1. Conservation of Momentum in One –Dimension (along a line)

2. What is the impulse equation J = ?

3. Conservation of Momentum in One –Dimension (along a line) J = ∆p

4. Conservation of Momentum in One –Dimension (along a line) J = ∆p But what is J in terms of F net and ∆t ?

5. Conservation of Momentum in One –Dimension (along a line) J = ∆p But what is J in terms of F net and ∆t ? F net ∆t = ∆p

6. Conservation of Momentum in One –Dimension (along a line) J = ∆p But what is J in terms of F net and ∆t ? F net ∆t = ∆p Let’s switch left side and right side

7. Conservation of Momentum in One –Dimension (along a line) J = ∆p But what is J in terms of F net and ∆t ? F net ∆t = ∆p Let’s switch left side and right side ∆p = F net ∆t

8. Conservation of Momentum in One –Dimension (along a line) J = ∆p But what is J in terms of F net and ∆t ? F net ∆t = ∆p Let’s switch left side and right side ∆p = F net ∆t The above equation works for one particle, but it also works for a system of particles, so we can modify the equation like this:

9. Conservation of Momentum in One –Dimension (along a line) J = ∆p But what is J in terms of F net and ∆t ? F net ∆t = ∆p Let’s switch left side and right side ∆p = F net ∆t The above equation works for one particle, but it also works for a system of particles, so we can modify the equation like this: ∆p system = F net system ∆t

10. Conservation of Momentum in One –Dimension (along a line) ∆p system = F net system ∆t

11. Conservation of Momentum in One –Dimension (along a line) ∆p system = F net system ∆t In physics, we can have an “isolated” system of particles in which the external, unbalanced or net force on that system of particles is zero.

12. Conservation of Momentum in One –Dimension (along a line) ∆p system = F net system ∆t In physics, we can have an “isolated” system of particles in which the external, unbalanced or net force on that system of particles is zero. Therefore, for an isolated system of particles, we know that F net system = 0. How will this change our equation above?

13. Conservation of Momentum in One –Dimension (along a line) ∆p system = F net system ∆t In physics, we can have an “isolated” system of particles in which the external, unbalanced or net force on that system of particles is zero. Therefore, for an isolated system of particles, we know that F net system = 0. If F net system =0, then ∆p system = 0

14. Conservation of Momentum in One –Dimension (along a line) ∆p system = F net system ∆t In physics, we can have an “isolated” system of particles in which the external, unbalanced or net force on that system of particles is zero. Therefore, for an isolated system of particles, we know that F net system = 0. If F net system =0, then ∆p system = 0 If ∆p system = 0, then what can be said about p system ?

15. Conservation of Momentum in One –Dimension (along a line) ∆p system = F net system ∆t In physics, we can have an “isolated” system of particles in which the external, unbalanced or net force on that system of particles is zero. Therefore, for an isolated system of particles, we know that F net system = 0. If F net system =0, then ∆p system = 0 If F net system =0, p system stays constant !!

16. Conservation of Momentum in One –Dimension (along a line) If F net system =0, then ∆p system = 0 Or If F net system =0, p system stays constant !!

17. Conservation of Momentum in One –Dimension (along a line) If F net system =0, then ∆p system = 0 Or If F net system =0, p system stays constant !! Another term for “staying constant” in physics is called “ ____________ “

18. Conservation of Momentum in One –Dimension (along a line) If F net system =0, then ∆p system = 0 Or If F net system =0, p system stays constant !! Another term for “staying constant” in physics is called “ conservation “.

19. Conservation of Momentum in One –Dimension (along a line) If F net system =0, then ∆p system = 0 Or If F net system =0, p system stays constant !! Another term for “staying constant” in physics is called “ conservation “. So, for an isolated system, where F net system =0, the total momentum of the system, p system, is conserved.

20. Conservation of Momentum in One –Dimension (along a line) If F net system =0, then ∆p system = 0 Or If F net system =0, p system stays constant !! Another term for “staying constant” in physics is called “ conservation “. So, for an isolated system, where F net system =0, the total momentum of the system, p system, is conserved. This is called the “_______________ of _____________ principle.

