Block A of mass m is hanging from a vertical | vedclass

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(B) $1$. Initial state: The spring is extended by $x_0 = \frac{mg}{k}$ due to the weight of block $A$.$2$. Collision: Block $B$ strikes block $A$ and

Vedclass · Accepted Answer

$g\sqrt{\frac{6m}{k}}$

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(B) $1$. Initial state: The spring is extended by $x_0 = \frac{mg}{k}$ due to the weight of block $A$.$2$. Collision: Block $B$ strikes block $A$ and sticks to it. By conservation of linear momentum,$mv = (m+m)V$,where $V$ is the velocity of the combined mass immediately after collision. Thus,$V = \frac{v}{2}$.$3$. Motion: The combined mass $(2m)$ moves upward. The spring is initially extended by $x_0 = \frac{mg}{k}$. For the spring to just attain its natural length,the combined mass must rise by a distance $x_0$ and come to rest momentarily.$4$. Energy Conservation: Let the position just after collision be the reference level for potential energy. The total energy just after collision is the kinetic energy of the combined mass plus the elastic potential energy of the spring. The total energy at the highest point (natural length) is the gravitational potential energy gained by the combined mass.$\frac{1}{2}(2m)V^2 + \frac{1}{2}k x_0^2 = (2m)g x_0$Substituting $V = \frac{v}{2}$ and $x_0 = \frac{mg}{k}$:$\frac{1}{2}(2m)(\frac{v}{2})^2 + \frac{1}{2}k(\frac{mg}{k})^2 = (2m)g(\frac{mg}{k})$$\frac{mv^2}{4} + \frac{m^2g^2}{2k} = \frac{2m^2g^2}{k}$$\frac{mv^2}{4} = \frac{3m^2g^2}{2k}$$v^2 = \frac{6mg^2}{k} \implies v = g\sqrt{\frac{6m}{k}}$.

Block $A$ of mass $m$ is hanging from a vertical spring of spring constant $k$ and is at rest. Block $B$ of mass $m$ strikes block $A$ with velocity $v$ and sticks to it. The velocity $v$ for which the spring just attains its natural length is:

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