Two masses M and m are connected by the arrangement shown in figure. What is the downward acceleration of mass M.

(\frac{2 M - m}{3 M + m}) g

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(\frac{2 M - m}{4 M + m}) 2 g

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(\frac{M - m}{M + m}) g

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(\frac{M}{2 M + m}) g

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Solution

The correct option is D $(\frac{2 M - m}{4 M + m}) 2 g$
Let $m_{A} = m_{1}$ and $m_{B} = m_{2}$
Force balance
$m_{2} a = m_{2} g - T$
By tension rule
for a is a by 2
$m_{1} \frac{a}{2} = 2 T - m_{1} g$
From above equations we get
$2 (m_{2} g - m_{2} a) - m_{1} g$
$m_{1} \frac{a}{2} + {2 m}_{2} a = (2 m_{2} - m_{1})$
$\frac{2 m_{2} - m_{1}}{\frac{m_{1}}{2} + 2 m_{2}} g$
On putting the values of $m_{1} = m$ and $m_{2} = M$ we get
$\frac{2 M - m}{4 M + m} 2 g$

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Two masses M and m are connected by the arrangement shown in figure. What is the downward acceleration of mass M.

The correct option is D (2M−m4M+m)2gLet mA=m1 and mB=m2Force balancem2a=m2g−TBy tension rule for a is a by 2 m1a2=2T−m1gFrom above equations we get 2(m2g−m2a)−m1gm1a2+2m2a=(2m2−m1)2m2−m1m12+2m2gOn putting the values of m1=m and m2=M we get2M−m4M+m2g