/s just before landing on the cart.



After the collision, the speed of the cart and clay is closest to:

1. 60 cm/s 2. 50 cm/s 3. 40 cm/s 4. 30 cm/s 5. 20 cm/s 6. 10 cm/s 7. 0 cm/s 8. Impossible to determine

Commentary

Purpose: To hone your understanding of momentum and momentum conservation, especially its vector nature.

Discussion: Momentum is conserved in a collision when no net impulse is delivered by external forces. (Internal forces cannot deliver a net impulse.)

The momentum of the cart-and-clay system is not conserved for this collision: the clay was moving verti-cally before the collision, and is moving horizontally afterwards. The vertical momentum of the system was nonzero and downward before the collision, and zero afterward. This change in momentum occurred because of the impulse delivered by the fl oor to the cart.

However, momentum is a vector quantity, and no external forces deliver an impulse with any horizon-tal component to the system. Thus, the system’s horizontal momentum is conserved, and this allows us to solve for the fi nal velocity of the combined cart and clay. The cart’s initial horizontal momentum of 1 60kg cm s× plus the clays initial horizontal momentum of zero must equal the fi nal system’s combined horizontal momentum of 1 5. kg final× v so the fi nal speed must be 40 cm/s.

Key Points:

• Momentum is conserved in a collision whenever external forces deliver no net impulse to the system.

• Momentum is a vector quantity, and one component may be conserved while another is not.


Some students will view this problem one-dimensionally, consider only the horizontal components, and arrive at the correct answer without ever entertaining the impulse delivered by the fl oor or realizing that total momentum is not conserved. This is neither incorrect nor undesirable, but having these students explicitly consider the vertical aspects of the problem will enhance their overall understanding of momen-tum and impulse.

Common mistakes include estimating the fi nal speed without taking into account the given masses of the cart and clay, and reversing the two masses in the calculation to yield 20 cm/s.

If students add the momenta of the clay and the cart as scalars, they will get 50 cm/s as the fi nal speed.

Productive discussion can be stimulated by asking students to identify the impulse delivered to the cart by the clay and to determine both components (downwards and leftwards) of it.


252 Chapter 6

Additional Questions:

1. (a) What is the net impulse delivered to the clay during the collision? (b) What agent(s) are responsible for this impulse? Only the cart is touching the clay, so it delivers an impulse up and to the right equal to the change in the clay’s momentum. 2. (a) What is the net impulse delivered to the cart during the collision? (b) What agent(s) are responsible for this impulse? Both the clay and the fl oor deliver impulses, but the net impulse is to the left. The downward component of the impulse delivered by the clay balances the upward impulse delivered by the fl oor.

Question D1.04a

Description: Developing understanding of momentum conservation.

Question

Cart A with momentum p (in magnitude) collides with cart B at rest and rebounds. What can you say about the momentum of cart B after the collision?

p

1. It is smaller than p. 2. It is equal to p. 3. It is larger than p. 4. It is impossible to predict what the momentum of cart B will be after the collision.

Commentary


Discussion: Momentum is conserved in a collision when no net impulse is delivered by external forces. (Internal forces cannot deliver a net impulse.) Even though each cart receives an impulse from the other, the net force on the system is zero, so the total momentum of the two-cart system is constant.

Cart A rebounds after the collision. This means that its momentum is initially positive and ends up negative, so the magnitude of its change in momentum and thus of the impulse it receives is larger than p. Therefore, cart B also receives an impulse larger than p. Since it was not moving before the collision, its fi nal momen-tum must be larger than p as well.

For example, if cart A began with 3 units of momentum and ended up with –1 unit, it would have received an impulse of –4 units. Cart B would therefore receive an impulse of +4 units, so it would have 4 units of momentum after the collision.

Note that cart A must have a smaller mass than cart B for this collision to occur. Also note that even though cart B ends up with more momentum than cart A started with, it has less kinetic energy. Thus, momentum is not “transferred” from one object to another the way energy is.



Key Points:


• Momentum is a vector quantity. When calculating changes in momentum, you must consider its direction as well as its magnitude for each object.

• When an object reverses direction, the magnitude of its change in momentum is larger than the magnitude of its initial momentum.


This is the fi rst of two questions exploring simple reasoning with two-body collisions.

Many students will think that there is not enough information because they cannot compute the fi nal momentum of cart B. Part of this question’s intent is to demonstrate how many conclusions can be drawn from reasoning even when a computation is not possible.

Other students may think that momentum is transferred the same way energy is transferred: that is, that an object cannot give another object any more momentum than it has originally. Similarly, some students may treat momentum as a scalar quantity and think that the change in momentum of cart A is smaller than p: it has some momentum after the collision (direction does not matter), so its change is smaller than p.

Question D1.04b

Description: Developing understanding of momentum conservation.

Question

Cart A with momentum p (in magnitude) and speed 2v collides with cart B at rest. After the collision, cart A has speed v and cart B has speed 3v, as shown. What can you say about the momentum of cart B after the collision?

p, 2v

v 3v

1. It is smaller than p. 2. It is equal to p. 3. It is larger than p. 4. It is impossible to predict what the momentum of cart B will be after the collision.

Commentary



254 Chapter 6

Discussion: Momentum is conserved in a collision when no net impulse is delivered by external forces. (Internal forces cannot deliver a net impulse.) Even though both carts receive an impulse from the other, the net force on the system is zero, so the total momentum of the two-cart system is constant.

Cart A continues in the same direction after the collision. This means that its the magnitude of its change in momentum, and thus of the impulse it receives, is smaller than p. (Otherwise, it would change direction.) Therefore, cart B also receives an impulse smaller than p. And since it was not moving before the collision, its fi nal momentum must be smaller than p as well.

It is irrelevant that cart B is moving at 3v after the collision. Though its velocity is greater, its momentum must be smaller, so we can deduce that its mass must be smaller as well.

Key Points:


• Momentum depends on both mass and velocity. A large speed does not always indicate a large momentum.


This is the second of two questions exploring simple reasoning with two-body collisions.

Students may think that there is not enough information because they cannot compute the fi nal momentum of cart B. Part of this question’s intent is to demonstrate how many conclusions can be drawn from reason-ing even when a computation is not possible.

Some students will choose the correct answer without fully appreciating the situation. They might believe that cart B simply cannot ever end up with more momentum than cart A, so they will always say that the momentum of B is smaller than p. Explicitly or implicitly, they are treating momentum as a scalar quantity. Thus, it is imperative to fi nd out why students are choosing the responses they do during the discussion phase of the question.

Question D1.05a

Description: Understanding inelastic collisions, and integrating force and kinematic ideas.

Question

A 500-g block traveling at 60 cm/s slides onto a 1-kg cart as shown. The block stops sliding before reaching the end of the cart.

As the block is sliding on the cart, which object has the larger net force on it?

1. The cart 2. The block 3. Neither; the net force is the same. 4. Impossible to determine



Commentary

Purpose: To explore a totally inelastic collision through kinematics and Newton’s laws.

Discussion: There are three forces on the block: (1) gravitation due to the Earth, down; (2) normal due to the cart, up; and (3) kinetic friction due to the cart, left. The gravitational force and the normal force bal-ance, so the net force is the kinetic friction force exerted by the cart.

There are four forces on the cart: (1) gravitation due to the Earth, down; (2) normal due to the fl oor, up; (3) normal due to the block, down; and (4) kinetic friction due to the block, right. (We are ignoring any frictional effects in the wheels of the cart.) The normal force exerted by the fl oor balances the weight of the cart and the normal force exerted by the block, so the net force is the kinetic friction force exerted by the block.

The friction forces exerted by the block and cart on each other are an action-reaction pair, which means they are equal in magnitude and opposite in direction (as required by Newton’s third law). Thus, the net forces on the two objects are equally large, though in opposite directions.

Note that the friction force on the block slows it down, but the friction force on the cart speeds it up. When the two are moving at the same speed, there is no longer any friction force on either one.

Key Points:

• When two objects interact, the forces they exert on each other have the same magnitude and opposite direction (Newton’s third law).

• You do not need to determine the magnitude of the friction forces to know that they are equal.


This is the fi rst of three questions analyzing a totally inelastic collision via kinematics and Newton’s laws. The set helps to set up a later presentation of momentum ideas, develops students’ understanding of Newtonian’s laws by reasoning with them in a new context, and develops their understanding of inelastic collisions. Students do not need to know anything about momentum to answer any of the questions in the set.

It is common for students who recognize that the block has a larger acceleration than the cart to reason that the force on the block must also be larger. The following question asks specifi cally about the accelerations to help resolve this misconception.

It is also common for students to think the question is impossible to answer without knowing the coeffi cient of kinetic friction.


256 Chapter 6

Question D1.05b


Question


As the block is sliding on the cart, which object has the larger acceleration?

1. The cart 2. The block 3. Neither; the acceleration is the same. 4. Impossible to determine

Commentary


Discussion: The net forces on the block and cart are the same (as discussed in the previous question), though we do not know what the actual value is. According to Newton’s second law, if each experiences the same net force but the cart is twice as massive, it must have half the acceleration.

Note that the block must be sliding on the cart for the accelerations to be different. As soon as their veloci-ties are the same, the block stops sliding, the friction force becomes zero, and the accelerations are the same (i.e., zero).

Key Points:

• Newton’s second law relates the acceleration of an object to the net force it experiences.

• Two objects experiencing the same net force can have different resulting accelerations, if their masses are different.


This is the second of three questions analyzing a totally inelastic collision via kinematics and Newton’s laws. The set helps to set up a later presentation of momentum ideas, develops students’ understanding of Newton’s laws by reasoning with them in a new context, and develops their understanding of inelas-tic collisions. Students do not need to know anything about momentum to answer any of the questions in the set.

This question, in conjunction with the previous, can help determine whether any of your students still have lingering misunderstandings about Newton’s second law and the distinction between force and acceleration.



Question D1.05c


Question


The fi nal speed of the cart is closest to:

1. 0 2. 10 cm/s 3. 20 cm/s 4. 30 cm/s 5. 40 cm/s 6. 50 cm/s 7. 60 cm/s 8. Impossible to determine

Commentary


Discussion: The net forces on the block and the cart are the same, which means the block has an accelera-tion twice as large as that of the cart (as discussed in the previous two questions). Even though we do not know the actual value of the accelerations, we can still predict what the fi nal speed of the cart will be.

The block will come to rest relative to the cart after some amount of time spent sliding. Since the accelera-tion of the block is twice as large as the acceleration of the cart, its velocity will change twice as much as the cart’s in that time period. (Acceleration is the rate of change of velocity.) So, for every 1 cm/s the cart gains, the block must lose 2 cm/s.

Eventually, the cart and the block are traveling to the right at the same speed. If the block’s speed has changed twice as much as the cart’s, then the fi nal speed must be 20 cm/s: the block loses 40 cm/s while the cart gains 20 cm/s.

This situation is, basically, a collision in which two objects stick together. We call this a totally inelastic collision. It is “inelastic” because kinetic energy is not conserved. It is “total” or “complete” because the maximum amount of kinetic energy is lost and the two objects colliding stick together and move at the same velocity.


258 Chapter 6

Key Points:

• Even without knowing actual numerical values for the coeffi cient of friction, forces, accelerations, or time spent sliding, you can reason your way to the answer.

• Collisions can be analyzed with only kinematics and Newton’s laws (though momentum and energy ideas make it easier).


This is the third of three questions analyzing a totally inelastic collision via kinematics and Newton’s laws. The set helps to set up a later presentation of momentum ideas, develops students’ understanding of Newton’s Laws by reasoning with them in a new context, and develops their understanding of inelastic col-lisions. Students do not need to know anything about momentum to answer any of the questions in the set, but they need to know Newton’s laws and kinematics. Graphs are useful as well. They will learn that with some basic reasoning with force and motion ideas they can predict what the fi nal speed of the block-cart system will be.

Students may try to use momentum ideas, even if you have not introduced them yet. They should be encouraged to use force and motion ideas, building on the previous two questions in this set.

A graph of velocity vs. time might help some students. We do not know the exact slopes, but we do know that the slope for the block (though negative) is twice as large as that for the cart. When the two graphs meet, the slopes become zero, because when the block stops sliding, there is zero net force on each. It does not matter whether the coeffi cient of friction is small or large; the fi nal speed will be the same: 20 cm/s. The size of the coeffi cient of friction merely affects how long it will take for the “collision” to take place.

Time

Vel

ocit

y (c

m/s

)

0

20

40

60

Small m

Large m

Some students may choose answer (4), 30 cm/s, because think that the fi nal velocity of the block is zero: because it has “stopped sliding.” They should be reminded that they must view the entire process within one inertial reference frame, e.g., the fl oor. If the block is at rest relative to the fl oor, and the cart is moving to the right, then the block is sliding on the cart. It is impossible for the block to be moving slower than the cart (relative to the fl oor).



Question D1.06

Description: Integrating ideas about mass, velocity, momentum, and impulse in the context of a collision.

