1. Turn in Homework!

2. Momentum Objectives: You will be able to: Describe the momentum of an object. Determine the impulse given to an object.

3. Bell Ringer: answer on the sheet of binder paper from yesterday. Write a sentence to explain:  Newton’s First Law of Motion  Newton’s Third Law of Motion

4. Pre-test! This will allow me to: set you a realistic target grade. Plan lessons to prevent repetition and focus on areas of new knowledge.

5. Impulse & Momentum Impulse is another way to describe Newton’s second law of motion. Newton’s second law of motion F = ma, can be rewritten as: F ∆ t = m ∆ v F ∆ t is the impulse. m ∆ v is the momentum of the object. Momentum is a quantity defined as the product of the mass and velocity of an object. Or, think of it as the inertia of an object in motion. Linear Momentum = mass x velocity (p = mv).

6. To stop an object that has momentum, an impulse will have to act on an object to change its state of motion. Both momentum and impulse are vector quantities – they have magnitude and direction. Impulse points in the same direction as the change in momentum. Impulse is the product of the average force on an object and the time interval it acts. Measured in N.s FΔ t

7. Impulse-Momentum Theorem Momentum and impulse are related. The impulse-Momentum theorem is FΔ t = p f – p i p f = mv f and p i = mv i So FΔ t = mv f - mv i Khan Academy video

8. Guided Practice 1 p. 231 List the variables Write out the formula Set up the formula with the correct variables Solve for the base-ball’s final velocity.

9. Group Practice Work through the worked problem in your textbook with your group. p. 232 Make sure everyone in the group can solve the problem.

10. Guided Practice 2 A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?

11. Solution 2 p = mv p = (2250) (25) p = 56250 kg. m/s or 56000 kg. m/s to the correct number of sig figs or 5.6 x 10 4 kg. m/s in scientific notation

12. Guided Practice 3 A 1400 kg car moving west with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.

13. Solution 3 Use F ∆ t = mv f - mv i F = mv f – mv i ∆ t F = (1400 x 0) – (1400 x 15) 0.30 F = - 7.0 x 10 4 kg. m/s 2 or - 7.0 x 10 4 N to the East.

14. Independent Practice 1. Calculate the momentum of the following objects. a) a 75 kg speed skater moving forward at 16 m/s. b) a 135 kg ostrich running north at 16.2 m/s. c) a 5.0 kg baby on a train moving eastward at 72 m/s. d) a seated 48.5 kg passenger on a train that has stopped. 2. Calculate the velocity of a 0.8 kg kitten with a momentum of 5 kg.m/s forward.

15. Solutions 1. a) 1200 kg.m/s b) 2187 kg. m/s c) 360 kg. m/s d) 0 kg.m/s 2. 6.3 m/s

16. Homework: Pre AP Complete on binder paper & turn in Monday: p. 233 Questions 1 - 5.