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  4. 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 />

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Q. 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

NTA AbhyasNTA Abhyas 2020

Solution:

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