1. 1 Do Now: What makes the shuttle go UP? Objectives: Utilize IMPULSE to calculate: Force – time – change in velocity Home work: Page 233: #’s 1 – 5 all Page 250-52: #’s 47, 51, 55, 60, 64 Momentum LESSON ONE

2. Momentum ρ = mv You must know how to do these actions: 1. Calculate time needed to slow or accelerate an object to a certain velocity using applied Force with known masses. IMPULSE (F)(Δt) = Δρ 2. Find velocities of objects that collide in inelastic collisions 1. (stick together, car crash, clay and balls etc.) 3. Find velocities of objects that collide in elastic collisions ( 1. do NOT stick together, billiard balls) 4. Using the conservation of momentum, find velocities of objects using just masses, velocities. 5. Find velocities and directions of objects that collide in two dimensions.

3. 3 1.IMPULSE (F)(Δt) = Δρ = mΔv  Δm Δv

4. 4 Before the collision, the baseball moves toward the bat. During the collision, the baseball is squashed against the bat. After the collision, however, the baseball moves at a higher velocity away from the bat, and the bat continues in its path, but at a slower velocity. How are the velocities of the ball, before and after the collision, related to the force acting on it? Newton’s second law describes how the velocity of an object is changed by a net force acting on it. The change in velocity of the ball must have been caused by the force exerted by the bat on the ball. The force changes over time.

5. 5 Newton’s second law of motion, F = ma, can be rewritten by using the definition of acceleration as the change in velocity divided by the time needed to make that change. It can be represented by the following equation: F = ma = m Multiplying both sides of the equation by the time interval,  t, results in the following equation: F  t = m  v Impulse, or F  t, is the product of the average force on an object and the time interval over which it acts. Impulse is measured in newton-seconds.

6. 6 What does an air bag do from an “IMPULSE” point of view? IMPULSE LESSON ONE SUMMARY

7. 7 Using the Impulse-Momentum Theorem to Save Lives A large change in momentum occurs only when there is a large impulse. A large impulse can result either from a large force acting over a short period of time or from a smaller force acting over a long period of time. What happens to the driver when a crash suddenly stops a car? An impulse is needed to bring the driver’s momentum to zero. An air bag reduces the force by increasing the time interval during which it acts. It also exerts the force over a larger area of the person’s body, thereby reducing the likelihood of injuries.

8. Momentum LESSON TWO J Elastic events 8 Do Now: What is the impulse needed to accelerate a 5.00 Kg mass from rest to 10.0 m/s? Objectives: Utilize ELASTIC to calculate: Force – time – change in velocity Home work: Due Tuesday Page 233: #’s 1 – 25 all Page 250-52: #’s 47, 51, 55, 60, 64 50, 70, 72, 73, 74

9. 9 Two-Particle Collisions Although it would be simple to consider the bat as a single object, the ball, the hand of the player, and the ground on which the player is standing are all objects that interact when the baseball player hits the ball. A much easier system is the collision of two balls of unequal mass and velocity. During the collision of the two balls, each briefly exerts a force on the other. Despite the differences in sizes and velocities of the balls, the forces they exert on each other are equal and opposite, according to Newton’s third law of motion. These forces are represented by the following equation. F B on A = -F A on B Momentum LESSON TWO

10. 10 Elastic Collisions two masses stay separate Kinetic Energy is conserved Momentum LESSON TWO

11. 11 Use the result of Newton’s third law of motion –F A on B = F B on A. P A2 = -F A on B Δt + P A1 Add the momenta of the two balls. THIS IS IT!!!! m 1 v 1 + m 2 v 2 = m 1 v ' 1 + m 2 v ' 2 This shows that the sum of the momenta of the balls is the same before and after the collision. That is, the momentum gained by ball 2 is equal to the momentum lost by ball 1. If the system is defined as the two balls, the momentum of the system is constant. Momentum LESSON TWO

12. What is happening?? Newton’s Cradle: Dominique Toussaint on wikipedia

13. 13 1. 2. 3. 4. 5. 6. 7. List ELASTIC “Events”

14. 14 F Δt = Δm Δv F = Δm Δv / Δt = 3.95 (7.91) / 1.00 = 31.2445 N Wt = mg = 3.95 * 9.8 = 38.71 N ….. It won’t get off the ground! LESSON THREE DO NOW: A 4.00 Kg rocket launches from rest. 0.05 Kg of fuel is exhausted @ – 625 m/s for 1.00 sec. How far up does the rocket travel? Home work: Due MONDAY Jan 14 Page 233: #’s 1 – 5 all WX:3.3 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0 Page 240: #’s 19 – 21 all WX:3.6 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0 Page 250-52: #’s 47, 51, 55, 60, 64 WX:3.3 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0 50, 70, 72, 73, 74 WX:3.3 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0

15. 15 IN-Elastic Event Momentum LESSON THREE Jan 16 Do Now: Two balls collide. The 1 st ball has a mass of 3.00 Kg and an initial velocity of 2.00 m/s @ 0.00º The 2 nd ball has a mass of 9.00 Kg and an initial velocity of 0.00 m/s After the collision: The 1 st ball has a velocity of 0.5 m/s @ 0.00º What is the velocity of the second ball if it too moves off @ 0.00º ? Objectives: Utilize IN-ELASTIC to calculate: Force – time – change in velocity Home work: Page 238: #’s 13 – 18 all Page 243 #22 Page 252-53: #’s: 76, 77 (add part “b” the bullet imbeds itself in the block) Page 252-53: #’s: 86, 90 Page 255: # 2- 5, 7 m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v ‘ 3

