A block A of mass 2m is hanging from a vertical massless spring of spring constant k and is in equilibrium. Another block B of mass m strikes the block A with velocity u and sticks to it as shown in the figure. The magnitude of the acceleration of the combined system of the blocks just after the collision is

Question

A block A of mass 2m is hanging from a vertical massless spring of spring constant k and is in equilibrium. Another block B of mass m strikes the block A with velocity u and sticks to it as shown in the figure. The magnitude of the acceleration of the combined system of the blocks just after the collision is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-jn7lxqcihty4.png" /> (A) g/2 (B) g/3 (C) g (D) zero

Tardigrade Admin · Accepted Answer

For block A to be in equilibrium, the spring force acting on it should be equal to its weight. This means Fspring=2mg Just after the collision, the combined mass of the system is 3m and the spring force is the same, hence a=(3 m g - 2 m g/3 m)=(g/3)

$g /2$

$g /3$

$g$

zero

For block $A$ to be in equilibrium, the spring force acting on it should be equal to its weight. This means
$F_{s p r in g} = 2 m g$
Just after the collision, the combined mass of the system is $3 m$ and the spring force is the same, hence
$a = \frac{3 m g - 2 m g}{3 m} = \frac{g}{3}$

g/2

g/3

g

zero

For block A to be in equilibrium, the spring force acting on it should be equal to its weight. This means Fspring​=2mg Just after the collision, the combined mass of the system is 3m and the spring force is the same, hence a=3m3mg−2mg​=3g​

$g /2$

$g /3$

$g$

For block $A$ to be in equilibrium, the spring force acting on it should be equal to its weight. This means
$F_{s p r in g} = 2 m g$
Just after the collision, the combined mass of the system is $3 m$ and the spring force is the same, hence
$a = \frac{3 m g - 2 m g}{3 m} = \frac{g}{3}$