In a simple Atwood machine, two unequal masses $m_{1}$ and $m_{2}$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement shown in figure, $m_{1} = 300 g$ and $m_{2} = 600 g$ . The system is released from the rest.Find the distance traveled by the first block in the first two seconds.

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Solution

$m_{1} = 0.3 k g, m_{2} = 0.6 k g$
$T - (m_{1} g + m_{2} a) = 0$ ... (i)
$\Rightarrow T = m_{1} g + m_{1} a$
$T + m_{2} a - m_{2} g = 0$ .... (ii)
$\Rightarrow T = m_{2} g - m_{2} a$
From equation (i) and equation (ii)
$m_{2} g + m_{1} a + m_{2} a - m_{2} g = 0$ ,
from $(i)$
$\Rightarrow a (m_{1} + m_{2}) = g (m_{2} - m_{1})$
$\Rightarrow a = f (\frac{m_{2} - m_{1}}{m_{1} + m_{2}}) = 9.8 (\frac{0.6 - 0.3}{0.6 + 0.3}) = 3.266 m s^{- 2}$
a) $t = 2 s e c$ acceleration $= 3.266 m s^{- 2}$
Initial velocity $u = 0$
So, distance travelled by the body is $S = u t + 1 / 2 a t^{2} \Rightarrow 0 + 1 / 2 (3.266) 2^{2} = 6.5 m$

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In a simple Atwood machine, two unequal masses $m_{1}$ and $m_{2}$ are connected by a string going over a clamped light smooth pulley. In a typical arrangement (Figure) $m_{1} = 300$ g and $m_{2} = 600$ g. The system is released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string. (c) Find the force exerted by the clamp on the pulley.