For elas6c collisions we have an addi1onal equa1on

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Unformatted text preview: gt;>m1 and v2i = 0 (massive target at rest) v1 f ≈ − m2 v1i = − v1i m2 v2 f ≈ 2 m1 v1i m2 i.e., light objects bounces off, heavy target moves very slowly. 4. If m1>>m2 and v2i = 0 (massive projec1le) v1 f ≈ v1i , v2 f ≈ 2 v1i i.e., projec6le just ‘plows through’, light target bounces off at twice the speed of the projec6le. Example: Ballis6c pendulum How to measure the speed of a bullet (m=9.5g)? Answer: Shoot bullet into a suspended block (M=5.4 kg), in which the bullet becomes lodged. If the block is ini;ally at rest and swings up a ver;cal distance of 6.3 cm when struck by the bullet, how fast was the bullet going ? Energy conserva6on (aber bullet strikes block): ΔU = ( m + M ) gh = 1 ( m + M )v 2 , 2 − ΔK = f 2 ⇒ v f , 2 = 2 gh = 1.1m/s Momentum conserva6on: Pi = Pf pm, i + p M , i = pm, f + p M , f → mvi ,1 + M ⋅ 0 = ( m + M )v f ,1 ⇒ vi ,1 = m+ M 5.4kg v f ,1 = (1.1m/s) = 632m/s = 1413mph m 9.5 × 10-3k...
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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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