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KD5 Conservation of momentum and energy

Chapter 7; Chapter 8. KD5 Conservation of momentum and energy. Review: Momentum. Momentum - Mass in motion momentum = mass X velocity Vector quantity (magnitude and direction). Review: Momentum. What are the 2 ways to increase momentum?

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KD5 Conservation of momentum and energy

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  1. Chapter 7; Chapter 8 KD5 Conservation of momentum and energy

  2. Review: Momentum • Momentum - Mass in motion • momentum = mass X velocity • Vector quantity (magnitude and direction)

  3. Review: Momentum • What are the 2 ways to increase momentum? • If the velocity of an object doubles, what happens to the momentum? • Which has more momentum a car moving down a hill or a roller skate moving with the same speed? • Which has more momentum a skateboard moving at 2 m/s or a truck at a red light? Increase mass OR increase velocity Momentum Doubles Car due to more mass with the same velocities Skateboard due to the Truck has no velocity Thus no momentum

  4. Review: Momentum • Calculate the momentum of a 90 kg football player running at 6 m/s. 540 kg m/s

  5. Law of Conservation • Physical properties in an isolated system don’t change • Just transferred from 1 object to another • Examples: • Law of Conservation of Momentum • Law of Conservation of Energy • Law of Conservation of Mass

  6. Law of Conservation of Momentum • Example: 2 cars collide • The momentum of the 2 vehicles before the collision is equal to the total momentum of the 2 vehicles after the collision • The momentums of each car might change but the total momentum remains the same

  7. Law of Conservation of Momentum • The total momentum of any closed, isolated system does not change • Momentum is NEVER lost! • Just transferred

  8. Law of Conservation of Momentum • p = p or • m1v1 + m2v2 = m1v1 + m2v2

  9. Example • A ball with a mass of 0.10 kg moves to the right with a speed of 2.0 m/s. It hits a .040 kg ball which is standing still. After the balls hit, the 2nd ball moves to the right with a speed of 0.80 m/s. What is the velocity of the 1st ball?

  10. Answer • A ball with a mass of 0.10 kg moves to the right with a speed of 2.0 m/s. It hits a .040 kg ball which is standing still. After the balls hit, the 2nd ball moves to the right with a speed of 0.80 m/s. What is the velocity of the 1st ball? m(ball 1) v(ball 1) + m(ball 2) v(ball 2) = m(ball 1) v(ball 1) + m(ball 2) v(ball 2) .1kg (2.0 m/s) + .04kg (0 m/s) = .1 kg (v) + .04 (.8 m/s) .2 + 0 = .1 (v) + .032 .168 = .1 (v) 1.68 = v

  11. Collisions • p (before a collision or explosion) = p (after a collision or explosion)

  12. 2 Types of Collisions • 1. Elastic Collision-Objects hit and bounce off • Both momentum and kinetic energy are conserved • Ex) Pool balls collide

  13. 2 Types of Collisions • 2. Inelastic Collision-Objects hit and stick together • Only momentum is conserved • Ex) 2 cars collide and stick together

  14. Energy • Energy-The ability to do work • Forms of energy: • Electrical • Chemical • Solar • Thermal • Mechanical

  15. Mechanical Energy • 1. Kinetic energy • 2. Gravitational Potential Energy • 3. Elastic Potential Energy

  16. Mechanical Energy: Kinetic • Kinetic Energy-Energy of motion • KE = ½ mv² • m=mass (kg) • v=velocity (m/s) • KE=kinetic energy (J)

  17. Mechanical Energy: Gravitational • Gravitational Potential Energy-Stored energy of position • GPE = m g h • m=mass (kg) • g=9.8 m/s² • h=height (m) • GPE=gravitational potential energy (J)

  18. Mechanical Energy: Elastic • Elastic Potential Energy-Stored energy of stretch or compression • EPE = ½ k x² • k=elastic constant (N/m) • x=distance stretched/compressed from rest (m) • EPE=elastic potential energy (J)

  19. Elastic Potential Energy

  20. Law of Conservation of Energy • The total energy of a system remains constant • Energy is neither created or destroyed (only transferred to another form) • Total Energy Before = Total Energy After • KE + GPE + EPE = KE´ + GPE´ + EPE´

  21. “The Process” to Solve Energy Problems • 1. Sketch a picture of the problem • 2. Identify the types of mechanical energy present at the important stages of the problem • 3. Set the before energies equal to the after • 4. Substitute • 5. Solve and Label

  22. Example: • Bill throws a 0.1 kg ball straight up with a speed of 7.5 m/s. How high did the ball go?

  23. Remember • Bill throws a 0.1 kg ball straight up with a speed of 7.5 m/s. How high did the ball go? • KE + GPE + EPE = KE´ + GPE´ + EPE´ ½ mv2 + mgh + 1/2 kx2 = ½ mv2 + mgh + 1/2 kx2 Only velocity to start with (only KE) = Only height to end with (only GPE) ½ m v2 = mgh ½ (.1 kg) (7.5 m/s)2 = .1kg (9.8 m/s2) v2

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