Laws of MotionMedium
Question
In the figure a block ′A′ of mass ′m′ is attached at one end of a light spring and the other end of the spring is connected to another block ′B′ of mass 2m through a light string. ′A′ is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is ′a′. In the next case, B is held and A is in static equilibrium. Now when B is released, its acceleration immediately after the release is ′b′. The value of a/b is :
(Pulley, string and the spring are massless)
(Pulley, string and the spring are massless)
Options
A.0
B.1/2
C.2
D.undefined
Solution
For first case tension in spring will be
Ts = 2mg just after ′A′ is released.
2mg − mg = ma ⇒ a = g
In second case Ts = mg
2mg − mg = 2mb
b = g/2
a/b = 2
Ts = 2mg just after ′A′ is released.
2mg − mg = ma ⇒ a = g
In second case Ts = mg
2mg − mg = 2mb
b = g/2
a/b = 2
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