Question

Two masses $M_1=5kg$ and $M_2=10kg$ are connected at the ends of an inextensible string passing over a frictionless pulley as shown. When the masses are released, then the acceleration of the masses will be

• A. g
• B. g/2
• C. g/3
• D. g/4

Answer 1

Answer: (C)

Since $M_2>M_1$, therefore $M_2$ moves downwards and $M_1$ moves upwards with an acceleration a as shown in the figure.

Free body diagram of $M_1$

The equation of motion for $M_1$ is
$T-M_{1}g=M_{1}a\dots$(i)
Free body diagram of $M_2$

The equation of motion for $M_2$ is
$M_{2}g-T=M_{2}a\dots \dots (ii)$
Adding (i) and (ii), we get
$a=\frac{\left(M_2-M_1\right)g}{M_2+M_1}$
$=\frac{(10-5)g}{(10+5)}=\frac{g}{3}$