SHMMedium
Question
Two block A and B each of mass m are connected by a masses spring of natural length L and spring constant k. The blocks are initially resting 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, moving on the floor with a speed v along the the line joining A and B collides elastically with A. Then -
Options
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 of the spring is 
C.The maximum compression of the spring is v 
D.The maximum compression of the spring is v
Solution
At maximum compression 
& kinetic energy of A - B system will be minimum
so vA = vB =
KAB = 
mv2
From energy conservation
kxm2 ⇒ xm = 
& kinetic energy of A - B system will be minimum so vA = vB =
KAB = mv2From energy conservation
kxm2 ⇒ xm = Create a free account to view solution
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