JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Two masses \(A\) and \(B ,\) each of mass \(M\) are fixed together by a massless spring. \(A\) force acts on the mass \(B\) as shown in figure. If the mass \(A\) starts moving away from mass \(B\) with acceleration \('a',\) then the acceleration of mass \(B\) wil be :-
- A \(\frac{ Ma - F }{ M }\)
- B \(\frac{ MF }{ F + Ma }\)
- C \(\frac{ F + Ma }{ M }\)
- D \(\frac{ F - Ma }{ M }\)
Answer & Solution
Correct Answer
(D) \(\frac{ F - Ma }{ M }\)
Step-by-step Solution
Detailed explanation
\(a _{ cm }=\frac{ m _{1} a _{1}+ m _{2} a _{2}}{ m _{1}+ m _{2}}\) \(\frac{ F }{2 M }=\frac{ Ma + Ma _{ B }}{2 M }\) \(a _{ B }=\frac{ F - Ma }{ M }\)
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