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Elastic and Inelastic Collisions

Elastic and Inelastic Collisions. Chapter 6, Section 3 Pg. 222-229. Inelastic collisions. 1. 2. Inelastic (or “hit and stick”) collisions occur when two objects stick together and move with the same velocity after colliding. m 1 ν 1,i + m 2 ν 2,i = m 1 ν 1,f + m 2 ν 2,f.

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Elastic and Inelastic Collisions

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  1. Elastic and Inelastic Collisions Chapter 6, Section 3 Pg. 222-229

  2. Inelastic collisions 1 2 Inelastic (or “hit and stick”) collisions occur when two objects stick together and move with the same velocity after colliding. m1ν1,i + m2ν2,i = m1ν1,f + m2ν2,f ν1,f = ν2,f m1ν1,i + m2ν2,i = (m1 + m2)νf

  3. Inelastic collisions and Kinetic Energy Kinetic Energy does not appear to be conserved in inelastic collisions! Some energy converted to: Sound Thermal Energy Energy loss also due to deformation

  4. Sample Problem (Pg. 226 #1) A 0.25 kg arrow with a velocity of 12.0 m/s to the east strikes and pierces the center of a 6.80 kg target. If the velocity of the arrow and target combined is 0.43 m/s, what is the decrease in kinetic energy during the collision?

  5. m1 = 0.25 kg m2 = 6.80 kg vf = 0.43 m/s v2,i = 0 m/s v1,i = 12.0 m/s Starting KE KEi = KE1,i + KE2,i KEi = ½ m1v1,i² KEi = ½ (0.25 kg) (12.0 m/s)² = 18.0 J

  6. m1 = 0.25 kg m2 = 6.80 kg vf = 0.43 m/s v1,i = 12.0 m/s v2,i = 0 m/s Final KE KEf = KE1,f + KE2,f KEf = ½ (m1 + m2) vf² KEf = ½ (0.25 kg + 6.80 kg) (0.43 m/s)² KEf = 0.65 J Decrease in KE = 18.0 J – 0.65J = 17.4 J

  7. 1 2 Elastic Collisions A collision between two objects in which both objects do not stick together! Kinetic energy is conserved in perfect elastic collisions! m1ν1,i + m2ν2,i = m1ν1,f + m2ν2,f ½ m1v1,i² + ½ m2v2,i² = ½ m1v1,f² + ½ m2v2,f²

  8. Sample Problem (Pg. 229 #3) A 4.0 kg ball sliding to the right at 8.0 m/s has an elastic collision with another 4.0 kg ball at rest. If the final velocity of the second ball after the collision is 8.0 m/s to the right and the first ball’s new velocity is 0.0 m/s, show that the kinetic energy remains constant from start to finish.

  9. 1 2 m1 = 4.0 kg m2 = 4.0 kg V1,f = 0.0 m/s v1,i = 8.0 m/s v2,i = 0 m/s V2,f = 8.0 m/s ½ m1v1,i² + ½ m2v2,i² = ½ m1v1,f² + ½ m2v2,f² ½ (4.0 kg)(8.0 m/s) ² + ½ (4.0 kg)(0.0)² = 128 J ½ (4.0 kg)(0.0 m/s) ² + ½ (4.0 kg)(8.0)² = 128 J

  10. Assignment • Practice Problems Pg. 224 #1 Pg. 226 #2 Pg. 229 # 2 Pg. 233 # 27

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