1. C OLLISION Reporters: Cabag Andaluz Diaz Janier

2. CRASHES BILLIARDS

3. The act or process of colliding; a crash or conflict. A brief dynamic event consisting of the close approach of two or more particles, such as atoms, resulting in an abrupt change of momentum or exchange of energy.

4. M OMENTUM Involves motion -to have momentum, an object must be moving at a certain velocity Involves mass -the more massive, the more difficult to change your state immediately -is the measure of one’s motion, equivalent to the product of one’s mass and velocity.

5. E XPRESS IN : p= m v Where p is the momentum ain kg-m/s

6. S AMPLE CONDITIONS : 1. If you have a VW Bug going 50 km/h and a truck going at the same velocity, which has more momentum? 2. What about two trucks, one moving at 25 km/h and the other moving 50 km/h. Which has more momentum? 3.

7. Example: Calculate the momentum of 110 – kg missile traveling at 100 m/s eastward. Given: Mass (m) – 110 kg of missile Velocity (v) – 100 m/s Momentum - ?

8. Solution: Direct substitution p = mv = 110 kg (100 m/s) = 11000 kg – m/s, eastward

10. I MPULSE Force (F) applied by one object to another object within a given time interval ∆t Express in: F∆t= ∆p Where F∆t is the impulse in newton-second (N-s)

11. D ERIVATION :

12. Transporting ∆t to the other side of equation: F ∆t = m ∆v F ∆t= ∆p

13. C ONSERVATION OF M OMENTUM All collisions have to conserve momentum.

15. Elastic Collision in One Dimension

19. Stops

21. E XAMPLE ( CASE 1)

22. E XAMPLE ( CASE 2)

24. CASE 3

25. Mass of moving object less than the mass of the stationary object (m 1 < m 2 ) (m 1 - m 2 ) v’ = (m 1 + m 2 ) v1 m1 Moves. But, in the opposite direction. 2 m1 v’= (m 1 + m 2 ) v1 m2 Moves. V 2 has the same direction as V 1

26. Example : Again, m 2 is stationary on the track while the other m 1 rolls at 1 m/s towards the right. This time, the mass of m 2 is 2 kg while the mass of m 1 is 1 kg. What happens when the balls collide elastically?

27. Since the masses are not equal, we solve again for v 1 ’ and v 2 ’ So (m 1 - m 2 ) 1 kg – 2 kg v’ = (m 1 + m 2 ) v1 = 2 kg + 1kg = 1 m/s = -(1/3) m/s NEGATIVE OPPOSITE DIRECTION 2 m1 2 kg v2’= (m 1 + m 2 ) v1 = 3 kg = 1 m/s = (2/3) m/s

28. Before Collision m1 1m/s m2 1kg 2kg After Collision m1 m2 (1/3) m/s 2/3 m/s

29. T OTAL INELASTIC COLLISION IN ONE DIMENSION

30. If in an elastic collision the masses maintain their mass state after collision, a total inelastic collision produces a combined mass after collision

31. Before Collision 1 m/s After Collision Vf Formula: m1v1 + m2v2 = (m1 + m2) vf m1 m2 m1 +

32. If m2 is stationary, m2v2 = 0 FORMULA: m1 Vf = v1 ( m1 + m2 )

33. Example: Colliding with a much heavier object may produce an inelastic collision. A head-on collision between a car and a train may drag the car along the path of the train. Assume that 1000-kh car stalled on the railroad tracks is smashed by a 50,000-kg train travelling at 8.33 m/s (about 30 km/hr). What is the final velocity after a total inelastic collision?

34. m1 Vf = v1 ( m1 + m2 ) = 8.33 m/s 50,000 kg 51,000 kg = 8.17 m/s

35. C OLLISIONS IN TWO DIMENSIONS

36. When we move to collisions in two dimensions, we can separately balance momentum in the x- direction and in the y-direction BeforeAfter p1’ p2’ a 1a2 ß1 ß2 p 1 p 2

37. Using the conservation of energy: Along the x-axis: ∑p x before collision = ∑p x after collision (Eqn. 7.16) p 1 cos a 1 + p 2 cos a 2 = p 1’ cos ß 1 + p 2’ cos ß 2 Along the y-axis: ∑p y before collision = ∑p y after collision (Eqn. 7.17) p 1 sin a 1 + p 2 sin a 2 = p 1’ sin ß 1 + p 2’ sin ß 2

38. Example A pool ball weighing 2 kg is traveling at 30◦ at 0.8 m/s hits another ball moving at 0.5 m/s at 180◦. If the second ball leaves the collision at 0◦ and the first moves away at 150◦, find the final velocity vectors of the balls. Before After v’ 1 0.5 m/s ß 1 0.8 m/s v’ 2 30◦ ß 2 = 0

39. Answer Given mass of the pool balls m 1, m 2 2kg Before Collision velocity of the first pool ball v 1 0.8m/s angle of the first pool ball a 1 30◦ Velocity of second pool ball v 2 0.5 m/s Angle of second pool ball a 2 180◦ After collision Angle of first pool ball ß 1 150◦ Angle of second pool ball ß 2 0◦ Find: Final velocities of the balls ( v 1’ and v 2’ )

40. Solution Along the x-axis: ∑px before collision = ∑px after collision p 1 cos a 1 + p 2 cos a 2 = p 1’ cos ß 1 + p 2’ cos ß 2 m(v 1 cos a 1 + v 2 cos a 2 ) + m(v 1’ cos ß 1 = v 2’ cos ß 2 ) v 1 cos a 1 + v 2 cos a 2 = v1’ cos ß 1 + v 2 cos ß 2 0.8 m/s (cos 30◦) + 0.5 m/s (cos 180◦) = v 1’ cos 150◦ + v 2’ cos 0◦ 0.8m/s (0.866) + 0.5 m/s (-1) = v 1’ (-0.866) + v 2’ (1) 0.70 m/s – 0.5m/s = -0.866 v 1’ + v 2’ Expressing v 2’ in terms of v 1’ v 2’ = 0.866 v 1’ + 0.20m/s

41. Along the y-axis: ∑py before collision = ∑py after collision p 1 sin a 1 + p 2 sin a 2 = p 1’ sin ß 1 + p 2’ sin ß 2 m( v 1 sin a 1 + v 2 sin a 2 ) = m(v 1’ sin ß 1 + v 2’ sin ß 2 ) v 1 sin a 1 v 2 sin a 2 = v 1’ sin ß 1 + v 2’ sin ß 2 0.8 m/s (sin 30◦) + 0.5m/s (sin 180◦) = v 1’ sin 150◦ + v 2 sin 0◦ 0.8m/s (0.5) + 0.5m/s (0) = v 1’ (0.5) + v 2’ (0) 0.4m/s = 0.5 v 1’ 0.4m/s / 0.5 = v 1’ v 1’ = 0.m/s < 150 ◦ Solving for v 2’ v 2’ = 0.866 v 1’ + 0.20 m/s = 0.866 (0.8m/s) + 0.20 m/s = 0.9 m/s <0 ◦