Question
Two blocks of masses 2 kg and 1 kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is $\_\_\_\_$ m.
(Take $g = 10\text{ }m/s^{2}$ )
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Solution
Sol. [Key Concept: Find each block's acceleration, then $a_{cm} = \frac{m_{1}a_{1} + m_{2}a_{2}}{m_{1} + m_{2}}$; $S = \frac{1}{2}a_{cm}t^{2}$.]
$${\ a = \frac{\left( m_{2} - m_{1} \right)}{m_{1} + m_{2}}\text{ }g = \frac{(2 - 1)}{3} \times 10 = \frac{10}{3}\text{ }m/s^{2} }{a_{cm} = \frac{(1)(10/3) - (2)(10/3)}{3} = \frac{- 10}{9}\text{ }m/s^{2}( \downarrow ) }{S_{cm} = u_{cm}t + \frac{1}{2}a_{cm}t^{2} }$$$= 0 + \frac{1}{2}\left( \frac{10}{9} \right)(4) = \frac{20}{9}\text{ }m($ downward $)$
$S_{\text{COM}\text{~}} = \frac{20}{9}($ Downward $) = 2.22$
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