Simple Harmonic Motion on an air track

In summary, the problem involves a 0.20-kilogram mass colliding with and sticking to a 1.3-kilogram mass on a horizontal, frictionless air track. The 1.3-kilogram mass is connected to a massless spring with a spring constant of 100 Newtons per meter. The problem asks for the linear momentum and kinetic energy of the masses before and after the collision, as well as the amplitude and period of the resulting harmonic motion. The final values for momentum and kinetic energy are 0.6 kg m/s and 0.12 Joules respectively, and the amplitude is 0.0015 meters. The period can be calculated using the formula T = 2π
  • #1
CloCon
8
0

Homework Statement



A 0.20-kilogram mass is sliding on a horizontal, frictionless air track with a speed of 3.0 meters per second when it instantaneously hits and sticks to a 1.3-kilogram mass initially at rest on the track. The 1.3-kilogram mass is connected to one end of a massless spring, which has a spring constant of 100 Newtons per meter. The other end of the spring is fixed.

Homework Equations


F = -kx
KE = 1/2m(v)squared
momentum = mass x velocity

The Attempt at a Solution



The first couple parts of the problem ask you to solve for the linear momentum and kinetic energy of the masses before and immediately after collision. I interpreted "immediately after the impact" to indicate that I'm supposed to calculate the linear momentum/KE of the masses without taking the spring into account.

a. momentum before impact = 0.6 kg m/s
KE before impact = 0.9 Joules
b. momentum after impact = 0.6 kg m/s
KE after impact = 0.3 J

c. Determine the amplitude of the harmonic motion.
d. Determine the period of the harmonic motion.


I don't really know where to start on parts C and D- I thought to use F=mA to determine the spring force (F=-kX) needed to stop the motion of the mass since the k value is provided (100 n/m), but there's no A value to plug into F=mA. I suspect that KE initial + PE initial = KE final + PE final might factor in because at maximum amplitude PE = 0.3 J, but I don't know where. How might I solve this?

EDIT:

Took another crack at it; am I on the right track?

PE final = 0.3 Joules
PE = (1/2)k(x)squared
0.3 J= (1/2)*(100 N/m)*x squared
x = .0015 m
Would that be the amplitude? If so, how might I find the period?

EDIT 2:

Went Wikipedia hunting and came up with the formula T = 2pi ROOT(m/k). Plugged in all the numbers and came out with .76 seconds. I haven't a clue if that formula applies for a mass on an air track as it applies for a hanging mass, though, so please tell me whether I've got it right or all wrong.

In case you're curious, this problem is from the 1995 Physics AP free response section.
 
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  • #2
Are you sure about that 0.3J for final kinetic energy? I get something different...
CloCon said:
EDIT:

Took another crack at it; am I on the right track?

PE final = 0.3 Joules
PE = (1/2)k(x)squared
0.3 J= (1/2)*(100 N/m)*x squared
x = .0015 m
Would that be the amplitude?
That is the right procedure for finding the amplitude.

CloCon said:
EDIT 2:

Went Wikipedia hunting and came up with the formula T = 2pi ROOT(m/k). Plugged in all the numbers and came out with .76 seconds. I haven't a clue if that formula applies for a mass on an air track as it applies for a hanging mass, though, so please tell me whether I've got it right or all wrong.
That formula applies for any mass attached to a spring, whenever the sum of all other forces on the mass (except for the spring force) is a constant. (Zero is a constant, of course)
 
  • #3
Thanks for the confirmation on the amplitude and period, diazona. I redid the calculation and discovered that I'd neglected to square the velocity to find KE- is this more in line with your calculations?

Momentum final = mV
0.6 kg m/s = 1.5 kg * v
V final = 0.4 m/s

KE final = 1/2 (mass 1 + mass 2) * Vf squared
KE final = 1/2 (1.5 kg) * (0.4 m/s)^2
KE final = .12 Joules
 

1. What is simple harmonic motion on an air track?

Simple harmonic motion on an air track refers to the back-and-forth movement of an object along a straight line, known as oscillation, caused by a restoring force that is proportional to the object's displacement from its equilibrium position. This type of motion is characterized by a constant frequency and amplitude.

2. How does an air track create simple harmonic motion?

An air track consists of a smooth, low-friction surface that is supported by a cushion of air. When an object is placed on the air track and given a push, it will experience minimal friction and continue to move back and forth in a straight line, creating simple harmonic motion.

3. What factors affect the period of simple harmonic motion on an air track?

The period, or time it takes for one complete cycle of oscillation, is affected by the mass of the object, the spring constant of the air track, and the amplitude of the oscillation. Increasing the mass or the spring constant will result in a longer period, while increasing the amplitude will result in a shorter period.

4. How is simple harmonic motion on an air track related to Hooke's Law?

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. This is also true for the restoring force in simple harmonic motion on an air track, where the force is provided by the spring in the track. As the object moves away from its equilibrium position, the spring exerts a force that pulls it back towards the equilibrium position.

5. What are some real-life applications of simple harmonic motion on an air track?

Simple harmonic motion on an air track is commonly used in educational settings to demonstrate concepts such as oscillation, frequency, and amplitude. It is also used in research and development to study the behavior of objects in a controlled, low-friction environment. Additionally, simple harmonic motion is found in many mechanical systems, such as pendulums and springs, and understanding this type of motion is essential in engineering and design.

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