Linear momentum and its conservation by Victor R. Oribe

Lessons about: MOMENTUM, IMPULSE and LAW OF CONSERVATION OF MOMENTUM AND COLLISION

1. Linear Momentum and its Conservation By: Victor Rea Oribe

2. Complete the Matrix below by providing the needed information Newton’s Laws of Motion Statement of the law Illustration or example or Diagram Misconceptions or difficulties in understanding the Laws

3. Momentum Momentum describes an object’s resistance to stopping. The magnitude of momentum (p) is measured as the product of the object’s mass (m in kg) and velocity (v in m/s). Momentum = mass x velocity p = m x v kg.m /s = kg x m/s

4. Velocity is a vector quantity. Vector - has magnitude and direction. Therefore, Momentum is also a vector quantity

5. Sample Problem # 1 A 295-kg motorcycle moves at a velocity of 80 km/h. What is its momentum? Given: m = 295-kg v = 80 km/h 80 km 1 h x 1 h 3,600 s x 1000 m 1 km = 80000 m 3600 s v = 22.22 m/s p = ? p = m v p = (295 kg) ( 22.22 m/s) p = 6, 554.9 kg.m /s

6. Sample Problem # 2 What is the velocity of a 1,200kg car that has a momentum of 30,000 kg.m /s? Given: m = 1,200 kg p = 30,000 kg.m /s v = ? p = mv v = p m v = 30,000 kg . m/s 1,200 kg = 25 m/s

7. Sample Problem # 2 If a car has a momentum of 30,000 kg.m /s and its velocity is 25m/s, what is its mass? Given: p = 30,000 kg.m /s v = 25 m/s m = ? m = p v m = 30,000 kg m/s 25 m/s = 1,200 kg

8. Sample Exercises: A 45.9-g golf-ball is initially at rest on the tee. As the club hits the ball, the ball acquires a speed of 76.0 m/s. What is the change in momentum of the golf balls? Given: m = 45.9 g m = kg m = 45.9 g = 1 kg 1000g m = 45.9 g . kg 1000 g m = 0.0459 kg v = 76.0 m/s p = ? p = mv p = (0.0459 kg)(76.0 m/s) p = 3.5 kg . m/s

9. Sample Exercises: 2. What should be the velocity of a 2600 kg delivery truck to have the same momentum as the race car with a mass of 1,100 kg moving with a velocity of 110 km/h? For the Race Car m = 1,100 kg v = 110 km/h = 30.55 m/s p = mv = (1,100kg)(30.55 m/s) = 33,605 kg . m/s For the Delivery Truck m = 2,600 kg p = 33, 605 kg.m /s v = ? v = p/m v = 33,605 kg.m /s/2,600 kg v = 12.925 m/s

10. Research Work: Research on the maximum speeds and masses of the objects below Arrange the following according to the increasing magnitude of momentum Assume that all of them are moving in their maximum speed. Car 6. ant Bus 7. horse Bullet 8. motorcycle Airplane 9. cheetah Turtle 10. ostrich

11. Impulse

12. If a soccer ball is kicked and attains a great momentum, it will be difficult to stop it To stop the ball, it is necessary to use force against its momentum for a given period of time. The greater the momentum of an object, the greater the force, or the longer the time or both is needed to stop the ball.

13. What is the effect of the force and time in the motion of the object? According to Newton’s Second Law of Motion, an UNBALANCED FORCE causes an object to accelerate (speed up) or decelerate (slow down). If an unbalanced forced acts on an object against its motion, speed – the object slows down If the force acts in the same direction of the motion – the objects speeds up .

14. Therefore: Unbalanced force causes change in momentum of an object. To illustrate the effect, recall the Newton’s second Law of Motion Force = mass ( m ) x acceleration ( a ) = kg x m/s 2 = Newton N = kg.m /s 2

15. Acceleration ( a ) is mathematically defined as a = v – v t Relating the two equations: F = m a a = v – v t F = m ( v – v ) t F t = mv – mv Since ∆ v = v – v Then: F t = m ∆ v

16. F t = m ∆ v The right side of the equation represents the change in momentum caused by the unbalanced force acting at a certain period of time. It can be read as the “force ( F ) times the time ( t ) equals the mass ( m ) times the change (delta) in velocity ( ∆ v )

