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AP EAMCET20224 Jul 2022Evening ShiftPhysicsLaws of MotionActual

Two masses M 1 and M 2 are arranged as shown in the figure. Let ' a ' be the magnitude of the acceleration of the mass M 1 . If the mass of M 1 is doubled and that of M 2 is halved, then the acceleration of the system is (Treat all surfaces as smooth : masses of pulley and rope are negligible )

Options

  1. AM 1 + M 2 4 M 1 + M 2 a
  2. B2 M 1 + M 2 4 M 1 + M 2 a
  3. CM 1 + 2 M 2 4 M 1 + 2 M 2 a
  4. DM 1 + 2 M 2 M 1 + M 2 a

Correct answer

A. M 1 + M 2 4 M 1 + M 2 a

Step-by-step solution

Free body diagrams of M 1 and M 2 are shown below. From the free body diagram of M 2 M 2 g sin θ - T = M 2 a And from the free body diagram of M 1 T = M 1 a Therefore, a = M 2 g sin θ M 1 + M 2 When mass of M 1 is doubled and mass of M 2 is halved, a ' = M 2 2 g sin θ 2 M 1 + M 2 2 ⇒ a ' = M 2 g sin θ 4 M 1 + M 2 = M 1 + M 2 4 M 1 + M 2 a

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