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A system is composed of 2 blocks of mass m1,m2 connected by a massless spring with  spring constant k.The blocks slide on a frictionless plane.The unstretched length of spring is l.Initially m2 is held so that the spring is compressed to l/2 and m1 is forced against a stop.M2 is released at t=0.Find the motion of the center of mass of the system as a function of time.

11 years ago

till spring assumes its stress free length, m2 executes SHM and m1 is stationary. the angular frequency is ω=√(k/m2),

the equation of motion of m2 in the SHM is x=l-(l/2)cos(ωt)

centre of mass=m2l(1-(cos(ωt)/2))/(m1+m2) for 0≤t≤(π/2)√(m2/k)

after that kinetic energy of centre of mass =kl2/8=(m1+m2)v2/2, v=(l/2)√(k/(m1+m2))

centre of mass =l+(t-(π/2)√(m2/k))(l/2)√(k/(m1+m2)) for t>(π/2)√(m2/k)

3 years ago
since, initially m1 is rest untill the spring come to it original length l
at this point of time the system all potential energy is goes to the kinetic energy of the m2.
from which we can calculate the speed of m2 which is v=(k/4m)^1/2*l
their is no externel force will acting on the system .
therefore system total momentum is conserve
which is p=m2*v=m2*(k/4m)^1/2*l.
therefore velocity of CM Vcm=(m2*(k/4m)^1/2*l)/(m1+m2).
therefore position  of CM at any time is =((m2*(k/4m)^1/2*l)/(m1+m2))*t +(m2*l/2)/(m1+m2) wrt to wall