1983 Exam Question w/SHM and conservation of momentum

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SUMMARY

The discussion focuses on a physics problem involving conservation of momentum and simple harmonic motion (SHM) with two blocks of mass M and 2M. The speed of the blocks immediately after impact is calculated using the conservation of momentum, yielding a result of 2/3Vo. The period of the subsequent SHM is derived using the formula T = 2π√(m/k), where the combined mass is 3M. The main challenge discussed is determining the maximum compression of the spring, which involves equating the initial kinetic energy of the blocks to the potential energy stored in the spring.

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rvhockey
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A block of mass M is resting on a horizontal, frictionless table and is attached to a relaxed spring of spring constant k. A second block of mass 2M and initial speed vo collides with and sticks to the first block. Develop expressions for the following quantities in terms of M, k, and Vo.

a) v, the speed of the blocks immediately after impact
b) x, the maximum distance the spring is compressed
c) T, the period of the subsequent simple harmonic motion




m1v1+m2v2=m1v1'+m2v2'
EPE=.5kx2
KE = .5mv2
T = 2pi*sqrt(m/k)




I'm only having trouble with part b. I knew to use conservation of momentum for a and get 2/3vo. c was easy as you could just use the last equation and plug in 3M and k. But for b I'm not sure if I'm doing it right. What I'm doing it plugging in 3M and 2/3vo to solve for the initial KE and setting it equal to .5kx2 to get x. Is this right?
 
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Looks right to me. All that's happening is the kinetic energy of the blocks is transforming into the potential energy of the spring.
 

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