4 if m1m2 and v2i 0 massive projec1le v1 f v1i

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Unformatted text preview: 2 2 2 The last equa6on can be wri]en as: m1 ( v12i − v12f ) = m2 ( v2 f − v2 i ) ...........(2 ) The first equa6on can be wri]en as: m1 v1i − v1 f = m2 v2 f − v2 i ...........( 3) ( ) ( ) Dividing (3) by (2) and and transposing the terms we get: v1i − v2 i = − v1 f − v2 f .............( 4 ) ( ) Solving the equa6ons (1) and (2) simultaneously we obtain v1 f = v2 f m1 − m2 2 m2 v1i + v2 i m1 + m2 m1 + m2 2 m1 m − m1 = v1i + 2 v2 i m1 + m2 m1 + m2 Elas6c collisions (contd…) Some special cases: 1. m1 = m2  ­> v1 f = v2 f v1 f = v2 i v2 f = v1i m1 − m2 2 m2 v1i + v2 i m1 + m2 m1 + m2 2 m1 m − m1 = v1i + 2 v2 i m1 + m2 m1 + m2 i.e., the par6cles swap veloci6es. In billiard ball case, one observes that the cue ball stops and the struck ball moves away with the same velocity the cue ball had before the collision. 2. If par1cle 2 is ini1ally at rest (v2i = 0) v1 f = m1 − m2 v1i m1 + m2 v2 f = 2 m1 v1i m1 + m2 3. If m2&...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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