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Two masses $m_1 = 5\, kg$ and $m_2 = 4.8\, kg$ tied to a string are hanging over a light frictionless pulley. ............ $m/s^2$ is the acceleration of the masses when they are free to move . $(g = 9.8\, m/s^2)$
$0.2$
$9.8 $
$5$
$4.8$
Solution
On release, the motion of the system will be according to the figure. $\mathrm{m}_{1} \mathrm{g}-\mathrm{T}=\mathrm{m}_{1} \mathrm{a} \ldots \ldots$ $(i)$ and
$\mathrm{T}-\mathrm{m}_{2} \mathrm{g}=\mathrm{m}_{2} \mathrm{a} \ldots \ldots$ $(ii)$
On solving$:$
$\mathrm{a}=\left(\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right) \mathrm{g} \ldots \ldots \ldots$ (iii)
Here, $m_{1}=5 \mathrm{kg}, \mathrm{m}_{2}=4.8 \mathrm{kg}, \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}$
$a=\left(\frac{5-4.8}{5+4.8}\right) \times 9.8$
$=\frac{0.2}{9.8} \times 9.8$
$=0.2 \mathrm{m} / \mathrm{s}^{2}$
