# Simple Harmonic Motion on an air track

1. May 8, 2010

### CloCon

1. The problem statement, all variables and given/known data

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.

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

3. 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.

Last edited: May 8, 2010
2. May 9, 2010

### diazona

Are you sure about that 0.3J for final kinetic energy? I get something different...
That is the right procedure for finding the amplitude.

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. May 9, 2010

### CloCon

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