Question

Find the acceleration of mass $m_1$ and $m_2$ in the arrangement shown in figure, if mass $m_2$ is twice that of mass $m_1$, and the angle that the inclined plane forms with the horizontal is equal to ${30}^{\circ }$. Mass of the pulley and string are negligible and friction is absent everywhere.

• A. $(a{)}_{m_2}=\frac{2g}{18}$
• B. $(a{)}_{m_2}=\frac{7g}{9}$
• C. $(a{)}_{m_1}=\frac{7g}{18}$
• D. $(a{)}_{m_1}=\frac{7g}{9}$

Answer 1

Answer: (B), (C)

$(a{)}_{m_1}=\frac{7g}{18}$

Here pulley $1$ is movable and if it moves by $x$ downwards, then total length of string slackened is $2x$.
However, one end of the string is fixed so in order to maintain string length constant, block $m_2$ will move downwards by $2x$ and block $m_1$ will move $x$ upwards along the incline.

Differentiating the displacement twice w.r.t time:
$a_1=a$ for $m_1$, and $a_2=2a$ for $m_2$ respectively.


Applying Newton's $2^{\text{nd}}$ law for $m_1$ and $m_2$ along the direction of acceleration:
$2T-m_{1}g\sin \theta =m_{1}a...\left(i\right)$
$m_{2}g-T=m_2\left(2a\right)...\left(ii\right)$

Here $m_2=2m_1,\theta ={30}^{\circ }...\left(iii\right)$

Now, adding $\left(ii\right)\times 2+\left(i\right)$,
$\Rightarrow 2m_{2}g-m_{1}g\sin {30}^{\circ }=a(m_1+4m_2)$
$\Rightarrow 4m_{1}g-\frac{m_{1}g}{2}=a(m_1+8m_1)$
$[\because m_2=2m_1]$
$\Rightarrow \frac{7m_{1}g}{2}=a\left(9m_1\right)$
$\therefore a=\frac{7g}{18}$
Therefore,
Acceleration of $m_1=a=\frac{7g}{18}$
Acceleration of $m_2=2a=2\times \frac{7g}{18}=\frac{7g}{9}$