A body of mass 1.0kg 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 of 3.0 m//"sec" and gets embedded in it. Find the amplitude of oscillation

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

1.	A body of mass 1.0kg 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 of 3.0 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` `implies V= (mv)/(m+M)= (0.5 xx 3)/((1+0.5))= 1m//"sec"` Now just after collision the SYSTEM will have `KE= (1)/(2)(m+M)V^(2)` at equilibrium POSITION. So after collosion by conservation of mechanical energy `KE_("max")= PE_("max")` `(1)/(2)(m+M)v^(2)= (1)/(2)KA^(2)` `implies A= Vsqrt((m+M)/(k))= 1sqrt((1.5)/(600))= (1)/(20)m= 5cm`