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sample2_extra_exam_soln

# sample2_extra_exam_soln - Solutions to sample exam 2 TAM...

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Solutions to sample exam 2 TAM 212, Fall, 2002 Solution for sample problem #1 1 : The change in kinetic energy of the plunger = work done by external forces: Work E K = . . ( ) ( ) 2 2 2 / 1 2 / 1 . . initial plunger final plunger v m v m E K = 0 = initial v , so ( ) ( ) ( ) 2 2 / 32 / 5 2 / 1 . . v s ft lb E K = , where v is its speed just before it hits the button. The external forces include that due to gravity and that due to the spring. ( )( ) lb in lb in h mg Work down gravity = = = 75 . 8 75 . 1 5 ( ) ( ) 2 2 2 / 1 final initial sping k Work δ δ = To find the δ ’s we need to solve for it in the equilibrium position: ( ) in lb in lb F F ma F gravity spring y y 5 . 0 5 / 10 0 = δ = δ = + = so ( )( ) ( ) ( ) ( ) lb in in in in lb Work sping = == 1875 . 2 75 . 0 1 / 10 2 / 1 2 2 ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 / 667 . 11 12 / 1 / 140 1875 . 2 75 . 8 / / 078125 . s ft in ft x s ft in v lb in v s ft lb = = + = s ft v / 42 . 3 = (down) 2 Let m be the mass of the plunger and M be the mass of the button / base combination. During the collision, external forces are negligible, so conservation of momentum gives: button plunger plunger Mv mv Mx mv , 2 , 2

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sample2_extra_exam_soln - Solutions to sample exam 2 TAM...

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