17. In the expression: F t = mv – mv mv represents the final momentum p and mv represents the initial momentum p of an object The equation can now be written as: F t = p – p or F t = ∆ p Again, this equation tells that the product of the net or unbalanced force, F and time, t when the force act is equal to change in momentum, p . This is known as the IMPULSE- MOMENTUM THEOREM

18. In Physics, F t is equal to the concept called IMPULSE ( I ) It is describes as the effect of the unbalanced force on the object. It is represented by the equation: I , Impulse = F t Since : F t = ∆ p Then : I = ∆ p

19. Example # 1 A 430-kg tricycle travelling at 30 km/h slows down to 10 km/h in 5 s as it approaches a road intersection. What is the force applied on the brakes of the tricycle? What is the impulse exerted on the vehicle? Given: m = 430 kg v = 10 km/h = 2.78 m/s v = 30 km/h = 8.3 m/s t = 5 s Find: (a) F (b) I

20. Solutions: Find the initial momentum p o = mv p = (430kg) (8.3 m/s) = 3,569 kg . m/s Find the final momentum p = mv p = (430 kg) ( 2.78 m/s) = 1195.4 kg . m/s

21. Find the Force ( F ) Ft = ∆p ∆p = p –p = 1195.4 kg . m/s – 3569 kg .m/s = - 2373.6 kg . m/s Ft = ∆p F (5s) = -2373.6 kg . m/s F = - 2373.6 kg . m/s 5 s F = 474.72 kg . m/s 2 or - 475N Note: T he negative sign indicates F is a retarding force

22. Find the Impulse I = ∆p I = -2373.6 kg . m/s

23. Example # 1 What force is needed to stop a 4,500 kg tourist bus travelling at 90 km/h in 18 s? Given: m = 4,500 kg v = 90 km /h = 25 m/s v = 0 t = 18 s Find: F

24. Solutions: A. Compute the initial momentum p = mv = (4500 kg) ( 25 m/s) = 112, 500 kg . m/s B. Compute the final momentum p = mv = (4500 kg) ( 0 m/s) = 0 C. Find the Force (F) Ft = ∆p F = ∆p (p – p ) t F = 0 – 112500 kg.m /s 18 s F = -6250 kg.m /s 2 or N

25. Practice Exercise # 1: A 12.0-kg hammer hits a nail at a velocity of 13.2 m/s in 0.005 s. What is the average force exerted on the nail? Given: m = 12.0kg v = o v = 13.2 m/s t = 0.005s Solution: Compute the initial momentum p = mv = (12.0 kg) (13.2 m/s) = 158.4 kg.m /s B. Compute the final momentum p = mv = (12.0 kg) (0) = 0 C. Compute the Force Ft = ∆p F = ∆p / t F = (0- 158.4 kg.m /s) / 0.005 S F = - 31,680 kg. m/s 2

26. Practice Exercise # 2: What is the velocity of a 348-kg motorcylce which has a momentum of 4833.33 kg.m /s? How much force is needed to stop it in 5 s. What is the impulse. Given: v = ? v = ? m = 348 kg p = 4833.33 kg. m/s F =? t = 5 s I = ? v = ? p = mv 4833.33 kg.m /s = (348 kg) (v) v = 4833.33 kg.m /s / 348 kg v = 13.9 m/s Final momentum p = mv = (348kg) (0) p = 0 Ft = ∆p F = (4833.33kg.m/s – 0) / 5s F =

27. Quiz (understanding 30%) What is the momentum of an object with m=2.00 kg and v= 40.0 m/s? 2. A car weighing 15,680 N and moving at 20 m/s is acted upon by a force of 640 N until it stops. Find:(a) the mass of the car 3. A car weighing 15,680 N and moving at 20 m/s is acted upon by a force of 640 N until it stops. Find: (b) the car’s initial momentum 4. A force of 30000 N j is exerted for 4.00 s, on a 95,000 kg mass. What is the impulse of the force for this 4.00 s? 

28. Quiz 1. What is the momentum of an object with m=2.00 kg and v= 40.0 m/s? Solution: ρ = mv   ρ = (2.00 kg)(40.0 m/s) ρ = 80.0 kg-m/s Given: m = 2.00-kg v = 40.0 m/s ρ = ?

