Chapter 7: Linear Momentum (p)

1. Chapter 7:Linear Momentum (p)

2. (Linear) Momentum (p) • Linear Momentum (p) is defined as the product of mass (m) and velocity (v): p = m v • SI Units of linear momentum = kg.m/s • p is a vector quantity: specified by magnitude and direction. • The direction of p is the same as the direction of velocity v.

3. Example:A car of mass 1,500 kg is traveling at a constant speed of 12 m/s due north. What is its linear momentum? p = mv= (1,500 kg) x (12 m/s) = 18,000kg.m/s due north • Constant momentum: Means both the mass (m) and velocity (v) is not changing. Usually, mass stays constant during motion. • Some exceptions: • A rocket traveling: total mass changes as it burns fuel. • An open freight car being loaded while in motion. Its mass will change as it moves

4. Example: 1. A car of mass 1,500 kg is traveling at a constant speed of 12 m/s in a circular path. Is its linear momentum constant? No: Because direction of velocity in a circular motion changes. 2. A ball of mass 0.2 kg is thrown horizontally at a wall. If it hits the wall at 25 m/s and bounces back with the same speed, what is the change in its momentum? (A) 0 kg-m/s (B) 5 kg.m/s (C) -5 kg.m/s (D) 10 kg.m/s (E) –10 kg.m/s

5. Example: 3. A ball of mass 0.2 kg is thrown at an angle of 30o above the horizontal with a speed of 5 m/s and travels as a projectile. • What is the y-component of its momentum at the instant it is thrown? • At its maximum height, what is its momentum? 4. True/False? An object traveling with larger velocity must have larger momentum than another one traveling with smaller velocity.

6. TOTAL MOMENTUM

7. Example 1: Car (1) is moving due east (+) at 30 m/s. Another car (2) is moving due east (+) at 30 m/s. For these two cars each of mass 1,500 kg, find their (a) total kinetic energy. (b) total momentum.

8. Example 2: Car (1) is moving due east (+) at 30 m/s. Another car (2) is moving due west (-) at 30 m/s. For these two cars each of mass 1,500 kg, find their (a) total kinetic energy. (b) total momentum.

9. Example 3: Yes/No? Is it possible to have a system of objects where the total momentum is zero but the total kinetic energy is not zero?

10. Conservation of Linear Momentum Consider two objects with mass mA and mB moving towards each other with initial velocities v1(A) and v1(B) V1(B) V1(A) mB mA Total initial momentum: pi = mAV1(A) + mBV1(B)

11. If these two objects collide and later their velocities after impact are V2(A) and V2(B): V2(B) V2(A) mB mA Total final momentum: pf = mAV2(A) + mBV2(B)

12. It has been established that on condition that no net external forces act on any system of colliding objects, the total momentum of the system will always remain conserved. ie, pi = pf OR: mAV1(A) + mBV1(B) mA = mAV2(A) +mBV2(B)

13. Law of Conservation of Linear Momentum: Momentum is “Conserved” means it can not be created nor destroyed • Can be transferred from one object to another • Total Momentum does not change with time. • Total momentum “before” = total momentum “after.” • This is a BIG deal! • In science, any law of conservation is a very powerful tool in understanding the physical universe.

14. Law of Conservation of Energy • In any natural process, total energy is always “conserved”, i.e. energy can not be created nor destroyed. • Can be transformed from one form to another. • Can be transferred from one system to another. In science, any law of conservation is a very powerful tool in understanding the physical universe.

15. Example: Jane and Fred are on skates facing each other. Jane then pushes Fred so he is going 2.0 m/s. If Fred is twice as heavy as Jane, how fast does Jane end up moving? • Spinitial = Spfinal • 0 = mFred VFred + mJane VJane • VJane = mFred VFred / mJane= 4 m/s

16. Example: Car (1) is moving due east at 30 m/s. Another car (2) is moving due west at 30 m/s. For these two cars each of mass 1,500 kg, find • Total kinetic energy. • Total momentum. • Yes/No? Is it possible to have a system of objects where the total momentum is zero but the total kinetic energy is not zero?

17. A bullet of mass 20 grams initially traveling at a speed of 200 m/s lodges in a block of wood of mass 2 kg at rest on a frictionless floor. What is the velocity with which the bullet and block of wood travel after impact? A 15,000 kg open box-car is moving at 7 m/s on a level road. 3,000 kg of water then falls straight down into the box-car. The speed of the box-car now with the water in it is

18. Example 7.1: A car w/ mass 1200 kg is driving north at 30 m/s, and turns east driving 13.6 m/s. What is the magnitude of the car’s change in momentum? pinitial = m vinitial = (1200 Kg) x 30 m/s = 36000 kg m/s North pfinal = m vfinal = (1200 Kg) x 13.6 m/s = 16320 kg m/s East North-South: pfinal – pinitial = (0 – 36000) = -36000 kg m/s v2 = 13.6 m/s East-West: pfinal – pinitial = (16320 - 0) = +16320 kg m/s v1 = 30 m/s Magnitude : Sqrt(p2North – p2East ) = 39526 kg m/s

