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borishaifa
3 months ago
7

For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia

n ham. The slices of ham are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m . The slices of ham 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.What is the amplitude of oscillation A of the scale after the slices of ham land on the plate?Express your answer numerically in meters and take free-fall acceleration to be g= 9.80m/s^2What is the period of oscillation T of the scale?Express your answer numerically in seconds.
Physics
1 answer:
serg [3.5K]3 months ago
8 0

Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

A slice is dropped onto the plate from a height of 0.250 m,

therefore the speed of the slice upon impact is calculated as

v = \sqrt{2gh}

We know that

v = \sqrt{2(9.81)(0.250)}

v = 2.21 m/s

Now applying the conservation of momentum:

mv = (m + M)v_f

m = 0.300 kg

M = 0.400 kg

From this equation, we find:

0.300 (2.21) = (0.300 + 0.400) v_f

v_f = 0.95 m/s

0.400 (9.81) = 200 x_1

When the slice rests on the plate, the new mean position can be expressed as

x_1 = 0.01962 m

(0.300 + 0.400)9.81 = 200 x_2

We also determine that the speed of SHM is represented as

x_2 = 0.0343 m

Here, we derive values from

v = \omega\sqrt{A^2 - x^2}

\omega = \sqrt{\frac{k}{m + M}}

\omega = \sqrt{\frac{200}{0.300 + 0.400}}

\omega = 16.9 rad/s

a = x_2 - x_1 = 0.0343 - 0.01962 = 0.0147 m

Using the previous formula gives:

0.95 = 16.9\sqrt{A^2 - 0.0147^2}

A = 0.0581 m

Part b)

The time period for the scale is computed as

T = \frac{2\pi}{\omega}

T = \frac{2\pi}{16.9}

T = 0.37 s

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