Question

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 Italian 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.

Answer 1

Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

Explanation:

Slice is dropped on the plate from height 0.250 m

so the speed of the slice while it hit the plate is given as

$v = \sqrt{2gh}$

here we know that

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

$v = 2.21 m/s$

now from momentum conservation

$mv = (m + M)v_f$

$m = 0.300 kg$

$M = 0.400 kg$

from the above equation we have

$0.300 (2.21) = (0.300 + 0.400) v_f$

$v_f = 0.95 m/s$

Now the initial mean position while it will not hit the plate is given as

$0.400 (9.81) = 200 x_1$

$x_1 = 0.01962 m$

now when slice is placed on the plate then the new mean position will be given as

$(0.300 + 0.400)9.81 = 200 x_2$

$x_2 = 0.0343 m$

now we know that speed of SHM is given as

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

here we have

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

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

$\omega = 16.9 rad/s$

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

now from above formula

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

$A = 0.0581 m$

Part b)

Time period of scale is given as

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

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

$T = 0.37 s$