Question

Compare two collisions that are perfectly inelastic. In case (A) a car traveling with speed V collides head-on with a sports car having half the mass and traveling in the opposite direction with twice the speed. In case (B) an identical car traveling with the same speed V collides head-on with a light truck having twice the mass and traveling in the opposite direction with half the speed.

A)

M,V

M,V

M/2, 2V

2M, V/2

B)

In which case is the impulse delivered to the car during the collision larger?

1. A 2. B 3. Both the same 4. Cannot be determined

Commentary

Purpose: To integrate ideas about mass, velocity, momentum, and impulse.

Discussion: Does the sports car in A deliver the larger impulse, because it is traveling so much faster than the truck in B? Or does the truck deliver the larger impulse, because it is so much heavier than the sports car?

By defi nition, the impulse delivered to the car is a function of the force exerted by the sports car or truck and how this force varies with time. Since we can never know this information, we can’t use the defi nition of impulse here.

Another way of fi nding the impulse is via the impulse–momentum theorem, which states that the net impulse delivered to an object is equal to the change in its momentum. So, if we can determine the fi nal velocity of the car, we can fi nd its change in momentum and determine the situation in which the car expe-riences the larger impulse.

A critical bit of information is that the collisions are completely inelastic. This means that the vehicles stick together after the collision. This allows us to determine the fi nal velocity of the car in each case. Before the collision in case A, the car has a momentum of MV and the sports car has a momentum of –MV, so the total momentum of the pair is zero. Thus, after the collision, the stuck-together vehicles will be at rest. (We are choosing a coordinate system in which the positive direction points to the right.) Similarly, the truck has an initial momentum of –MV, so the vehicles are at rest after the crash in case B also.

In both cases the car comes to rest, so in both cases its change in momentum is the same (–MV ). Thus, it must experience the same impulse.


260 Chapter 6

Key Points:

• The impulse–momentum theorem is often an easier way to determine impulse than the defi nition of impulse, since the defi nition requires knowing how the forces vary with time. The impulse–momentum theorem requires knowing only the change in momentum.

• The net impulse exerted on something is equal to the change in its momentum, which depends on both velocity and mass.

• Because momentum is a vector quantity, the total momentum of a system of moving objects can be zero.


Students often think that either the mass or the speed is more critical for determining momentum, changes in momentum, or impulse. It can be hard for them to sort out the effect of increasing one and decreasing the other. By directing their attention to momentum, however, they have only one variable to compare.

Note that if the total momentum of the situations were not zero, the answer to this question would depend on whether the total momentum were positive (to the right) or negative (to the left). For example, if the ini-tial speed of the sports car is 4V instead of 2V and that of the truck is V instead of V/2, the total momentum in each system is –MV, and the fi nal speed of the two vehicles would be –2V/3 for case A and –V/3 for case B, which means the car experiences the larger impulse in case A. But if the sports car and truck are at rest before the collision, the total momentum in the system is MV, and the fi nal speed of the two vehicles would be 2V/3 for case A and V/3 for case B, which means the impulse is larger in case B.

Additional Questions:

1. How would your answer change if the sports car and truck are moving twice as fast before the col-lision? That is, what if the sports car is moving at 4V initially and the truck is moving at V initially? Explain.

2. How would your answer change if the sports car and the truck are not moving before the collision? Explain.

3. How would your answer change if the car is moving twice as fast initially (2V instead of V )? Explain. 4. How would your answer change if the car is at rest before the collision? Explain.

Question D1.07

Description: Developing a general, abstract understanding of energy and momentum conservation in colli-sions.

Question

A cart of mass m is moving with speed v. Is it possible for the cart’s kinetic energy or the magnitude of its momentum to be larger after it collides with something?

m v



1. No; it is impossible. 2. Yes; the kinetic energy can be larger. 3. Yes; the momentum can be larger. 4. Yes; both can be larger.

Commentary

Purpose: To generalize your understanding of collisions and the effects they can have on one of the objects (in terms of energy and momentum).

Discussion: The cart’s momentum has magnitude mv and its kinetic energy is 12 mv2. If the cart’s speed

increases, its momentum and kinetic energy will increase. It is not possible for the kinetic energy but not the magnitude of momentum to increase, or vice versa. Thus, answers (2) and (3) are impossible.

But is it possible for the speed to increase, thus causing momentum and kinetic energy to increase? Yes! For example, the cart could be rammed from behind by something moving faster. Not all collisions are head-on!

Or, it could collide head-on with an object having a much larger momentum. Even if it sticks to the object, its fi nal speed can be larger than its initial speed (though in the opposite direction). For instance, if the cart collides head-on with a cart of mass 3m and speed v moving to the left, and energy is conserved, then the 3m cart is at rest after the collision, and the m cart is moving to the left with speed 2v.

You may be thinking that momentum is always conserved in a collision, and kinetic energy is either conserved or lost depending on whether the collision is elastic or inelastic. This is true, but for the entire system, not for each individual object involved.

Key Points:

• If an object’s kinetic energy increases, the magnitude of its momentum must also increase.

• Energy and momentum can be transferred between objects during a collision, and sometimes one object ends up with more than it had to start.

• When trying to fi gure out whether some general statement about physics is possible, think of extreme examples.


Most collision problems students encounter in physics are head-on and in many cases cause both objects to slow down, so students might erroneously generalize. This question helps to remedy that. Ask your students how many considered a rear-end (or even side-on) collision.

Another common mistake is for students to automatically apply “momentum is conserved” or “energy is conserved or lost” ideas without thinking through what system or bodies these apply to.

Some confusion is possible about whether “larger,” for momentum, depends on the sign as well as the mag-nitude. We intend this question to be about the magnitude only, but as with all ambiguities, the best result is to have the confusion articulated and discuss how the choice of interpretation would affect students’ answers.


262 Chapter 6

Question D1.08

Description: Integrating momentum and energy ideas and developing problem-solving skills.

Question

Two blocks, M = 2m, sit on a horizontal frictionless surface with a compressed massless spring between them.

m M

After the spring is released M has velocity v. The total energy initially stored in the spring was:

1. mv2 2. 2 mv2 3. 3 mv2 4. 4 mv2 5. 5 mv2

6. None of the above 7. Cannot be determined

Commentary

Purpose: To integrate energy and momentum ideas and develop problem solving skills.

Discussion: Momentum and energy are both conserved in this situation, since no outside forces act on the mass-spring-mass system except for gravity and the normal force of the table, which balance each other out. The initial momentum is zero, and the initial energy is potential energy stored in the spring. The fi nal momentum must also be zero, and the total kinetic energy of the blocks must be equal to the initial energy stored in the spring.

Mass M has a fi nal kinetic energy of 12

2M v . Since M = 2m, this is equal to mv2. To conserve momentum, mass m must be moving twice as fast as mass M, so its fi nal kinetic energy is 1

222m v( ) = 2 mv2. Therefore,

the total kinetic energy of the two blocks is 3 mv2, which must be the total amount of energy initially stored in the spring.

Key Points:

• If the sum of all external forces on a system is zero, the system’s total momentum remains constant.

• If the net work done on a system by external forces is zero, the system’s total mechanical energy remains constant.

• Some questions require the use of more than one principle.




Some students may answer (6) because they have confused M and m.

It is important to determine the reasons that any student might select (7). They might be unwilling to assume that the system is initially at rest. Students taking this perspective should not be disconfi rmed, but congratulated for making a critical interpretation of the wording.

QUICK QUIZZES

1. (b). The relation between the kinetic energy of an object and the magnitude of the momentum of that object is KE p m= 2 2 . Thus, when two objects having masses m m1 2< have equal kinetic energies, we may write

p

m

p

m12

1

22

22 2= so p

p

m

m1

2

1

2

1= < and p p1 2<

2. (c). Because the momentum of the system (boy + raft) remains constant with zero magnitude, the raft moves towards the shore as the boy walks away from the shore.

3. (c). The total momentum of the car-truck system is conserved. Hence, any change in momentum of the truck must be counterbalanced by an equal magnitude change of opposite sign in the momentum of the car.

4. (a). The total momentum of the two-object system is zero before collision. To conserve momentum, the momentum of the combined object must be zero after the collision. Thus, the combined object must be at rest after the collision.

5. (a) Perfectly inelastic. Any collision in which the two objects stick together afterwards is perfectly inelastic.

(b) Inelastic. Both the Frisbee and the skater lose speed (and hence kinetic energy) in this collision. Thus, the total kinetic energy of the system is not conserved.

(c) Inelastic. The kinetic energy of the Frisbee is conserved. However, the skater loses speed (and hence, kinetic energy) in this collision. Thus, the total kinetic energy of the system is not conserved.

6. (a). If all of the initial kinetic energy is transformed, then nothing is moving after the collision. Consequently, the fi nal momentum of the system is necessarily zero. Because momentum of the system is conserved, the initial momentum of the system must be zero meaning that the two objects must have had equal magnitude momenta in opposite directions before the collision.


264 Chapter 6

ANSWERS TO MULTIPLE CHOICE QUESTIONS

1. The magnitude of the impulse is

I p p p m m mf i f i f i= = − = − = −( )∆ v v v v

or

I = ( ) −( ) = ⋅0 450 12 8 3 20 4 32. . . .kg m s m s kg m s

making (b) the correct choice.

2. The impulse given to the ball is I F t m m mf i f i= = − = −av ( ) ( )∆ v v v v . Choosing the direction of the fi nal velocity of the ball as the positive direction, this gives

Fm

tf i

av

kg m s=

−( )( ) =

×( ) + − −−v v

∆57 0 10 25 0 213. . ..

. .0

43 7 43 7m s

0.060 skg m s N2( )⎡⎣ ⎤⎦ = ⋅ =

and the correct choice is (c).

3. Assuming that the collision was head-on so that, after impact, the wreckage moves in original direction of the car’s motion, conservation of momentum during the impact gives

m m m m m mc t f c c t t c t+( ) = + = + ( )v v v v0 0 0

or

v v vv

fc

c t

m

m m

m

m m=

+⎛⎝⎜

⎞⎠⎟

=+

⎛⎝

⎞⎠ =

2 3

showing that (c) is the correct choice.

4. The mass in motion after the rice ball is added to the bowl is twice the original moving mass. Therefore, to conserve momentum, the speed of the (rice ball + bowl) after the event must be one half of the initial speed of the bowl (i.e., v vf i= 2). The fi nal kinetic energy is then

KE m m mf fi= +( ) = ( )⎛

⎝⎞⎠ =1

2

1

22

2

122

ball bowl bowlvv

22

1

2 22m

Eibowlv

⎛⎝

⎞⎠ =

and the correct choice is (c).



5. Billiard balls all have the same mass and collisions between them may be considered to be elastic. The dual requirements of conservation of kinetic energy and conservation of momentum in a one-dimensional, elastic collision are summarized by the two relations:

m m m mi i f f1 1 2 2 1 1 2 2v v v v+ = + [1]

and

v v v v1 2 1 2i i f f− = − −( ) [2]

In this case, m m1 2= and the masses cancel out of the fi rst equation. Call the cue ball #1 and the red ball #2 so that v v1 3i = − , v v2i = + , v v1 f = cue, and v v2 f = red. Then, the two equations become:

− + = +3v v v vcue red or v v vcue red+ = −2 [1]

and

− − = − −( )3v v v vcue red or v v vcue red− = 4 [2]

Adding the fi nal versions of these equations yields 2 2v vcue = , or v vcue = . Substituting this result into either [1] or [2] above then yields v vred = −3 . Thus, the correct response for this question is (c).

6. We choose the original direction of motion of the cart as the positive direction. Then, vi = 6 m s and v f = −2 m s. The change in the momentum of the cart is

∆ p m m mf i f i= − = −( ) = ( ) − −( ) = −v v v v 5 2 6 40kg m s m s kkg m s⋅

and choice (c) is the correct answer.

7. As in question 5 above, the requirements of conserving both momentum and kinetic energy in this one-dimensional, elastic collision are summarized by the equations

m m m mi i f f1 1 2 2 1 1 2 2v v v v+ = + [1]

and

v v v v1 2 1 2i i f f− = − −( ) [2]

Choosing eastward as the positive direction, we have m1 0 10= . kg, v1 0 20i = + . m s, m2 0 15= . kg, and v2 0i = . The general equations then become

0 10 0 20 0 15 0 0 10. . . .kg m s kg kg( ) +( ) + ( )( ) = ( )v11 20 15f f+ ( ). kg v

or, after simplifying,

v v1 21 50 0 20f f+ =. . m s [1]

and

0 20 0 1 2. m s − = − +v vf f or v v2 1 0 20f f= + . m s [2]

Substitute the fi nal version of [2] into the fi nal version of [1] to obtain 2 50 0 101. .v f = − m syielding

v1 0 040f = − . m s or v1 0 040f = . m s westward

This means that (d) the correct choice for this question.


