Lecture18-2A

# Lecture18-2A - Today's Lecture Lecture 18 Chapter 11...

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Today Today s Lecture s Lecture Lecture 18: Chapter 11, Impulse, Conservation of Momentum, Inelastic and Elastic Collisions

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I Δ P P f P i F ave I Δ t
Example: Example: Impluse Impluse A 59g tennis ball is thrown up and at the peak of its trajectory it is hit by a racket with a horizontal force given by F = at – bt 2 where t is measured in milliseconds, a = 1200N / ms , and b = 400N / ms 2 . The ball and racket separate after 3ms . Find (a) the impulse and (b) the average impulsive force, and (c) the speed of the ball as it leaves the racket. (a) The impulse is the integral: (b) The average impulsive force: (c) The speed of the ball: I Fdt 0 3 ms at bt 2 dt I 600 N / ms 3 ms 2 400 3 N / ms 2 3 ms 3 5.4 3.6 1.8 kgm / s F I / Δ t 1.8/ 3 10 3 600 N v P / m I / m 1.8/.059 30.5 m / s

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Inelastic Collisions in Two Inelastic Collisions in Two - - Dimensions Dimensions Example: Elementary Particles Example: Elementary Particles Two deuterium nuclei combine to form an a particle. One of the incident nuclei is detected moving at 3.5 Mm/s and the a particle moving at 2.3 Mm/s at an angle q = 21 o to the x axis. Find the initial velocity of the second deuteron. This is a perfectly inelastic collision so only momentum is conserved. Solving for : v 2 m d v 2 m d v 1 m v f v 2 m m d v f v 1 Considering each component we find: v 2 x m m d v fx v 1 x 2 2.3cos21 3.5 .79 Mm / s v 2 y m m d v fy 2 2.3sin21 1.65 Mm / s

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Inelastic Collisions in Two Inelastic Collisions in Two - - Dimensions Dimensions Example: Ballistic Pendulum Example: Ballistic Pendulum If a bullet with mass m strikes the wooden block with mass M , find the initial velocity of the bullet if the block rises to a height h after impact. Initially momentum is conserved so that: mv M m V Conserving energy as the block (and bullet) rise in the gravitational field we have: 1 2 M m V 2 M m gh V 2 gh Solving for v : v M m m 2 gh
Elastic Collisions in One Elastic Collisions in One - - Dimension Dimension For an elastic collisions in one-dimension both energy and momentum are conserved: m 1 v 1 i m 2 v 2 i m 1 v 1 f m 2 v 2 f 1 2 m 1 v 1 i 2 1 2 m 2 v 2 i 2 1 2 m 1 v 1 f 2 1 2 m 2 v 2 f 2 Rearranging and simplifying: m 1 v 1 i v 1 f m 2 v 2 f v 2 i m 1 v 1 i 2 v 1 f 2 m 2 v 2 f 2 v 2 i 2 Now we can divide the second equation by the first and then rearrange: m 1 v 1 i m 2 v 2 i m 2 v 2 f m 1 v 1 f v 1

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Lecture18-2A - Today's Lecture Lecture 18 Chapter 11...

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