Two mass `m_(1) and m_(2)` are attached to a flexible inextensible massless rope, which passes over a frictionless and massless pully. Find the accelerations of the masses and tension in the rope.

Explanation

Fix an inertial frame to the ground to observe the motion of the masses.
Let the tension in the rope be T(in fact, tension is the property of a point of the rope). In this case with ideal pulley (massless and frictionless) and ideal rope(inextensible,massless, and flexible), the tension will remain constant throughout the rope. Let the acceleration of `m_(2)` be vertically downwards, and accelerations on `m_(1)` also be a, vertically upwards. This is because the rope is inextensible, during motion, the length of the rope must not change, and the rope must not slacken either.
From the above statement, you must not conclude that the accelerations of the masses connected by a rope are always equal. The relationshop between the accelerations of the masses depends on the configuration of the pulley-rope system , which can be obtained from the fact that the length of an ideal rope must not change and the rope must not slacken.
Using eq. `Sigma vec(F )=m vec(a)` for the force diagram of `m_(1) and m_(2)`
`T-m_(1)g=m_(1)a` ...(i)
`m_(2)g-T=m_(2)a` ...(ii)
Adding (i) and (ii) , `m_(2)g-m_(1)g=m_(1)a+m_(2)a`
`a=((m_(2)-m_(1))/(m_(2)+m_(1)))g`
substituting this value of a in (i) we get
`T=m_(1)g+m_(1)((m_(2)-m_(1))/(m_(2)+m_(1)))g-(2m_(1) m_(2))/((m_(1)+m_(2)))g`...(iv).