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Two blocks of masses m_(1) = 2 kg and m_(2) = 5 kg are moving in the same direction along a frictionless surface with speeds 10 m//s and 3 m//s, respectively, m_(2) being ahead of m_(1). An ideal spring with k = 1120 N//m is attached to the back side of m_(2). Find the maximum compression of the spring when the blocks collide. What are the final velocities of the blocks when they separate?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :Form conservation of momentum <br/> `v=(m_(1)v_(1)+m_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)v_(2))/(m_(1)+m_(2))=5m//s` <br/> conservationof energy <br/> `1/2m_(1)v_(1)^(2)+1/2m_(2)v_(2)^(2)=1/2(m_(1)+m_(2))v^(2)+1/2kx^(2)` <br/> `implies kx^(2)=(m_(1)v_(1)^(2)+m_(2)v_(2)^(2)-(m_(1)-m_(2))v^(2))/k` <br/> `implies x=sqrt(40/1120)=0.25m` <br/> when the blocks have equa <a href="https://interviewquestions.tuteehub.com/tag/speeds-650188" style="font-weight:bold;" target="_blank" title="Click to know more about SPEEDS">SPEEDS</a>, spring has maximum commpression. after his instasnt,the spring againexpands, and after sometimes `m_(1)` loses contact with the spring. <br/> Let `v_(1)` and `v_(2)` be the velocities of the blocks after they lose contact. <br/> `m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)u_(2)` <br/> and `v_(1)-v_(2)=-1(u_(1)-u_(2))` <br/> `[e=1,` because there is no loss in `KE`] <br/> Solving for `v_(1)` and `v_(2)` we get <br/> `v_(1)=0m//s` and `v_(2)=7m//s`</body></html>


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