Question

In the figure mass $m_1=4$ times mass of $m_2$. The pulleys are smooth and light and thread is also light. At time $t=0$, the system is at rest in the positions of the masses as in the figure. If the system is released, find the maximum height reached by the mass $m_2$. $(g=10m{s}^{-2})$.

Answer 1

Given,

$m_1=4m_2$

Let acceleration of system of masses is a.

Here , $m_1$ is greater than $m_2$ then, acceleration of mass $m_1$ in downward and acceleration of mass $m_2$ is upward.

when the body of mass $m_1$goes upward for distance '$x$', it will free the string for mass '$m_1$' upto length '$2x$'.

Therefore let acceleration of body $m_1=a$

So, acceleration of body of mass $m_2=2a$

Writing force balance equation for each mass we get,

$m_{1}g-2T=m_{1}a$........(i)

$T-m_{2}g=2m_{2}a$..........(ii)

Multiplying equation (ii) by 2 and adding with equation (i) we get,

$m_{1}g-2m_{2}g=m_{1}a+4m_{2}a$

$a=\frac{(m_1-2m_2)g}{m_1+m_2}=\frac{(4m_2-2m_2)\times 10}{4m_2+4m_2}=\frac{2\times 10}{8}=\frac{10}{4}m/s^2$

Time taken by $m_1$ to reach the floor is given by,

$s_1=ut+\frac{1}{2}{a}_1{t}^2$

$0.2=0\times t+\frac{1}{2}\times \frac{10}{4}\times t^2$

$t^2=\frac{0.2\times 2\times 4}{10}\Rightarrow t=0.4s$

Thus, the vertical distance covered by $m_2$ in $t=0.4s$ is given by,

$s_2=ut+\frac{1}{2}{a}_2{t}^2$

$s_2=0\times 0.2+\frac{1}{2}\times \frac{20}{4}\times (0.4{)}^2$

$s_2=0.4m$.