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Two blocks A and B, each of mass m, are ...

Two blocks A and B, each of mass m, are connected by a masslesss spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in fig. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then

A

the kinetic energy of the `A-B` system, at maximum compression of the spring, is zero

B

the kinetic energy of the `A-B` system, at maximum compression of the spring, is `(m v^(2))/(4)`

C

the maximum compression of the spring is `v sqrt(((m)/(K)))`

D

the maximum compression of the spring is `vsqrt((m)/(2K))`

Text Solution

Verified by Experts

The correct Answer is:
B, D

After collision between `C` and `A, C` stops while `A` moves with speed of `C` i.e., `v` [in head on elastic collision, two equal masses exchange their velocities]. At maximum compression, `A` and `B` will move with same speed `(v)/(2)` (From conservation of linear momentum).

Let `x` be the maximum compression in this position.
`KE` of `A-B` system at maximum compression
`= (1)/(2) (2m) ((v)/(2))^(2)` or `K_("max") = (m v^(2))/(4)`
From conservation of mechanical energy in two position shown in figure.
or `(1)/(2)m v^(2) = (1)/(4) m v^(2) + (1)/(2) Kx^(2)`
`(1)/(2)Kx^(2) = (1)/(4)m v^(2), x = v sqrt((m)/(2K))`
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