# The device shown here is called an Atwood's machine. m$_1$ and m$_2$ are suspended at the ends of a string of negligible mass passing over a friction-less pulley as shown here. How far will m$_1$ fall in time $t$ after the system is released?

Answer: $\large\frac{1}{2}$$\large\frac{m_1-m_2}{m_1+m_2}$$gt^2$
When the bodies are released, the heavier one moves downward and the lighter one moves upward.
Net force in the direction of motion of $m_1$ is $F_1 = m_1g - T$ which is equal to $m_1a$ as per Newton's second law.
$\Rightarrow m_1g - T = m_1 a \quad (i)$
Net force in the direction of motion of $m_2$ is $F_2 = T-m_2g$ which is equal to $m_2a$ as per Newton's second law.
$\Rightarrow T-m_2g = m_2a \quad (ii)$
Solving $(i)$ and $(ii)$ for acceleration, $a \rightarrow a = \large\frac{m_1-m_2}{m_1+m_2}$$g Now, from Kinematics we know that a = \large\frac{2d}{t^2}$$ \rightarrow d = \large\frac{1}{2}$$at^2 \Rightarrow distance d = \large\frac{1}{2}$$\large\frac{m_1-m_2}{m_1+m_2}$$gt^2$
edited Aug 19, 2014