Forces and Motion In this lesson: 1.Newtons Second Law 2.Momentum & Impulse

Forces and Motion

In this lesson:1. Newton’s Second Law2. Momentum & Impulse

Newton’s Second LawThe 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.

Newton’s Second LawAcceleration is directly proportional to Force

This means a large Force causes

A large acceleration

Newton’s Second LawAcceleration is inversely proportional to Mass

This means a large Mass results in A small

acceleration

Newton’s Second LawThis relationship is written mathematically as:

F=ma

Newton’s Second Law

t

Va

This relationship is written mathematically as:

A useful form of Newton's Second Law requires the substitution of the acceleration formula below.

F=ma

Newton’s Second LawThe substitution results in the following formula:

The result is two new concepts:

Impulse & Momentum

Concept CheckQuestion (True or False) Answer

1. Acceleration depends on net force, the mass, and shape of the object.

2. If I triple the net force on an object the acceleration will triple.

3. A ball with a mass of 0.25kg that is thrown with 4N of force will accelerate at 1 m/s/s.

4. If I triple the mass of an object the acceleration will triple.

5. When two objects that were moving at a constant speed collide force is created.

Click to check your answers

Concept CheckQuestion (True or False) Answer

1. Acceleration depends on net force, the mass, and shape of the object. False

2. If I triple the net force on an object the acceleration will triple. True

3. A ball with a mass of 0.25kg that is thrown with 4N of force will accelerate at 1 m/s/s. True

4. If I triple the mass of an object the acceleration will triple. False

5. When two objects that were moving at a constant speed collide force is created. True

Click to check your answers

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.

Solve the following momentum problems:

Problem Answer

1. A moving car has momentum. If it moves twice as fast, its momentum is ___________ as much.

2. A steel ball whose mass is 2.0 kg is rolling at a rate of 2.8m/s. What is its momentum?

3. A marble is rolling at a velocity of 1.5 m/s with a momentum of 0.10 kgm/s. What is its mass?

4. Two cars, one twice as heavy as the other, move down a hill at the same speed. Compared to the lighter car, the momentum of the heavier car is ____________ as much.

5. A 5100-kg freight truck accelerates from 4.2 m/s to 7.8 m/s what is it’s change in momentum?

Solve the following momentum problems:

Problem Answer

1. A moving car has momentum. If it moves twice as fast, its momentum is ___________ as much.

twice

2. A steel ball whose mass is 2.0 kg is rolling at a rate of 2.8m/s. What is its momentum?

5.6kg·m/s

3. A marble is rolling at a velocity of 1.5 m/s with a momentum of 0.10 kgm/s. What is its mass?

0.067kg or 67g

4. Two cars, one twice as heavy as the other, move down a hill at the same speed. Compared to the lighter car, the momentum of the heavier car is ____________ as much.

Twice

5. A 5100-kg freight truck accelerates from 4.2 m/s to 7.8 m/s what is it’s change in momentum?

18 000 kg·m/s

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.

vmtF

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

• Forces are transferred– Action reaction forces happen

• 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

Total momentum

before an interaction

This means that… ip

Total momentum

after an interaction

fp=Σ is the Greek letter Sigma and mean “sum” or “total”

So, this equation would read…

“total momentum initial equals total momentum final”

More ways to represent

conservation of momentum:

During an Impact

fi pp

ffiipppp 2121

ffiivmvmvmvm 22112211

“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 ”

1) Example - Meteorite• A meteorite breaks up into

two 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 before the break up to the momentum after the break up.

1)Compare the momentum before the break up to the momentum after the

break up.Record what you know.

Known

Possible formulas

Show Work

Final Answer

Unknown

1) Compare the momentum before the break up to the momentum after the break up.