16. In - Elastic 16 m 1 v 1 + m 2 v 2 = (m 1 + m 2 )v ‘ 3 #1 Two cars collide…and they stick together. The 1 st car has a mass of 1875 Kg and an initial velocity of 23.00 m/s @ 0.00º The 2 nd car has a mass of 1025 Kg and an initial velocity of 17.00 m/s @ 0.00º After the collision: What is the velocity of the two cars if they both move off @ 0.00º ? #2 Two cars collide…and they stick together. The 1 st car has a mass of 1875 Kg and an initial velocity of 23.00 m/s @ 0.00º The 2 nd car has a mass of 1025 Kg and an initial velocity of 17.00 m/s @ 180.0º After the collision: What is the velocity (speed and direction) of the two cars? #3 If the cars in #2 took 1.00 seconds to collide and stick together, What was the force on the SECOND Car? What was the acceleration felt by a 60.00 Kg passenger in car number 2? Momentum LESSON THREE QUIZ

17. 17 1. 2. 3. 4. 5. 6. 7. List IN - ELASTIC “Events”

18. 18 Momentum LESSON Four

19. 19 Two- Dimensional Collisions Up until now, you have looked at momentum in one dimension only. The law of conservation of momentum holds for all closed systems with no external forces. It is valid regardless of the directions of the particles before or after they interact. Now you will look at momentum in two dimensions. For example, billiard ball A strikes stationary ball B. Consider the two balls to be the system. The original momentum of the moving ball is p A1 ; the momentum of the stationary ball is zero. Therefore, the momentum of the system before the collision is equal to p A1. Momentum LESSON Four Home work Page 242: Example Problem # 4 Page 243: #’s 22– 25 all Do Now: What is the initial momentum of ball #1 if the mass is 10.4Kg and the velocity is + 12.45 m/s?

20. 20 After Collision m1m1 m2m2 v1'v1' v2'v2' After the collision, both balls are moving and have momenta. According to the law of conservation of momentum, the initial momentum equals the vector sum of the final momenta, so p A1 = p A2 + p B2. The equality of the momentum before and after the collision also means that the sum of the components of the vectors before and after the collision must be equal. Therefore, the sum of the final y-components must be zero. p A2y + p B2y = 0 They are equal in magnitude but have opposite signs. The sum of the horizontal components is also equal. p A1 = p A2x + p B2x

21. 21 DO NOW  In Class: A 2.00-kg ball, A, is moving at a speed of 5.00 m/s. It collides with a stationary ball, B, of the same mass. After the collision, A moves off in a direction 30.0º to the left of its original direction. Ball B moves off in a direction 90.0º to the right of ball A's final direction. How fast are they moving after the collision? PHET Momentum Billiard Balls Solution: 1. Sketch the before and after states Before AB v A1 v B1 =0 A B After v A2 v B2

22. 22 2. Draw a momentum vector diagram. Note that p A2 and p B2 form a 90º angle. Initial Note that ALL the Initial momentum is in the X DIRECTION 90ºp A2 30º p B2 p2p2 3. Perform calculations: Initial Final Determine Initial momenta: Final momenta p IAx + p IAy + p IBx + p IBy = p FAx + p FAy + p FBx + p FBy p IAx = m a v IAX = (2.00 kg)(5.00m/s) = 10.0 kgm p FAx = m a v FAx Cos(30) p IAy = 0p FBx = m a v FBx Cos(-60) p IBx = 0p FAy = m b v FAy SIN (30) p IBy = 0p FBy = m b v FBy SIN (-60) p A1 p B1

23. 23 2. Two Equations  Two Variables 3. Solve for v FA in terms of v FB p Ix = 10 = m a v FAx Cos(30) + m a v FBx Cos(-60) p Iy = 0 = m b v FAy SIN (30) + m b v FBy SIN (-60)

24. 24 HOMEWORK TEST  Due NOW In Class: A 5.00-kg ball, A, is moving at a speed of 10.00 m/s. It collides with a stationary ball, B, of the same mass. After the collision, A moves off in a direction 60.0º to the left of its original direction. Ball B moves off in a direction 30.0º to the right of ball A's final direction. How fast are they moving after the collision? PHET Momentum Billiard Balls Solution: 1. Sketch the before and after states Before AB v A1 v B1 =0 A B After v A2 v B2

25. 25 2. Draw a momentum vector diagram. Note that p A2 and p B2 form a 90º angle. 3. Perform calculations: Determine initial momenta: p A1 = m a v a1 = (5.00 kg)(10.00m/s) = 50.0 kgm/s {X-Dir} p B1 = 0 Find p 2 : p A ‘ + p B ‘ = 50.0 kgm/s X direction + 0 Y direction p A1 p B1 60 deg 30 deg

26. 26

27. Two Dimensional IN CLASS “Initial”ρ x = M 1 (V 1 )(Cos{Θ }) + M 2 (V 2 )(Cos{Θ }) “Initial”ρ y = M 1 (V 1 )(Sin{Θ }) + M 2 (V 2 )(Sin{Θ }) Two Equations two Unknowns Page 253 # 79 27