# Preview Section 1 Momentum and Impulse

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Preview Section 1 Momentum and Impulse
Section 2 Conservation of Momentum Section 3 Elastic and Inelastic Collisions Section 4 Extra Questions

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.

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.

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.

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.

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

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.

Classroom Practice Problems
A 1350 kg car has a velocity of 22.0 m/s to the north. When braking rapidly, it stops in 4.50 s. What was the momentum of the car before braking? What is the magnitude of the force required to stop the car? Answers: 2.97 x 104 kg • m/s to the north 6.60 x 103 N 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.

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.

Classroom Practice Problems
A 65 kg passenger in a car travels at a speed of 8.0 m/s. If the passenger is stopped by an airbag in 0.75 s, how much force is required? Answer: 6.9 x 102 N If the car does not have an air bag and the passenger is instead stopped in s when he strikes the dashboard, by what factor does the force increase? Answer: F = 2.0 x 104 N so it is 29 times greater 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.

Now 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? Students should list the many safety features. They should now realize that the similarity between them is extending the time of impact in order to reduce the force required to stop.

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)

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).

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.

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

Classroom Practice Problems
A 62.0 kg astronaut on a spacewalk tosses a kg baseball at 26.0 m/s out into space. With what speed does the astronaut recoil? Step 1: Find the initial momentum of both astronaut and baseball. Answer: zero because vi = 0 for both Step 2: Since pi = 0, then pf, astronaut= -pf, baseball Step 3: Substitute and solve for vf,astronaut Answer: m/s or cm/s Does a pitcher recoil backward like the astronaut when throwing the ball? Explain. 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 dealth 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.

Classroom Practice Problem
Gerard is a quarterback and Tyler is a defensive lineman. Gerard’s mass is 75.0 kg and he is at rest. Tyler has a mass of 112 kg, and he is moving at 8.25 m/s when he tackles Gerard by holding on while they fly through the air. With what speed will the two players move together after the collision? Answer: 4.94 m/s Initial momentum: zero for Gerard and 924 kg•m/s for Tyler Both players are combined after the collision (m = 187 kg) and move with the same speed (v). Initial momentum = Final momentum 924 kg•m/s = (187 kg)(v) Remind students that they did not need to know anything about the force. They did not need to know how much force was present, or if it was constant, or how long the force lasted.

Now 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? Scenario 1: The forces are equal and the velocities are equal. Scenario 2: The forces are still equal (3rd law) but the velocities are not equal. The girl with more mass will have a smaller velocity. However, when calculating the momentum (mv) for each girl, they will be the same as before. Scenario 3: The forces are still equal (3rd law) and the total momentum of the two will be equal to the momentum of the moving girl prior to the collision. It is impossible to describe the motion of either one after the collision without more information.

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.

Classroom Practice Problems
An 2.0 x 105 kg train car moving east at 21 m/s collides with a 4.0 x 105 kg fully-loaded train car initially at rest. The two cars stick together. Find the velocity of the two cars after the collision. Answer: 7.0 m/s to the east Now calculate the kinetic energy of the two cars before and after the collision. Was kinetic energy conserved? Answer: KEbefore= 4.4 x 107 J, KEafter= 1.5 x 107 J KE is not conserved. It is less after the collision. 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.

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.

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.

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.

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.

Types of Collisions Click below to watch the Visual Concept.

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.

Now what do you think? 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 Perfectly inelastic means they stick together, so the modeling clay, football players, baseball and glove and possibly the cars would be the least elastic. The most elastic is the billiard ball collision, followed by the toy rubber ball collision and bat and ball collision. The only way to get a perfectly accurate ranking is to measure the KE before and after each collision. The greater the % lost, the less elastic the collision.

Preview Multiple Choice Short Response Extended Response

Multiple Choice, continued
2. The vector below represents the momentum of a car traveling along a road. The car strikes another car, which is at rest, and the result is an inelastic collision. Which of the following vectors represents the momentum of the first car after the collision? F. G. H. J.

Multiple Choice, continued
3. What is the momentum of a kg baseball thrown with a velocity of 35 m/s toward home plate? A. 5.1 kg • m/s toward home plate B. 5.1 kg • m/s away from home plate C. 5.2 kg • m/s toward home plate D. 5.2 kg • m/s away from home plate

Multiple Choice, continued
Use the passage below to answer questions 4–5. After being struck by a bowling ball, a 1.5 kg bowling pin slides to the right at 3.0 m/s and collides head-on with another 1.5 kg bowling pin initially at rest. 4. What is the final velocity of the second pin if the first pin moves to the right at 0.5 m/s after the collision? F. 2.5 m/s to the left G. 2.5 m/s to the right H. 3.0 m/s to the left J. 3.0 m/s to the right

Multiple Choice, continued
Use the passage below to answer questions 4–5. After being struck by a bowling ball, a 1.5 kg bowling pin slides to the right at 3.0 m/s and collides head-on with another 1.5 kg bowling pin initially at rest. 5. What is the final velocity of the second pin if the first pin stops moving when it hits the second pin? A. 2.5 m/s to the left B. 2.5 m/s to the right C. 3.0 m/s to the left D. 3.0 m/s to the right

Multiple Choice, continued
6. For a given change in momentum, if the net force that is applied to an object increases, what happens to the time interval over which the force is applied? F. The time interval increases. G. The time interval decreases. H. The time interval stays the same. J. It is impossible to determine the answer from the given information.

Multiple Choice, continued
8. Two shuffleboard disks of equal mass, one of which is orange and one of which is yellow, are involved in an elastic collision. The yellow disk is initially at rest and is struck by the orange disk, which is moving initially to the right at 5.00 m/s. After the collision, the orange disk is at rest. What is the velocity of the yellow disk after the collision? Think energy! F. zero G m/s to the left H m/s to the right J m/s to the right Talk about Conservation of energy.

Multiple Choice, continued
Use the information below to answer questions 9–10. A kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. 9. What is the large bead’s velocity after the collision? A cm/s to the right B cm/s to the right C cm/s to the right D cm/s to the right

Multiple Choice, continued
Use the information below to answer questions 9–10. A kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The bead collides elastically with a larger kg bead that is initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. 10. What is the total kinetic energy of the system after the collision? F  10–4 J G  10–4 J H  10 –4 J J  10 –4 J

Short Response 11. Is momentum conserved when two objects with zero initial momentum push away from each other?

Short Response, continued
An 8.0 g bullet is fired into a 2.5 kg pendulum bob, which is initially at rest and becomes embedded in the bob. The pendulum then rises a vertical distance of 6.0 cm. 340 m/s 13. What was the initial speed of the bullet? Show your work.

Short Response, continued
Base your answers to questions 13–14 on the information below. An 8.0 g bullet is fired into a 2.5 kg pendulum bob, which is initially at rest and becomes embedded in the bob. The pendulum then rises a vertical distance of 6.0 cm. 1.5 m/s 14. What will be the kinetic energy of the pendulum when the pendulum swings back to its lowest point? Show your work.

Extended Response 15. An engineer working on a space mission claims that if momentum concerns are taken into account, a spaceship will need far less fuel for the return trip than for the first half of the mission.Write a paragraph to explain and support this hypothesis.