266 Chapter 6

8. First, consider the motion of the block, with embedded bullet, from just after impact until it comes to rest. During this time, the only force doing work on the block is the friction force between it and the horizontal surface. The work–energy theorem then gives

W f x m mk f inet total totalº= ( ) = −cos1801

2

1

22 2∆ v v

Since v f = 0 and f n m gk k k= = ( )µ µ total , this becomes

− ( ) = −µk im g x mtotal total∆ 01

22v or vi k g x= ( )2µ ∆

and the speed of the (block with embedded bullet) just after impact is

vi = ( )( )( ) =2 0 400 9 80 8 00 7 92. . . .m s m m s2

Conservation of momentum from just before impact to just after gives m m ibullet totalv v0 = , so the speed of the bullet before impact is

v v0

0 200 0 004

0=

⎛⎝⎜

⎞⎠⎟

= +m

m itotal

bullet

kg kg. .

...

0047 92 404

kgm s m s

⎛⎝⎜

⎞⎠⎟

( ) =

which is seen to be choice (d).

9. With the kinetic energy written as KE p m= 2 2 , we solve for the magnitude of the momentum

as p m KE= ( )2 . The ratio of the fi nal momentum of the rocket to its initial momentum is then given by

p

p

m KE

m KE

m

m

KE

KEf

i

f f

i i

f

i

f

i

=( )( )

=⎛⎝⎜

⎞⎠⎟

( )( ) =

2

2

11

28 2⎛

⎝⎞⎠ = or p pf i= 2

and the correct choice is (a).

10. The kinetic energy of a particle may be written as

KEm m

m

m

m

p

m= = = ( )

=v v v2 2 2 2 2

2 2 2 2.

The ratio of the kinetic energies of two particles is then

KE

KE

p m

p m

p

p

m

m

( )( ) = =

⎛⎝⎜

⎞⎠⎟

⎛⎝⎜

⎞2

1

22

2

12

1

2

1

2

1

2

2

2 ⎠⎠⎟

We see that, if the magnitudes of the momenta are equal p p2 1=( ), the kinetic energies will be equal only if the masses are also equal. The correct response is then (c).



11. Expressing the kinetic energy as KE p m= 2 2 (see questions 9 and 10), we see that the ratio of the magnitudes of the momenta of two particles is

p

p

m KE

m KE

m

m

KE

KE2

1

2 2

1 1

2

1

2

1

2

2=

( )( )

=⎛⎝⎜

⎞⎠⎟

( )( )

Thus, we see that if the particles have equal kinetic energies KE KE( ) = ( )[ ]2 1 , the magnitudes of their momenta are equal only if the masses are also equal. However, momentum is a vector quantity and we can say the two particles have equal momenta only it both the magnitudes and directions are equal, making choice (d) the correct answer.

12. Consider the sketches at the right. The leftmost sketch shows the rocket immediately after the engine is fi red (while the rocket’s velocity is still essentially zero). It has two forces acting on it, an upward thrust F exerted by the burnt fuel being ejected from the engine, and a downward force of gravity. These forces produce the upward acceleration a of the rocket according to Newton’s second law:

ΣF F F May g= − =

Since F Mgg = , the thrust exerted on the rocket by the ejected fuel is

F F Ma M a gg= + = +( )

The rightmost part of the sketch shows a quantity of burnt fuel that was initially at rest within the rocket, but a very short time ∆t later is moving downward at speed v. As this material is ejected, it exerts the upward thrust F on the rocket. By Newton’s third law, the rocket exerts a downward force of equal magnitude on this burnt fuel. This force imparts an impulse I F t p m= ( ) = = −( )∆ ∆ ∆ v 0 to the ejected material. Thus, the rate the rocket is burning and ejecting fuel must be

∆∆m

t

F M a g=

−=

+( ) =×( ) +( )

v v0

3 00 10 36 0 9 805. . .kg mm s

4.50 10 m skg s

2

3

⎡⎣ ⎤⎦×

= ×3 05 103.

and we see that choice (a) is the correct response.

Note: Failure to include the gravitational force in this analysis will lead some students to incor-rectly select choice (b) as their answer.


268 Chapter 6

ANSWERS TO EVEN-NUMBERED CONCEPTUAL QUESTIONS

2. The glass, concrete, and steel were part of a rigid structure that shattered upon impact of the airplanes with the towers and upon collapse of the buildings as the steel support structures weakened due to high temperatures of the burning fuel. The sheets of paper fl oating down were probably not in the vicinity of the direct impact, where they would have burned after being exposed to very high temperatures. The papers were most likely situated on desktops or open fi le cabinets and were blown out of the buildings as they collapsed.

4. No. Only in a precise head-on collision with equal and opposite momentum can both objects end up at rest. Yes. In the second case, assuming equal masses for the two objects, if object 2, originally at rest, is struck head-on by object 1, object 2 will depart with the original velocity of object 1. Then object 1 is left at rest.

6. Since the total momentum of the skater-Frisbee system is conserved, the momentum transferred to the skater equals the magnitude of the change in the Frisbee’s momentum. This is greatest when the skater throws the Frisbee back after catching it.

8. A certain impulse is required to stop the egg. But, if the time during which the momentum change of the egg occurs is increased, the resulting force on the egg is reduced. The time is increased when the sheet billows out as the egg is brought to a stop. The force is reduced low enough so that the egg will not break.

10. The resulting collision is intermediate between an elastic and a completely inelastic collision. Some energy of motion is transformed as the pieces buckle, crumple, and heat up during the colli-sion. Also, a small amount is lost as sound. The most kinetic energy is lost in a head-on collision, so the expectation of damage to the passengers is greatest.

12. The passenger must undergo a certain momentum change in the collision. This means that a certain impulse must be exerted on the passenger by the steering wheel, the window, an air bag, or something. By increasing the time during which this momentum change occurs, the resulting force on the passenger can be decreased.

14. Its speed decreases as its mass increases. There are no external horizontal forces acting on the box, so its momentum cannot change as it moves along the horizontal surface. As the box slowly fi lls with water, its mass increases with time. Because the product mv must be constant, and because m is increasing, the speed of the box must decrease.

PROBLEM SOLUTIONS

6.1 Use p m= v:

(a) p = ×( ) ×( ) = ×− −1 67 10 5 00 10 8 35 1027 6 21. . .kg m s kgg m s⋅

(b) p = ×( ) ×( ) = ⋅−1 50 10 3 00 10 4 502 2. . .kg m s kg m s

(c) p = ( )( ) = ⋅7 1 7505.0 kg 0.0 m s kg m s

(d) p = ×( ) ×( ) = × ⋅5 98 10 2 98 10 1 78 1024 4 29. . .kg m s kg mm s



6.2 From the impulse–momentum theorem,

F t p m mf iav ∆ ∆( ) = = −v v

Thus,

Fm

tf i

av

35 10 kg ft s=

−( )( ) =

×( ) × −( )−v v

∆5 2 0 10 02.

00.0020 s 0

m s

ft skN

−⎛⎝⎜

⎞⎠⎟

=1

3 2811 7

..

6.3 (a) If p pball bullet= , then

vv

ballbullet bullet

ball

kg= =

×( )−m

m

3 00 10 1 53. . 00 10

0 14531 0

3×( )=

m s

kgm s

..

(b) The kinetic energy of the bullet is

KE mbullet bullet bullet

3.00 10 kg= =

×( )−1

2

3 12v

.550 10

23 38 10

3 2

3×( )

= ×m s

J.

while that of the baseball is

KE mball ball ball

.145 kg 1.0 m s= =

( )( ) =1

2

0 3

22

2

v 669 7. J

The bullet has the larger kinetic energy by a factor of 48.4.

6.4 (a) Since the ball was thrown straight upward, it is at rest momentarily (v = 0) at its maximum height. Therefore, p = 0 .

(b) The maximum height is found from v vy y ya y202 2= + ( )∆ with vy = 0.

0 202= + −v y g y( )( )∆ max. Thus,

∆ygy( ) =

max

v02

2

We need the velocity at ∆ ∆y y gy= =( )max 2 402v ; thus v vy y ya y2

02 2= + ( )∆ gives

v vv v

y yy ygg

202 0

202

24 2

= + −( )⎛

⎝⎜⎞

⎠⎟= , or v

vy

y= =0

2

15

2

m s

Therefore,

p m y= =( )( ) = ⋅v0 10 15

21 1

..

kg m skg m s upward

6.5 (a) I F t p m f i= ( ) = = − = ( ) − =av kg m s∆ ∆ v v 84 0 0 6 70 563. . kkg m s⋅

(b) FI

tav

kg m s

sN= = ⋅ =

∆563

0 750751

.

6.6 KEm m

m

m

m

p

m= = = ( )

=v v v2 2 2 2 2

2 2 2 2


270 Chapter 6

6.7 From problem 6.6, KE p m= 2 2 , and hence, p m KE= ( )2 . Thus,

mp

KE=

⋅=

⋅( )( ) =

2 2

2

25 0

2 2751 14

..

kg m s

Jkg

and

v = =( )

= ( )= ( )

=p

m

m KE

m

KE

m

2 2 2 275

1 1422 0

J

kgm s

..

6.8 (a) The impulse delivered by a force is equal to the area under the force versus time curve. From the fi gure at the right, this is seen to be a triangular area having a base of 1 50 1 50 10 3. .ms s= × − and altitude of 18 000 N . Thus,

I = ×( )( ) = ⋅−1

21 50 10 18 000 13 53. .s N N s

(b) FI

tav 3

N s

1.50 10 sN= = ⋅

×= × =−∆

13 59 00 10 9 003.. . kN

6.9 (a) We choose the positive direction to be the direction of the fi nal velocity of the ball.

I p m f i= = −( ) = ( ) + − −( )∆ v v 0 280 22 0 15 0. . .kg m s m s⎡⎡⎣ ⎤⎦

or

I = + ⋅ = ⋅10 4 10 4. .kg m s kg m s in the direction off the final velocity

(b) The average force the player exerts on the ball is

FI

tav

kg m s

sN= = ⋅ =

∆10 4

0 060 0173

.

.

By Newton’s third law, the ball exerts a force of equal magnitude back on the player’s fi st.

6.10 (a) FI

tav =∆

, where I is the impulse the man must deliver to the child: I m f= −child v v0 .

Fm

tf

av

child kg mi h

0.10 s=

−=

( ) −v v0 12 0 0 120 0

∆. ..

.447

16 4 103m s

mi hN

⎛⎝⎜

⎞⎠⎟

= ×

or

Fav Nlb

Nlb= ×( )⎛

⎝⎞⎠ = ×6 4 10

0 224 8

11 4 103 3.

..

(b) It is unlikely that the man has suffi cient arm strength to guarantee the safety of the child during a collision. The violent forces during the collision would tear the child from his arms.

(c) The laws are soundly based on physical principles: always wear a seat belt when in a car.

3210

5000

10000

15000

20000

F (N) F�18000 N

t (ms)3210

5000

10000

15000

20000

F (N) F�18000 N

t (ms)



6.11 The velocity of the ball just before impact is found from v vy y ya y202 2= + ∆ as

v v1 02 2 0 2 9 80 1 25 4= − + = − + −( ) −( ) = −y ya y∆ . . .m s m2 995 m s

and the rebound velocity with which it leaves the fl oor is

v v22 2 0 2 9 80 0 960 4= + − = + − −( ) +( ) = +f ya y∆ . . .m s m2 334 m s

The impulse given the ball by the fl oor is then

I F v v v� �� = = ( ) = −( )

= ( ) +

∆ ∆t m m 2 1

0 150 4 34. .kg mm s m s N s 1.39 N s upwar− −( )⎡⎣ ⎤⎦ = + ⋅ = ⋅4 95 1 39. . dd

6.12 Take the direction of the ball’s fi nal velocity (toward the net) to be the +x-direction.

(a) I p m f i= = −( ) = ( ) − −( )∆ v v 0 0600 40 0 50 0. . .kg m s m s⎡⎡⎣ ⎤⎦, giving

I = + ⋅ =5 40. kg m s 5 40. N s toward the net⋅ .

(b) Work KE m f i= = −( )

=( ) ( )

∆ 1

2

0 0600 40 0

2 2

2

v v

. .kg m s −− ( )⎡⎣ ⎤⎦ = −50 0

227 0

2.

.m s

J

6.13 I F t p m= ( ) = = ( )av ∆ ∆ ∆v

Thus, I m= = ( ) −( ) = ⋅∆v 70 0 5 20 0 364. .kg m s kg m s , and

FI

tav2kg m s

skg m s= = ⋅ = ⋅

∆364

0 832438

.

or

�Fav N directed forward= 438

6.14 Choose toward the east as the positive direction.

(a) The impulse delivered to the ball as it is caught is

I p v v� � � �= = − = − ( ) +( ) = −∆ m mf i 0 0 500 15 0 7. . .kg m s 550 kg m s⋅

or

I�

= ⋅7 50. kg m s westward

(b) The average force exerted by the ball on the receiver is the negative of the average force exerted by the receiver on the ball, or

F FI��

av receiver av ball

k( ) = −( ) = − = − −∆t

7 50. gg m s

sN

⋅⎛⎝

⎞⎠ = +

0 0200375

.