Write a equation that shows the momentum before = the momentum after

Known

Possible formulas

Show Work

Final Answer

Unknown

Beforem1 = 16 kgv1=12 m/s

Afterm2= 8.0 kgv2=12 m/sm3 = 8.0 kgv3=12 m/s

mvp

1) Compare the momentum before the break

up to the momentum after the break up.Replace p with the momentum

formulaKnown

Possible formulas

Show Work

Final Answer

Unknown



mvp

Momentum before = sum of the momentums after p1 = p2+p3

= +


Plug in your numbers

Known

Possible formulas

Show Work

Final Answer

Unknown



mvp


m1v1= m2v2+m3v3


Solve each side of the equation to confirm if it is true

Known

Possible formulas

Show Work

Final Answer

Unknown



mvp


m1v1= m2v2+m3v3

16 kg(12m/s) = 8.0kg(12m/s)+ 8.0kg(12m/s)

1) Compare the momentum before the break up to the momentum

after the break up.

Known

Possible formulas

Show Work

Final Answer

Unknown



mvp


m1v1= m2v2+m3v3

16 kg(12m/s) = 8.0kg(12m/s)+ 8.0kg(12m/s) 192 kg·m/s = 96 kg·m/s + 96 kg·m/s 192 kg·m/s = 192 kg·m/s

They are equal!!

Total Momentum

• 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.

2) 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?

2) What is the speed of the car and brick after?

WAIT!! We have 2 variables V1 and V2

Known

Possible formulas

Show Work

Final Answer

Unknown

Beforem1 = 1.0 kgv1= 60.0 m/s

Afterm2= 1.0 kgv2=?m3 = 2.0 kgv3=?

Stupid problem can’s be solve. Ms Schwartz is just trying to kill my brain


WAIT!! We have 2 variables V1 and V2

Known

Possible formulas

Show Work

Final Answer

Unknown


Afterm2= 1.0 kgv2=?m3 = 2.0 kgv3=?

Stupid problem can’s be solve. Ms Schwartz is just trying to kill my brain

It is solvable but we need to assume something.

What can assume?


The Cart and brick stick together they have the same speed v2=v3

Now Set up your equationKnown

Possible formulas

Show Work

Final Answer

Unknown

Afterm2= 1.0 kgv2= v3

m3 = 2.0 kg



Plug in your numbers and Solve

Known

Possible formulas

Show Work

Final Answer

Unknown


Afterm2= 1.0 kgv2=?m3 = 2.0 kgv3=?

mvp

Momentum before = sum of the momentums after

p1 = p2+p3

m1v1 = m2v2+ m3v3


And the Answer is? Known

Possible formulas

Show Work

Final Answer

Unknown

Beforem1 = 1.0 kgv1= 60 m/s

Afterm2= 1.0 kgv2=?m3 = 2.0 kgv3=?

mvp


p1 = p2+p3

m1v1 = m2v2+ m3v3

1.0kg(60.0 m/s) = 1.0kg(v)+ 2.0 kg(v) 60 kg·m/s = 3 kg (v)


And the Answer is?

Known

Possible formulas

Show Work

Final Answer 20.0 m/s

Unknown

Beforem1 = 1.0 kgv1= 60 m/s

Afterm2= 1.0 kgv2=?m3 = 2.0 kgv3=?

mvp


p1 = p2+p3

m1v1 = m2v2+ m3v3

1.0kg(60 m/s) = 1.0kg(v)+ 2.0 kg(v) 60 kg·m/s = 3.0 kg (v) 60 kg·m/s/3kg = (v)

20.0 m/s = v

Watch it change

What do we know? And Set up the equation

3) A 3000. kg truck travelling at 20.0 m/s hits a 1000. kg car and they

stick together. What is the speed of each after the impact?

3) What is the speed of the car and truck after?

And the Answer is?

Known

Possible formulas

Show Work

Final Answer

Unknown

Beforem1 = 3000. kgv1= 20.0 m/sm2= 1000. kgv2= 0 m/s

Afterm3= 3000. kgv3=?m4 = 1000. kgv4=?

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

3000.kg(20.0m/s)+1000kg(0m/s) = 3000.kg(v)+1000kg(v)

3) What is the speed of the car and truck after?

