A block of mass m₁ rests on a smooth horizontal table. It is connected by a light inextensible string passing over a smooth pulley at the edge of the table to a hanging block of mass m₂ . When the system is released from rest, the tension in the string is found to be one-fourth of the weight of the block of mass m₁ . The ratio of the masses m₁ m₂ is

Step-by-step solution

Let the acceleration of the system be a . The equations of motion for the blocks are: For mass m₁ on the table: T = m₁ a For hanging mass m₂ : m₂ g - T = m₂ a Adding these equations gives the acceleration: a = m₂ g m₁ + m₂ Substituting a back into the first equation, the tension is: T = m₁ m₂ g m₁ + m₂ We are given that the tension is one-fourth of the weight of m₁ : T = m₁ g 4 Equating the two expressions for tension: m₁ m₂ g m₁ + m₂ = m₁ g 4 Cancelling m₁ g from both sides: m₂ m₁ + m₂ = 1 4 Cross-multiplying yiel