1. Preview Section 1 Momentum and Impulse
Section 2 Conservation of Momentum
Section 3 Elastic and Inelastic Collisions

2. What do you think?
Imagine an automobile collision in which an older model car from the 1960s collides with a car at rest while traveling at 15 mph. Now imagine the same collision with a 2007 model car. In both cases, the car and passengers are stopped abruptly.
List the features in the newer car that are designed to protect the passenger and the features designed to minimize damage to the car.
How are these features similar?
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.
Students will mention seat belts and air bags. Hopefully some will be aware of collapsible steering columns, shock absorbing bumpers, fenders and car body parts that collapse easily, padded dashboards, and so on. If you can get a fairly complete list, they may see the common thread in these features. That thread is the ability to extend the amount of time required for the car and the passenger to stop.

3. What do you think? What are some common uses of the term momentum?
Write a sentence or two using the term momentum.
Do any of the examples provided reference the velocity of an object?
Do any of the examples reference the mass of an object?
This slide provides an opportunity for students to assess their own understanding of momentum. There may be some references to motion, but more likely are comments about how the momentum shifted in a soccer match or other sporting event.

4. Momentum Momentum (p) is proportional to both mass and velocity.
A vector quantity
SI Units: kg • m/s
Momentum is in the same direction as the velocity.
Students may associate momentum only with velocity, so discuss the fact that a slow moving truck has as much momentum as a car moving at much greater speeds.

5. Momentum and Newton’s 2nd Law
Prove that the two equations shown below are equivalent.
F = ma and F = p/t
Newton actually wrote his 2nd Law as F = p/t.
Force depends on how rapidly the momentum changes.
Have students show that the two equations are equivalent.
To do so, substitute v/t for a in the equation F = ma.
It will now read F = mv/t
Next, point out to the students that the mass of an object does not change while it is accelerating (except in cases such as rockets taking off, where the fuel is being expelled so the mass is decreasing). Therefore, mv = (mv) = p.

6. Classroom Practice Problems
A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?
Answers:
5.63 x 104 kg • m/s to the east
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

7. Classroom Practice Problems
What velocity must a car with a mass of 1210 kg have in order to have the same momentum as the pickup truck in the previous problem?
Answers:
46.49 m/s to the east
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

8. Classroom Practice Problems
An ostrich with a mass of 146 kg is running to the right with a velocity of 17 m/s. Find the momentum of the ostrich.
Answers:
2.48 x 103 kg • m/s to the right
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

9. Classroom Practice Problems
A 21 kg child is riding a 5.9 kg bike with a velocity of 4.5 m/s to the NW.
What is the total momentum of the child and the bike? What is te momentum of the child? What is the momentum of the bike?
Answers:
kg • m/s to the NW
94.50 kg • m/s to the NW
26.55 kg • m/s to the NW
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

10. Impulse and Momentum The quantity Ft is called impulse.
SI units: N•m or kg•m/s
Impulse equals change in momentum.
Another version of Newton’s 2nd Law
Changes in momentum depend on both the force and the amount of time over which the force is applied.

11. Impulse-Momentum Theorem
Click below to watch the Visual Concept.
Visual Concept

12. Changing momentum
Greater changes in momentum(p) require more force (F) or more time (t) .
A loaded truck requires more time to stop.
Greater p for truck with more mass
Same stopping force
Discuss the problem stopping vehicles with larger masses. The stopping force of friction between the road and the tires is limited, so the only way to stop a truck with more momentum is with a greater amount of time.

13. Classroom Practice Problems
A 0.50 kg football is thrown with a velocity of 15 m/s to the right. A stationary receiver catches the ball and brings it to rest in 0.020s. What is the force exerted on the receiver?
Answers:
380 N to the right
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

14. Classroom Practice Problems
A 0.40 kg soccer ball approaches a player horizontally with a velocity of 18 m/s to the north. The player strikes the ball and causes it to move in the opposite direction with a velocity of 22 m/s. What impulse was delivered to the ball by the player?
Answers:
16.0 kg • m/s to the south
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

