Q. Two masses $m_{1}$ and $m_{2\ }(m_{1} > m_{2})$ are connected by massless flexible and inextensible string passed over massless and frictionless pulley. The acceleration of centre of mass is
A) $\left( \frac{m_{1} - m_{2}}{m_{1} + m_{2}} \right)^{2}g$
B) $\frac{m_{1} - m_{2}}{m_{1} + m_{2}}g$
C) $\frac{m_{1} + m_{2}}{m_{1} - m_{2}}g$
D) Zero
Explanation
(a) The free body diagram showing the various forces acting on the pulley mass are as follows
[image:5_image209.jpeg]
Equating the vertical forces, we have
$m_{1}\ g - T = m_{1}a$ …(i)
$T - m_{2}g = m_{2}a$ …(ii)
From Eqs. (i) and (ii), we get
$a = \frac{m_{1}g - m_{2}g}{m_{1} + m_{2}}\ \ \ \ldots.(iii)$
The acceleration of centre of mass is
$a_{CM} = \frac{m_{1}\ a - m_{2}a}{m_{1} + m_{2}}$
Putting the value of $a$ from Eq. (iii), we get
$a_{CM} = \frac{{(m_{1} - m_{2})}^{2}}{{(m_{1} + m_{2})}^{2}}g$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.