Q.
Two blocks A and B, each of mass m, are connected by a massless spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at ists 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
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a
The kinetic energy of the A-B system, at maximum compression of the spring, is zero.
b
The kinetic energy of A-B system, at maximum compression of the spring, is mv2/4.
c
The maximum compression of the spring is v(m/k)
d
The maximum compression of the spring is v(m/2K)
answer is X.
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Detailed Solution
mv=mv'=mv'⇒v'=v2∴ K.E. of the system in situation 3 is12mv' 2+12mv' 2=mv' 2=mv24 (∵v'=v2)This is the kinetic energy possessed by A-B system (since, C is the rest). Let x be the maximum compression of the spring applying energy conservation12mv2=12mv' 2+12+Kx2⇒12mv2=14mv2+12Kx2⇒12Kx2=14mv2∴x=vm2K
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