19. A ball is projected straight up. Which graph shows the linear momentum of the ball as a function of time? (D) (C) (B) (A) t t t t (E) (F) t t

20. IMPULSE • Objects A and B colliding: The force of impact A exerts on B = FBA. This causes velocity of B to change from v1(B) to v2(B) • FBA = maB = m[v2(B) – v1(B)]/t • OR FBAt = m(v2(B) – v1(B)) = p • The quantity FBAt is called impulse (of a force). • Impulse = F t = p Unit = N.s • Change in momentum requires force acting over a time duration.

21. IMPULSE Force F (N) Time t (s) Impulse = area under the graph

22. Force F (N) 200 0 Time t (s) 0.2 0.4 0.6 • Calculate the impulse • If this impulse was applied on a 3 kg mass at rest, what would its final velocity be?

23. Ft = p p = (mv) = m(v) and v = p/m F = ma and a = F/m • Example: • A force of 30 N is applied for 5 s to each of two objects of mass m and M (m < M). Which of the masses experience the greater • Momentum change? • Velocity change? • Acceleration?

24. Why do we flex our knees when when jumping? • Increases the time of contact for the ground to bring you to rest. • In turn reduces force exerted on your body.

25. A 160-gram baseball with a velocity of 20 m/s is hit by a bat and leaves at 25 m/s in the opposite direction. If the contact lasted for only 0.012 s, what was the magnitude of the average force on the ball? [1,000 grams = 1 kg]

26. Example on conservation of momentum: A bullet of mass 200 g traveling at a speed of 150 m/s hits a 3 kg block of wood at rest on a frictionless table. If the bullet lodges inside the block, with what speed will the bullet-block composite travel after impact?

27. Elastic and Inelastic Collisions 1. Elastic Collisions: Collisions in which the total kinetic energy is conserved. SKinitial = SKfinal 2. Inelastic Collisions: Collisions in which the total kinetic energy is NOT conserved. SKinitial SKfinal

28. Before: V1(B) V1(A) mB mA Ki = ½ mAv21(A) + ½ mBv21(B) After: V2(B) V2(A) mB mA Kf = ½ mAv22(A) + ½ mBv22(B) For Elastic Collision: Ki = Kf OR ½ mAv21(A) + ½ mBv21(B) = ½ mAv22(A) + ½ mBv22(B)

29. In most cases, collisions occur inelastically. • Part of the total initial kinetic energy is converted to other forms of energy such as light, heat, sound, etc. • Ki = Kf + heat + sound + light, etc • However, total energy and total linear momentum are still conserved even in inelastic collisions. • Elastic collision is an ideal case. Collision of billiard balls when no heat is produced is the closest approximation to elastic collision.

30. Center of Mass Center of Mass = Balance point of a large object = Balance point of a number of discrete objects For an object with a regular shape (sphere, cylinder, cube etc, CM is located at its geometric center.

31. Example m m x xCM = (0 + mL)/2m = L/2 m 5m x xCM = (0 + 5mL)/6m = 5L/6 X = L X = 0

32. Example A 55-kg man walks his 5-kg dog using a 3 meter long lease. Where is the center of mass of the man-dog system?

33. Example: Find the center of mass. y 7 kg 8 1 kg 6 4 3 kg 2 x 0 4 6 8 2

34. Collisions in Two Dimensions When dealing with collisions in 2-D, momentum conservation is applied separately to the x and y components of the total momentum: Spx (Before) = Spx (After) -----[x-components] And Spy (Before) = Spy (After) -----[y-components]

35. After y m1 60o x 30o m2 Before 8 m/s v1 At rest m1 0.1 kg m2 0.4 kg v2 • Find the speeds v1 and v2 after the collision. • Is the collision elastic or inelastic?

36. A large seed initially at rest explodes into two pieces which move off. Which of these could be possible paths the two pieces would take? (II) (III) (I)

37. Two objects with different masses (m and M with m < M) have the same kinetic energy. Which has the larger magnitude of momentum? [Hint: K = ½ p2/m]

38. A ball is projected straight up. Which graph shows the total energy of the ball as a function of time? (D) (C) (B) (A) t t t t (E) (F) t t

39. P7.35 A BMW of mass 2.0 x 103 kg is traveling at 42 m/s. It approaches a 1.0 x 103 kg VW going 25 m/s in the same direction and strikes it in the rear. Neither driver applies the brakes. Neglect frictional forces due to the road and air resistance. If the collision slows the BMW down to 33 m/s, what is the speed of the VW after collision?

40. A 75-kg person jumps off a table and lands on the ground with a speed of 3.5 m/s. By flexing his knees, he comes to rest in 0.40 seconds. Determine the average force exerted on his body in this process.