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