Conservation of Linear Momentum (spring)

[songsteel]

Homework Statement

A mass 3M moving horizontally with velocity v_{0} on a frictionless surface, strikes head-on and sticks to a horizontal spring system of natural length l and spring constant k with masses M at each end. The spring has negligible mass.

Determine the maximum compression of the spring.

Homework Equations

Conservation of Linear Momentum:
Initial Linear Momentum = Final Linear Momentum
Conservation of Energy:
Initial Kinetic Energy + Initial Elastic Potential Energy = Final Kinetic Energy + Final Elastic Potential Energy

The Attempt at a Solution

3Mv_{0} = (3M + M)Vfinal
Vfinal = \frac{3}{4}v_{0}

At Maximum compression: all the Kinetic Energy would have been converted to Elastic Potential Energy; let x be the maximum compression.

\frac{1}{2}(4M)(\frac{3}{4}v_{0})^{2} = \frac{1}{2}(k)(x)^{2}

\frac{9}{8}Mv_{0}^{2} = \frac{kx^{2}}{2}

x = √(\frac{9Mv_{0}^{2}}{4k})

According to the answersheet I was given, the final answer should be x = √(\frac{9Mv_{0}^{2}}{20k}) instead.

Would really appreciate it if anyone could help me out. Thanks!

 

[voko]

Your assumption that at max compression KE = 0 has no basis. In fact, it is not compatible with conservation of momentum.

 

[songsteel]

I see, would it be correct to write

\frac{1}{2}(4M)(\frac{3}{4}v_{0})^{2} = \frac{1}{2}(k)(x)^{2} + \frac{1}{2}(5M)(V1^{2})

If it is then what method should I use to find v1?

 

[voko]

The latter equation is meaningless, because 5M is the mass of the entire system, but V1, even if it was the velocity of the center of mass, is not the only contribution to kinetic energy.

Perhaps the easiest approach here would to use the center of mass frame. Find its velocity. Find the kinetic energy of the center of mass. Then, the internal energy in the center of mass frame is the original energy minus the center of mass energy. In the center of mass frame, the system undergoes simple harmonic motion, whose energy is known.

 

[songsteel]

I think I got it. Thanks!