Question 199: Laws of Motion

Question

Two identical blocks each of mass ' M ' attached to the ends of a massless inextensible string which passes over a pulley with a fixed axis as shown below. A small mass ' $m$ ' is now placed on the block B. The acceleration with which the two blocks move together is
[g = gravitational acceleration]

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& (\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{T}=(\mathrm{M}+\mathrm{m}) \mathrm{a} \
& \mathrm{~T}-\mathrm{Mg}=\mathrm{Ma}
\end{aligned}\
&	ext { Adding the above equations, }\
&\begin{aligned}
& (M+m) g-M g=(2 M+m) a \
	herefore \quad & a=\frac{m g}{(2 M+m)}
\end{aligned}
\end{aligned}$$

Answer

$\frac{m g}{2 M+m}$

Answer

$\frac{M g}{M+2 m}$

Answer

$\frac{\mathrm{Mg}}{2 \mathrm{M}+\mathrm{m}}$

Answer

$\frac{m g}{M+2 m}$

Laws of Motion

Two identical blocks each of mass ' M ' attached to the ends of a massless inextensible string which passes over a pulley with a fixed axis as shown below. A small mass ' m ' is now placed on the block B. The acceleration with which the two blocks move together is [g = gravitational acceleration]

Choose the correct answer: