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

Two blocks `A` and `H`. each of mass `m`, are connected by a massless spring of natural length `I`. and spring constant `K`. The blocks are initially resting in 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

C

the maximum compression of the spring is `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
In situation i mass `C` is moving towards right with velocity `v, A` and `B` are at rest. In situation ii, which is just after the collision of `C` with `A, C` stops and `A` acquires a velocity `v`. when `A` starts moving towards right, the spring suffers a compression due ot which `B` also starts moving towards right. The compression of the spring continues till there is a relative velocity between `A` and `B`. Once this relative velocity becomes zero, both `A` and `B` move with the same velocity `v'` and the spring is in a state of maximum compression.

Applying momentum conservation in situations ii and iii
`mv=mv'+mv'impliesv'=v/2`
Therefore, `KE` of the system in situation iii is
`1/2mv^('2)+1/2mv^('2)=mv^('2)=(mv^(2))/4`
Applying energy conservation we get
`1/2mv^(2)=1/2mv^('2)+1/2Kx^(2)`
solve to get `x=vsqrt(m/(2K))`
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