Question

Two masses $m_1=1kg$ and $m_2=2kg$ are connected by a light inextensible string and suspended by means of a weightless pulley as shown in figure.

Assuming that both the masses start from rest, the distance travelled by $2kg$ mass in $2s$ is (given, $g=10m{s}^{-2}$ )

• A. $\frac{20}{9}m$
• B. $\frac{40}{9}m$
• C. $\frac{20}{3}m$
• D. $\frac{1}{3}m$

Answer 1

Answer: (C)

Given, $m_1=1kg,m_2=2kg,t=2s$ and $g=10m{s}^{-2}$
Net force on system $=m_{2}g-m_{1}g=\left(m_2-m_1\right)g$
Net mass $=\left(m_1+m_2\right)$
$\therefore a=\frac{\text{ Net force }}{\text{ Net mass }}=\left(\frac{m_2-m_1}{m_1+m_2}\right)g$
$=\left(\frac{2-1}{1+2}\right)10=\frac{10}{3}$
Distance travelled by $2kg$ mass in $2s,s=\frac{1}{2}\times a\times t^2$
$=\frac{1}{2}\times \frac{10}{3}\times 4=\frac{20}{3}m$Given, $m_1=1kg,m_2=2kg,t=2s$ and $g=10m{s}^{-2}$
Net force on system $=m_{2}g-m_{1}g=\left(m_2-m_1\right)g$
Net mass $=\left(m_1+m_2\right)$
$\therefore a=\frac{\text{ Net force }}{\text{ Net mass }}=\left(\frac{m_2-m_1}{m_1+m_2}\right)g$
$=\left(\frac{2-1}{1+2}\right)10=\frac{10}{3}$
Distance travelled by $2kg$ mass in $2s,s=\frac{1}{2}\times a\times t^2$
$=\frac{1}{2}\times \frac{10}{3}\times 4=\frac{20}{3}m$