# Mastering Physics: 'weighing lunch': spring oscillation

1. Jun 7, 2010

### louza8

1. The problem statement, all variables and given/known data

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100 {\rm kg}of turkey. The slices of turkey are weighed on a plate of mass 0.400 {\rm kg} placed atop a vertical spring of negligible mass and force constant of 200 {\rm N/m}. The slices of turkey are dropped on the plate all at the same time from a height of 0.250 {\rm m}. They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.

What is the amplitude of oscillations A of the scale after the slices of turkey land on the plate?

Express your answer numerically in meters and take free-fall acceleration to be g = 9.80{\rm m/s}^2.

2. Relevant equations

F=-kx
deltaW=0.5kx^2
deltaW=0.5mv^2
mv+mv=(m+m)vf (inelastic)

3. The attempt at a solution

m=0.1kg
m2=0.4k
k=200N/m
h=0.25

Initial equilibrium with 0.4kg plate:

ma=kx
(0.4*9.81)/200 = x
x=0.0196m

New equilibrium position with turkey

Need velocity of turkey+plate

v^2=u^2 + 2ash

=sqrt(2*9.81*0.25)
v=2.21472m/s

momentum after inelastic collision

mv=(m+m)v_f

(0.1*2.215)/0.5=0.442m/s

equate energies

0.5(m+m)v^2 + 0.5kx^2 = 0.5kA^2

sqrt([0.5(0.5)0.442^2 + 0.5(200)(0.0196^2)]/100)

A=0.029462m

so this says kinetic energy of turkey+plate + energy in initial displacement = final energy

answer is 0.027 i'm not sure where I stuffed up, i think in the initial displacement maybe?

no turkies were harmed in writing this question

2. Jun 8, 2010

### Frostfire

With out going through it step by step, it looks right and your answer is close enough to reason that, try carrying the values to an extra digit to remove possible rounding errors.