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I had done this for the more general case of three different masses and got ##\frac{m_1m_2}{Mm_1+Mm_2+m_1m_2}g##. This reduces to that expression when ##m_1=m_2##. Note the curious fact that my expression is symmetric in ##m_1, m_2##.kuruman said:A lot has been said already, I agree with most but not all of it. When I first solved this problem I went back to basics to answer the original question, "Find the acceleration of the big block immediately after the system is released." To do that, one needs to draw a free body diagram (FBD) and then apply Newton's second law. Here is my FBD with the assumption that immediately after release, the hanging part of the string is vertical and the tension is ##T=mg##. I did not include a normal force exerted by the block on the hanging mass because that force is zero at ##t=0## and remains zero as the block accelerates to the left at ##t>0##.
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This FBD and the ##T=mg## assumption do not result in the purported answer. Based on the above, one can find the horizontal acceleration is one's head. Has anyone obtained the answer ##a=\dfrac{mg}{(2M+m)}##? Because of PF rules, I cannot post my derivation here, but I will be happy to send it to you by PM. However, if you find the error of my ways on the basis of what I have already said, I will welcome your reply here. As some of you already know, I have an occasional blind spot.
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