10. Motion of two bodies connected by a string Case (A) : Motion of unequal masses suspended from a light frictionless pulley:
A and B are two bodies of mass m1and m2 respectively suspended by means of a light string passing over a smooth pulley P.

Let m₂ > m₁. If the string is light and continuous a tension T exists all along the string. The forces acting on A and B are clearly shown. Let A moves up with an acceleration a and B move down with the same acceleration.
For the motion of A
T – m₁g = m₁a ...(i)
For the motion of B
m₂g – T = m₂a ...(ii)
Solving,
Case (B) : Let us consider the case of a body of mass (m₁), to which a light and string is attached rests on a smooth horizontal plane. The string passes over a frictionless pulley fixed at the end of plane. Another end of the string carries a mass ( m₂ ) as shown in fig. Our aim is to calculate the acceleration of the system and tension in the string :

Here We have
(m₂g – T ) = m₂a ....(1)
and T = m₁a ....(2)
Solving these equations we have,

Case (C) : Here we shall consider the above case with a difference that (m₁) placed on smooth inclined plane making an angle(θ) with horizontal as shown in fig. :

In this case
T – m₁g sin θ = m₁a
and (m₂g – T) = m₂a
solving we get,

Case (D) : Let us consider the case when masses (m₁) and (m₂) are on inclined plane making angles (α) and (β) with horizontal respectively as shown in figure :

We have, m₁g sin α – T = m₁a
and T – m₂ g sin β = m₂ a
solving these equations we get,