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Forces and Motion

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

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Forces and Motion

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

  2. 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. Newton’s Second Law

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

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

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

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

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

  8. Concept Check Click to check your answers

  9. Concept Check Click to check your answers

  10. Momentum of an object is simply defined as the mass times the velocity. It is usually abbreviated as “ρ”. Momentum 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:

  12. Solve the following momentum problems:

  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 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. During an Impact Total momentum after an interaction Total momentum before 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. 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. Known Show Work Unknown Possible formulas Final Answer 1)Compare the momentum before the break up to the momentum after the break up.Record what you know.

  21. Known Show Work Unknown Possible formulas Final Answer 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 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. Known Show Work Unknown Possible formulas Final Answer 1) Compare the momentum before the break up to the momentum after the break up.Replace p with the momentum formula 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. Known Show Work Unknown Possible formulas Final Answer 1) Compare the momentum before the break up to the momentum after the break up.Plug in your numbers 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. Known Show Work Unknown Possible formulas Final Answer 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 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. Known Show Work Unknown Possible formulas Final Answer 1) Compare the momentum before the break up to the momentum after the break up. 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. 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

  27. 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. Known Show Work Unknown Possible formulas Final Answer 2) What is the speed of the car and brick after?WAIT!! We have 2 variables V1 and V2 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. Known Show Work Unknown Possible formulas Final Answer 2) What is the speed of the car and brick after?WAIT!! We have 2 variables V1 and V2 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. Known Show Work Unknown Possible formulas Final Answer 2) What is the speed of the car and brick after?The Cart and brick stick together they have the same speed v2=v3Now Set up your equation After m2= 1.0 kg v2= v3 m3 = 2.0 kg Before m1 = 1.0 kg v1= 60.0 m/s

  31. Known Show Work Unknown Possible formulas Final Answer 2) What is the speed of the car and brick after?Plug in your numbers and Solve 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. Known Show Work Unknown Possible formulas Final Answer 2) What is the speed of the car and brick after?And the Answer is? 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. Known Show Work Unknown Possible formulas Final Answer 20.0 m/s 2) What is the speed of the car and brick after?And the Answer is? 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. 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= 3000. kg v3=? m4 = 1000. kg v4=? Before m1 = 3000. kg v1= 20.0 m/s m2= 1000. 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 = 3000. kg v1= 20.0 m/s m2= 1000. kg v2= 0 m/s After m3= 3000. kg v3=? m4 = 1000. 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. 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 = 1000. kg v1= 20.0 m/s m2= 3000. kg v2= 0 m/s After m3= 1000. kg v3=? m4 = 3000. 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 = 1000. kg v1= 20.0 m/s m2= 3000. kg v2= 0 m/s After m3= 1000. kg v3=? m4 = 3000. 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. 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 = 1000. kg v1= 20.0 m/s m2= 3000. kg v2= 20.0 m/s After m3= 1000. kg v3=? m4 = 3000. kg v4=? Unknown Possible formulas Final Answer

  46. 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= 1000. kg v3=? m4 = 3000. kg v4=? Before m1 = 1000. kg v1= 20.0 m/s m2= 3000. kg v2= -20.0 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)+ 3000 kg (v) Unknown Possible formulas Final Answer

  48. 5) What is the speed of the car and Truck after? Known Show Work Before m1 = 1000. kg v1= 20.0 m/s m2= 3000. kg v2= -20.0 m/s After m3= 1000. kg v3=? m4 = 3000. 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)+ 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 Possible formulas Final Answer -10.0 m/s

  49. Watch What happens.

  50. 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

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