Download presentation

Presentation is loading. Please wait.

Published byLondon Emans Modified over 3 years ago

1
**In this lesson: Newton’s Second Law Momentum & Impulse**

Forces and Motion In this lesson: Newton’s Second Law Momentum & Impulse

2
**The common definition of a force is any push or pull.**

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.

3
**Newton’s Second Law Acceleration is directly proportional to Force**

A large acceleration This means a large Force causes

4
**Newton’s Second Law Acceleration is inversely proportional to Mass**

This means a large Mass results in A small acceleration

5
**This relationship is written mathematically as:**

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

6
**This relationship is written mathematically as:**

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

7
**Newton’s Second Law Impulse & Momentum**

The substitution results in the following formula: The result is two new concepts: Impulse & Momentum

8
**Click to check your answers**

Concept Check Question (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

9
**Click to check your answers**

Concept Check Question (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. 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

10
**It is usually abbreviated as “ρ”.**

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.

11
**Solve the following momentum problems:**

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?

12
**Solve the following momentum problems:**

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? kg·m/s

13
**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.

14
Conservation of

15
**Presentation Goals Explain the concept of conservation of momentum**

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.

16
**During an Impact In any interaction between two or more objects:**

Forces are transferred Action reaction forces happen Objects undergo acceleration The velocity changes Each object’s momentum changes 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

17
**= This means that… During an Impact Total momentum**

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

18
**More ways to represent conservation of momentum:**

During an Impact More ways to represent conservation of momentum: “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 ”

19
**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.

20
**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

21
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 After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s Before m1 = 16 kg v1=12 m/s

22
1) Compare the momentum before the break up to the momentum after the break up. Replace p with the momentum formula Known Possible formulas Show Work Final Answer Unknown After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s Before m1 = 16 kg v1=12 m/s Momentum before = sum of the momentums after p1 = p2+p3 = +

23
**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 After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s Before m1 = 16 kg v1=12 m/s Momentum before = sum of the momentums after p1 = p2+p3 m1v1= m2v2+m3v3

24
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 After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s Before m1 = 16 kg v1=12 m/s 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)

25
**1) Compare the momentum before the break up to the momentum after the break up.**

Known Possible formulas Show Work Final Answer Unknown After m2= 8.0 kg v2=12 m/s m3 = 8.0 kg v3=12 m/s Before m1 = 16 kg v1=12 m/s 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!!

26
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.

27
**2) A 1. 0 kg moving cart (velocity= 60. 0 m/s) catches a 2. 0 kg brick**

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?

28
**2) What is the speed of the car and brick after. WAIT**

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 After m2= 1.0 kg v2=? m3 = 2.0 kg v3=? Before m1 = 1.0 kg v1= 60.0 m/s Stupid problem can’s be solve. Ms Schwartz is just trying to kill my brain

29
**2) What is the speed of the car and brick after. WAIT**

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 After m2= 1.0 kg v2=? m3 = 2.0 kg v3=? Before m1 = 1.0 kg v1= 60.0 m/s 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?

30
**2) What is the speed of the car and brick after**

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 equation Known Possible formulas Show Work Final Answer Unknown After m2= 1.0 kg v2= v3 m3 = 2.0 kg Before m1 = 1.0 kg v1= 60.0 m/s

31
**2) What is the speed of the car and brick after**

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 After m2= 1.0 kg v2=? m3 = 2.0 kg v3=? Before m1 = 1.0 kg v1= 60.0 m/s Momentum before = sum of the momentums after p1 = p2+p3 m1v1 = m2v2+ m3v3

32
**2) What is the speed of the car and brick after? And the Answer is?**

Known Possible formulas Show Work Final Answer Unknown After m2= 1.0 kg v2=? m3 = 2.0 kg v3=? Before m1 = 1.0 kg v1= 60 m/s 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)

33
**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 After m2= 1.0 kg v2=? m3 = 2.0 kg v3=? Before m1 = 1.0 kg v1= 60 m/s 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

34
Watch it change

35
**3) A 3000. kg truck travelling at 20. 0 m/s hits a 1000**

3) A kg truck travelling at 20.0 m/s hits a kg car and they stick together. What is the speed of each after the impact? What do we know? And Set up the equation

36
**3) What is the speed of the car and truck after? And the Answer is?**

Known Show Work After m3= kg v3=? m4 = kg v4=? Before m1 = kg v1= 20.0 m/s m2= kg v2= 0 m/s 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) Unknown Possible formulas Final Answer

