Question

In the figure shown below, $m_a=3\text{kg}$ and $m_b=1\text{kg}$. String is inextensible and light. Find the acceleration of centre of mass of the system.

• A. $-\frac{g}{4}\hat{i}{\text{m/s}}^2$
• B. $-\frac{g}{2}\hat{j}{\text{m/s}}^2$
• C. $-\frac{g}{4}\hat{j}{\text{m/s}}^2$
• D. $\frac{g}{6}\hat{j}{\text{m/s}}^2$

Answer 1

The correct option is C $-\frac{g}{4}\hat{j}{\text{m/s}}^2$


From F.B.D of block $a$
$m_{a}g-T=m_{a}a$
$3g-T=3a\dots \dots \left(i\right)$
For block $b$
$T-m_{b}g=m_{b}a$
$T-g=a\dots \dots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$
$2g=4a\Rightarrow a=\frac{g}{2}{\text{m/s}}^2$
$a_a=\text{acc. of}3\text{kg is}\frac{g}{2}$ (downwards)
$=-\frac{g}{2}\hat{j}{\text{m/s}}^2$
$a_b=\text{acc. of}1\text{kg is}\frac{g}{2}$ (upwards)
$=\frac{g}{2}\hat{j}{\text{m/s}}^2$
$a_{\text{com}}=\frac{m_a{a}_a+m_b{a}_b}{m_a+m_b}\Rightarrow \frac{3\times -\frac{g}{2}\hat{j}+1\times \frac{g}{2}\hat{j}}{3+1}$
$a_{\text{com}}=-\frac{g}{4}\hat{j}{\text{m/s}}^2$ (downwards)