A smooth massless string passes over a smooth fixed pulley. Two masses m1 and `m_(2) (m_(1) gt m_(2))` are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is

Options

A. `(m_(1)+m_(2))g`
B. `((m_(1)-m_(2))^(2))/(m_(1)+m_(2))g`
C. `(m_(1)-m_(2))g`
D. `((m_(1)+m_(2))^(2))/(m_(1)-m_(2))g`

Answer

B

Explanation

Let a be the common acceleration of the system and T be the tension in the string.
The equations of motion of two masses are
`m_(1)g-T=m_(1)a` …(i)
`T-m_(2)g=m_(2)a` …(ii)
Adding eqn. (i) and eqn. (ii), we get
`a=((m_(1)-m_(2)))/(m_(1)+m_(2))g` ...(iii)
Acceleration of centre of mass of the system is
`a_(CM)=(m_(1)a_(1)+m_(2)a_(2))/(m_(1)+m_(2))=(m_(1)a-m_(2)a)/(m_(1)+m_(2))`
(`becausea_(1)` and `a_(2)` are equal in magnitude but opposite in direction)
`a_(CM)=((m_(1)-m_(2))/(m_(1)-m_(2)))a=((m_(1)-m_(2))/(m_(1)+m_(2)))((m_(1)-m_(2))/(m_(1)+m_(2)))g` (Using (iii))
`=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g`
The total external force acting on the two masses is
`F_("ext")=(m_(1)+m_(2))a_(CM)=(m_(1)+m_(2))((m_(1)+m_(2))/(m_(1)+m_(2)))^(2)g`
`=((m_(1)-m_(2))^(2))/(m_(1)+m_(2))g`