# Momentum

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Category: Education

## Presentation Description

Covers Newton's second Law, Momentum and Impulse, and conservation of momentum.

## Comments

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## Presentation Transcript

### Forces and Motion :

Forces and Motion In this lesson: Newton’s Second Law 3. Conservation of Momentum Momentum & Impulse

### Slide 2:

Newton’s Second Law Acceleration is directly proportional to Force This means a large Force causes A large acceleration

### Slide 3:

Newton’s Second Law Acceleration is inversely proportional to Mass This means a large Mass results in A small acceleration

### Slide 4:

Newton’s Second Law This relationship is written mathematically as: F=ma

### Slide 5:

Newton’s Second Law The common definition of a force is any push or pull. A more interesting and useful definition is any interaction between two (or more) objects. Newton’s second law can explain that interaction and the resulting change in motion.

### Slide 6:

Newton’s Second Law This relationship is written mathematically as: A useful form of Newton's Second Law requires the substitution of the acceleration formula below. F=ma Which can also be written as…

### Slide 7:

Newton’s Second Law The substitution results in the following formula: The result is two new concepts: Impulse & Momentum

### Concept Check :

Concept Check Click to check your answers

### Concept Check :

Concept Check Click to check your answers

### Slide 10:

Momentum Momentum of an object is simply defined as the mass times the velocity. It is usually abbreviated as “ρ”. Mass measured in kg times velocity measured in m/s results in a momentum with units of kg m/s.

### Slide 11:

Solve the following momentum problems:

### Slide 12:

Solve the following momentum problems:

### Take Home Points: :

Take Home Points: Forces result from interactions of objects. Acceleration is directly proportional to force and indirectly proportional to mass. Newton’s second law can be written as mv is called momentum; mΔv is change in momentum.

Conservation of

### Presentation Goals :

Presentation Goals Following this presentation you should be able to: Explain the concept of conservation of momentum Apply the conservation of momentum in real world situations to predict outcomes of interactions. Solve conservation of momentum problems using mathematical models.

### During an Impact :

During an Impact Action reaction forces happen Momentum is transfered Objects undergo acceleration The velocity changes Each object’s momentum changes In any interaction between two or more objects: But the Total Momentum of a system* remains constant. *System: group of interacting, interrelated, or interdependent elements or parts that function together as a whole

### During an Impact :

During an Impact Total momentum before an interaction This means that… Total momentum after an interaction is the Greek letter Sigma and means “sum” or “total” So, this equation would read… “total momentum initial equals total momentum final” =

### During an Impact :

More ways to represent conservation of momentum: During an Impact “total momentum initial equals total momentum final” “initial momentum of object #1 plus the initial momentum of object #2 equals final momentum of object #1 plus the final momentum of object #2 ” “mass of object #1 times the initial velocity of object #1 plus mass of object #2 times the initial velocity of object #2 equals mass of object #1 times the final velocity of object #1 plus mass of object #2 times the final velocity of object #2 ”

### Example #1- Meteorite :

Example #1- Meteorite A meteorite entering the earths atmosphere breaks up into two equal pieces. The mass of the original meteorite is 16 kg and travels at a rate of 12 m/s. The two pieces each have a mass of 8.0 kg. Newton’s 1st law says that unless an outside force is present the speed will remain constant. So the speed of each piece is 12 m/s. Compare the momentum of the meteorite before the break up to the momentum of the pieces after the break up as a proof of conservation.

### First Step: Record what you know. :

First Step: Record what you know. So from the problem we know…. There are 3 objects; the meteorite (object #1) and the two pieces of the meteorite (object #2 and object #3). We know each objects mass and velocity. Let’s enter this information into our scaffold: Before m1 = 16 kg v1=12 m/s After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s

### Step 2: Select a equation to use :

For 3 objects; the meteorite (object #1) and the two pieces of the meteorite (object #2 and object #3), We need a variation of: Let’s enter this information into our scaffold: Step 2: Select a equation to use Before m1 = 16 kg v1=12 m/s After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s

### Step 3: Manipulate & Substitute :

After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s Using our known information manipulate the equation to fit our problem. Then substitute our data. Let’s enter this information into our scaffold: Step 3: Manipulate & Substitute Before m1 = 16 kg v1=12 m/s

### Step 4: Solve :

After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s Step 4: Solve Now solve for total momentum before and after the collision. Let’s enter this information into our scaffold: Before m1 = 16 kg v1=12 m/s =190kg·m/s 96kg·m/s + 96kg·m/s =190kg·m/s 190kg·m/s =190kg·m/s

### Slide 24:

Even if the meteorite broke up into a thousand little pieces the momentum of all the pieces added together would still equal the momentum of the original meteorite. Total Momentum

### A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? :

A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? What do you know? Sample Problem #1

### A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? :

Before m1 = 1.0 kg v1= 60.0 m/s m2 = 2.0 kg v2= 0m/s In this problem we have 2 objects, the cart (object 1) and the brick (object 2). Initially the brick isn’t moving so it’s momentum (p2i) is zero. The cart and brick are moving together after the collision so their velocities are the same so we can call them just vf (vf =v1f=v2f). Our known variables and unknown variable are: A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? After m1 = 1.0 kg v1= ? m2 = 2.0 kg v2= ? vf =v1f=v2f

### A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? :

There are several possible formulas to use but they are all variations of:. The variations are: Before m1 = 1.0 kg v1= 60.0 m/s m2 = 2.0 kg v2= 0m/s A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? After m1 = 1.0 kg v1= ? m2 = 2.0 kg v2= ? vf =v1f=v2f

### A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? :

Before m1 = 1.0 kg v1= 60.0 m/s m2 = 2.0 kg v2= 0m/s We will use the last equation then substitute and solve for the final velocity (vf). The substitution and solution is: A 1.0 kg moving cart (velocity= 60.0 m/s) catches a 2.0 kg brick. What is the speed of the car and brick after? After m1 = 1.0 kg v1= ? m2 = 2.0 kg v2= ? vf =v1f=v2f 20m/s

### Take a look at the animation again. Watch the values change as the interaction occurs. :

Take a look at the animation again. Watch the values change as the interaction occurs. There are a lot of variables in these problems. It is essential to use a strategy like the one presented here to keep track of them all.

