1. Momentum and Its Conservation LEQ: What is Momentum?

2. Momentum: The velocity of a moving mass. p = m∆v Unit is Kg m/s For a constant velocity; an increase in mass will result in an increase in momentum.

3. Sampler A 2800 Kg car is parked on an incline. The driver steps out without setting the brake. The car begins to roll at 0.2 m/s. What is the momentum of the car? ans: p = m∆v, p = 2800 Kg (0.2 m/s) p = 560 Kg m/s

4. Sampler How would the momentum change with a car at twice the mass, 5600 Kg ? ans: p = 5600 Kg ( 0.02 m/s) = 1120 Kg m/s Or twice the mass = twice the momentum.

5. Problems 1.What is the momentum of a 146 Kg mass rolling at 17 m/s? ans: 2482 Kg m/s 2. A 21 Kg boy runs an jumps onto a sled that has a mass of 5.9 Kg and begins to slid down the hill at 4.5 m/s/ What is the total momentum? ans: 121 Kg m/s 3. The boy rolls off of the sled in #2. If he maintains the same velocity, what is the boys instantaneous momentum? ans: 94.5 Kg m/s

6. 4. If a 1210 Kg mass has a momentum of 5.6 x 10 4 Kg m/s, What is the velocity of this mass ? ans: 46.3 m/s 5. You are driving your car that has a mass of 2750 Kg at 38 mi/h in a 25 mi/h zone, what is your momentum? 38 mph = 16.98 m/s ans: 46695 Kg m/s 6. You apply a force of 2.5 x 10 3 N for 2.25 s to the car above. What is the change in the momentum and velocity in the car above?

7. Impulse Impulse: force exerted in time Impulse = F∆t Impulse-Momentum Theorem: Impulse is equal to the change in the momentum of a mass. F∆t = p f – p i = mv f - mv i

8. Sampler 1.How much force must be applied to stop a 2.8 Kg ball traveling at 5.6 m/s in 0.05 s? ans: F∆t = m∆v ; F = 313.6 N 2. When 2.5 x 10 4 N of force is applied to a 1200 Kg mass slows from 14 m/s to 6 m/s. How much time is required to slow the mass? ans: -0.384 s, 3. What does the negative mean in #2? ans: the negative mean in the opposite direction of the motion.

9. Sampler You are driving your car that has a mass of 2750 Kg at 38 mi/h in a 25 mi/h zone, what is your momentum? 3. How must force must be applied to your car in the above situation to stop in 0.05 s? ans: -933900 N

10. Homework Page 233 #1-4 Page 251 #56-69 odd

11. Conservation of Momentum Total momentum within a system is a constant for an elastic collision. Σp i = Σp f m 1i v 1i + m 2i v 2i = m 1f v 1f + m 2f v 2f

12. Sampler: Conservation A 15 kg ball at rest is hit by a 25 kg ball traveling at 4 m/s west. After the collision, the 15 kg ball moves at 2.3 m/s west. What is the velocity of the 25 kg ball? ans: 2.62 m/s

13. Sampler: Conservation Two cart both with a mass of 25 kg collide and stick together. One cart had a velocity of 2 m/s while the other is at rest before the collision. What is the final velocity of the carts? ans: 1 m/s HW: page 238 and 240, #13-21

14. Kinetic Energy and Momentum Kinetic Energy is NOT conserved during an inelastic collision. An inelastic collision occurs when two objects a. collide and deform moving away separately. b. collide and stick together. Energy is loss due to sound and internal transfer of energy.

15. Elastic: NO transfer of energy Inelastic: transfer of energy. To find the amount of Kinetic Energy loss: ∆KE = KE f – KE i Where KE i = KE 1i + KE 2i and KE f = KE 1f + KE 2f Remember KE = ½ mv 2

16. Sampler Two clay balls collide in an inelastic collision. The first ball has a mass of 0.5 Kg and an initial velocity of 4.0 m/s to the right. The mass of the second ball is 0.25 Kg and it initial velocity is 3.0 m/s to the left. After the collision they stick, what is the velocity of the composite ball? How much has the kinetic energy decreases? ans: 1.67 m/s right ; -4.07 J