A mass of 500 g is resting on a vertical spring with a force constant of 55.0 N/m. A mass of 250 g is dropped from a height of 12.0 cm onto the larger mass and sticks to it. What is the amplitude of the resulting oscillations?
A) 4.71 cm
B) 6.34 cm
C) 9.90 cm
D) 11.4 cm
E) 5.97 cm

Question

Quizplus · Accepted Answer

First, we need to find the initial potential energy of the 250 g mass before it is dropped onto the 500 g mass:

U = mgh = (0.25 kg)(9.81 m/s^2)(0.12 m) = 0.2943 J

When the 250 g mass is dropped onto the 500 g mass, their total mass becomes 750 g. The spring is compressed, and their potential energy is converted to spring potential energy:

0.2943 J = (1/2)(0.055 kg)(x^2)

where x is the amplitude of the resulting oscillations.

Solving for x, we get:

x = 0.0597 m = 5.97 cm

Therefore, the answer is E.

A Mass of 500 G Is Resting on a Vertical