29. Quiz 2. A car weighing 15,680 N and moving at 20 m/s is acted upon by a force of 640 N until it stops. Find : (a) the mass of the car Solution: W = mg 15,680 kg-m/s 2 = (m) (9.8 m/s 2 ) m = 15,680kg-m/s 2 ÷ 9.8 m/s 2 m = 1,600-kg Given: W = 15,680 N g = 9.8 m/s 2 m = ?

30. Quiz 3. A car weighing 15,680 N and moving at 20 m/s is acted upon by a force of 640 N until it stops. Find : (b) the car’s initial momentum  Solution: ρ = mv  = (15,680 N ÷ 9.8 m/s 2 ) 20 m/s = 1,600 kg x 20 m/s = 32,000 kg-m/s Given: W = 15,680 N g = 9.8 m/s 2 m = ? v = 20 m/s F = 460N

31. Quiz 4. A force of 30000 N is exerted for 4.00 s, on a 95,000 kg mass. What is the impulse of the force for this 4.00 s? Solution: Impulse = Ft = 30,000 N ( 4.00 s) I = 120,000 N-s Given : F = 30,000 N t = 4.00 s m = 95,000 kg I = ?

32. Law of Conservation of Momentum and Collisions Collision is the interaction that occurs when two or more objects hit each other. When two objects collide, each object exerts a force on another for a short amount of time. This force imparts an impulse, or changes the momentum of each of the colliding objects.

33. Experiments proved that momentum can be conserved. This means that the total momentum of any set of objects remains the same unless outside forces act on the object. Therefore the momentum of each object involved in the collision changes, the total momentum of the system remains constant. P before = p after

34. For the collision of two objects Total momentum before collision = Total momentum after collision P 1 + p 2 = p 1’ + p 2’ Where: p 1 = momentum of the first object before collision p 2 = momentum of the second object before collision p 1’ = momentum of the first object before collision p 2’ = momentum of the second object after collision

35. Collision may be classified as a) elastic collision – collision in which both momentum and kinetic energy are conserved.

36. Before collision m 1 v 1 + m 2 v 2 (1,000 kg) (20 m/s) + (3,000 kg) (0 m/s) = 20,000 kg-m/s After collision m 1 v 1 ’ + m 2 v 2 ’ (1,000 kg) (-10 m/s) + (3,000 kg) (10 m/s) -10,000 kg-m/s + 30,000 kg-m/s = 20,000 kg-m/s

37. An analysis of the kinetic energy of the two objects reveals that the total system kinetic energy before the collision is 800,000 Joules (200,000 J for the car plus 600,000 J for the truck). After the collision, the total system kinetic energy is 800000 Joules (800,000 J for the car and 0 J for the truck). The total kinetic energy before the collision is equal to the total kinetic energy after the collision.

38. m = 1.2 kg m 2 = 1.2 kg v 1 = 2.0 m/s v 2 = 0 m = 1.2 kg m 2 = 1.2 kg v’ 1 = 0 m/s v 2 ‘ = 2.0 m/s Total Momentum Before the Collision P total = p 1 + p 2 = m 1 v 1 + m 2 v 2 = (1.2kg)(2.0m/s) + (1.2kg)(0) = 2.4 kg.m /s Total Momentum After the Collision P total = p 1 ‘ + p 2 ‘ = m 1 v 1 ‘ + m 2 v 2 ’ = (1.2kg)(0m/s) + (1.2kg)(0) = 2.4 kg.m /s Total Kinetic Energy Before the Collision KE total = m 1 v 1 2 / 2 + m 2 v 2 2 / 2 = (1.2kg) (2.0m/s) 2 / 2 + (1.2kg)(0)/2 = 2.4 kg.m 2 /s 2 + 0 = 2.4 N Total Kinetic Energy After the Collision KE total = m 1 v 1 ‘ 2 / 2 + m 2 v 2 ‘ 2 / 2 = (1.2kg) (0m/s) 2 /2 + (1.2kg)(2.0m/s)2 = 0 + 2.4kg.m 2 s 2 = 2.4 N

39. b) Perfectly inelastic collision – occurs when the two bodies join together during collision and move together as one body after collision.

40. The total momentum is conserved but the total kinetic energy before is not equal to the kinetic energy after collision. the kinetic energy is lost in the deformation caused by the collision or conversion to other forms of energy like sound or thermal energy.