15. Classroom Practice Problems
A 0.50 kg object is at rest. A 3.00 N force to the right acts on the object during a time interval of 1.50s.
What is the velocity of the object at the end of this interval? At the end of this interval, a constant force of 4.00N to the left is applied for 3.00s. What is the velocity at the end of the 3.00s?
Answers:
9.0 kg • m/s to the right
15 kg • m/s to the left
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

16. Classroom Practice Problems
An 82 kg man drops from rest on a diving board 3.0m above the surface of the water and comes to rest 0.55s after reaching the water. What force does the water exert on him?
Answers:
1100 N to the upwards
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
(1) p = mv
(2) F = p/t
Note that the change in momentum is negative or to the south, because the momentum changes from north to zero.
You might point out that the amount of force braking can provide is limited by the brakes and the force of friction between the tires and the road.

17. Stopping Time Ft = p = mv
When stopping, p is the same for rapid or gradual stops.
Increasing the time (t) decreases the force (F).
What examples demonstrate this relationship?
Air bags, padded dashboards, trampolines, etc
Decreasing the time (t) increases the force (F).
Hammers and baseball bats are made of hard material to reduce the time of impact.
Keep reminding students that the change in momentum is the same no matter how much time is required to stop. The object loses all of its momentum.
Additional examples:
(1) Catching a baseball - the player lets his hand move back with the ball to extend the amount of time and reduce the force.
(2) Shock absorbers on car bumper
(3) Boxing gloves as opposed to bare fists
(4) Boxers move their head back when being hit (Muhammad Ali turned this into an art) in order to reduce the force by extending the time.

18. Classroom Practice Problems
A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20.0 m/s by a 6250 N braking force acting opposite the car’s motion. Use the impulse-momentum theorem to answer the following question.
What is the car’s velocity?
Answer: 14 m/s to the north
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
F = p/t or F = (mv)/t
Students can get the second answer without calculating the force by simply determining the ratio between the two times (0.75/0.026). It is still beneficial to actually calculate the force.

19. Classroom Practice Problems
A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20.0 m/s by a 6250 N braking force acting opposite the car’s motion. Use the impulse-momentum theorem to answer the following question.
How far does the car move during 2.50 s?
Answer: 42 m
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
F = p/t or F = (mv)/t
Students can get the second answer without calculating the force by simply determining the ratio between the two times (0.75/0.026). It is still beneficial to actually calculate the force.

20. Classroom Practice Problems
A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20.0 m/s by a 6250 N braking force acting opposite the car’s motion. Use the impulse-momentum theorem to answer the following question.
How long does it take the car to come to a complete stop?
Answer: F = 8.0s
Work through these problems with the students slowly and ask questions about the quantities and the units so they become more familiar with the concepts.
F = p/t or F = (mv)/t
Students can get the second answer without calculating the force by simply determining the ratio between the two times (0.75/0.026). It is still beneficial to actually calculate the force.

21. Now what do you think? How is momentum defined?
How is Newton’s 2nd Law written using momentum?
What is impulse?
What is the relationship between impulse and momentum?
mass  velocity
F = p/t
Ft
Impulse = change in momentum, or Ft = (mv)

22. What do you think?
Two skaters have equal mass and are at rest. They are pushing away from each other as shown.
Compare the forces on the two girls.
Compare their velocities after the push.
How would your answers change if the girl on the right had a greater mass than her friend?
How would your answers change if the girl on the right was moving toward her friend before they started pushing apart?
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.

23. Momentum During Collisions
When the bumper cars collide,
F1 = -F2 so F1t = -F2t, and therefore p1 = -p2 .
The change in momentum for one object is equal and opposite to the change in momentum for the other object.
Total momentum is neither gained not lost during collisions.
Newton’s 3rd Law and 2nd Law lead to the idea of momentum being conserved in collisions.
The 3rd Law is used in the first step (F1 = -F2).
The 2nd Law is used in the last step (F=p/t or Ft = p).

24. Conservation of Momentum
Total momentum remains constant during collisions.
The momentum lost by one object equals the momentum gained by the other object.
Conservation of momentum simplifies problem solving.
Using conservation of momentum makes it possible to determine the effect of a force without knowing how much force was involved. Also, the force need not be constant.

