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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
Beforem1 = 16 kgv1=12 m/s
Afterm2= 8.0 kgv2=12 m/sm3 = 8.0 kgv3=12 m/s
mvp
Momentum before = sum of the momentums after p1 = p2+p3
= +
1) Compare the momentum before the break up to the momentum after the break up.
Plug in your numbers
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
Momentum before = sum of the momentums after p1 = p2+p3
m1v1= m2v2+m3v3
1) Compare the momentum before the break up to the momentum after the break up.
Solve each side of the equation to confirm if it is true
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
Momentum before = sum of the momentums after p1 = p2+p3
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
Beforem1 = 16 kgv1=12 m/s
Afterm2= 8.0 kgv2=12 m/sm3 = 8.0 kgv3=12 m/s
mvp
Momentum before = sum of the momentums after p1 = p2+p3
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
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
It is solvable but we need to assume something.
What can assume?
2) What is the speed of the car and brick after?
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
Beforem1 = 1.0 kgv1= 60.0 m/s
2) What is the speed of the car and brick after?
Plug in your numbers and Solve
Known
Possible formulas
Show Work
Final Answer
Unknown
Beforem1 = 1.0 kgv1= 60.0 m/s
Afterm2= 1.0 kgv2=?m3 = 2.0 kgv3=?
mvp
Momentum before = sum of the momentums after
p1 = p2+p3
m1v1 = m2v2+ m3v3
2) What is the speed of the car and brick after?
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
Momentum before = sum of the momentums after
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)
2) What is the speed of the car and brick after?
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
Momentum before = sum of the momentums after
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
Momentum before = sum of the momentums after
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
Final Answer 15.0 m/s
Unknown
Beforem1 = 3000. kgv1= 20.0 m/sm2= 1000. kgv2= 0 m/s
Afterm3= 3000. kgv3=?m4 = 1000. kgv4=?
mvp
Momentum before = sum of the momentums after
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
Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= 0 m/s
Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?
mvp
Momentum before = sum of the momentums after
p1+p2 = p3+p4
m1v1 +m2v2= m3v3 + m4v4
1000.kg(20.0m/s)+3000kg(0m/s) = 1000.kg(v)+3000kg(v)
4) What is the speed of the car and Truck after?
Known
Possible formulas
Show Work
Final Answer 5.0 m/s
Unknown
Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= 0 m/s
Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?
mvp
Momentum before = sum of the momentums after
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?
5) What is the speed of the car and Truck after?
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
5) What is the speed of the car and Truck after?
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
Momentum before = sum of the momentums after
p1+p2 = p3+p4
m1v1 +m2v2= m3v3 + m4v4
1000.kg(20.0m/s)+3000kg(-20.0m/s) = 1000.kg(v)+ 3000 kg (v)
Unknown
5) What is the speed of the car and Truck after?
Known
Possible formulas
Show Work
Final Answer -10.0 m/s
Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= -20.0 m/s
Afterm3= 1000. kgv3=?m4 = 3000. kgv4=?
mvp
Momentum before = sum of the momentums after
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
Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= -20.0 m/s
Afterm3= 1000. kgv3=?m4 = 3000. kgv4= 0 m/s
mvp
Momentum before = sum of the momentums after
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
Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= -20.0 m/s
Afterm3= 1000. kgv3=?m4 = 3000. kgv4= 0 m/s
mvp
Momentum before = sum of the momentums after
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
Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= 0.0 m/s
Afterm3= 1000. kgv3=?m4 = 3000. kgv4= -10.0 m/s
mvp
Momentum before = sum of the momentums after
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
Beforem1 = 1000. kgv1= 20.0 m/sm2= 3000. kgv2= 0.0 m/s
Afterm3= 1000. kgv3=?m4 = 3000. kgv4= -10.0 m/s
mvp
Momentum before = sum of the momentums after
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.
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