1983 Exam Question w/SHM and conservation of momentum

[rvhockey]

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 v_o 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 2/3v_o. 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 2/3v_o to solve for the initial KE and setting it equal to .5kx² to get x. Is this right?

 

[Kruum]

Looks right to me. All that's happening is the kinetic energy of the blocks is transforming into the potential energy of the spring.

 

[nlightner]

Yes, you are on the right track. To solve for the maximum distance the spring is compressed, you need to use the conservation of energy principle. Initially, the system has only kinetic energy (KE) due to the initial speed of the second block. After the collision, the system has both kinetic and potential energy (PE) due to the compressed spring. Therefore, the initial KE must be equal to the final KE and PE. This can be represented by the equation:

.5m1v1^2 + .5m2v2^2 = .5kx^2

Substituting in the values of m1, m2, and v2 (which is equal to 2/3vo), we get:

.5(M)(0)^2 + .5(2M)(2/3vo)^2 = .5kx^2

Solving for x, we get:

x = (2/3vo)^2 / k

Therefore, the maximum distance the spring is compressed is dependent on the mass of the block and the initial speed, but is inversely proportional to the spring constant.