A body of mass 1.0 kg is suspended from a weightless spring having force constant 600 N/m. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of3.0 m/sec and gets embedded in it. Find the amplitude of oscillation.

A body of mass 1.0 kg is suspended from a weightless spring having for

1.	A body of mass 1.0 kg is suspended from a weightless spring having force constant 600 N/m. Another body of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of3.0 m/sec and gets embedded in it. Find the amplitude of oscillation.
Answer» SOLUTION :By conservation of LINEAR MOMENTUM in the collision `mv=(m+M)V` `impliesV=(mv)/(m+M)=(0.5xx3)/((1+0.5))=1` m/sec Now just after collision the system will have `KE=1/2(m+M)V^(2)` at equiibrium 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`