Work, Power and EnergyMedium

Question

Two blocks A and B, of masses M and 2M, respectively, are connected by a massless spring of natural length L0 and spring constant K. The blocks are initially at rest on a smooth horizontal floor with spring at its natural length L0. A third block C of mass M, identical to that of block A, moves on the floor with speed v along the line joining A and B and collides with A elastically. In the subsequent motion :-

Options

A.the spring will be compressed to a maximum when at a length of v√(M/(3K))
B.the kinetic energy of A and B together, when the spring is compressed to the maximum, is Mv2/6
C.the blocks A and B stop for a moment when the spring is at the maximum compression.
D.the time required to reach the maximum compression from the normal length is (π/2)√(2M/(3K))

Solution

[Key Concept: At maximum compression the two blocks share the centre-of-mass velocity; the relative motion is SHM with the reduced mass.]

Step 1: Equal-mass elastic collision: C stops and A moves with v.

Step 2: At maximum compression A and B move together with v/3, so KE = (1/2)(3M)(v/3)2 = Mv2/6. Option B true; Option C false.

Step 3: (1/2)Mv2 − Mv2/6 = (1/2)Kx2 gives xmax = v√(2M/(3K)), not v√(M/(3K)), and this is a compression, not a length. Option A false.

Step 4: Reduced mass μ = M(2M)/(3M) = 2M/3; time = T/4 = (π/2)√(2M/(3K)). Option D true.

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