Question

Weighing Lunch

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100 kg of turkey. The slices of turkey are weighed on a plate of mass 0.400 kgplaced atop a vertical spring of negligible mass and force constant of 200 N/m . The slices of turkey are dropped on the plate all at the same time from a height of 0.250 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.

Part A

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 m/s2 .

A = ?? m

Part B

What is the period of oscillation T of the scale?

Express your answer numerically in seconds.

T = ?? s

Answer 1

a)

here

The scale spring had already been compressed before the meat was dropped on it. Its initial compression was such that F = mg = kx, or
0.4kg * 9.8m/s² = 200N/m * x
x = 0.0196 m
U = ½kx² = ½ * 200N/m * (0.0196m)² = 0.04 J
The new equilibrium point is lower after the meat is added:
0.1kg * 9.8m/s² = 200N/m * x
x = 0.0049 m
Let's pick the point of maximum compression of the scale as our reference point.
initial E = PE + U = mg(0.25m + 0.0049m + A) + 0.04J
PE = 0.1kg * 9.8m/s² * (0.2549m + A) + 0.04J = 0.2898 + 0.98A
where A is the amplitude of motion.

At maximum compression,
E = U = ½k(0.0196m + 0.0049m + A)² = ½ * 200N/m * (0.0245 + A)²
E = U = 100 * (0.0245² + 0.049A + A²) = 0.06 + 4.9A + 100A²

These two energy levels are equal:
0.2898 + 0.98A = 0.06 + 4.9A + 100A²
100A² + 3.92A - 0.2298 = 0
A = -0.07 m, 0.032 m ← take positive root as amplitude

b)

T = 2π/ω = 2π√(m/k) = 2π√(0.5kg / 200kg/s²) = 0.31 s