Two masses M 1 and M 2 are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M 2 is twice that of M 1 . the acceleration of the system

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Two masses M 1 and M 2 are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass M 2 is twice that of M 1 . the acceleration of the system is a 1 . When the mass M 2 is thrice that of M 1 . The acceleration of The system is a 2 . The ratio a 1 a 2 will be

TestHub · Accepted Answer

In both cases m 1 > m 2 . Therefore, m 1 will accelerate downward and m 2 will accelerate upward. For m 1 , m 1 g - T = m 1 a For m 2 , T - m 2 g = m 2 a Adding the above equations we get, a = m 1 - m 2 m 1 + m 2 g ⇒ a 1 = m 1 - m 2 m 1 + m 2 g = 2 m 2 - m 2 2 m 2 + m 2 g = g 3 and ⇒ a 2 = m 1 - m 2 m 1 + m 2 g = 3 m 2 - m 2 3 m 2 + m 2 g = g 2 ⇒ a 1 a 2 = 2 3

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