25. Conservation of Momentum
Click below to watch the Visual Concept.
Visual Concept

26. Classroom Practice Problems
A 63.0 kg astronaut is on a spacewalk when the tether line to the shuttle breaks. The astronaut is able to throw a 10.0 kg oxygen tank in the direction away from the shuttle with a speed of 12.0 m/s, propelling the astronaut back to the shuttle
Assuming that the astronaut starts from res, find the final speed of the astronaut after throwing the tank
Answer: 1.90 m/s
Show students that, when the initial momentum = 0, the equation for conservation of momentum simplifies to pf,a= -pf,b.
Note that the recoil speed is quite small, because the mass of the astronaut is much greater than that of the baseball. However, a speed of roughly 6 cm/s is measurable. Ask students what the recoil speed would be if he threw another 62.0 kg astronaut.
Students may have trouble with the idea of recoiling when you throw something because they have not experienced it. You could ask them what happens if they are standing on skates or a skateboard and thrown something forward.
For the pitcher, several factors come into play. First of all, he is moving forward (not at rest) as he throws the ball so the recoil would simply slow him down. Secondly, his feet are pushing against Earth so the force would be transferredto Earth, and the entire Earth would recoil (very slowly).
If the pitcher were on ice skates (frictionless surface) and at rest when throwing, he would recoil backward just as the astronaut does.
Other examples of recoil include firing a rifle and shooting a cannon. Ask students how the old pirate ships dealt with the recoil of the cannon, which was pretty significant. They may recall that the cannons were mounted in such a way that they could slide or roll back after firing. They were not permanently fixed to the boat or the entire boat would have recoiled slightly. More likely, it would have damaged the boat where it was attached.

27. Classroom Practice Problems
A 63.0 kg astronaut is on a spacewalk when the tether line to the shuttle breaks. The astronaut is able to throw a 10.0 kg oxygen tank in the direction away from the shuttle with a speed of 12.0 m/s, propelling the astronaut back to the shuttle
Determine the maximum distance the astronaut can be from the craft when the line breaks in order to return to the craft within 60.0 s.
Answer: 114 m
Show students that, when the initial momentum = 0, the equation for conservation of momentum simplifies to pf,a= -pf,b.
Note that the recoil speed is quite small, because the mass of the astronaut is much greater than that of the baseball. However, a speed of roughly 6 cm/s is measurable. Ask students what the recoil speed would be if he threw another 62.0 kg astronaut.
Students may have trouble with the idea of recoiling when you throw something because they have not experienced it. You could ask them what happens if they are standing on skates or a skateboard and thrown something forward.
For the pitcher, several factors come into play. First of all, he is moving forward (not at rest) as he throws the ball so the recoil would simply slow him down. Secondly, his feet are pushing against Earth so the force would be transferredto Earth, and the entire Earth would recoil (very slowly).
If the pitcher were on ice skates (frictionless surface) and at rest when throwing, he would recoil backward just as the astronaut does.
Other examples of recoil include firing a rifle and shooting a cannon. Ask students how the old pirate ships dealt with the recoil of the cannon, which was pretty significant. They may recall that the cannons were mounted in such a way that they could slide or roll back after firing. They were not permanently fixed to the boat or the entire boat would have recoiled slightly. More likely, it would have damaged the boat where it was attached.

28. Classroom Practice Problems
An 85.0 kg fisherman jumps from a doc into a 135 kg rowboat at rest on the west side of the dock. If the velocity of the fisherman is 4.30 m/s to the west as he leaves the dock, what is the final velocity of the fisherman and the boat?
Answer: 1.66 m/s to the west
Show students that, when the initial momentum = 0, the equation for conservation of momentum simplifies to pf,a= -pf,b.
Note that the recoil speed is quite small, because the mass of the astronaut is much greater than that of the baseball. However, a speed of roughly 6 cm/s is measurable. Ask students what the recoil speed would be if he threw another 62.0 kg astronaut.
Students may have trouble with the idea of recoiling when you throw something because they have not experienced it. You could ask them what happens if they are standing on skates or a skateboard and thrown something forward.
For the pitcher, several factors come into play. First of all, he is moving forward (not at rest) as he throws the ball so the recoil would simply slow him down. Secondly, his feet are pushing against Earth so the force would be transferredto Earth, and the entire Earth would recoil (very slowly).
If the pitcher were on ice skates (frictionless surface) and at rest when throwing, he would recoil backward just as the astronaut does.
Other examples of recoil include firing a rifle and shooting a cannon. Ask students how the old pirate ships dealt with the recoil of the cannon, which was pretty significant. They may recall that the cannons were mounted in such a way that they could slide or roll back after firing. They were not permanently fixed to the boat or the entire boat would have recoiled slightly. More likely, it would have damaged the boat where it was attached.

