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Announcements Test corrections due today. Today’s topics – Conservation of momentum

Announcements Test corrections due today. Today’s topics – Conservation of momentum Notion of impulse Analysis of collisions Elastic Inelastic. Linear momentum: p = m v. Generalization for a composite system:. Conservation of momentum:.

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Announcements Test corrections due today. Today’s topics – Conservation of momentum

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  1. Announcements • Test corrections due today. • Today’s topics – • Conservation of momentum • Notion of impulse • Analysis of collisions • Elastic • Inelastic PHY 113 -- Lecture 10

  2. Linear momentum: p = mv Generalization for a composite system: Conservation of momentum: Energy may also be conserved (“elastic” collision) or not conserve (“inelastic” collision) PHY 113 -- Lecture 10

  3. Snapshot of a collision: Pi Impulse: Pf PHY 113 -- Lecture 10

  4. Example from HW: Plot shows model of force exerted on ball hitting a wall as a function of time. Given information about piand pf , you are asked to determine Fmax. PHY 113 -- Lecture 10

  5. Notion of impulse: FDt = Dp Example: • Dp = mvf - mvi • Dp=(-mvf sin qf - mvi sin qi) i • +(-mvf cos qf + mvi cos qi) j m vi qi qf vf -pi pf PHY 113 -- Lecture 10

  6. Notion of impulse: causeeffect • Dp = mvf - mvi • Dp=(-mvf sin qf - mvi sin qi) i • +(-mvf cos qf + mvi cos qi) j Example: m vi qi Dp = pf - pi = F Dt qf vf -pi pf PHY 113 -- Lecture 10

  7. Example: Suppose that you (mg= 500 N) jump down a distance of 1m, landing with stiff legs during a time Dt=0.1s. What is the average force exerted by the floor on your bones? PHY 113 -- Lecture 10

  8. PHY 113 -- Lecture 10

  9. Analysis of before and after a collision: One dimensional case: Conservation of momentum: m1v1i + m2v2i = m1v1f+m2v2f If energy (kinetic) is conserved, then: ½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2+ ½ m2v2f2 Extra credit: Show that PHY 113 -- Lecture 10

  10. More examples in one dimension: vi vf before totally inelastic collision Conservation of momentum: m1vi = (m1+m2) vf Energy lost: PHY 113 -- Lecture 10

  11. Elastic collision in 2-dimensions Statement of conservation of momentum: If mechanical (kinetic) energy is conserved, then:  4 unknowns, 3 equations -- need more information PHY 113 -- Lecture 10

  12. Example: Conservation of momentum Detector measures v2f , f Note: energy is not conserved in this case. PHY 113 -- Lecture 10

  13. Elastic collision v1i q1 q2 Special case: m1 = m2 v1f v2f q1 + q2 = 90o PHY 113 -- Lecture 10

  14. Other examples: h2 Suppose you are given h1, m1, m2 -- find h2 (assume elastic collision) PHY 113 -- Lecture 10

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