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 11/22/2018 PHY 113 -- Lecture 10

Generalization for a composite system: Linear momentum: p = mv Generalization for a composite system: Conservation of momentum: Energy may also be conserved (“elastic” collision) or not conserve (“inelastic” collision) 11/22/2018 PHY 113 -- Lecture 10

Snapshot of a collision: Pi Impulse: Pf 11/22/2018 PHY 113 -- Lecture 10

Example from HW: Plot shows model of force exerted on ball hitting a wall as a function of time. Given information about pi and pf , you are asked to determine Fmax. 11/22/2018 PHY 113 -- Lecture 10

Dp=(-mvf sin qf - mvi sin qi) i +(-mvf cos qf + mvi cos qi) j m vi qi 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 11/22/2018 PHY 113 -- Lecture 10

Dp=(-mvf sin qf - mvi sin qi) i +(-mvf cos qf + mvi cos qi) j Example: Notion of impulse: cause effect 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 11/22/2018 PHY 113 -- Lecture 10

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? 11/22/2018 PHY 113 -- Lecture 10

11/22/2018 PHY 113 -- Lecture 10

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 11/22/2018 PHY 113 -- Lecture 10

More examples in one dimension: vi vf before totally inelastic collision Conservation of momentum: m1vi = (m1+m2) vf Energy lost: 11/22/2018 PHY 113 -- Lecture 10

Elastic collision in 2-dimensions Statement of conservation of momentum: If mechanical (kinetic) energy is conserved, then:  4 unknowns, 3 equations -- need more information 11/22/2018 PHY 113 -- Lecture 10

Conservation of momentum Detector measures v2f , f Example: Conservation of momentum Detector measures v2f , f Note: energy is not conserved in this case. 11/22/2018 PHY 113 -- Lecture 10

Elastic collision v1i q1 q2 Special case: m1 = m2 v1f v2f q1 + q2 = 90o 11/22/2018 PHY 113 -- Lecture 10

Suppose you are given h1, m1, m2 -- find h2 (assume elastic collision) Other examples: h2 Suppose you are given h1, m1, m2 -- find h2 (assume elastic collision) 11/22/2018 PHY 113 -- Lecture 10