Blocks of mass m is hanging from a vertical spring of force constant k. Another identical block B strikes the block A with velocity v and sticks to it. The value of v for which the spring just attains

Question

Blocks of mass m is hanging from a vertical spring of force constant k. Another identical block B strikes the block A with velocity v and sticks to it. The value of v for which the spring just attains natural length is :-

TestHub · Accepted Answer

The first block stretches the spring by

$x_{0} = \frac{m g}{k}$

The blocks stick together, so momentum conservation gives

$v_{0} = \frac{v}{2}$

For the spring to reach natural length, the final speed is zero.

Initial kinetic energy:

$K_{i} = \frac{1}{2} (2 m) (\frac{v}{2})^{2} = \frac{m v ^{2}}{4}$

Initial spring energy:

$U_{s, i} = \frac{1}{2} k (\frac{m g}{k})^{2}$

Rise in gravitational potential energy:

$Δ U_{g} = 2 m g (\frac{m g}{k})$

Energy conservation:

$\frac{m v ^{2}}{4} + \frac{m ^{2} g ^{2}}{2 k} = \frac{2 m ^{2} g ^{2}}{k}$

$v^{2} = \frac{6 m g ^{2}}{k}$

$v = g \frac{6 m}{k}$

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