29. Classroom Practice Problems
Each croquet ball in a set has a mass of 0.50 kg. the green ball, traveling at 12.0 m/s, strikes the blue ball, which is at rest. Assuming that all collisions are head-on, find the final speed of the blue ball in each of the following situations
The green ball stops moving after it strikes the blue ball.
Answer: 12.0 m/s
Show students that, when the initial momentum = 0, the equation for conservation of momentum simplifies to pf,a= -pf,b.
Note that the recoil speed is quite small, because the mass of the astronaut is much greater than that of the baseball. However, a speed of roughly 6 cm/s is measurable. Ask students what the recoil speed would be if he threw another 62.0 kg astronaut.
Students may have trouble with the idea of recoiling when you throw something because they have not experienced it. You could ask them what happens if they are standing on skates or a skateboard and thrown something forward.
For the pitcher, several factors come into play. First of all, he is moving forward (not at rest) as he throws the ball so the recoil would simply slow him down. Secondly, his feet are pushing against Earth so the force would be transferredto Earth, and the entire Earth would recoil (very slowly).
If the pitcher were on ice skates (frictionless surface) and at rest when throwing, he would recoil backward just as the astronaut does.
Other examples of recoil include firing a rifle and shooting a cannon. Ask students how the old pirate ships dealt with the recoil of the cannon, which was pretty significant. They may recall that the cannons were mounted in such a way that they could slide or roll back after firing. They were not permanently fixed to the boat or the entire boat would have recoiled slightly. More likely, it would have damaged the boat where it was attached.

30. Classroom Practice Problems
Each croquet ball in a set has a mass of 0.50 kg. the green ball, traveling at 12.0 m/s, strikes the blue ball, which is at rest. Assuming that all collisions are head-on, find the final speed of the blue ball in each of the following situations
The green ball continues moving after the collision at 2.4 ms in the same direction.
Answer: 9.6 m/s
Show students that, when the initial momentum = 0, the equation for conservation of momentum simplifies to pf,a= -pf,b.
Note that the recoil speed is quite small, because the mass of the astronaut is much greater than that of the baseball. However, a speed of roughly 6 cm/s is measurable. Ask students what the recoil speed would be if he threw another 62.0 kg astronaut.
Students may have trouble with the idea of recoiling when you throw something because they have not experienced it. You could ask them what happens if they are standing on skates or a skateboard and thrown something forward.
For the pitcher, several factors come into play. First of all, he is moving forward (not at rest) as he throws the ball so the recoil would simply slow him down. Secondly, his feet are pushing against Earth so the force would be transferredto Earth, and the entire Earth would recoil (very slowly).
If the pitcher were on ice skates (frictionless surface) and at rest when throwing, he would recoil backward just as the astronaut does.
Other examples of recoil include firing a rifle and shooting a cannon. Ask students how the old pirate ships dealt with the recoil of the cannon, which was pretty significant. They may recall that the cannons were mounted in such a way that they could slide or roll back after firing. They were not permanently fixed to the boat or the entire boat would have recoiled slightly. More likely, it would have damaged the boat where it was attached.