F��

av receiverN eastward( ) = 375


272 Chapter 6

6.15 (a) The impulse equals the area under the F versus t graph. This area is the sum of the area of the rectangle plus the area of the triangle. Thus,

I = ( )( ) + ( )( ) = ⋅2 0 3 01

22 0 2 0 8 0. . . . .N s N s N s

(b) I F t p m f i= ( ) = = −( )av ∆ ∆ v v

8 0 1 5 0 5 3. . , .N s kg giving m s⋅ = ( ) − =v vf f

(c) I F t p mI

mf i f i= = = − = +av , so( ) ( )∆ ∆ v v v v

v f = − + ⋅ =2 08 0

1 53 3.

.

..m s

N s

kgm s

6.16 (a) Impulse = area under curve = (two triangular areas of altitude 4.00 N and base 2.00 s) + (one rectangular area of width 1.00 s and height of 4.00 N.) Thus,

I =( )( )⎡

⎣⎢⎤⎦⎥

+ ( )( )24 00 2 00

24 00 1 00

. .. .

N sN s == ⋅12 0. N s

(b) I F t p mI

mf i f i= ( ) = = −( ) = +av , so∆ ∆ v v v v

N s

.00 kgm sv f = + ⋅ =0

12 0

26 00

..

(c) v vf i

I

m= + = − + ⋅ =2 00

12 04 00.

..m s

N s

2.00 kgm s

6.17 (a) The impulse is the area under the curve between 0 and 3.0 s. This is

I = = ⋅( . )( . )4 0 3 0 12N s N s

(b) The area under the curve between 0 and 5.0 s is

I = + − = ⋅( . )( . ) ( . )( . ) .4 0 3 0 2 0 2 0 8 0N s N s N s

(c) I F t p mI

mf i f i= ( ) = = −( ) = +av , so∆ ∆ v v v v

At 3.0 s: v vf i

I

m= + = + ⋅ =0

128 0

N s

1.50 kgm s.

At 5.0 s: v vf i

I

m= + = + ⋅ =0

85 3

.0 N s

1.50 kgm s.



6.18 Fp��

av = ∆∆t

so Fp

txx

av( ) = ∆∆

and Fp

ty

yav( ) =

∆∆

Fm

t

my

y f y i

av

º( ) =( ) − ( )⎡

⎣⎢⎤⎦⎥ = −v v v v

∆cos . cos60 0 660 0

0. º[ ] =

∆t

Fm

t

mx

x f x i

av

º( ) =( ) − ( )⎡

⎣⎤⎦ =

−( ) − +v v v v∆

sin . s60 0 iin .60 0º( )[ ]∆t

= − ° =− ( )( )2 60 0 2 3 00 10 0 60 0m

t

vsin . . . sin .

∆kg m s °°

= −0.200 s

N260

Thus, F��

av N in the negative -direction or= 260 x pperpendicular to the wall .

6.19 (a) ∆ ∆ ∆t

x x

f i

= =( )+

= ( )+

=v v vav

m

m s

2 2 1 20

0 25 09 60

.

.. ×× −10 2 s

(b) Fp

t

m

tav

kg m s= =

( ) =( )( )

× −∆∆

∆∆

v 1 400 25 0

9 60 10

.

. 2253 65 10

sN= ×.

(c) atav

2 2m s

sm s m s= =

×= =−

∆∆

v 25 0

9 60 10260 2602

.

.(( )⎛

⎝⎜⎞⎠⎟

=1

9 8026 6

m s2

gg

..

6.20 Choose the positive direction to be from the pitcher toward home plate.

(a) � � � � �I F p v v= ( ) = = −( ) = ( ) −(av kg m s∆ ∆t m f i 0 15 22. )) − ( )⎡⎣ ⎤⎦20 m s

� �I F= ( ) = − ⋅ ⋅av kg m s or kg m s toward th∆t 6 3 6 3. . ee pitcher

(b) �

�F

Iav

kg m s

sN= = − ⋅

×= − ×−∆t

6 3

2 0 103 2 103

3.

..

or

�Fav N toward the pitcher= ×3 2 103.

6.21 Requiring that total momentum be conserved gives

m m m mfclub club ball ball club club ball bv v v v+( ) = + aall( )i

or

200 40 46 200 5 0g m s g g 5 m sball( )( ) + ( ) = ( )( ) +v

and

vball m s= 65

�y

60.0°

60.0°

v

v

�x

�y

60.0°

60.0°

v

v

�x


274 Chapter 6

6.22 (a) The mass of the rifl e is

mw

g= = = ⎛

⎝⎞⎠

30 30

9 8

N

9.80 m skg2 .

We choose the direction of the bullet’s motion to be negative. Then, conservation of momentum gives

m m mfrifle rifle bullet bullet rifle riflev v v+( ) = ++( )m

ibullet bulletv

or

30 9 8 5 0 10 3003. .( )⎡⎣ ⎤⎦ + ×( ) −( )−kg kg m sriflev == +0 0

and

vrifle

kg m s

kgm=

×( )( )=

−9 8 5 0 10 300

300 49

3. .. ss

(b) The mass of the man plus rifl e is

m = =73074 5

N

9.80 m skg2 .

We use the same approach as in (a), to fi nd

v = ×⎛⎝⎜

⎞⎠⎟

( ) = ×−

−5 0 10

74 5300 2 0 10

32.

..

kg

kgm s m s

6.23 The velocity of the girl relative to the ice, vGI, is v v vGI GP PI= + where vGP = velocity of girl relative to plank, and vPI velocity of plank relative to ice.= Since we are given that vGP m s= 1 50. , this becomes

v vGI PIm s= +1 50. [1]

(a) Conservation of momentum gives

m mG GI P PIv v+ = 0, or v vPIG

PGI= −

⎛⎝⎜

⎞⎠⎟

m

m [2]

Then, Equation [1] becomes

v vGIG

PGIm s= −

⎛⎝⎜

⎞⎠⎟

1 50.m

m or 1 1 50+

⎛⎝⎜

⎞⎠⎟

=m

mG

PGI m sv .

giving

vGI

m s

kg

50 kg

m s=+

⎛⎝⎜

⎞⎠⎟

=1 50

145 0

1

1 15.

..

(b) Then, using [2] above,

vPI

kg

50 kgm s m s= −

⎛⎝⎜

⎞⎠⎟

( ) = −45 0

11 15 0 346

.. .

or vPI m s directed opposite to the girl’= 0 346. ss motion .



6.24 We shall choose southward as the positive direction.

The mass of the man is

mw

g= = =730

74 5N

9.80 m skg2 .

Then, from conservation of momentum,

we fi nd

m m m mfman man book book man man book bookv v v v+( ) = +( ))i

or

74 5 1 2 5 0 0 0. . .kg kg m sman( ) + ( ) −( ) = +v and vman m s= × −8 1 10 2.

Therefore, the time required to travel the 5.0 m to shore is

t

x= =×

=−∆vman

m

m ss

5 0

8 1 10622

.

.

6.25 (a) Using subscript a for the astronaut and t for the tank, conservation of momentum gives m m m ma af t tf a ai t tiv v v v+ = + . Since both astronaut and tank were initially at rest, this becomes

m ma af t tfv v+ = +0 0 or v vaft

atf

m

m= −

⎛⎝⎜

⎞⎠⎟

The mass of the astronaut alone (after the oxygen tank has been discarded) is ma = 75 0. kg. Taking toward the spacecraft as the positive direction, the velocity imparted to the astronaut is

vaf = −⎛⎝⎜

⎞⎠⎟

−( ) = +12 08 00 1 28

.. .

kg

75.0 kgm s m ss

and the distance she will move in 2.00 min is

d taf= = ( )( ) =v 1 28 120 154. m s s m

(b) By Newton’s third law, when the astronaut exerts a force on the tank, the tank exerts a force back on the astronaut. This reaction force accelerates the astronaut towards the spacecraft.

6.26 (a) Using subscript c for the (fl atcar + cannon) and p for the projectile, conservation of

momentum in the horizontal direction gives m m m mc cf x p pf x c ci x p pi xv v v v( ) + ( ) = ( ) + ( ) .

Assuming that the fl atcar, cannon, and projectile were initially at rest, v vci pi= = 0, giving the initial recoil speed of the (fl atcar + cannon) as

v vcf x

p

cpf x

m

m( ) = ( ) = ⎛

⎝⎞⎠ ×1 00

1 00.

.ton

36.0 ton110 30 0 24 13 m s º m s( ) =cos . .

(b) The fl atcar, cannon, and Earth undergo a change in momentum in the y-direction that is equal an opposite the vertical component of momentum imparted to the projectile. Because of the great mass of Earth, however, the effect goes unnoticed.


276 Chapter 6

6.27 Consider the thrower fi rst, with velocity after the throw of vthrower. Applying conservation of momentum yields

65 0 0 045 0 30 0 65 0. . . .kg kg m sthrower( ) + ( )( ) =v kg kg .50 m s+( )( )0 045 0 2.

or vthrower m s= 2 48. .

Now, consider the (catcher + ball), with velocity of vcatcher after the catch. From momentum conservation,

60 0 0 045 0 0 045 0 30 0. . . .kg kg kgcatcher+( ) = ( )v m s kg( ) + ( )( )60 0 0.

or

vcatcher m s= × −2 25 10 2.

6.28 (a) B exerts a horizontal force on A.

(b) A exerts a force on B that is opposite in direction to the force B exerts on A.

(c) The force on A is equal in magnitude to the force on B, but is oppositely directed.

(d) Yes. The momentum of the system (the two skaters) is conserved because the net external force on the system is zero (neglecting friction).

(e) ∆ ∆ ∆p p p m mx x A x B A A B( ) = ( ) + ( ) = ⇒ −( ) +system

0 0v vvB −( ) =0 0

or v vAB

AB

B

B

m

m

m

m= −

⎛⎝⎜

⎞⎠⎟

= −⎛

⎝⎜

⎞

⎠⎟ ( ) =

0 9002 00

.. m s −− 2 22. m s

� �v vA m s in the direction opposite to= 2 22. BB

6.29 (a) Manm1

Before impact

Wifem2

Couplem1�m2

v1 v2 vf

After impact

(b) The collision is best described as perfectly inelastic , because the skaters remain in contact after the collision.

(c) m m m m f1 1 2 2 1 2v v v+ = +( ) (d) vv v

fm m

m m= +

+1 1 2 2

1 2

(e) vf =( )( ) + ( )( )70 0 8 00 50 0 4 00

70

. . . .

.

kg m s kg m s

00 50 06 33

kg kgm s

+=

..



6.30 Consider a system consisting of arrow and target from the instant just before impact until the instant after the arrow emerges from the target. No external horizontal forces act on the system, so total horizontal momentum must be conserved, or

m m m ma a t t f a a t t iv v v v+( ) = +( )

Thus,

vv v v

a f

a a i t t i t t f

a

m m m

m( ) =

( ) + ( ) − ( )

=( ) +22 5 35. g .. .

..

0 300 2 50 0

22 51 67

m s g m s

gm s

( ) + ( ) −( ) −=

6.31 When Gayle jumps on the sled, conservation of momentum gives

50 0 5 00 50 0 4 00 02. . . .kg kg kg m s+( ) = ( )( ) +v

or the speed of Gayle and the sled as they start down the hill is v2 3 64= . m s.

After Gayle and the sled glide down 5.00 m, conservation of mechanical energy (taking y = 0 at the level of the top of the hill) gives

1

255 0 55 0 9 80 5 003

2. . . .kg kg m s m2( ) + ( )( ) −(v )) = ( )( ) +1

255 0 3 64 0

2. .kg m s

so Gayle’s speed just before the brother hops on is v3 10 5= . m s.

After her Brother jumps on, conservation of momentum yields

55 0 30 0 55 0 10 50 04. . . .kg kg kg m s+( ) = ( )( ) +v

and the speed of Gayle, brother, and sled just after brother hops on is v4 6 82= . m s.

After all slide an additional 10.0 m down (to a level 15.0 m below the level of the hilltop), conservation of mechanical energy from just after brother hops on to the end gives the fi nal speed as

1

285 0 85 0 9 80 15 05

2. . . .kg kg m s m2( ) + ( ) ( ) −(v ))

= .1

285 0 kgg .82 m s kg m s m2( ) ( ) + ( ) ( ) −(6 85 0 9 80 5 00

2. . . ))

or v5 15 6= . m s .

6.32 For each skater, the impulse–momentum theorem gives

Fp

t

m

tav

kg m s

0.100 s= = =

( )( ) =∆∆

∆∆

v 75 0 5 003

. ..775 103× N

Since Fav N< 4500 , there are no broken bones .


278 Chapter 6

6.33 (a) If M is the mass of a single car, conservation of momentum gives

3 3 00 2 1 20M M Mf( ) = ( ) + ( )( )v . .m s m s , or vf = 1 80. m s

(b) The kinetic energy lost is KE KE KEi flost = − , or

KE M M Mlost m s m s= ( ) + ( )( ) − (1

23 00

1

22 1 20

1

23

2 2. . ))( )1 80

2. m s

With M = ×2 00 104. kg, this yields KElost J= ×2 16 104. .

