A block of mass 2.0 kg is fixed to the top of a light vertical spring of force constant 800 N/m…

Step 1. The impact (a). The putty arrives at $u = \sqrt{2gh} = \sqrt{2 \times 10\ \text{m/s}^2 \times 0.45\ \text{m}} = 3.0\ \text{m/s}$. Momentum is conserved:

Step 2. The new centre. Before the impact the spring was squeezed by $Mg/k = 2.50\ \text{cm}$. With the putty it must be squeezed by $(M + m)g/k$, which is $mg/k = 0.625\ \text{cm}$ more. So at the impact the body sits $0.625\ \text{cm}$ above the centre of the new motion.

Step 3. Amplitude (b) and period (c). $\omega = \sqrt{k/(M + m)} = \sqrt{320}\ \text{rad/s} = 17.889\ \text{rad/s}$. At the start the displacement is $0.625\ \text{cm}$ and the speed is $0.60\ \text{m/s}$, so

and $T = 2\pi/\omega = 0.351\ \text{s}$.

Step 4. Greatest compression (d). The lowest point is $A$ below the new centre: $2.50\ \text{cm} + 0.625\ \text{cm} + 3.41\ \text{cm} = 6.54\ \text{cm}$.

Step 5. Time to the lowest point (e). Measure $y$ downwards from the new centre: $y = A\sin(\omega t - \phi)$ with $\sin\phi = 0.625/3.41$, so $\phi = 0.1842\ \text{rad}$, and $y$ starts negative and growing. The lowest point is where $\omega t - \phi = \pi/2$: