Question

Two particles having mass ratio $n:1(n>1)$ are interconnected by a light, inextensible string that passes over a smooth pulley. If the system is released, then the acceleration of the centre of mass of the system is
($g=$acceleration due to gravity)

• A. $(n-1{)}^{2}g$
• B. ${\left(\frac{n+1}{n-1}\right)}^{2}g$
• C. ${\left(\frac{n-1}{n+1}\right)}^{2}g$
• D. $\left(\frac{n+1}{n-1}\right)g$

Answer 1

The correct option is C ${\left(\frac{n-1}{n+1}\right)}^{2}g$

Let the mass of the second particle is $m_2=m$. So, the mass of the first particle will be, $m_1=nm$.

As $m_1>m_2$, so acceleration of $\text{1}$ is downward and $\text{2}$ is upward.


The net pulling force on the system is,

$F=(m_1-m_2)g=(n-1)mg$

The total mass being pulled is,

$m_{total}=nm+m=(n+1)m$

So,

$a=\frac{\text{Net pulling force}}{\text{Total mass}}=\frac{(n-1)mg}{(n+1)m}$

$a=\frac{(n-1)g}{(n+1)}.......\left(1\right)$

Now,

$a_{\text{COM}}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$

For downward motion, assume acceleration is positive and for upward motion, is acceleration negative. i.e, $a_1=a$ and $a_2=-a$

$\therefore a_{\text{COM}}=\frac{\left(mn\right)\left(a\right)-\left(m\right)\left(a\right)}{(n+1)m}=\left(\frac{n-1}{n+1}\right)a$

Using equation $\left(1\right)$, we have

$a_{\text{COM}}={\left(\frac{n-1}{n+1}\right)}^{2}g$

Hence, option $\left(C\right)$ is the correct answer.