A block of mass 200 g is suspended from a vertical spring. At...

Mass of the block, $M = 200 \text{ g} = 0.2 \text{ kg}$
Initial extension, $x_0 = 1.0 \text{ cm} = 0.01 \text{ m}$
Spring constant, $k = \dfrac{Mg}{x_0} = \dfrac{0.2 \times 10}{0.01} = 200 \text{ N/m}$

Velocity of the particle of mass $m = 120 \text{ g} = 0.12 \text{ kg}$ just before impact:
$v = \sqrt{2gh} = \sqrt{2 \times 10 \times 0.45} = 3 \text{ m/s}$

By conservation of linear momentum during the inelastic collision:
$mv = (M + m)V$
$0.12 \times 3 = (0.2 + 0.12)V \Rightarrow V = \dfrac{0.36}{0.32} = \dfrac{9}{8} \text{ m/s}$

The new equilibrium position of the spring is:
$x_{eq} = \dfrac{(M + m)g}{k} = \dfrac{0.32 \times 10}{200} = 0.016 \text{ m} = 1.6 \text{ cm}$

The angular frequency of the oscillation is:
$\omega = \sqrt{\dfrac{k}{M + m}} = \sqrt{\dfrac{200}{0.32}} = 25 \text{ rad/s}$

The amplitude $A$ of the oscillation can be found using the velocity at the initial position $x_0$:
$V = \omega \sqrt{A^2 - (x_0 - x_{eq})^2}$
$\dfrac{9}{8} = 25 \sqrt{A^2 - (0.01 - 0.016)^2}$
$\dfrac{9}{200} = \sqrt{A^2 - (-0.006)^2}$

Squaring both sides and converting to cm:
$A^2 = (4.5)^2 + (-0.6)^2 = 20.25 + 0.36 = 20.61 \text{ cm}^2$
$A = \sqrt{20.61} \approx 4.54 \text{ cm}$

The maximum extension of the spring is:
x_\max = x_{eq} + A = 1.6 + 4.54 = 6.14 \text{ cm} \approx 6.1 \text{ cm}

Answer: $6.1 \text{ cm}$