Question

Two masses $m_{1}=1\, kg$ and $m_{2}=2\, kg$ 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 $2\, kg$ mass in $2\, s$ is (given, $g=10\, ms ^{-2}$ )

• 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, $m_{1}=1\, kg , m_{2}=2\, kg , t=2\, s$ and $g=10\, ms ^{-2}$
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 $2\, kg$ mass in $2 s , s=\frac{1}{2} \times a \times t^{2}$
$=\frac{1}{2} \times \frac{10}{3} \times 4=\frac{20}{3} m$Given, $m_{1}=1\, kg , m_{2}=2\, kg , t=2\, s$ and $g=10\, ms ^{-2}$
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 $2\, kg$ mass in $2 s , s=\frac{1}{2} \times a \times t^{2}$
$=\frac{1}{2} \times \frac{10}{3} \times 4=\frac{20}{3} m$