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.

Explanation

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`