1. Momentum & Impulse Day #1: Introduction HW #7

2. Momentum & Collisions: Define Momentum: Momentum, p, is defined as the product of mass and velocity. Units: momentum = (mass) (velocity) Momentum is a vector, and the direction of the momentum is the same as the velocity.

3. Define Impulse and Relation to Newton’s 2 nd Law: Newton’s original statement of his 2 nd law of motion: The time rate of change of the momentum of an object equals the net force impressed upon the object.

4. Is this the same as F = ma? Approximately true for real differences, considered true when used in calculus. For all the situations so far, mass has been constant. In that limit, F = ma. When mass changes, the problems are considerably harder.

5. Impulse is defined as the change of momentum: Units: impulse = (force) (time) The two units for momentum are equivalent, and both are used.

6. Example #1: (a) What is the momentum of a 10.0 kg mass that travels at 20.0 m/s? (b) What would be the speed of a 40.0 kg mass with the same momentum?

7. (c) Which object has the larger kinetic energy? In general: For a given momentum, a smaller mass will have greater KE

8. Example #2: A 255 gram baseball is thrown at 40.0 m/s towards the right. The ball is struck by a bat and subsequently the ball travels at 50.0 m/s towards the left. The ball is in contact with the bat for 2.00 ms (millisecond). (a) What is the impulse delivered to the ball? Take towards the right as the positive direction. Impulse =

9. (b) What is the net force on the ball?

10. Application of Momentum: Collisions. Initially, each mass has velocity and momentum. F A on B = Applied Force F B on A = Applied Force During the collision, masses exert equal but opposite forces on one another. Finally, each mass has velocity and momentum.

11. Each mass will have an initial momentum: = initial momentum of mass m 1. = initial momentum of mass m 2. Each mass will have an final momentum: = final momentum of mass m 1. = final momentum of mass m 2.

12. During the collision, each mass exerts a force on the other. By Newton’s 3 rd law, these are equal and opposite. Impact takes a time  t. The impulse on each mass is: With a little rearranging, get: This is known as Conservation of Total Momentum

13. Example #3: A 534 kg cannon fires an 6.39 kg projectile at 1342 m/s. (a) What is the total initial momentum? Before the projectile is fired, everything is at rest. Thus the total momentum is zero.

14. (b) Determine the recoil velocity of the cannon after the projectile is fired. The ( – ) sign means the cannon recoils in the direction opposite to the projectile.

15. Momentum & Impulse Day #2: Collisions Conservation of Momentum HW #7

16. Example #4: Four railroad cars, each of mass 2.50 × 10 4 kg, are coupled together and coasting along horizontal tracks at speed v o toward the south. A very strong but foolish movie actor riding on the second car uncouples the front car and gives it a big push, increasing its speed to 4.00 m/s south. The remaining three cars continue moving south, now at 2.00 m/s. (a) Find the initial speed of the cars. before after

17. Momentum Conservation:

18. (b) How much work did the actor do? Work done = change of KE:

19. Example #5: A 10.0 kg mass travels towards the right at a speed of 6.00 m/s and collides with and sticks to a 20.0 kg initially at rest. (a) What is the speed of the combined object after the collision?

21. (b) What is the amount of kinetic energy after the collision? Write it as a percentage of the initial kinetic energy.

22. Example #6: A 10.0 kg mass travels towards the right at a speed of 6.00 m/s and collides with and sticks to a 20.0 kg moving initially towards the left at 4.50 m/s. (a) What is the speed of the combined object after the collision?

24. (b) What is the amount of kinetic energy after the collision? Write it as a percentage of the initial kinetic energy.

25. Example #7: A mass of 5.00 kg slides towards the left at a speed of 10.0 m/s. At the same time, a mass of 15.0 kg slides towards the right at 5.00 m/s. The two masses collide with one another and stick together after the collision. (a) Determine the velocity (speed and direction) of the combined mass after the impact.