31. Classroom Practice Problems
Each croquet ball in a set has a mass of 0.50 kg. the green ball, traveling at 12.0 m/s, strikes the blue ball, which is at rest. Assuming that all collisions are head-on, find the final speed of the blue ball in each of the following situations
The green ball continues moving after the collision at 0/3 m/s in the same direction.
Answer: 11.7 m/s
Show students that, when the initial momentum = 0, the equation for conservation of momentum simplifies to pf,a= -pf,b.
Note that the recoil speed is quite small, because the mass of the astronaut is much greater than that of the baseball. However, a speed of roughly 6 cm/s is measurable. Ask students what the recoil speed would be if he threw another 62.0 kg astronaut.
Students may have trouble with the idea of recoiling when you throw something because they have not experienced it. You could ask them what happens if they are standing on skates or a skateboard and thrown something forward.
For the pitcher, several factors come into play. First of all, he is moving forward (not at rest) as he throws the ball so the recoil would simply slow him down. Secondly, his feet are pushing against Earth so the force would be transferredto Earth, and the entire Earth would recoil (very slowly).
If the pitcher were on ice skates (frictionless surface) and at rest when throwing, he would recoil backward just as the astronaut does.
Other examples of recoil include firing a rifle and shooting a cannon. Ask students how the old pirate ships dealt with the recoil of the cannon, which was pretty significant. They may recall that the cannons were mounted in such a way that they could slide or roll back after firing. They were not permanently fixed to the boat or the entire boat would have recoiled slightly. More likely, it would have damaged the boat where it was attached.

32. Classroom Practice Problems
Each croquet ball in a set has a mass of 0.50 kg. the green ball, traveling at 12.0 m/s, strikes the blue ball, which is at rest. Assuming that all collisions are head-on, find the final speed of the blue ball in each of the following situations
The blue ball and the green ball move in the same direction after the collision; the green ball has a speed of 1.6 m/s.
Answer: 10.4 m/s
Show students that, when the initial momentum = 0, the equation for conservation of momentum simplifies to pf,a= -pf,b.
Note that the recoil speed is quite small, because the mass of the astronaut is much greater than that of the baseball. However, a speed of roughly 6 cm/s is measurable. Ask students what the recoil speed would be if he threw another 62.0 kg astronaut.
Students may have trouble with the idea of recoiling when you throw something because they have not experienced it. You could ask them what happens if they are standing on skates or a skateboard and thrown something forward.
For the pitcher, several factors come into play. First of all, he is moving forward (not at rest) as he throws the ball so the recoil would simply slow him down. Secondly, his feet are pushing against Earth so the force would be transferredto Earth, and the entire Earth would recoil (very slowly).
If the pitcher were on ice skates (frictionless surface) and at rest when throwing, he would recoil backward just as the astronaut does.
Other examples of recoil include firing a rifle and shooting a cannon. Ask students how the old pirate ships dealt with the recoil of the cannon, which was pretty significant. They may recall that the cannons were mounted in such a way that they could slide or roll back after firing. They were not permanently fixed to the boat or the entire boat would have recoiled slightly. More likely, it would have damaged the boat where it was attached.

33. What do you think?
Collisions are sometimes described as elastic or inelastic. To the right is a list of colliding objects. Rank them from most elastic to most inelastic.
What factors did you consider when ranking these collisions?
A baseball and a bat
A baseball and a glove
Two football players
Two billiard balls
Two balls of modeling clay
Two hard rubber toy balls
An automobile collision
When asking students to express their ideas, you might try one of the following methods.
(1) You could ask them to write their answers in their notebook and then discuss them.
(2) You could ask them to first write their ideas and then share them with a small group of 3 or 4 students. At that time you can have each group present their consensus idea. This can be facilitated with the use of whiteboards for the groups.
The most important aspect of eliciting student’s ideas is the acceptance of all ideas as valid. Do not correct or judge them. You might want to ask questions to help clarify their answers. You do not want to discourage students from thinking about these questions and just waiting for the correct answer from the teacher. Thank them for sharing their ideas. Misconceptions are common and can be dealt with if they are first expressed in writing and orally.
Answers will vary, with the toy balls generally near the top of the list and the football players near the bottom of the list. The point is to get students to think about the meaning of these terms in an everyday context before hearing the definitions. Some students may realize that billiard ball collisions are nearly perfectly elastic. Listen to their ideas and try to see if there is any consensus among the students without interjecting the “correct” answers. In order to determine an exact order for the list, we would need to measure the loss in KE for each of these collisions.

