A body of mass 1kg is suspended from a weightless spring having force constant 600 N/m Another ody of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of 3m/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 600 N/m Another ody of mass 0.5 kg moving vertically upwards hits the suspended body with a velocity of 3m/s and gets embedded in it. Find amplitude of oscillation.
Answer» <html><body><p></p>Solution :By <a href="https://interviewquestions.tuteehub.com/tag/conservation-929810" style="font-weight:bold;" target="_blank" title="Click to know more about CONSERVATION">CONSERVATION</a> of linear momentum in the collision `mv=(m+M)vimpliesV=(mv)/(m+M)=(0.5xx3)/((1+0.5))=1m//s` <br/> Now just after collisions the <a href="https://interviewquestions.tuteehub.com/tag/system-1237255" style="font-weight:bold;" target="_blank" title="Click to know more about SYSTEM">SYSTEM</a> will have `KE=1/2(m+M)V^(2)` at equilibrium position. <br/> So after collision by conservation of mechanical <a href="https://interviewquestions.tuteehub.com/tag/energy-15288" style="font-weight:bold;" target="_blank" title="Click to know more about ENERGY">ENERGY</a> `KE_("<a href="https://interviewquestions.tuteehub.com/tag/max-546895" style="font-weight:bold;" target="_blank" title="Click to know more about MAX">MAX</a>")=PE_("max")` <br/> `1/2(m+M)V^(2)=1/2KA^(2)impliesA=Vsqrt(((m+M)/K))=1sqrt(1.5/600)=1/20m=5cm`</body></html>