Final Exam Review

1. Final Exam Review Conservation of Energy Impulse & Momentum Conservation of Momentum

2. Conservation of Energy • Def: Total energy in a system remains constant. Energy is not created or destroyed. • Key Point: Energy can change it’s form, but it never really disappears. (Rabbit into the hat)

3. How to find the total Energy • 2 Ways to determine the total energy in a system: • 1) Calculate energy when it’s ALL Potential • 2) Calculate energy when it’s ALL Kinetic

4. Example • A 1000kg roller coaster ride begins at the top of a 50m hill and has a drop to ground level before continuing onto the rest of the ride. • A) What is the Potential Energy at the beginning? • B) What is the Kinetic Energy at the bottom of the first hill? • C) What is the total energy in this system?

5. Example 1000 kg PE = mgh 50m

6. Example 1000 kg PE = mgh = 490,000 J 50m KE = ?

7. Example 1000 kg PE = mgh = 490,000 J 50m KE = 490,000J

8. Example PE = mgh = 490,000 J ALL energy is PE here. Total energy in the system is 490,000 J All energy is KE here. 50m KE = 490,000J

9. Example PE = mgh = 490,000 J What is the PE here? What is the KE? 50m KE = 490,000J 25m Total energy in the system is 490,000 J

10. Example PE = mgh = 490,000 J What is the PE here? PE=mgh=(1000)(9.8)(25)=245,000J What is the KE? 50m KE = 490,000J 25m Total energy in the system is 490,000 J

11. Example PE = mgh = 490,000 J What is the PE here? PE=mgh=(1000)(9.8)(25)=245,000J What is the KE? KE=Total-PE= 490,000-245,000=245,000J 50m KE = 490,000J 25m Total energy in the system is 490,000 J

12. Practice • Pg. 177 Practice E • Do 1,2 &4 ONLY.

13. Momentum • Formula: p = mv momentum(kg*m/s) = mass(kg) * velocity(m/s)

14. Impulse • Def: The applied force needed to change the momentum of an object. • Formula: FΔt = m(vf-vi) Impulse=Δp

15. Practice • Pg 199- Practice A • #1 ONLY • Pg 201- Practice B • #1 & 2 ONLY

16. Conservation of Momentum • Key Point: The total momentum between two objects always remains constant.

17. Example • A 3 kg ball rolling to the right at 5m/s hits a 2kg ball at rest. The 3kg ball comes to rest after the collision. What is the velocity of the ball initially at rest? 3kg 2kg 0m/s 5m/s

18. Example • A 3 kg ball rolling to the right at 5m/s hits a 2kg ball at rest. The 3kg ball comes to rest after the collision. What is the velocity of the ball initially at rest? Total momentum: (3)(5) + (2)(0) = 15kgm/s 3kg 2kg 0m/s 5m/s

19. Example • After collision: Total momentum: 15kgm/s 3kg 2kg ? m/s 0m/s Total momentum: 15 = (3)(0) + (2)(v) 15 = 2v 7.5 m/s = v

20. Practice • Pg. 211 – Section Review • #3 ONLY.