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. [image:5_image131.jpeg] Assuming that both the masses start from rest, the distance travelled by the centre of mass in 2 s is $(take\ g = 10m/s^{2})$

Question

Dr. Abubacker · Accepted Answer

(a) $Given,\ m_{1} = 1kg,\ m_{2} = 2\ kg\ and\ g = 10m/s^{2}$
$a = \left( \frac{m_{2} - m_{1}}{m_{1} + m_{2}} ight)g$
$= \left( \frac{2 - 1}{1 + 2} ight)10 = \frac{10}{3}$
$S = \frac{1}{2}a\ t^{2}$
$= \frac{1}{2} 	imes \frac{10}{3} 	imes 4 = \frac{20}{3}$
$m = \frac{2 	imes \frac{20}{3} - 1 	imes \frac{20}{3}}{3} = \frac{20}{9}$

Q. 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 the centre of mass in 2 s is $(take\ g = 10m/s^{2})$

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. 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 the centre of mass in 2 s is $(take\ g = 10m/s^{2})$

Explanation

Reviewed by Dr. Abubacker

Want more practice?

Q. 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 the centre of mass in 2 s is $(take\ g = 10m/s^{2})$