21. Conservation of Momentum in One –Dimension (along a line) If F net system =0, then ∆p system = 0 Or If F net system =0, p system stays constant !! Another term for “staying constant” in physics is called “ conservation “. So, for an isolated system, where F net system =0, the total momentum of the system, p system, is conserved. This is called the “conservation” of “momentum” principle.

22. Conservation of Momentum in One –Dimension (along a line) Another way of stating the conservation of momentum principle is … For an isolated system, where F net system = 0, p system (before interaction) = p system (during interaction) = p system (after interaction)

23. Conservation of Momentum in One –Dimension (along a line) Another way of stating the conservation of momentum principle is … For an isolated system, where F net system = 0, p system (before interaction) = p system (during interaction) = p system (after interaction) Use for… Rockets explosionscollisions

24. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A

25. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = ?

26. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s

27. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s a) conservation of mo says … p system (before collision) = p system (after collision)

28. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s a) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = m a v a2 + m b v b2

29. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s a) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = m a v a2 + m b v b2 Because motion is 1-dimensional, East = + West = -

30. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s a) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = m a v a2 + m b v b2 Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = (4250 ) v a2 + 8500 (3.500 )

31. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s a) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = m a v a2 + m b v b2 Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = (4250 ) v a2 + 8500 (3.500 ) 21250 + 0 = (4250 ) v a2 + 29750

32. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s a) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = m a v a2 + m b v b2 Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = (4250 ) v a2 + 8500 (3.500 ) 21250 + 0 = (4250 ) v a2 + 29750 21250 – 29750 = (4250 ) v a2

33. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s a) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = m a v a2 + m b v b2 Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = (4250 ) v a2 + 8500 (3.500 ) 21250 + 0 = (4250 ) v a2 + 29750 21250 – 29750 = (4250 ) v a2 -8500 = (4250 ) v a2 v a2 = - 2.00 m/s = 2.00 m/s [W]

34. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision)

35. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = ( m a + m b ) v combined cars Because motion is 1-dimensional, East = + West = -

36. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = ( m a + m b ) v combined cars Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = ( 4250 + 8500 ) v combined cars

37. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = ( m a + m b ) v combined cars Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = ( 4250 + 8500 ) v combined cars 21250 = 12750 v combined cars

38. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = ( m a + m b ) v combined cars Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = ( 4250 + 8500 ) v combined cars 21250 = 12750 v combined cars v combined car = 1.67 m/s [E]

39. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = ( m a + m b ) v combined cars Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = ( 4250 + 8500 ) v combined cars 21250 = 12750 v combined cars v combined car = 1.67 m/s [E] v a2 = ?

40. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = ( m a + m b ) v combined cars Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = ( 4250 + 8500 ) v combined cars 21250 = 12750 v combined cars v combined car = 1.67 m/s [E] v a2 = 1.67 m/s [E]

41. Example: A 4250.0 kg freight car A is moving at 5.00 m/s [E] on a straight, level railroad track of negligible friction. The car collides with another 8500.0 kg stationary freight car B on the same track. After the collision, what is the velocity of freight car A if freight car B … a) moves off at 3.500 m/s [E] b) locks together with car A Given: m a = 4250.0 kg m b = 8500.0 kg v a1 = 5.00 m/s [E] v b1 = 0 m/s b) conservation of mo says … p system (before collision) = p system (after collision) m a v a1 + m b v b1 = ( m a + m b ) v combined cars Because motion is 1-dimensional, East = + West = - ( 4250 ) ( +5 ) + ( 8500 ) ( 0 ) = ( 4250 + 8500 ) v combined cars 21250 = 12750 v combined cars v combined car = 1.67 m/s [E] v a2 = 1.67 m/s [E] coupled collision

42. Try this example! Happy is a 50.0 g frog. Jumpy is a 75.0 g frog. Together they sit at rest on a 100.0 g lily-pad. Suddenly, Happy jumps off at a velocity of 4.0 m/s [W] relative to the water. At the same time, Jumpy jumps off at 5.0 m/s [E] relative to the water. With what velocity does the lily- pad move with respect to the water after the frogs jump off? Assume negligible water friction. We will take-up in class.

43. Homework Do handout 1-d conservation of momentum, impulse… New textbook: Read p228-p231 Q1,2 p231 check answers same page Q2 –Q7 p232 check answers p717 Old textbook: Read p239 –p243 Do q3 to q9 p243 check answers same page