CP A 10.0 .0 -kg mass is traveling to the right with a speed of 2.00 m / s on a smooth horizonta

step 1 So, here we can apply the conservation of linear momentum. Linear momentum to the collision and

CP A 10.0 .0 -kg mass is traveling to the right with a speed...

CP A 10.0 .0 -kg mass is traveling to the right with a speed of 2.00 $\mathrm{m} / \mathrm{s}$ on a smooth horizontal surface when it collides with and sticks to a second 10.0 -kg mass that is initially at rest but is attached to a light spring with force constant 110.0 $\mathrm{N} / \mathrm{m}$ . (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

CP A 10.0 .0 -kg mass is traveling to the right with a speed of 2.00 $\mathrm{m} / \mathrm{s}$ on a smooth horizontal surface when it collides with and sticks to a second 10.0 -kg mass that is initially at rest but is attached to a light spring with force constant 110.0 $\mathrm{N} / \mathrm{m}$ . (a) Find
the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

Key Concepts

Conservation of Momentum

In a collision process, especially when external forces are negligible during the brief collision interval, the total linear momentum of the system is conserved. This principle allows us to compute the common velocity of two objects after they stick together in an inelastic collision by equating the total momentum before the collision to the total momentum after.

Inelastic Collisions

An inelastic collision is one in which the colliding objects stick together after impact, resulting in a loss of mechanical energy. However, while kinetic energy is not conserved, momentum remains conserved. Such collisions are significant in problems that combine collisions with other mechanics, such as oscillations.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) describes the oscillatory behavior of systems in which the restoring force is directly proportional to the displacement from equilibrium. After the collision, when the combined mass interacts with a spring, the system's motion can be modeled as SHM, allowing the use of formulas for amplitude, frequency, and period.

Angular Frequency in Mass-Spring Systems

For a mass-spring system undergoing simple harmonic motion, the angular frequency is defined as ? = sqrt(k/m), where k is the spring constant and m is the total mass attached to the spring. Understanding this relationship is crucial for determining the oscillation frequency and period, as it connects the physical properties of the system to its dynamic behavior.

Transcript

00:01 So, here we can apply the conservation of linear momentum.

00:10 Linear momentum to the collision and the conservation of energy to the motion after collision.

00:16 So, conservation of momentum is there to the collision and conservation of energy of energy after collision.

00:37 This is the key thing that we have to see.

00:40 Now, frequency we know equal to 1 upon 2 .5 root of.

00:45 Of k upon m and time period equal to one upon frequency.

00:53 Now according to the question, the object returns to the equilibrium position in time t by two.

01:04 In time t by two object returns to equilibrium.

01:16 So in the first part, the a part, the momentum conservation during collision momentum conservation during pollution we have to see this m v0 equal to 2m into v and capital v equal to half b0 that is half into 2 .00 meter per second so that is v equal to 1 .00 meter per second.

02:13 Now energy conservation after collision you have to see energy conservation after pollution that is half mv square equal to minus half k x square from here we can find out x that is equal to mv square upon k that is equal to root of 20 .0 kg into 1 .00 meter per second the whole square from where did we get? we got it from here.

03:02 Okay? now, this is in root itself upon 1 .10 .0 newton per meter.

03:13 So when we calculate, we are going to get the answer to be 0 .4 to 6 meter and this is our amplitude.

03:29 Now, omega or the angular frequency is equals to 2 pi f, okay? that is k by m because f equal to 1 by 2 pi root of k by m.

03:51 Now putting the value, what we are going to get, f equal to 1 by 2 pi root of 1 .10 .0 newton upon meter upon 20 .0 kg that is equal to 0 .373 hertz.

04:18 This is 0 .0 .0 .373 hertz...

Please give Ace some feedback