41. Before: m = 460.0 kg m 2 = 460.0 kg v 1 = 100.0 m/s v 2 = 0 After m = 460.0 kg m = 460.0 kg v’ = v 1 ’ = 5.0 m/s Perfectly inelastic collision between equal masses occur when two bodies join together during collision and move together as one body after collision.

42. Before: m = 460.0 kg m 2 = 460.0 kg v 1 = 10.0 m/s v 2 = 0 After m = 460.0 kg m = 460.0 kg v’ = v 1 ’ = 5.0 m/s Total Momentum Before the Collision P total = p 1 + p 2 = m 1 v 1 + m 2 v 2 = (460.0kg)(10.0m/s) + (460.0kg)(0) = 4,600.0 kg.m /s Total Momentum After the Collision P total = p 1 ‘ + p 2 ‘ = (m 1 + m 2 ) v’ = (460.0kg + 460.0kg)(5.0 m/s) = 4,600.0 kg.m /s The total momentum is conserved before and after the collision

43. Before: m = 460.0 kg m 2 = 460.0 kg v 1 = 100.0 m/s v 2 = 0 After m = 460.0 kg m = 460.0 kg v’ = v 1 ’ = 5.0 m/s The total kinetic energy before is not equal to the kinetic energy after the collision. Total Kinetic Energy Before the Collision KE total = m 1 v 1 2 / 2 + m 2 v 2 2 / 2 = (460.0kg) (10.0m/s) 2 / 2 + (460.0kg)(0) = 32,000 N.m or Joule Total Kinetic Energy After the Collision KE total =( m 1 m 2 ‘) (v 2 ‘) 2 / 2 = (460.0kg + 460.0kg) (5.0m/s) 2 /2 = (920.0kg) (25m 2 /s 2 ) 2 = 11,500 N.m or J

44. c) Inelastic collision Objects of equal masses move off separately at different speeds in the same direction after collision. The figure shows that in an inelastic collision, object of equal masses move off separately at different speeds in the same direction after collision

45. Total Momentum Before the Collision P total = p 1 + p 2 = m 1 v 1 + m 2 v 2 = (460.0kg)(10.0m/s) + (460.0kg)(0) = 4,600.0 kg.m /s Total Momentum After the Collision P total = p 1 ‘ + p 2 ‘ = (m 1 + m 2 ) v’ = (460.0kg + 460.0kg)(5.0 m/s) = 4,600.0 kg.m /s 460.0kg 460.0kg 10.0m/s V 2 = 0m/s 2.5 0 m /s 7.5 0 m/s

46. Total Kinetic Energy Before the Collision KE total = m 1 v 1 2 / 2 + m 2 v 2 2 / 2 = (460.0kg) (10.0m/s) 2 / 2 + (460.0kg)(0) /2 = (460kg)(100.0m 2 /s 2 /2 + 0 = 23,000 N.m or J Total Kinetic Energy After the Collision KE total = m 1 v 1 ‘ 2 / 2 + m 2 v 2 ‘ 2 / 2 = (460.0kg) (2.5m/s) 2 /2 + (460.0kg)(7.5.0m/s) 2 / 2 = (460.0kg)(6.25m 2 /s 2 /2 + (460.0kg)(56.26m 2 /s 2 /2 = 2,875.0kg.m 2 /s 2 / 2 + 25,879.6kg.m 2 /s 2 /2 = 1,437.5kg.m 2 /s 2 + 12,939.8kg.m 2 /s 2 = 14,377.3 N.m or J. The total kinetic energy before is not equal to the kinetic energy after collision. The kinetic energy is lost in the deformation caused by the collision or conversion to other forms of energy like sound or thermal(heat) 460.0kg 460.0kg 10.0m/s V 2 = 0m/s 7.5 0 m/s 2.5 0 m /s

47. Sample Problem # 1: Consider the perfectly elastic collision between two balls of masses 120g (m 1 ) and 240g (m 2 ). If ball 1 is moving with a velocity of 32.0 cm/s to the right and ball 2 with a ve3locity of 25.0 cm/s in the same direction. Find the velocity of ball 1, if ball 2 has a final velocity of 28.0 cm/s.

