Melvin H. answered • 10/07/14
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When the two blocks collide momentum is conserved. But because the two blocks stick together, some initial kinetic energy is lost (to energy needed to bind the blocks together). One can write for the conservation of momentum
mv = (m + M)V where m = 2 kg, v = 5 m/s and M = 6 kg and V is the velocity of the "stuck-together" block.
Solving for V gives V = 1.25 m/s
[Comment: Conservation of energy would require KEi = KEf + BE (the "binding energy"). We could find the value of BE if we wanted, but we don't need to for this problem.]
The (remaining) kinetic energy of the combined block is gradually shifted to the compressing spring until the block comes momentarily to rest. We can (again) use the conservation of energy to write
½(m + M)V2 = ½kA2 where A is the amplitude of the spring's subsequent motion, which is also its maximum compression in this problem.
a) One can now solve for A. (I got A = 0.289 meters)
b) On can now solve for the frequency f =(1/2π) (√(k/(m+M)) (I got f = 0.689 osc/sec)
c) The period T = 1/f (I got T = 1.45 seconds)
Janine T.
Interesting. I figured that there were two masses and the question was asking about the question after the collision, so I added them together. Using the equations Bryon provided, I got the same for the amplitude and the values were reversed for "b" and "c". What value did you use for "n" in part B? Thank you.
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10/07/14
Byron S.
10/07/14