### Practice Problems :

Use the problem scaffold to solve the following conservation of momentum sample problems. Practice Problems The solutions are provided on the subsequent slides.

### Practice Problems :

A 3000.kg truck travelling at 20.0m/s hits a 1000.kg car and they stick together. What is the velocity of each vehicle after the impact? Practice Problems Solution on next slide

### A 3000.kg truck travelling at 20.0m/s hits a 1000.kg car and they stick together. What is the final velocity of the vehicles? :

Before m1 = 3000. kg v1= 20.0 m/s m2 = 1000. kg v2= 0m/s A 3000.kg truck travelling at 20.0m/s hits a 1000.kg car and they stick together. What is the final velocity of the vehicles? After m1 = 3000.kg v1= ? m2 = 1000. kg v2= ? vf =v1f=v2f 15m/s

### Watch the momentum and velocity change during the interaction. :

Watch the momentum and velocity change during the interaction.

### Practice Problems :

2. This time the 1000.kg car travelling at 20.0m/s hits the 3000.kg truck! What is the velocity of each vehicle after the impact? Practice Problems Solution on next slide

### This time the 1000.kg car travelling at 20.0m/s hits the 3000.kg truck! What is the final velocity :

Before m1 = 1000. kg v1= 20.0 m/s m2 = 3000. kg v2= 0m/s This time the 1000.kg car travelling at 20.0m/s hits the 3000.kg truck! What is the final velocity After m1 = 1000.kg v1= ? m2 = 3000. kg v2= ? vf =v1f=v2f 5.00m/s

### Watch the momentum and velocity change during the interaction. :

Watch the momentum and velocity change during the interaction.

### Practice Problems :

3. A 1000.kg car is travelling at 20.0m/s. A 3000kg truck is travelling in the opposite direction at 20.0m/s. After the collision they stick together. At what speed and in which direction do they go? What is the velocity of each vehicle after the impact? Practice Problems Solution on next slide Remember: Sign of the velocity shows direction!

### A 1000.kg car is travelling at 20.0m/s. A 3000kg truck is travelling in the opposite direction at 20.0m/s. :

Before m1 = 1000. kg v1= +20.0 m/s m2 = 3000. kg v2= -20.0m/s A 1000.kg car is travelling at 20.0m/s. A 3000kg truck is travelling in the opposite direction at 20.0m/s. After m1 = 1000.kg v1= ? m2 = 3000. kg v2= ? vf =v1f=v2f -10.0m/s

### Watch the momentum and velocity change during the interaction. :

Watch the momentum and velocity change during the interaction.

### Practice Problems :

4. Objects can bounce rather then stick together. This time all the momentum of the 3000. kg truck travelling at 20.0 m/s gets passed to the 1000. kg car that is travelling at 20.0 m/s in the opposite direction. What is the velocity of the car after the impact? Practice Problems Solution on next slide Remember: Sign of the velocity shows direction!

### The momentum of the 3000.kg truck travelling at 20.0m/s gets passed to the 1000.kg car that is travelling at 20.0m/s in the opposite direction. :

Before m1 = 1000. kg v1= +20.0 m/s m2 = 3000. kg v2= -20.0m/s The momentum of the 3000.kg truck travelling at 20.0m/s gets passed to the 1000.kg car that is travelling at 20.0m/s in the opposite direction. After m1 = 1000.kg v1= ? m2 = 3000. kg v2= 0m/s vf =v1f=v2f -40.0m/s

### Watch the momentum and velocity change during the interaction. :

Watch the momentum and velocity change during the interaction.

### Practice Problems :

5. A1000. kg car travelling at 20.0m/s hits a stationary 3000.kg truck. The truck starts to move at a rate of 10.0m/s. What is the velocity of the car after the impact? Practice Problems Solution on next slide Remember: Sign of the velocity shows direction!

### A1000. kg car travelling at 20.0m/s hits a stationary 3000.kg truck. The truck starts to move at a rate of 10.0m/s. :

Before m1 = 1000. kg v1= +20.0 m/s m2 = 3000. kg v2= 0.0m/s A1000. kg car travelling at 20.0m/s hits a stationary 3000.kg truck. The truck starts to move at a rate of 10.0m/s. After m1 = 1000.kg v1= ? m2 = 3000. kg v2= 10.0m/s vf =v1f=v2f -10.0m/s

### Watch the momentum and velocity change during the interaction. :

Watch the momentum and velocity change during the interaction.

### Slide 46:

Take home points The total momentum before a collision must equal the total momentum after the collision. This is known as the Law of Conservation of Momentum This is mathematically represented: Using this concept we can calculate the velocity or mass of an object before or after a collision.