48. Sample Problem #1: Consider the perfectly elastic collision between two balls of masses 120g (m 1 ) and 240g (m 2 ). If ball 1 is moving with a velocity of 32.0 cm/s to the right and ball 2 with a ve3locity of 25.0 cm/s in the same direction. Find the velocity of ball 1, if ball 2 has a final velocity of 28.0 cm/s. Before collision m 1 = 120g m 2 = 240g v 1 = 32.0cm/s v 2 = 25.0cm/s p total = m 1 v 1 + m 2 v 2 = 120g(32.0cm/s) + 240g (25.0 cm/s) = 9840 g.cm/s After collision m 1 = 120g m 2 = 240g v 1’ = 32.0cm/s v 2’’ = 28.9cm/s P’ total = m 1 v 2’ + m 2 v 2’ = 120g(v 1’ ) + 240g (28.9cm/s) = 120g (v 1’ ) + 6936 g.cm/s

49. After collision m 1 = 120g m 2 = 240g v 1’ = 32.0cm/s v 2’’ = 28.9cm/s P’ total = m 1 v 2’ + m 2 v 2’ = 120g(v 1’ ) + 240g (28.9cm/s = 120g (v 1’ ) + 6936 g.cm/s 9840 g. cm/s = 120g (v‘ 1 ) + 6936 cm/s 9840 g.cm/s – 6936 cm/s = 120g (v’ 1 ) 120g (v’ 1 ) = 2,904 g.cm/s v’ 1 = 2,904 g.cm/s / 120g v’ 1 = 24.2 cm/s Before collision m 1 = 120g m 2 = 240g v 1 = 32.0cm/s v 2 = 25.0cm/s p total = m 1 v 1 + m 2 v 2 = 120g(32.0cm/s) + 240g (25.0 cm/s) = 9840 g.cm/s

50. Sample Problem #2: A 70.0-kg ice skater moving on ice at 4.62 m/s hits from behind another 70.9 kg skater, who is initially at rest. After collision, the skaters move in the same direction at different speeds. The first skater moves at 1.30 m/s. What is the speed of the second skater? What is the total momentum of the skaters before collision? Is the collision elastic or inelastic?

51. Sample Problem # 2: A 70.0-kg ice skater moving on ice at 4.62 m/s hits from behind another 50.9 kg skater, who is initially at rest. After collision, the skaters move in the same direction at different speeds. The first skater moves at 1.30 m/s. What is the speed of the second skater? What is the total momentum of the skaters before collision? Is the collision elastic or inelastic? m 1 = 70.0kg v 1 = 4.62m/s m 2 = 50.9kg v 2 = 0 m/s m 1 = 70.0kg v‘ 1 = 1.30 m/s m 2 = 50.9kg v ‘ 2 = ?

52. Solutions: Given: m 1 = 70.0 kg m 2 = 50.9 kg v‘ 1 = 1.30 m/s v 1 = 4.62m/s v‘ 2 = ? p total = ? p total = m 1 v 1 + m 2 v 2 = (70.0kg)(4.62m/s) + (50.0kg)(0m/s) = 323.4kg.m/s + 0 = 323.4 kg . m/s p‘ total = m 1 v‘ 1 + m 2 v‘ 2 = (70.0 kg) (1.30 m/s) + (50.9 kg) (v’ 2 ) = 91 kg . m/s + (50.9 kg) (v’ 2 ) 323.4 kg.m /s = 91 kg.m /s + (50.9-kg) (v’ 2 ) 323.4 kg.m /s – 91 kg.m /s = (50.0-kg) (v’ 2 ) 232.4 kg.m /s = (50.0-kg) (v’ 2 ) (50.0-kg)(v’ 2 ) = 232.4 kg.m /s v’ 2 = 232.3 kg.m /s ÷ 50.0 kg v’ 2 = 4.65 m/s

53. Sample Problem # 3: One rainy day, a car with a mass of 1250-kg moving at 20.0 m/s hits the rear end of another car with a mass of 1610-kg moving at 8.0 m/s in the same direction. What is the final velocity of the two cars if they stick together. What is the change in kinetic energy of the system? What type of collision occurred in the system?

54. Sample Problem # 3: One rainy day, a car with a mass of 1250-kg moving at 20.0 m/s hits the rear end of another car with a mass of 1610-kg moving at 8.0 m/s in the same direction. What is the final velocity of the two cars if they stick together. What is the change in kinetic energy of the system? What type of collision occurred in the system? m 1 = 1250-kg m 2 = 1610-kg v 1 = 20.0 m/s v 2 = 8.0 m/s v ‘ = ?