37
**3) What is the speed of the car and truck after?**

Known Show Work Before m1 = kg v1= 20.0 m/s m2= kg v2= 0 m/s After m3= kg v3=? m4 = kg v4=? 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 Unknown Possible formulas Final Answer 15.0 m/s

38
Watch it happen

39
**4) This time the 1000. kg car travelling at 20. 0 m/s hits the 3000**

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

40
**4) What is the speed of the car and Truck after? And the Answer is?**

Known Show Work Before m1 = kg v1= 20.0 m/s m2= kg v2= 0 m/s After m3= kg v3=? m4 = kg v4=? 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) Unknown Possible formulas Final Answer

41
**4) What is the speed of the car and Truck after?**

Known Show Work Before m1 = kg v1= 20.0 m/s m2= kg v2= 0 m/s After m3= kg v3=? m4 = kg v4=? 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 Unknown Possible formulas Final Answer 5.0 m/s

42
Watch it happen

43
**Remember sign of velocity indicates direction**

44
**5) A 1000. kg car is travelling at 20. 0 m/s**

5) A 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?

45
**5) What is the speed of the car and Truck after?**

Wait a sec. Something is not right with the signs in the known. What is it? Known Show Work Before m1 = kg v1= 20.0 m/s m2= kg v2= 20.0 m/s After m3= kg v3=? m4 = kg v4=? Unknown Possible formulas Final Answer

46
**Remember sign of velocity indicates direction**

Remember sign of velocity indicates direction. And the truck is going in the opposite direction. Set up the problem

47
**5) What is the speed of the car and Truck after? And the Answer is?**

Known Show Work After m3= kg v3=? m4 = kg v4=? Before m1 = kg v1= 20.0 m/s m2= kg v2= m/s 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) kg (v) Unknown Possible formulas Final Answer

48
**5) What is the speed of the car and Truck after?**

Known Show Work Before m1 = kg v1= 20.0 m/s m2= kg v2= m/s After m3= kg v3=? m4 = kg v4=? 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) kg (v) -40,000 kg·m/s = 4000.kg (v) -40,000 kg·m/s /4000.kg = v -10.0 m/s = v Unknown Possible formulas Final Answer m/s

49
Watch What happens.

50
**6) Objects can bounce rather then stick together.**

This time all the momentum of the kg truck travelling at 20.0 m/s gets passed to the 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

51
**6) What is the speed of the car after? And the Answer is?**

Known Show Work After m3= kg v3=? m4 = kg v4= 0 m/s Before m1 = kg v1= 20.0 m/s m2= kg v2= m/s 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) kg (0) Unknown Possible formulas Final Answer

52
**6) What is the speed of the car after?**

Known Show Work After m3= kg v3=? m4 = kg v4= 0 m/s Before m1 = kg v1= 20.0 m/s m2= kg v2= m/s 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) kg (0) -40,000 kg·m/s = 1000.kg (v) -40,000 kg·m/s /1000.kg = v m/s = v Unknown Possible formulas Final Answer

53
Watch What happens

54
**7) A1000. kg car travelling at 20. 0 m/s hits a stationary 3000**

7) A1000. kg car travelling at 20.0 m/s hits a stationary 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.

55
**7) How fast does the car go? And the Answer is?**

Known Show Work After m3= kg v3=? m4 = kg v4= m/s Before m1 = kg v1= 20.0 m/s m2= kg v2= 0.0 m/s 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 Possible formulas Final Answer

56
**7) How fast does the car go?**

Known Show Work After m3= kg v3=? m4 = kg v4= m/s Before m1 = kg v1= 20.0 m/s m2= kg v2= 0.0 m/s 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 Possible formulas Final Answer

57
Watch what happens.

58
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.

59
Now Complete the additional Conservation of Momentum problems in your packet.

Similar presentations

OK

© Houghton Mifflin Harcourt Publishing Company The student is expected to: Chapter 6 Section 1 Momentum and Impulse TEKS 6C calculate the mechanical energy.

© Houghton Mifflin Harcourt Publishing Company The student is expected to: Chapter 6 Section 1 Momentum and Impulse TEKS 6C calculate the mechanical energy.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on acid-base indicators experiment Ppt on national symbols of india Ppt on reuse of waste materials Ppt on floppy disk Ppt on different sectors of economy Ppt on production planning and control Ppt on power system stability using facts devices Ppt on security attacks in information security Ppt on water activity of foods Class 7 science ppt on light