6.34 (a) From conservation of momentum,

3 21 2M M Mf( ) = + ( )v v v

or

v v vf = +( )1

321 2

(b) The kinetic energy before is

KE M MM

i = + ( ) = +( )1

2

1

22

221

222

12

22v v v v

After collision:

KE MM M

f f= ( ) =+( )⎡

⎣⎢⎢

⎤

⎦⎥⎥

= +1

23

3

2

2

9 642 1 2

2

12v

v vv vv v v1 2 2

24+( )

or

KEM M M

f = + +6

2

3

2

312

1 2 22v v v v

The kinetic energy lost is

KE KE M M Mi f− = −⎛⎝

⎞⎠ + −⎛

⎝⎞⎠ −1

2

1

61

2

3

2

312

22

1 2v v v v

or

KE KEM M

i f− = + −( ) = −( )3

231

222

1 2 1 2

2v v v v v v

6.35 (a) Because momentum is conserved even in a perfectly inelastic collision such as this, the

ratio is p pf i = 1 .

(b) p p m m m mf i f i= ⇒ +( ) = + ( )1 2 1 1 2 0v v or vv

fim

m m=

+1 1

1 2

KE m m mi i i= + ( ) =1

2

1

20

1

21 12

2 1 12v v and KE m mf f= +( )1

2 1 22v

so KE

KE

m m

m

m m

m

mf

i

f

i i

i=+( )

=+( )1 2

2

1 12

1 2

1 12

12

1v

v v

v22

1 2

21

1 2m m

m

m m+( )=

+

MMM

M

v1

M M

After

Before

v2

vf

MMM

M

v1

M M

After

Before

v2

vf



6.36 Let us apply conservation of energy to the block from the time just after the bullet has passed through until it reaches maximum height in order to fi nd its speed V just after the collision.

1

2

1

22 2m mgy m mgyi i f fv v+ = + becomes

1

20 02mV mgyf+ = +

or

V g yf= = ( )( ) =2 2 9 80 0 120 1 53. . .m s m m s2

Now use conservation of momentum from before until just after the collision in order to fi nd the initial speed of the bullet, v.

7 0 10 0 1 5 1 53 7 0 103 3. . . .×( ) + = ( )( ) + ×− −kg kg m sv kg 00 m s( )( )2

from which v = ×5 3 102. m s .

6.37 The leftmost part of the sketch depicts the situation from when the actor starts from rest until just before he makes contact with his costar. Using conservation of energy over this period gives

KE PE KE PE i+( ) = +( )1

or

1

20 01 1

2m mgRv + = +

so his speed just before impact is

v1 2 2 9 80 3 75 8 57= = ( )( ) =gR . . .m s m m s2

Now, employing conservation of momentum from just before to just after impact gives

m m m m1 2 0 1 1 2 0+( ) = + ( )v v or vv

01 1

1 2

80 0 8 57

80 0 55 0=

+=

( )( )+

m

m m

. .

. .

kg m s

kg kgm s= 5 08.

Finally, using conservation of energy from just after impact to the end yields

KE PE KE PEf+( ) = +( )0 or 01

21 2 1 2 02+ +( ) = +( )m m gh m m v

and

hg

= =( )( ) =

v02 2

2

5 08

2 9 801 32

.

..

m s

m sm

2

m1

m1�m2m2

vi�0

vf �0

v1

v2�0

v0

R�3.75 m

R

Start

Just before Just after

End

h

y�0

m1

m1�m2m2

vi�0

vf �0

v1

v2�0

v0

R�3.75 m

R

Start

Just before Just after

End

h

y�0


280 Chapter 6

6.38

After

Before

37.0°53.0°

v1i�5.00 m/sv2i�0

v1i

v1f

v2f

�y

�xm m

m

m

Consider the sketches above which show the situation just before and just after collision.

Conserving momentum in y-direction: p p m myf yi f f= ⇒ − =º ºv v1 237 0 53 0 0sin . sin . , or

v v v2 1 1

37 0

53 00 754f f f= °

°⎛⎝

⎞⎠ =sin .

sin ..

Now, conserving momentum in the x-direction:

p p m m mxf xi f f i= ⇒ ° + ° = +v v v1 2 137 0 53 0 0cos . cos .

or

v v v1 1 137 0 0 754 53 0f f icos . . cos .º º+ ( ) =

and

vv

11

37 0 0 754 53 0

5 00

3fi=

+ ( ) =cos . . cos .

.

cosº º

m s

77 0 0 754 53 03 99

. . cos ..

º ºm s

+ ( ) =

Then,

v v2 10 754 0 754 3 99 3 01f f= = ( ) =. . . .m s m s

Now, we can verify that this collision was indeed an elastic collision:

KE mm

mi i= = ( ) = ( )1

2 25 00 12 51

2 2 2v . .m s m s2

and

KE m mm m

f f f= + = ( ) + (1

2

1

2 23 99

23 011

222 2v v . .m s m s)) = ( )2 212 5m . m s2

so KE KEf i= , which is the criteria for an elastic collision.



6.39 Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V = the initial velocity of the ball-bullet combination. Then, using conservation of momentum from just before to just after collision gives

M m V m Vm

M m+( ) = + =

+⎛⎝

⎞⎠v v0 or

Now, we use conservation of mechanical energy from just after the collision until the ball reaches maximum height to fi nd

01

20

2

1

22

2

+ +( ) = +( ) + = =M m g h M m V hV

g gmax maxormm

M m+⎛⎝

⎞⎠

22v

With the data values provided, this becomes

hmax 2m s

kg

kg kg= ( ) +

⎛⎝

1

2 9 80

0 030

0 15 0 030.

.

. .⎜⎜⎞⎠⎟

( ) =2

2200 m s 57 m

6.40 First, we will fi nd the horizontal speed, v0 x, of the block and embedded bullet just after impact. After this instant, the block-bullet combination is a projectile, and we fi nd the time to reach the

fl oor by use of ∆y t a ty y= +v012

2, which becomes

− = + −( )1 00 01

29 80 2. .m m s2 t , giving t = 0.452 s

Thus,

v0

2 004 43x

x

t= = =∆ .

.m

0.452 sm s

Now use conservation of momentum for the collision, with vb = speed of incoming bullet:

8 00 10 0 258 4 433. .×( ) + = ×( )( )− −kg 10 kg m s3vb , so

vb = 143 m s (about 320 mph)

6.41 First, we use conservation of mechanical energy to fi nd the speed of the block and embedded bullet just after impact:

KE PE KE PE m M Vs f s i+( ) = +( ) +( ) + = +becomes

1

22 0 0

1

222kx

and yields

Vkx

m M=

+=

( )( )+( )

2 2150 0 800

0 0120 0 100

N m m

k

.

. . ggm s= 29 3.

Now, employ conservation of momentum to fi nd the speed of the bullet just before impact: m M m M Vv + ( ) = +( )0 , or

v = +⎛⎝

⎞⎠ =

⎛⎝⎜

⎞⎠⎟

(m M

mV

0 11229 3

..

kg

0.0120 kgm s)) = 273 m s


282 Chapter 6

6.42 (a) Conservation of momentum gives m m m mT fT c fc T iT c icv v v v+ = + , or

v

v v vfT

T iT c ic fc

T

m m

m=

+ −( )

=( )( ) +9 000 20 0kg m s. 11 200 25 0 18 0

9 000

kg m s

kg

( ) −( )[ ]. .

v fT = 20 9. m s East

(b) KE KE KE m m mi f c ic T iT clost = − = +⎡⎣⎢

⎤⎦⎥

−1

2

1

2

1

22 2v v vv v

v v v

fc T fT

c ic fc T iT

m

m m

2 2

2 2 2

1

2

1

2

+⎡⎣⎢

⎤⎦⎥

= −( ) + −−( )⎡⎣ ⎤⎦

= ( ) −( )( ) +

v fT2

1

21 200 625 324 9 00kg m s2 2 00 400 438 2kg m s2 2( ) −( )( )⎡⎣ ⎤⎦.

KElost J, which becomes internal e= ×8 68 103. nnergy

Note: If 20.9 m �s were used to determine the energy lost instead of 20.9333 as the answer to Part (a), the answer would be very different. We have kept extra digits in all intermediate answers until the problem is complete.

6.43 (a) From conservation of momentum,

5 00 10 0 5 00 21 2. . .g g g 0.0 cm s( ) + ( ) = ( )( ) +v vf f 00

or

v v1 22 2f f+ = 0.0 cm s [1]

Also for an elastic, head-on, collision, we have v v v v1 2 1 2i f f i− = − −( ), which becomes 20 0 0 1 2. cm s − = − +v vf f , or

v v2 1 20 0f f= + . cm s [2]

Substituting equation [2] into [1] yields v v1 12 20 0 2f f+ +( ) =. cm s 0.0 cm s, or

3 20 01v f = − . cm s and v1 6 67f = − . cm s

Then [2] gives

v2 6 67 20 0 13 3f = − + =. . .cm s cm s cm s

(b) KE KE KEi i i= + = ×( )( ) +−1 2

3 21

25 00 10 0 200 0. .kg m s == × −1 00 10 4. J

KE mf f2 2 22 31

2

1

210 0 10 1= = ×( ) ×(− −v . kg 3.3 10 m s2 )) = × −2 58 84 10. J, so

KE

KEf

i

25

4

8 84 10

1 00 100 884= ×

×=

−

−.

..

J

J



6.44 (a) �y

m2

m3

m1

m1�48.0 kg

m2�62.0 kg

m3�112 kgq �x

v3

v1

v1�12.0 m/s

v2�15.0 m/s

v2

105°

(b) x-direction: Σ Σp p m m mxf xi= ⇒ + +º º1 1 2 2 3 30 105v v vcos cos coosθ = 0

y-direction: Σ Σp p m m myf yi= ⇒ + +º º1 1 2 2 3 30 105v v vsin sin siinθ = 0

(c) p mx1 1 1 0 48 0 12 0 1 576= = ( )( )( ) = ⋅v cos . .º kg m s kg mm s

p mx2 2 2 105 62 0 15 0 0 259= = ( )( ) −( ) =v cos . . .º kg m s −− ⋅241 kg m s

(d) p my1 1 1 0 48 0 12 0 0 0= = ( )( )( ) =v sin . .º kg m s

p my2 2 2 105 62 0 15 0 0 966= = ( )( ) +( ) =v sin . . .º kg m s 8898 kg m s⋅

(e) x-direction: 576 241 112 03kg m s kg m s kg⋅ − ⋅ + ( ) =v cosθ

y-direction: 0 898 112 03+ ⋅ + ( ) =kg m s kg v sinθ

(f ) x-direction: v3

576 241

112cosθ = − ⋅ + ⋅kg m s kg m s

kg or v3 2 99cos .θ = − m s

y-direction: v3

898

112sinθ = − ⋅kg m s

kg or v3 8 02sin .θ = − m s

Then, squaring and adding these results, recognizing that cos sin2 2 1θ θ+ = , gives

v32 2 2 2 2

2 99 8 02cos sin . .θ θ+( ) = −( ) + −( )m s m s and v3273 3 8 56= =. .m m2 s s

(g) v

v3

3

8 02

2 992 68

sin

costan

.

..

θθ

θ= = −−

=m s

m s so θ = ( ) + =−tan .1 2 68 180 250º º

Note that the factor of 180° was included in the last calculation because it was recognized that both the sine and cosine of angle θ were negative. This meant that θ had to be a third quadrant angle. Use of the inverse tangent function alone yields only the principle angles − ≤ ≤ +( )90 90º ºθ that have the given value for the tangent function.

(h) Because the third fragment must have a momentum equal in magnitude and opposite direction to the resultant of the other two fragments momenta,

all three pieces must travel in the same plaane .


284 Chapter 6

6.45 Conservation of momentum gives

25 0 10 0 25 0 20 01 2. . . .g g g cm s( ) + ( ) = ( )( ) +v vf f 110 0 15 0. .g cm s( )( )

or

2 50 65 01 2. .v vf f+ = cm s [1]

For head-on, elastic collisions, we know that v v v v1 2 1 2i i f f− = − −( ). Thus,

20 0 15 0 1 2. .cm s cm s +− = −v vf f or v v2 1 5 00f f= + . cm s [2]

Substituting equation [2] into [1] yields 3 50 60 01. .v f = cm s, or v1 17 1f = . cm s .

Equation [2] then gives v2 17 1 5 00 22 1f = + =. . .cm s cm s cm s .

6.46 First, consider conservation of momentum and write

m m m mi i f f1 1 2 2 1 1 2 2v v v v+ = +

Since m m1 2= , this becomes

v v v v1 2 1 2i i f f+ = + [1]

For an elastic head-on collision, we also have v v v v1 2 1 2i i f f− = − −( ), which may be written as

v v v v1 2 1 2i i f f− = − + [2]

Adding Equations [1] and [2] yields

v v2 1f i= [3]

Subtracting Equation [2] from [1] gives

v v1 2f i= [4]

Equations [3] and [4] show us that, under the conditions of equal mass objects striking one another in a head-on, elastic collision, the two objects simply exchange velocities. Thus, we may write the results of the various collisions as

(a) v1 0f = , v2 1 50f = . m s

(b) v1 1 00f = − . m s , v2 1 50f = . m s

(c) v1 1 00f = . m s , v2 1 50f = . m s



6.47 (a) Over a the short time interval of the collision, external forces have no time to impart signifi cant impulse to the players. The two players move together after the tackkle , so the collision is completely inelastic.