55. Solutions: Given: m 1 = 1250 kg v 1 = 20 m/s m 1 = 1610 kg v 2 = 8.0 m/s v’ = ? p total = m 1 v 1 + m 2 v 2 = (1250kg)(20m/s) + (1610kg)(8.0m/s) = 25,000 kg.m /s + 12,880 kg.m /s = 37,880 kg . m/s p‘ total = p’ 1 + p’ 2 = (m 1 + m 2 ) v’ = (1250-kg + 1610-kg) (v’) 37,880 kg.m /s = (1250-kg + 1610-kg) (v’) 37,880 kg.m /s = (2,860-kg) (v’) (2,860-kg) (v’) = 37,880 kg . m/s v’ = 37,880 kg . m/s ÷ 2,860-kg v’ = 13.2 m/s

56. Practice Exercise # A 4.0x10 3 kg.m /s 2 force acts on a 1,200-kg car to stop it in 8.0s. What is the initial velocity of the car in km/h? Given: F = 4.O X 10 3 kg.m /s m = 1,200-kg v = ? t = 8.0s F = m (v/t) 4.0 x 10 3 kg.m /s = 1,200-kg (v/8.0s) 4.0 x 10 3 kg.m /s = (1,200-kg) (v) / 8.0s 4.0 x 10 3 kg.m /s = (150-kg/s)(v) 150-kg/s (v) = 4.0 x10 3 kg.m /s v = 4.0 x10 3 kg.m /s ÷ 150 kg/s v = 27 m/s 2 27m/s x 1km/1000m x 3600s/1h = 97.2km/h

57. Practice Exercise # 2 What force is needed to stop a 500-kg tricycle in 6.0s if it is travelling at 60 km/h? Given: F = ? m = 500 kg t = 6.0 s v = 60 km/h x 1000m/1km x 1h/3600s = 17m/s F = m(v/t) = (500-kg) (17m/s ÷ 6.0s) = (500-kg) (2.8 m/s 2 ) = 1,400 kg . m/s 2

58. Practice Exercise # 3.a A player hits a football which is initially at rest and moves 22 m/s. Find the momentum of the football if it has mass of 0.48kg. Given: v = 0 v = 22 m/s p =? m = 0.48kg p = m(v – v ) = (0.48kg) (22 m/s – 0) = (0.48kg)(22m/s) = 10.56 kg . m/s

59. Practice Exercise # 3.b A player hits a football (0.48-kg) which is initially at rest and moves 22 m/s. Find the force exerted on it when the time of contact is 0.03 s. Given: m = 0.48-kg v = 0 v = 22 m/s F =? t = 0.03 s F = m (v –v ÷ t) = (0.48kg) (22m/s -0 ÷ 0.03s) = (0.48kg) (733 m/s 2 ) = 351.8 kg . m/s 2

60. Practice Exercise # 3.c A player hits a football (0.48-kg) which is initially at rest and moves 22 m/s. Find the Impulse in the football. Given: m = 0.48-kg v = o v = 22 m/s p = ? p =? p = mv = (0.48kg)(0) = 0 p = mv = (0.48kg)(22m/s) = 10.56 kg . m/s I = ∆p = p – p = 10.56 kg.m /s – 0 = 10.56 kg.m /s

61. Practice Exercise # 4 A 660-kg motorcycle moving 20m/s rear-ends a 1017-kg car travelling at 16 m/s on ca concrete highway in the same direction. Sketch the situations showing the before and after collision conditions for both vehicles. If the two vehicles stick together, how fast did they move immediately after collision? Given: m 1 = 660-kg v 1 = 20 m/s m 2 = 1017-kg v 2 = 16 m/s v’ = ?

62. Solution Given: m 1 = 660-kg v 1 = 20 m/s m 2 = 1017-kg v 2 = 16 m/s v’ = ? p total = p 1 + p 2 = m 1 v 1 + m 2 v 2 = (660kg)(20m/s) + (1017kg)(16 m/s) = 13,200kg.m/s + 16272kg.m/s = 29,472 kg . m/s p’ total = p 1 + p 2 = (m 1 + m 2 ) (v’) = (660kg + 1017kg) (v’) = (1,677 kg) (v’) p total = p’ total 29,472 kg . m/s = (1,677 kg) (v’) (1,677 kg)(v’) = 29,472 kg . m/s v’ = 29,472 kg . m/s ÷ 1,677 kg v’ = 17.6 m/s