26. (b) Determine the amount of kinetic energy that is lost in this collision. {Hint: Subtract the initial KE from the final KE.} Write the kinetic energy lost as a fraction of the initial kinetic energy. %KE lost = The ( – ) sign means KE is lost.

27. Example #8: A 2000 kg car was traveling towards the east when it collided with and stuck to a 4000 kg car initially traveling towards the north. The wreckage slid at an angle of 55.2 degrees NE for a distance of 18.15 meters before stopping. Officer Bogart determines the coefficient of friction between the types of tires on the vehicle and the asphalt to be 0.600 for the conditions present that day. Determine the speeds of the cars before the collision. Which, if any, were driving over the 40 mph speed limit? Find v f from energy:

28. Momentum conserved in each component direction.

29. Example #9: A ballistic pendulum is shown at right. A heavy mass (M = 2.00 kg) hangs vertically downwards at rest. A small projectile (m = 0.0400 kg) travels with initial speed vo and collides with and sticks to the heavy mass. After the collision, the heavy mass swings out and rises up vertically 12.0 cm. (a) What was the speed of the combined block and projectile immediately after the collision? Order of Events: Bullet embeds in block. Solve for final speed with momentum conservation. Final speed of collision becomes initial speed of block rising. Use energy conservation.

30. MIT Ballistic Pendulum

31. Start with energy conservation.

32. (b) What was the initial velocity of the projectile? Next use momentum conservation. masses stick together

33. Elastic Collisions Elastic Collisions, 1-dim HW #6: Handout #18 – 21, 24 – 26. {on the schedule…}

34. Elastic Collisions in One Dimension: Definition of Elastic Collisions: All collisions conserve momentum! Elastic collisions also conserve total kinetic energy. Momentum conservation is always true. KE is only conserved for elastic collisions. {common source of error!} Inelastic collisions only conserve momentum. Usually KE is lost. inelastic = not elastic

35. Equation for momentum conservation: Equation for kinetic energy conservation: Two equations means two unknowns!

36. Derivation of reduction of kinetic energy equation: Show Note: You would be asked to derive this only on a bonus question… Start with momentum: Eq. #1

37. Next KE: Note: The underlined parts are equal from equation #1.

38. Note: This is the simplified form of Conservation of Total Kinetic Energy. This is only used with Elastic collisions.

39. Example #1: A mass m 1 = m travels towards the right at a speed v 1i = v o and collides with an identical mass m 2 = m initially at rest. If the two masses collide elastically, what is the final velocity of each mass? Start with momentum: Next use KE:

40. Solve the simultaneous equations: The particles completely exchange velocities.

41. Example #2: A 10.0 kg mass (m1) travels towards the right at 15.0 m/s and collides elastically with a 20.0 kg mass (m2) initially at rest. (a) Determine the final velocity of each mass. Start with momentum: Next use KE:

42. Solve: Pick one equation to solve for v 1f :

43. (b) What percentage of the initial kinetic energy has been transferred to the second mass?

44. Example #3: A 20.0 kg mass (m1) travels towards the right at 15.0 m/s and collides elastically with a 10.0 kg mass (m2) initially at rest. Determine the final velocity of each mass. Start with momentum: Next use KE:

45. Solve: Pick one equation to solve for v 2f :

46. Example #4: A cart with a mass of 340 grams moving on a frictionless surface at initial speed of 1.2 m/s collides elastically with a second mass at rest. The mass of the second object is unknown. After the elastic collision, the first cart travels in its original direction at 0.66 m/s. Determine the mass of the unknown cart and the speed of the unknown cart after the collision. Start with the KE equation, only one unknown!

47. Next use momentum to solve for the mass:

48. Example #5: A steel ball of mass 0.500 kg is fastened to a cord 70.0 cm long and fixed at the far end, and is released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic. (a) Find the speed of the block after the collision. Start with energy conservation to find the speed of the steel ball at the bottom before the impact. Next solve the elastic collision to find the velocities of the two objects after the collision. Finally use energy conservation again to solve for the height to which the steel ball will rise.