�y (north)

m1�90.0 kgm2�95.0 kg

v1i�5.00 m/sv2i�3.00 m/s

�x (east)qv1i

vf

v2i

m1�m2

m1

m2

(b) p p m m mxf xi f i= ⇒ +( ) = +Σ 1 2 1 1 0v vcosθ

or

vv

fim

m mcos

. .

.θ =

+( ) =( )( )1 1

1 2

90 0 5 00

90 0

kg m s

kkg kg+ 95 0. and v f cos .θ = 2 43 m s

p p m m myf yi f i= ⇒ +( ) = +Σ 1 2 2 20v vsinθ

giving

vv

fim

m msin

. .

.θ =

+( ) =( )( )2 2

1 2

95 0 3 00

90 0

kg m s

kkg kg+ 95 0. and v f sin .θ = 1 54 m s

Therefore,

v vf f2 2 2 2 2 2

1 54 2 43sin cos . .θ θ+( ) = = ( ) + ( ) m s m s

and

v f = =8 28 2 882. .m m s2 s

Also,

tansin

cos

.

..θ

θ

θ= = =

v

vf

f

1 54

2 430 633

m s

m sand θ = ( ) = °−tan . .1 0 633 32 3

Thus,

v�

f = 2 88. m s at 32.3° north of east

(c) KE KE KE m m m mi f i ilost = − = + − +( )1

2

1

2

1

21 12

2 22

1 2v v v ff2

= ( )( ) + ( )( )⎡1

290 0 5 00 95 0 3 00

2 2. . . .kg m s kg m s⎣⎣ ⎤⎦ − ( )( ) =1

2185 2 88

2kg m s 785 J.

The lost kinetic energy is transformed into other forms of energy, such as thermal energy and sound.


286 Chapter 6

6.48 Consider conservation of momentum in the fi rst event (twin A tossing the pack), taking the direction of the velocity given the backpack as positive. This yields

m m m mfA A pack pack A packv v+ = +( )( ) =0 0

or

vv

Apack pack

A

kg

55.0 kgf

m

m=

−= −

⎛⎝⎜

⎞⎠⎟

+12 03 0

.. 00 0 655 m s m s ( ) = − . and vA m sf = 0 655.

Conservation of momentum when twin B catches and holds onto the backpack yields

m m m mB f+( ) = ( ) +pack B B pack packv v0

or

vv

Bpack pack

B pack

kg m s

5f

m

m m=

+=

( ) +( )12 0 3 00. .

55.0 kg kgm s

+=

12 00 537

..

6.49 Choose the +x-axis to be eastward and the +y-axis northward.

If vi is the initial northward speed of the 3 000-kg car, conservation of momentum in the y direction gives

0 3 3 2 5+ ( ) = +( ) ( )000 kg 000 kg 000 kg .22 m svi sin 440 0. º⎡⎣ ⎤⎦

or

vi = 5 59. m s

Observe that knowledge of the initial speed of the 2 000-kg car was unnecessary for this solution.

6.50 We use conservation of momentum for both northward and eastward components.

For the eastward direction: M M Vf1 2 55 03.0 m s( ) = °cos . .

For the northward direction: M M Vi fv2 2 55 0= °sin . .

Divide the northward equation by the eastward equation to fi nd

M

M

MV

MVi f

f

v2

13 0

2 55 0

2 55 0.

sin .

cos .m s( ) =°°

or v2 13 0 55 0i = ( ) °. tan .m s

yielding

v2 13 02 237

55i = ( )⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥.

.tanm s

mi h

1 m s.. .0 41 5º mi h=

Thus, the driver of the north bound car was untruthful.



6.51 Choose the x-axis to be along the original line of motion.

(a) From conservation of momentum in the x direction,

m m m f5 0 4 30 0 2.00 m s .33 m s º( ) + = ( ) +cos . cosv θ

or

v2 1 25f cos .θ = m s [1]

Conservation of momentum in the y direction gives

0 4 30 0 2= ( ) +m m f.33 m s ºsin . sinv θ , or v2 2 16f sin .θ = − m s [2]

Dividing equation [2] by [1] gives

tan.

..θ = − = −2 16

1 251 73 and θ = − °60 0.

Then, either [1] or [2] gives v2 2 50f = . m s, so the fi nal velocity of the second ball is

�v2 2 50f = − °. m s at 60.0 .

(b) KE m m mi i= + = ( ) = ( )1

20

1

25 00 12 51

2 2v . .m s m s2 2

KE m m

m m

f f f= +

= ( ) +

1

2

1

2

1

24

1

22

12

22

2

v v

.33 m s .50 m ss m s2 2( ) = ( )212 5m .

Since KE KEf i= , this is an elastic collision .

6.52 The recoil speed of the subject plus pallet after a heartbeat is

Vx

t= = × = ×

−−∆

∆6 00 10

3 75 105

4..

m

0.160 sm s

From conservation of momentum, m MVv − = +0 0, so the mass of blood leaving the heart is

m MV= ⎛

⎝⎞⎠ = ( ) ×⎛

⎝⎜⎞−

v54 0

3 75 10

0

4

..

kgm s

.500 m s ⎠⎠⎟= × −4 05 10 40 52. .kg= g


288 Chapter 6

6.53 Choose the positive direction to be the direction of the truck’s initial velocity.

Apply conservation of momentum to fi nd the velocity of the combined vehicles after collision:

4 000 800 4 000 8 00 800kg kg kg m s k+( ) = ( ) +( ) +V . gg m s( ) −( )8 00.

which yields V = + 5 33. m s.

Use the impulse–momentum theorem, I F t p m f i= ( ) = = −( )av ∆ ∆ v v , to fi nd the magnitude of the average force exerted on each driver during the collision.

Truck Driver:

Fm

t

f i

avtruck kg m s m s

0=

−=

( ) −v v

∆80 0 5 33 8 00. . .

..120 sN= ×1 78 103.

Car Driver:

Fm

t

f i

avcar

kg m s m s=

−=

( ) − −( )v v

∆80 0 5 33 8 00. . .

00.120 sN= ×8 89 103.

6.54 First, we use conservation of mechanical energy to fi nd the speed of m1 at B just before collision. This gives 1

2 1 12

10 0m m ghiv + = + ,

or

v12 2 2 9 80 5 00 9 90= = ( )( ) =g hi . . .m s m m s2

Next, we apply conservation of momentum and knowledge of elastic collisions to fi nd the velocity of m1 at B just after collision.

From conservation of momentum, with the second object initially at rest, we have

m m mf f i1 1 2 2 1 1 0v v v+ = + , or v v v21

21 1f i f

m

m= −( ) [1]

For head-on elastic collisions, v v v v1 2 1 2i i f f− = − −( ). Since v2 0i = in this case, this becomes v v v2 1 1f f= + and combining this with [1] above we obtain

v v v v1 11

21 1f i i f

m

m+ = −( ) or m m m mf i1 2 1 1 2 1+( ) = −( )v v

so

v v11 2

1 21

5 00 10 0

5 00 10 0f i

m m

m m= −

+⎛⎝⎜

⎞⎠⎟

= −+

⎛ . .

. .⎝⎝⎞⎠ ( ) = −9 90 3 30. .m s m s

Finally, use conservation of mechanical energy for m1 after the collision to fi nd the maximum

rebound height. This gives KE PE KE PEg f g i+( ) = +( )

or

01

201 1 1

2+ = +m gh m fmax v and hgf

max 2

m s

m sm= =

−( )( ) =

v12 2

2

3 30

2 9 800 556

.

..



6.55 Note that the initial velocity of the target particle is zero (that is, v2 0i = ).

From conservation of momentum,

m m mf f i1 1 2 2 1 1 0v v v+ = + [1]

For head-on elastic collisions, v v v v1 2 1 2i i f f− = − −( ), and with v2 0i = , this gives

v v v2 1 1f i f= + [2]

Substituting equation [2] into [1] yields

m m mf i f i1 1 2 1 1 1 1v v v v+ +( ) =

or

m m m mf i1 2 1 1 2 1+( ) = −( )v v and v v11 2

1 21f i

m m

m m= −

+⎛⎝⎜

⎞⎠⎟

[3]

Now, we substitute equation [3] into [2] to obtain

v v v2 11 2

1 21f i i

m m

m m= + −

+⎛⎝⎜

⎞⎠⎟

or v v21

1 21

2f i

m

m m=

+⎛⎝⎜

⎞⎠⎟

[4]

Equations [3] and [4] can now be used to answer both parts (a) and (b).

(a) If m m i1 2 12 0 1 0 8 0= = =. . .g, g, and m sv , then

v1

8

3f = m s and v2

32

3f = m s

(b) If m m i1 2 12 0 10 8 0= = =. .g, g, and m sv , we fi nd

v1

16

3f = − m s and v2

8

3f = m s

(c) The fi nal kinetic energy of the 2.0 g particle in each case is

Case (a): KE mf f1 1 12 3

21

2

1

22 0 10

8

3= = ×( )⎛

⎝⎞⎠ =−v . kg m s 7 1 10 3. × − J

Case (b): KE mf f1 1 12 3

21

2

1

22 0 10

16

3= = ×( ) −⎛

⎝⎞⎠ =−v . kg m s 2 8 10 2. × − J

Since the incident kinetic energy is the same in cases (a) and (b), we observe that

the incident particle loses more kinetic eneergy in case (a) .


290 Chapter 6

6.56 If the pendulum bob barely swings through a complete circle, it arrives at the top of the arc (having risen a vertical distance of 2 �) with essentially zero velocity.

From conservation of mechanical energy, we fi nd the minimum velocity of the bob at the bottom

of the arc as KE PE KE PEg g+( ) = +( )bottom top

, or 12

2 0 2M V M g= + ( )� . This gives V g= 2 � as the needed velocity of the bob just after the collision.

Conserving momentum through the collision then gives the minimum initial velocity of the bullet as

m M g mv

v2

2 0⎛⎝

⎞⎠ + ( ) = +� or v = 4 M

mg �

6.57 We fi rst fi nd the speed of the diver when he reaches the water by using v vy ya y2

02 2= + ( )∆ . This becomes

vy2 0 2 9 80 3 0= + −( ) −( ). .m s m2 , and yields vy = − 59 m s

The negative sign indicates the downward direction.

Next, we use the impulse–momentum theorem to fi nd the resistive force exerted by the water as the diver comes to rest.

I F t p m f i= ( ) = = −( )net ∆ ∆ v v or F w t m f iwater −( ) = −( )∆ v v

and

Fwater N s kg m s−( )( ) = ( ) − −( )⎡⎣ ⎤⎦784 2 0 80 0 59.

yielding

Fwater N N = N upwa= +⎛⎝⎜

⎞⎠⎟

×78480 59

2 01 1 103

.. rrd( )

Fwater→

80 kg

w�784 N

Fwater→

80 kg

w�784 N



6.58 Use conservation of mechanical energy, KE PE KE PEg B g A+( ) = +( ) , to fi nd the speed of

the bead at point B just before it collides with the ball. This gives 12 1

2 0 0m m g yi Av + = + ,

or

v1 2 2 9 80 1 50 5 42i Ag y= = ( )( ) =. . .m s m m s2

Conservation of momentum during the collision gives

0 400 0 600 0 400 5 421 2. . . .kg kg kg( ) + ( ) = ( )v vf f mm s( ) + 0

or

v v1 21 50 5 42f f+ =. . m s [1]

For a head-on elastic collision, we have v v v v1 2 1 2i i f f− = − −( ), and with v2 0i = , this becomes

v v v1 2 1f f i= − or v v1 2 5 42f f= − . m s [2]

Substitute equation [2] into [1] to fi nd the speed of the ball just after collision as

v v2 25 42 1 50 5 42f f− + =. . .m s m s or v2

2 5 42

2 504 34f =

( ) =.

..

m sm s

Now, we use conservation of the mechanical energy of the ball after collision to fi nd the maximum height the ball will reach. This gives

0 012 2

2+ = +m g y m fball max ballv or ygf

max 2

m s

m sm= =

( )( ) =

v22 2

2

4 34

2 9 800 961

.

..

6.59 From the instant it is released from rest, at 2.00 m above ground, until just before contact, the ball is a freely falling body with a gy = − . Its speed just before impact is given by v vy y ya y2

02 2= + ( )∆ as

v vy y ya y= + ( ) = + −( ) −( ) =02 2 0 2 9 80 2 00 6 2∆ . . .m s m2 66 m s

and its velocity immediately prior to impact is v�

i = −6 26. m s.

After the ball leaves the ground on the rebound (to a height of 1.40 m), it is again in free-fall and v vy y ya y2

02 2= + ( )∆ gives its rebound speed as

v v02 2 0 2 9 80 1 40 5 2y y ya y= − ( ) = − −( ) +( ) =∆ . . .m s m2 44 m s

and its velocity immediately after impact is v�

f = +5 24. m s.

