Manipal MET2016PhysicsLaws of Motion
In the system shown in figure M₁ M₂ and pulley and threads are ideal. System is held at rest by thread B C . Just after thread B C is burnt.
Options
- AAcceleration M₁ and M₂ will be upward
- BMagnitude of acceleration of both masses will be M₁-M₂ M₁+M₂ g
- CAcceleration of M₁ and M₂ will be equal to zero
- DAcceleration of M₁ will be equal to zero, which that of M₂ will be M₁-M₂ M₂ g upward
Correct answer
D. Acceleration of M₁ will be equal to zero, which that of M₂ will be M₁-M₂ M₂ g upward
Step-by-step solution
As M₁ M₂ , so tension in thread connecting the block B and support C is (M₁-M₂ ) g . When this thread is burnt, this tension disappears. In this case tension in spring is M₁ g and it is elongated. So, tension in string connecting A and B is M₁ g . Hence, resultant force on A remains zero because tension in string is balanced by tension in spring. The block B has a net upward force (M₁-M₂ ) g , so initial acceleration of block B is (M₁-M₂ ) g M₂ Thus, initial acceleration of M₁ is zero and that of M₂ is (M₁-M₂ ) g M