Known

Possible formulas

Show Work


Unknown


Afterm3= 3000. kgv3=?m4 = 1000. kgv4=?

mvp


p1+p2 = p3+p4 m1v1 +m2v2= m3v3 + m4v4

3000.kg(20.0m/s)+1000kg(0m/s) = 3000.kg(v)+1000kg(v) 60,000 kg·m/s = 4000.kg (v)60,000 kg·m/s /4000. kg = v 15.0 m/s = v

Watch it happen

4) This time the 1000. kg car travelling at 20.0 m/s hits the 3000.

kg truck!

Make a prediction: will the end speed be greater or less then 15 m/s?

What do you know? Set up the formula

4) What is the speed of the car and Truck after?

And the Answer is?

Known

Possible formulas

Show Work

Final Answer

Unknown


Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(0m/s) = 1000.kg(v)+3000kg(v)


Known

Possible formulas

Show Work


Unknown


Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(0m/s) = 1000.kg(v)+3000kg(v) 20,000 kg·m/s = 4000.kg (v)20,000 kg·m/s /4000.kg =v

5.0 m/s = v

Watch it happen

Remember sign of velocity indicates

direction

5) A 1000. kg car is travelling at 20.0 m/s. A 3000 kg truck is

travelling in the opposite direction at 20.0 m/s. After the collision they stick together. At what speed and

in which direction do they go?

What do you know?


Known

Possible formulas

Show Work

Final Answer

Unknown

Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= 20.0 m/s

Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?

mvp

Wait a sec. Something is not right with the signs in the known.

What is it?

Remember sign of velocity indicates

direction. And the truck is going in the

opposite direction.

Set up the problem


And the Answer is?

Known

Possible formulas

Show Work

Final Answer

Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= -20.0 m/s

Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(-20.0m/s) = 1000.kg(v)+ 3000 kg (v)

Unknown


Known

Possible formulas

Show Work

Final Answer -10.0 m/s


Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(-20.0m/s) = 1000.kg(v)+ 3000 kg (v) -40,000 kg·m/s = 4000.kg (v) -40,000 kg·m/s /4000.kg = v

-10.0 m/s = v

Unknown

Watch What happens.

6) 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 speed of the car?

Set up the problem

6) What is the speed of the car after?

And the Answer is?

Known

Possible formulas

Show Work

Final Answer


Afterm3= 1000. kgv3=?m4 = 3000. kgv4= 0 m/s

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(-20.0m/s) = 1000.kg(v)+ 3000 kg (0)

Unknown

6) What is the speed of the car after?

Known

Possible formulas

Show Work

Final Answer


Afterm3= 1000. kgv3=?m4 = 3000. kgv4= 0 m/s

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(-20.0m/s) = 1000.kg(v)+ 3000 kg (0)

-40,000 kg·m/s = 1000.kg (v) -40,000 kg·m/s /1000.kg = v -400.0 m/s = v

Unknown

Watch What happens

7) A1000. kg car travelling at 20.0 m/s hits a stationary 3000. kg truck. The truck starts to move at a rate of 10.0 m/s. How fast does the car

go? What do you know? Set up the equation.

7) How fast does the car go? And the Answer is?

Known

Possible formulas

Show Work

Final Answer


Afterm3= 1000. kgv3=?m4 = 3000. kgv4= -10.0 m/s

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(0 m/s) = 1000.kg(v)+3000 kg(10.0m/s)

Unknown

7) How fast does the car go?

Known

Possible formulas

Show Work

Final Answer


Afterm3= 1000. kgv3=?m4 = 3000. kgv4= -10.0 m/s

mvp


p1+p2 = p3+p4

m1v1 +m2v2= m3v3 + m4v4

1000.kg(20.0m/s)+3000kg(0 m/s) = 1000.kg(v)+3000 kg(10.0m/s) 20,000 kg·m/s = 1000.kg (v)+ 30,000 kg·m/s

20,000 kg·m/s- 30,000 kg·m/s = 1000.kg (v) -10,000 kg·m/s /1000.kg = v -10.0 m/s = v

Unknown

Watch what happens.

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• Using this concept we can calculate the

speed of an object after the collision.

Now

• Complete the additional Conservation of Momentum problems in your packet.

Documents

Forces and Motion In this lesson: 1.Newtons Second Law 2.Momentum & Impulse