The impulse–momentum theorem, F v v��

av ∆t m mf i( ) = − , then gives the average force acting on the ball during the impact as

Fv v��

av

kg m s=

−( )=

( ) + − −m

tf i

∆0 500 5 24 6 26. . . m s

sN

( )⎡⎣ ⎤⎦ = +0 080 0

71 9.

. or 71 9. N upward


292 Chapter 6

6.60 The mass of the third fragment must be

m m m m3 1 22717 5 0 8 4 10 3 6= − − = − −( ) × =−

nucleus kg. . . ×× −10 27 kg

Conserving momentum in both the x- and y-directions gives the following:

y-direction: m m my y y1 1 2 2 3 3 0v v v+ + =

or

vv v

31 1 2 2

3

27 65 0 10 6 0 10y

y ym m

m= −

+= −

×( ) ×−. .kg m ss

kgm s

( ) +×

= − ×−

0

3 6 10

30

3 61027

6

. .

x-direction: m m mx x x1 1 2 2 3 3 0v v v+ + =

or

vv v

31 1 2 2

3

27 60 8 4 10 4 0 10x

x xm m

m= − + = −

+ ×( ) ×−. .kg m s

kgm s

( )×

= − ×−3 6 10

34

3 61027

6

. .

and

v v v3 32

32 6 2

34 3 6 10 30 3 6 10= + = − ( ) ×( ) + − ( ) ×x y . .m s 66 2 61 3 10m s m s( ) = ×.

Also, θ =⎛⎝⎜

⎞⎠⎟

+ = ⎛⎝

⎞⎠ +− −tan tan1 3

3

118030

34180

v

vy

x

º º == × =2 2 10 2202. degrees º.

Therefore, v�

361 3 10 220= ×. m s at º counterclockwise from the + -axisx

Note that the factor of 180° was included in the calculation for θ because it was recognized that both v v3 3x yand were negative. This meant that θ had to be a third quadrant angle. Use of the inverse tangent function alone yields only the principle angles − ≤ ≤ +( )90 90º ºθ that have the given value for the tangent function.

6.61 The sketch at the right gives before and after views of the collision between these two objects. Since the collision is elastic, both kinetic energy and momentum must be conserved.

Conservation of Momentum:

m m m mf f i i1 1 2 2 1 1 2 2v v v v+ = +

m m m m1 2 1 0 2 00( ) + = + −( )v v v

or

v v= −⎛⎝⎜

⎞⎠⎟

m

m1

201 [1]

Since this is an elastic collision, v v v v1 2 1 2i i f f− = − −( ), and with the given velocities this becomes

v v v0 0 0− −( ) = − −( ) or v v= 2 0 [2]

m1

Before impact

m2v1i � v0 v2i ��v0

m1

After impact

m2

v1f � 0v2f � v

m1

Before impact

m2v1i � v0 v2i ��v0

m1

After impact

m2

v1f � 0v2f � v

continued on next page



(a) Substituting equation [2] into [1] gives

2 101

20v v= −

⎛⎝⎜

⎞⎠⎟

m

m or m m1 2 3=

(b) From equation [2] above, we have v v0 2= .

6.62 (a) Let v1i and v2i be the velocities of m1 and m2 just before the collision. Then, using

conservation of mechanical energy: KE PE KE PEg i g+( ) = +( )0, or 1

22

00 0m mghiv + = + ,

gives

v v1 2 02 2 9 80 5 00 9 90i i g h= − = = ( )( ) =. . .m s m m s2

and

v1 9 90i = + . m s while v2 9 90i = − . m s

(b) From conservation of momentum:

2 00 4 00 2 00 9 90 41 2. . . .g g g m s( ) + ( ) = ( )( ) +v vf f .. .00 9 90g m s( ) −( ).

or

v v1 22 00 9 90f f+ ( ) = −. . m s [1]

For an elastic, head-on collision, v v v v1 2 1 2i i f f− = − −( ), giving

+ − −( ) = − +9 90 9 90 1 2. .m s m s v vf f or v v2 1 19 8f f= + . m s [2]

Substituting equation [2] into [1] gives v v1 12 00 19 8 9 90f f+ ( ) +( ) = −. . .m s m s,

or

v1

9 90 39 616 5f = − − = −. .

.m s m s

3.00m s

Then, equation [2] yields v2 16 5 19 8 3 30f = − + = +. . .m s m s m s .

(c) Applying conservation of energy to each block after the collision gives

1

20

1

202 2m mgh m mgf( ) + = + ( )max v or h

gf

max =v2

2

Thus,

hgff

112 2

2

16 5

2 9 8013 9= =

−( )( ) =

v .

..

m s

m sm

2

and

hgff

222 2

2

3

2 9 800 556= =

( )( ) =

v .30 m s

m sm

2..


294 Chapter 6

6.63 (a) Use conservation of mechanical energy to fi nd the speed of m1 just before collision. Taking y = 0 at the tabletop level, this gives 1

2 1 12 1

2 1 1 10 0m mg m m ghiv + ( ) = ( ) + , or

v1 12 2 9 80 2 50 7 00i g h= = ( )( ) =. . .m s m m s2

Apply conservation of momentum from just before to just after the collision:

0 500 1 00 0 500 7 001 2. . . .kg kg kg m( ) + ( ) = ( )v vf f ss( ) + 0

or

v v1 22 7 00f f+ = . m s [1]

For a head-on elastic collision, v v v v1 2 1 2i i f f− = − −( ), and with v2 0i = , this becomes

v v v2 1 1f f i= + or v v2 1 7 00f f= + . m s [2]


v v1 12 7 00 7 00f f+ +( ) =. .m s m s and v1

7 00

32 33f = − = −..

m sm s

Then, from equation [2], v2 2 33 7 00 4 67f = − + =. . .m s m s m s .

(b) Apply conservation of mechanical energy to m1 after the collision to fi nd the rebound height of this object

1

20

1

201 1 1 1 1

2m m gh m mgf( ) + ′ = + ( )v or ′ = =−( )( ) =h

gf

112 2

2

2 33

2 9 800 277

v .

..

m s

m sm

2

(c) From ∆y t a ty y= +v012

2, with v0 0y = , the time for m2 to reach the fl oor after it fl ies horizontally off the table is

ty

ay

=( ) =

−( )−

=2 2 2 00

0 639∆ .

.m

9.80 m ss2

During this time it travels a horizontal distance

∆x tx= = ( )( ) =v0 4 67 0 639 2 98. . .m s s m

(d) After the 0.500 kg mass comes back down the incline, it fl ies off the table with a horizontal velocity of 2.33 m/s. The time of the fl ight to the fl oor is 0.639 s as found above and the horizontal distance traveled is

∆x tx= = ( )( ) =v0 2 0 639 1 49.33 m s s m. .



6.64 Conservation of the x-component of momentum gives

3 0 32 0 0m m mx( ) + = − + ( )v v v or v v2 0

2

3x = [1]

Likewise, conservation of the y-component of momentum gives

− + ( ) =m my yv v1 23 0 and v v1 23y y= [2]

Since the collision is elastic, KE KEf i( ) = ( ) , or

1

2

1

23

1

2

1

231

222

22

02

02m m m my x yv v v v v+ ( ) +( ) = + ( )

which reduces to

v v v v12

22

22

023 4y x y+ +( ) = [3]

Substituting equations [1] and [2] into [3] yields

9 34

942

202

22

02v v v vy y+ +⎛

⎝⎞⎠ = or v v2 0

2

3y =

(a) From equation [2], the particle of mass m has fi nal speed v v v1 2 03 2y y= = and the particle of mass 3m moves at

v v v v v v2 22

22

02

02

0

4

9

2

9

2

3= + = + =x y

(b) θ =⎛⎝⎜

⎞⎠⎟

=⎛⎝⎜

⎞⎠⎟

=− −tan tan tan1 2

2

1 0

0

2 3

2 3

v

vv

vy

x

−− ⎛⎝⎜

⎞⎠⎟ =1 1

235.3°

6.65 (a) The momentum of the system is initially zero and remains constant throughout the motion. Therefore, when m1 leaves the wedge, we must have m m2 1 0v vwedge block+ = , or

v vwedge block= −⎛⎝⎜

⎞⎠⎟

= − ⎛⎝

⎞⎠

m

m1

2

0 500

3 004 00

.

.. m s m s( ) = − 0 667.

(b) Using conservation of energy as the block slides down the wedge, we have

KE PE KE PEg i g f+( ) = +( ) , or

0

1

2

1

201 1 2+ = + +m gh m mv vblock

2wedge2

Thus,

hg

m

m= +

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

=

1

2

1

19 6

2

1

2v vblock2

wedge

. mm sm s m s2 4 00

3 00

0 5000 667

2 2.

.

..( ) + ⎛

⎝⎞⎠ −( )⎡

⎣⎢⎤⎤⎦⎥

= 0 952. m


296 Chapter 6

6.66 Choose the positive x-axis in the direction of the initial velocity of the cue ball. Let vci be the ini-tial speed of the cue ball, vcf be the fi nal speed of the cue ball, vTf be the fi nal speed of the target, and θ be the angle the target’s fi nal velocity makes with the x-axis.

Conservation of momentum in the x-direction, recognizing that all billiard balls have the same mass, gives

m m mTf cf civ v vcos cos .θ + = +30 0 0º or v v vTf ci cfcos cos .θ = − 30 0º [1]

To conserve momentum in the y-direction, recognize that the y-components of the fi nal velocities of the target and cue balls must have opposite signs. Thus, if the cue ball scatters at 30.0° below the x-axis, the target ball must scatter at angle θ above the x-axis. The conservation equation for momentum in the y-direction is:

m mTf cfv vsin sin .θ − = +30 0 0 0º or v vTf cfsin sin .θ = °30 0 [2]

Since this is an elastic collision, kinetic energy is conserved, giving

1

2

1

2

1

22 2 2m m mTf cf civ v v+ = or v v vTf ci cf

2 2 2= − [3]

(b) To solve, square equations [1] and [2] and add the results to obtain

v v v v vTf ci ci cf cf2 2 2 2 22 30 0cos sin cos .θ θ+( ) = − +º ccos . sin .2 230 0 30 0º °+( )

or v v v v vTf ci ci cf cf2 2 22 30 0= − ° +cos . .

Now, substitute this result into equation [3] to get

v v v v v vci ci cf cf ci cf2 2 2 22 30 0− + = −cos . º or 2 30 0 0v v vcf cf ci−( ) =cos . º

Since vcf ≠ 0, it is necessary that v vcf ci= = ( ) =cos . . cos . .30 0 4 00 30 0 3 46º m s º m s .

Then, equation [3] yields v v vTf ci cf= −2 2 , or

vTf = ( ) − ( ) =4 00 3 2 002 2

. .m s .46 m s m s

(a) With the results found above, equation [2] gives

sin sin ..

.θ =

⎛

⎝⎜⎞

⎠⎟=

⎛⎝⎜

v

vcf

Tf

30 03 46

2 00º

m s

m s

⎞⎞⎠⎟

=sin . .30 0 0 866º , or θ = °60 0.

Thus, the angle between the velocity vectors after collision is

φ = + =60 0 30 0 90 0. . .° ° °



6.67 (a) Use conservation of the horizontal component of momentum from just before to just after the cannon fi ring.

Σ Σp px f x i( ) = ( ) gives

m mshell shell cannon recoil°v vcos .45 0 0( ) + = ,

or

v vrecoilshell

cannonshell °= −

⎛⎝⎜

⎞⎠⎟

m

mcos .45 0

= −⎛⎝⎜

⎞⎠⎟

( ) = −200

5 000125 45 0 3 54

kg

kgm s ºcos . . m s

(b) Use conservation of mechanical energy for the cannon-spring system from right after the cannon is fi red to the instant when the cannon comes to rest.

KE PE PE KE PE PEg s f g s i+ +( ) = + +( )

0 01

2

1

20 02 2+ + = + +kx mmax cannon recoilv

x

m

kmaxcannon recoil kg 3.54 m s

2= =

( ) −( )v2 25 000

..00 10 N mm4×

= 1 77.

(c) F k xmax max42.00 10 N m m N= = ×( )( ) = ×1 77 3 54 104. .

(d) No. The rail exerts a vertical external force (the normal force) on the cannon and prevents it from recoiling vertically. Momentum is not conserved in the vertical direction. The spring does not have time to stretch during the cannon fi ring. Thus, no external horizontal force is exerted on the system (cannon plus shell) from just before to just after fi ring. Momentum is conserved in the horizontal direction during this interval.

45.0°45.0°


298 Chapter 6

6.68 Observe from Figure P6.68, the platform exerts a 0.60-kN to support the weight of the standing athlete prior to t = 0 00. s. From this, we determine the mass of the athlete:

mw

g g= = = =0 60 600

61. kN N

9.8 m skg2

For the interval t t= =0 00 1 0. .s to s, we subtract the 0.60-kN used to counterbalance the weight to get the net upward force exerted on the athlete by the platform during the jump. The result is shown in the force versus time graph at the right. The net impulse imparted to the athlete is given by the area under this graph. Note that this area can be broken into two triangular areas plus a rectangular area.

