Two Blocks and a Spring
Question:
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.
Answer:
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
=
202.5=75*vB2
vB2=2.7
vB=1.64m/s
vA=2.46m/s
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