34. Perfectly Inelastic Collisions
Two objects collide and stick together.
Two football players
A meteorite striking the earth
Momentum is conserved.
Masses combine.
Ask students to suggest additional examples of perfectly inelastic collisions.

35. Classroom Practice Problems
A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is initially at rest at a stoplight. The car and truck stick together and move together after the collisions. What is the final velocity of the two vehicle mass?
Answer: 3.8 m/s to the south
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

36. Classroom Practice Problems
A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is initially at rest at a stoplight. The car and truck stick together and move together after the collisions. What is the final velocity of the two vehicle mass?
Answer: 3.8 m/s to the south
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

37. Classroom Practice Problems
A 1500 kg car traveling at 15.0 m/s to the south collides with a 4500 kg truck that is initially at rest at a stoplight. The car and truck stick together and move together after the collisions. What is the final velocity of the two vehicle mass?
Answer: 3.8 m/s to the south
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

38. Classroom Practice Problems
A grocery shopper tosses a 9.0 kg bag of rice into a stationary 18.0 kg cart. The bag hits the cart with a horizontal speed of 5.5 m/s toward the front of the cart. What is the final speed of the cart and the bag?
Answer: 1.8 m/s forward
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

39. Classroom Practice Problems
A 15,000 kg railroad car moving at 7.00 m/s to the north collides with and sticks to another railroad car of the same mass that is moving in the same direction at 1.50 m/s. What is the velocity of the joined cars after the collision?
Answer: 4.0 m/s to the north
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

40. Classroom Practice Problems
A dry cleaner throws a 22 kg bag of laundry onto a stationary 9.0 kg cart. The cart and laundry bag begin moving at 3.0 m/s to the right. Find the velocity of the laundry bag before the collision?
Answer: 4.2 m/s to the right
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

41. Inelastic Collisions Kinetic energy is less after the collision.
It is converted into other forms of energy.
Internal energy - the temperature is increased.
Sound energy - the air is forced to vibrate.
Some kinetic energy may remain after the collision, or it may all be lost.
If the objects are still moving after the collision, then there is still some KE. If both objects have stopped, such as might occur in some head-on collisions, then all of the KE is converted into other forms of energy.

42. Classroom Practice Problems
A 0.25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target
What is the final velocity of the combined mass?
What is the decrease in kinetic energy during the collision?
Answer: 0.43 m/s to the west
Answer: 17 J
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

43. Classroom Practice Problems
During practice, a student kicks a 0.40 kg soccer ball with a velocity of 8.5 m/s to the south into a 0.15 kg bucket lying on its side. The bucket travels with the ball after the collision.
What is the final velocity of the combined mass?
What is the decrease in kinetic energy during the collision?
Answer: 6.2 m/s to the south
Answer: 3 J
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

44. Classroom Practice Problems
A 56 kg ice skater traveling at 4.0 m/s to the north suddenly grabs the hand of a 65 kg skater traveling at 12.0 m/s in the opposite direction as they pass. The two skaters continue skating together with joined hands..
What is the final velocity of the combined mass?
What is the decrease in kinetic energy during the collision?
Answer: 4.6 m/s to the south
Answer: 3900 J
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

45. Elastic Collisions Objects collide and return to their original shape.
Kinetic energy remains the same after the collision.
Perfectly elastic collisions satisfy both conservation laws shown below.
Billiard balls colliding are nearly perfectly elastic. Ideal gases undergo perfectly elastic collisions between molecules and between the walls of the container.

46. Elastic Collisions
Two billiard balls collide head on, as shown. Which of the following possible final velocities satisfies the law of conservation of momentum?
vf,A = 2.0 m/s, vf,B = 2.0 m/s
vf,A = 0 m/s, vf,B = 4.0 m/s
vf,A = 1.5 m/s, vf,B = 2.5 m/s
Answer: all three
m = 0.35 kg m = 0.35 kg
v = 4.0 m/s v = 0 m/s
This slide and the next slide are designed to allow students to see that conservation of momentum can occur in several ways. However, conservation of momentum and kinetic energy can only occur in one way. This will take some time, but should develop the idea that conservation of both KE and momentum limits the possible results in an elastic collision. See the next slide for further explanation.
Momentum before the collision = 1.4 kg•m/s.
Any of the three results shown yields a total momentum after the collision of 1.4 kg•m/s.

