A horizontal spring with a spring constant of 22 N/m has a 600 gram block attached to it and is at rest on a frictionless surface. A second block which has a mass of 220 grams is pushed toward the 600 gram block at a speed of 1.50 m/s. The second block collides with and sticks to the 600 gram block.
a. What is the amplitude of the subsequent oscillation?
b. What is the period of the subsequent oscillation?
To solve this problem, we'll need to consider conservation of energy and Hooke's Law. Here's how to find the answers:
a. Amplitude of the subsequent oscillation:
We can start by finding the total initial mechanical energy of the system before the collision. Since the surfaces are frictionless, the only form of energy is the kinetic energy of the second block. Using the formula for kinetic energy: KE = 1/2 * mass * velocity^2, we can calculate the initial kinetic energy of the second block.
KE_initial = 1/2 * (0.220 kg) * (1.50 m/s)^2
Next, the total mechanical energy of the system after the collision is the potential energy stored in the spring due to its compression. According to Hooke's Law, the potential energy of a compressed spring is given by: PE = 1/2 * k * x^2, where k is the spring constant and x is the displacement from the equilibrium position.
Since the system is at rest after the collision, all of the initial kinetic energy is transformed into potential energy, so we can equate the two:
KE_initial = PE
Solving for x, we get:
x = sqrt((2 * KE_initial) / k)
Substituting the values we know, we can calculate the amplitude of the subsequent oscillation.
b. Period of the subsequent oscillation:
The period (T) of an oscillation is the time it takes for one complete cycle. It depends on the mass (m) of the block and the spring constant (k).
The formula for the period is:
T = 2π * sqrt(m / k)
Using this formula, we can calculate the period of the subsequent oscillation.
Please provide the known values so that I can help you with the calculations.