A body of mass 1kg is suspended from a weightless spring having force constant 600 N/m Another ody of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of 3m/s and gets embedded in it. Find amplitude of oscillation.

A body of mass 1kg is suspended from a weightless spring having force

1.	A body of mass 1kg is suspended from a weightless spring having force constant 600 N/m Another ody of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of 3m/s and gets embedded in it. Find amplitude of oscillation.
Answer» Solution :By CONSERVATION of linear momentum in the collision `mv=(m+M)vimpliesV=(mv)/(m+M)=(0.5xx3)/((1+0.5))=1m//s` Now just after collisions the SYSTEM will have `KE=1/2(m+M)V^(2)` at equilibrium position. So after collision by conservation of mechanical ENERGY `KE_("MAX")=PE_("max")` `1/2(m+M)V^(2)=1/2KA^(2)impliesA=Vsqrt(((m+M)/K))=1sqrt(1.5/600)=1/20m=5cm`