The net upward impulse is then

I = ( )( ) + ( )( ) +1

20 50 100

1

20 50 300 0 50. . .s N s N NN N N s( )( ) = ⋅100 150

The upward velocity vi of the athlete as he lifts off of the platform (at t = 1 0. s) is found from

I p m m mI

mi i i= = − = − ⇒ = = ⋅ =∆ v v v v0 0150

2N s

61 kg..5 m s

The height of the jump can then be found from v vf i ya y2 2 2= + ∆ (with v f = 0) to be

∆ya

f i

y

=−

=− ( )−( ) =

v v2 2 2

2

0 2 5

2 9 80 31

.

..

m s

m sm

2

0.40

0.30

0.20

0.10

0.00�0.50 0.00 0.50 1.0

t (s)

Fnet (kN)0.40

0.30

0.20

0.10

0.00�0.50 0.00 0.50 1.0

t (s)

Fnet (kN)



6.69 Let particle 1 be the neutron and particle 2 be the carbon nucleus. Then, we are given that m m2 112= .

(a) From conservation of momentum m m mf f i2 2 1 1 1 1 0v v v+ = + .

Since m m2 112= , this reduces to

12 2 1 1v v vf f i+ = [1]

For a head-on elastic collision

v v v v1 2 1 2i i f f− = − −( ) Since v2 0i = , this becomes

v v v2 1 1f i f= + [2]

Substitute equation [2] into [1] to obtain 12 1 1 1 1v v v vi f f+( ) + = , or

13 111 1v vf i= − and v v1 1

11

13f i= −

Then, equation [2] yields

v v2 1

2

13f i=

The initial kinetic energy of the neutron is KE mi i112 1 1

2= v , and the fi nal kinetic energy of the carbon nucleus is

KE m m mf f i2 2 22

1 121

2

1

212

4

169

48

169

1

2= = ( )⎛

⎝⎞⎠ =v v 11 1

21

48

169v i iKE⎛

⎝⎞⎠ =

The fraction of kinetic energy transferred is KE

KEf

i

2

1

48

1690 28= = . .

(b) If KE i1131 6 10= × −. J, then

KE KEf i2 113 1448

169

48

1691 6 10 4 5 10= = ×( ) = ×− −. .J J

The remaining energy 1 6 10 4 5 10 1 1 1013 14 13. . .× − × = ×− − −J J J stays with the neutron.


300 Chapter 6

6.70 (a)

nA→

nB→

FBA→

MA

2M

B

FAB→

Mg→

2Mg→

(b) From Newton’s third law, the force F��

BA exerted by B on A is at each instant equal in magni-

tude and opposite in direction to the force F��

AB exerted by A on B.

(c) There are no horizontal external forces acting on system C which consists of both blocks. The forces F

��BA

and F��

AB are internal forces exerted on one part of system C by another part

of system C.

Thus,

Σ ∆∆

∆Fp

p��

�external = = ⇒ =C

Ct0 0

This gives

� � � �p p p pC f C i A i B i( ) = ( ) = ( ) + ( ) or M M V M+( ) = +( ) +2 0v

so the velocity of the combined blocks after collision is V = + v 3.

The change in momentum of A is then

∆� � �p p pA A f A i

MV M M M= ( ) − ( ) = − = −⎛⎝

⎞⎠ = −v

vv v

32 3

and the change in momentum for B is:

∆� � �

p p pB B f B iMV M M= ( ) − ( ) = − = +⎛

⎝⎞⎠ = +2 0 2

32 3

vv

(d) ∆KE KE KE KE MC f A B i= ( ) − ( ) + ( )⎡⎣ ⎤⎦ = ( )⎛

⎝⎞⎠ −1

2

213

3

v22

2 13

20M Mv v+⎡⎣ ⎤⎦ = −

Thus, kinetic energy is not conserved in this inellastic collision .



6.71 (a) The owner’s claim should be denied . Immediately prior to impact, the total momentum

of the two-car system had a northward component and an eastward component. Thus, after impact, the wreckage moved in a northeasterly direction and could not possibly have damaged the owner’s property on the southeast corner.

(b) From conservation of momentum:

p p m m mx x x i x( ) = ( ) ⇒ +( ) = ( )after before 1 2 1 1v v ++ ( )m i x2 2v

or

kg km

vv v

xi x i x

m m

m m=

( ) + ( )+

=( )1 1 2 2

1 2

1 300 30 0. hh + 0

kg kgkm h

( )+

=1 300 1100

16 3.

p p m m my y y i y( ) = ( ) ⇒ +( ) = ( )after before 1 2 1 1v v ++ ( )m i y2 2v

or

kg

vv v

y

i y i ym m

m m=

( ) + ( )+

=+ ( )1 1 2 2

1 2

0 1100 20 0. kkm h

kg kgkm h

( )+

=1 300 1100

9 17.

Thus, the velocity of the wreckage immediately after impact is

v v v= + =x y2 2 18 7. km h and θ =

⎛⎝⎜

⎞⎠⎟

= ( ) = °− −tan tan . .1 1 0 564 29 4v

vy

x

or �v = 18 7. km h at 29.4° north of east, consisteent with part (a) .

6.72 Ignoring the force of gravity during the brief collision time, we use the conservation of momentum to obtain:

0 45 60 0 45 25 6. .kg kg kg m s( ) + ( ) = ( ) −( ) +v vbf pf 00 4 0kg m s( )( ).

or

v vpf bf= − ×( )−3 8 7 5 10 3. .m s [1]

Also, elastic collision ⇒ − = − −( ) = − −v v v vbf pf bi pi 255 4 0m s m s−( ). , or

v vbf pf= +29 m s [2]


vbf = ++ ×

=−29 3 8

1 7 5 10333

m s m sm s

.

.

The average acceleration of the ball during the collision is

at

bf biav

m s m s

s=

−=

− −( )×

= ×−

v v

∆33 25

20 102 9 13 . 003 m s2 .


302 Chapter 6

6.73 (a) The speed vi of both balls just before the basketball reaches the ground may be found from v vy y ya y2

02 2= + ∆ as

v vi y ya y= + = + −( ) −( ) =02 2 0 2 9 80 1 20 4 85∆ . . .m s m s2 m s

(b) Immediately after the basketball rebounds from the fl oor, it and the tennis ball meet in an elastic collision. The velocities of the two balls just before collision are:

For the tennis ball: v v1 4 85i i= − = − . m s

For the basketball: v v2 4 85i i= + = + . m s

We determine the velocity of the tennis ball immediately after this elastic collision as follows:

Momentum conservation gives

57 0 590 57 0 4 85 51 2. . .g g g m s( ) + ( ) = ( ) −( ) +v vf f 990 4 85g m s( ) +( ).

which reduces to

v v22

14 38 9 66 10f f= − ×( )−. .m s [1]

Elastic Collision ⇒ − = − −( )v v v v1 2 1 2i i f f or v v v v1 2 1 2f f i i= − +

so

v v1 2 4 85 4 85f f= − −( ) +. .m s m s and v v1 2 9 70f f= + . m s [2]

Substituting equation [1] into [2] gives ( . ) . .1 9 66 10 4 38 9 7021+ × = +− v f m s m s, or

v1 12 84f = + . m s

The vertical displacement of the tennis ball during its rebound is given by v vy y ya y202 2= + ∆ as

∆ya

y y

y

=−

=− ( )

−( ) =v v2

02 2

2

0 12 84

2 9 808

.

..

m s

m s2336 m



6.74 The woman starts from rest ( )v0 0y = and drops freely with a gy = − for 2.00 m before the impact with the toboggan. Then, v v2

202 2i y ya y= + ( )∆ gives her speed just before impact as

v v2 02 2 0 2 9 80 2 00 6i y ya y= + ( ) = + −( ) −( ) =∆ . . .m s m2 226 m s

The sketches at the right show the situation just before and just after the woman’s impact with the toboggan. Since no external forces impart any signifi cant impulse directed parallel to the incline (+x-direction) to the system consisting of man, woman, and toboggan during the very brief duration of the impact, we will consider the total momentum parallel to the incline to be conserved. That is,

m m m m m mf i i x i i1 2 1 1 2 2 1 1 2 2 30 0+( ) = + ( ) = +v v v v v sin . °°

or the speed of the system immediately after impact is

vv v

fi im m

m m= +

+=

( )1 1 2 2

1 2

30 0 90 0 8 00sin . . .º kg m ss kg m s º

kg

( ) + ( )( )+

55 0 6 26 30 0

90 0 55 0

. . sin .

. . kgm s= 6 15.

6.75 First consider the motion of the block and embedded bullet from immediately after impact until the block comes to rest after sliding distance d across the horizontal table. During this time, a kinetic friction force f n M m gk k k= = +( )µ µ , directed opposite to the motion, acts on the block. The net work done on the (block plus bullet) during this time is

W f d KE KE M m Vk f inet º= ( ) = − = − +( )cos180 01

22

so the speed, V, of the block and embedded bullet immediately after impact is

Vf d

M m

M m gd

M mgdk k

k= −− +( ) =

+( )+

=2 22

µµ

Now, make use of conservation of momentum from just before to just after impact to obtain

p p m M m V M m gdxi xf k= ⇒ = +( ) = +( )v0 2µ

and the initial velocity of the bullet was

v0 2= +⎛⎝

⎞⎠

M m

mgdkµ

30.0°v2i

30.0°Just before impact Just after impact

m1

m2m1�mman�mtoboggan�90.0 kgm2�mwoman�55.0 kgv1i �8.00 m/s

v2i �6.26 m/s

�x

v1i

30.0°

�x

vf

m1 �m

2

30.0°v2i

30.0°Just before impact Just after impact

m1

m2m1�mman�mtoboggan�90.0 kgm2�mwoman�55.0 kgv1i �8.00 m/s

v2i �6.26 m/s

�x

v1i

30.0°

�x

vf

m1 �m

2


304 Chapter 6

6.76 (a) Apply conservation of momentum in the vertical direction to the squid-water system from the instant before to the instant after the water is ejected. This gives

m m m ms s w w s wv v+ = +( )( )0 or v vsw

sw

m

m= −

⎛⎝⎜

⎞⎠⎟

= −⎛⎝⎜

⎞⎠⎟

−0.30 kg

kgm

0 8520

.ss m s( ) = 7 1.

(b) Apply conservation of mechanical energy to the squid from the instant after the water is ejected until the squid reaches maximum height to fi nd:

0 12

2+ = +m gy m mgys f s s iv or ∆y y ygf is= − = =

( )( ) =v2 2

2

7 1

2 9 82 6

.

..

m s

m sm

2

6.77 (a)

Just after collisionJust before collision

At rest

v1i �xm1 m2

→53.0°

�y

�x

m1

m2

v1f→

v2f→

f

The situations just before and just after the collision are shown above. Conserving momentum in both the x- and y-directions gives

p p m my f y i f f( ) = ( ) ⇒ − =º1 1 2 253 0v vsin sinφ or m mf f2 2 1 1 53v vsin sinφ = ° [1]

p p m m mx f x i f f i( ) = ( ) ⇒ + = +°1 1 2 2 1 153 0v v vcos cosφ ,

or

m m mf i f2 2 1 1 1 1 53v v vcos cosφ = − ° [2]

Dividing equation [1] by [2] yields

tansin

cos

. sinφ =

°− °

=( ) °v

v v1

1 1

53

53

1 0 53

2f

i f

m s

.. . cos.

0 1 0 530 57

m s m s( ) − ( ) °= or φ = 30°

Equation [1] then gives

vv

21 1

2

53 0 20 1 0 5f

fm

m=

°=

( )( )sin

sin

. . sin

φkg m s 33

0 30 301 1

°( ) °

=. sin

.kg

m s

(b) The fraction of the incident kinetic energy lost in this collision is

∆KE

KE

KE KE

KE

KE

KEi

i f

i

f

i

=−

= − = −( )

1 10 20 1 01

2 . .kg m s kg m s

kg

( ) + ( )( )( )

2 12

2

12

0 30 1 1

0 20 2 0

. .

. . mm s( )2

∆KE

KEi

= 0 30. or 30%



6.78

12.0 g100 g 112 g 112 g

7.5 m

At the endImmediately after impactImmediately before impact

At restAt rest

v→ V

→

Using the work–energy theorem from immediately after impact to the end gives

W f s KE KEknet end after°= ( ) = −cos180

or

− +( )[ ] = − +( )µk M m g s M m V01

22 and V g sk= 2 µ

Then, using conservation of momentum from immediately before to immediately after impact gives m M m Vv + = +( )0 , or

v = +⎛⎝

⎞⎠ = +⎛

⎝⎞⎠ =

⎛⎝⎜

⎞⎠⎟

M m

mV

M m

mg sk2

112µ g

12.0 g22 0 650 9 80 7 5. . .( )( )( )m s m2

v = 91 m s



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Solucionario Fundamentos de Física 9na edición Capitulo 6

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