47. Elastic Collisions
Two billiard balls collide head on, as shown. Which of the following possible final velocities satisfies the law of conservation of kinetic energy?
vf,A = 2.0 m/s, vf,B = 2.0 m/s
vf,A = 0 m/s, vf,B = 4.0 m/s
vf,A = 1.5 m/s, vf,B = 2.5 m/s
Answer: only vf,A = 0 m/s, vf,B = 4.0 m/s
m = 0.35 kg m = 0.35 kg
v = 4.0 m/s v = 0 m/s
This slide and the previous slide are designed to allow students to see that conservation of momentum can occur in several ways. However, conservation of momentum and kinetic energy can only occur in one way. This will take some time, but should develop the idea that conservation of both KE and momentum limits the possible results in an elastic collision.
KE before the collision is 1/2(0.35 kg)(4.0 m/s)2 = 2.8 J.
The first choice for velocities produce a KE after the collision of 1/2(0.35 kg)(2.0 m/s)2 + 1/2(0.35 kg)(2.0 m/s)2 = 1.4 J (a decrease in KE).
Similarly, the last data points also show a decrease in KE.
Only the velocities of 0 m/s for A and 4.0 m/s for B produce no change in KE. Therefore, this is the only possible result if the collision is elastic.
An infinite number of possible velocities would satisfy the conservation of momentum. The only result possible in a perfectly elastic collision (for equal masses) is that one ball stops and the other continues with the original speed.
Many students have seen the sets of 5 stainless steel balls hanging from threads and allowed to swing back and forth as they collide. If one ball is dropped, only one goes up from the other side. This device behaves like it does because the collisions are nearly perfectly elastic. The steel balls are deformed very slightly, and spring back with little loss of energy.

48. Types of Collisions Click below to watch the Visual Concept.

49. Classroom Practice Problems
A kg marble moving to the right at 22.5 cm/s makes an elastic head-on collision with a kg marble moving to the left at 18.0 cm/s..
Find the velocity of the second marble after the collision.
Calculate the total kinetic energy before and after the collision.
Answer: .23m/s to the right
Answer: KEi = 6.20 X 10-4 J =KEf
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

50. Classroom Practice Problems
A 16.0 kg canoe moving to the left at 12 m/s makes an elastic head-on collision with a 4.0 kg raft moving to the right at 6.0 m/s. After the collision, the raft moves to the left at 22.7 m/s.
Find the velocity of the canoe after the collision.
Calculate the total kinetic energy before and after the collision.
Answer: 5 m/s to the left
Answer: KEi = 1.3 X 103 J , Kef = 1.2 X 103 J
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

51. Classroom Practice Problems
A 4.0 kg bowling ball moving to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially at rest. The first ball stops after the collision.
Find the velocity of the second ball after the collision.
Calculate the total kinetic energy before and after the collision.
Answer: 8 m/s to the right
Answer: KEi = 1.3 X 102 J = Kef
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

52. Classroom Practice Problems
A 25 kg bumper car moving to the right at 5.5 m/s overtakes and collides elastically with a 35 kg bumper car moving to the right at 2.0 m/s. After the collision, the 25 kg bumper car slows to 1.4 m/s to the right.
Find the velocity of the 35 kg bumper car after the collision.
Calculate the total kinetic energy before and after the collision.
Answer: 5.1 m/s to the right
Answer: KEi = 450 J; Kef =480 J
Momentum is conserved, and that is the basis for the first problem. (2.0 x 105 kg)(21 m/s) = (6.0 x 105 kg) (v)
Students should find that KE is not conserved. In fact, it is reduced significantly.

53. Types of Collisions
Recap the three types of collisions. Ask students to describe the ”what happens” and “conserved quantity” columns before you reveal the text.