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

Explanation

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`