Two Blocks and a Spring


Consider two blocks, A and B, of mass 40 and 60 kg respectively, connected by a spring with spring constant 160 N/m. The blocks are placed on a smooth table with the spring between them compressed 1.5 meters and released with no initial velocity. Determine the speed of each block at the instant the spring reaches its relaxed length.


The potential energy stored in the spring at the time of the release of the blocks is 1/2*k*x2 = 1/2*180*2.25 = 202.5 Joules. This energy will be converted to kinetic energy of the blocks at the time the spring is relaxed, to the total kinetic energy of the blocks at that time is 202.5 Joules. So 202.5=1/2*40*vA2+1/2*60*vB2. The total momentum of the system must remain zero so the velocity of each block will be inversely proportional to its mass. So vA=60/40*vB.

Substituting for vA in the energy equation gives us:
202.5=20*(60/40*vB)2+30*vB2 =
202.5=20*9/4*vB2+30*vB2 =

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