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Final Exam Review

Prepare for your physics final exam with this comprehensive review covering the principles of conservation of energy, impulse, and momentum. Learn the fundamental definitions, such as how total energy in a closed system remains constant and how energy can transition between kinetic and potential forms. Explore practical examples, including a roller coaster scenario to calculate potential and kinetic energy. Understand the formula for momentum and the concept of impulse that influences it. Engage with targeted practice problems to reinforce your understanding.

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Final Exam Review

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  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.

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