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A mass m which is attached to a spring constant k, oscillates on a horizontal table with amplitutde A.At the instant when the spring is stretched by 3^1/2A/2(Aroot3/2),a second mass m is dropped vertically onto the original mass and immediately sticks to it.What is the amplitude of the resulting motion
As there is no net external force acting on the system, the linear momentum will remain conserved.So, mv=2mv'v'=v/2Now, Energy will be conserved 1/2k2=1/2*2*m*(v')2+1/2*k*3A2/4As v=
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