A block of mass m slides with speed v on a fr | vedclass

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(B) When the spring is at maximum compression, both blocks move with the same velocity $v_{cm}$.By the law of conservation of linear momentum: $mv + m

Vedclass · Accepted Answer

$\sqrt{\frac{m}{2 k}} v$

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(B) When the spring is at maximum compression, both blocks move with the same velocity $v_{cm}$.By the law of conservation of linear momentum: $mv + m(0) = (m + m)v_{cm} \Rightarrow v_{cm} = \frac{v}{2}$.By the law of conservation of energy, the initial kinetic energy of the system equals the sum of the final kinetic energy and the potential energy stored in the spring at maximum compression $x$:$\frac{1}{2}mv^2 = \frac{1}{2}(2m)v_{cm}^2 + \frac{1}{2}kx^2$$\frac{1}{2}mv^2 = m(\frac{v}{2})^2 + \frac{1}{2}kx^2$$\frac{1}{2}mv^2 = \frac{1}{4}mv^2 + \frac{1}{2}kx^2$$\frac{1}{4}mv^2 = \frac{1}{2}kx^2$$x^2 = \frac{mv^2}{2k} \Rightarrow x = v\sqrt{\frac{m}{2k}}$.

$A$ block of mass $m$ slides with speed $v$ on a frictionless table towards another stationary block of mass $m$. $A$ massless spring with spring constant $k$ is attached to the second block as shown in the figure. The maximum distance the spring gets compressed is

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