Two masses m₁ and m₂ (with m₂ m₁) are connected by a massless rope hung over a frictionless light pulley. If the masses are released then the acceleration of m₂ in the downward direction is given by A) (m₂ - m₁) $\times$ g/(m₁ - m₂) B) (m₂ - m₁) $\times$ g/(m₁ + m₂) C) (m₁ + m₂) $\times$ g/(m₂ - m₁) D) m₂ $\times$ g/(m₁ + m₂) E) g

Question

Two masses m₁ and m₂ (with m₂ > m₁) are connected by a massless rope hung over a frictionless light pulley. If the masses are released then the acceleration of m₂ in the downward direction is given by A) (m₂ - m₁) $\times$ g/(m₁ - m₂) B) (m₂ - m₁) $\times$ g/(m₁ + m₂) C) (m₁ + m₂) $\times$ g/(m₂ - m₁) D) m₂ $\times$ g/(m₁ + m₂) E) g

Quizplus · Accepted Answer

By using Newton's second law for both masses, we have: m₁g - T = m₁a T - m₂g = m₂a where T is the tension in the rope and a is the common acceleration of both masses. Adding the two equations above, we get: (m₁ - m₂)g = (m₁ + m₂)a Dividing both sides by (m₁ + m₂), we get the expression for the acceleration: a = (m₂ - m₁)g / (m₁ + m₂) which matches choice B.

Two Masses M1 and M2 (With M2 > M1) Are ×\times×

Two Masses M₁ and M₂ (With M₂ > M₁) Are $\times$