Mastering Physics: 'weighing lunch': spring oscillation

Click For Summary
SUMMARY

The discussion centers on calculating the amplitude of oscillations (A) of a vertical spring system after an inelastic collision involving 0.100 kg of turkey and a 0.400 kg plate. The spring has a force constant of 200 N/m, and the turkey is dropped from a height of 0.250 m. The calculated amplitude of oscillation is approximately 0.029462 m, with the final answer being 0.027 m. The solution involves applying principles of momentum conservation and energy equations relevant to simple harmonic motion (SHM).

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with inelastic collisions and momentum conservation
  • Knowledge of energy conservation principles in mechanical systems
  • Basic proficiency in algebra and physics equations
NEXT STEPS
  • Review the principles of energy conservation in oscillatory systems
  • Study the derivation of the amplitude in simple harmonic motion
  • Learn about the effects of mass and spring constant on oscillation frequency
  • Explore advanced topics in dynamics involving collisions and energy transfer
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators and tutors looking for practical examples of SHM and collision problems.

louza8
Messages
58
Reaction score
0

Homework Statement



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.

Homework Equations



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

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
 
Physics news on Phys.org
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.
 

Similar threads

Replies
11
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
3
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
Replies
1
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 17 ·
Replies
17
Views
4K