A 0.150-kg frame, when suspended from a coil spring, stretches

Step 1
The mass of the frame is ${\displaystyle m_A=0.150kg}$, the mass of the putty is ${\displaystyle m_B=0.200kg}$, the spring is initially stretched a distance ${\displaystyle x_o=0.070m}$ and the putty is dropped from rest from a height ${\displaystyle h=0.030m}$.
Step 2
We know that, in collision of all kinds, since there is no external net force on the system, the total momentum before the collision equals the total momentum after the collision:
${\displaystyle {\overrightarrow{p}}_1={\overrightarrow{p}}_2}$
If the collision is between two particle (A and B), then we get:
${\displaystyle {\overrightarrow{p}}_{A1}+{\overrightarrow{p}}_{B1}={\overrightarrow{p}}_{A2}+{\overrightarrow{p}}_{B2}}$
Where the momentum of a particle is given by:
${\displaystyle \overrightarrow{p}=m\overrightarrow{v}}$
So we can write the collision equation in the following form:
${\displaystyle m_A{\overrightarrow{v}}_{A1}+m_B{\overrightarrow{v}}_{B1}=m_A{\overrightarrow{v}}_{A2}+m_B{\overrightarrow{v}}_{B2}}$ (1)
We also know that, when a spring is stretched (or compressed) a distance x, the restoring force is given by:
${\displaystyle F_x=kx}$
Where the elastic potential energy is given by:
${\displaystyle U_{el}=\frac{1}{2}m{x}^2}$ (2)
Also, the gravitational potential energy is given by:
${\displaystyle U_{grav}=mgy}$ (3)
And the kinetic energy is given by:
${\displaystyle K=\frac{1}{2}m{v}^2}$ (4)
Step 3
Calculations
We have three stages, the first stage is dropping the putty from its initial height until it reaches the frame, the second stage is the collision between the putty and the frame, and the third stage is the stretching of the spring to its maximum stretch point.
Step 4
The first stage:
When the putty is left to fall from its initial height, gravity is the only force that does work on it; So, from the conservation of energy, the gravitational potential energy at this height is totally converted into kinetic energy at the point where it meets the frame.
Therefore, by taking the zero potential energy level at the initial position of the frame and from equation (3) and (4), we get:
${\displaystyle m_{B}gh=\frac{1}{2}{m}_B{v}_{B1}^2}$
${\displaystyle gh=\frac{1}{2}{v}_{B1}^2}$
${\displaystyle v_{B1}=\sqrt{2gh}}$
Now, we plug our value for h, so we get:
${\displaystyle v_{B1}=\sqrt{2\times 9.8\times 0.300}=2.42\frac{m}{s}}$
This is the velocity of the putty just before the collision.
Step 5
The second stage:
When the putty reaches the frame. the net external force at the system is zero. and, at the moment of the collision, the spring is slightly stretched and we can ignore this change.
So, we can assume the conservation of momentum of the system during the collision.
Since the putty and the frame move together after the collision, they have the same velocity after the collision:
${\displaystyle v_{A2}=v_{B2}=v_2}$
Now, we plug our values for ${\displaystyle m_A,m_B\text{and}{v}_{B1}}$ into equation (1), so we get:
${\displaystyle \left(0.150\right)\cdot \left(0\right)+\left(0.200\right)\cdot \left(2.42\right)=(0.150+0.200)\cdot v_2}$
${\displaystyle v_2=\frac{\left(0.200\right)\cdot \left(2.42\right)}{0.150+0.200}=1.38\frac{m}{s}}$
This is the velocity of the system right after the collision.
Step 6
The third stage:
First, we need to calculate the force constant k of the spring; So, before the collision, we apply Newton's second law to the frame along the vertical direction, so we get:
${\displaystyle \sum F_y=k{x}_o-m_{A}g=0}$

The 0.15kg frame has a weight of 1.47N and this causes an extension of 0.07m.
This gives a spring constant of 21N/m
A 0.2kg lump of putty dropped from 0.30m will have ${\displaystyle KE(=PE)}$ of 0.589J
When the putty lands on the platform the spring will extend and this energy is stored in the spring.
The energy stored in a spring is 0.5F x ext or, since k(spring constant) ${\displaystyle =\frac{F}{e}}$
$F=k\times ex\text{then energy stored }=0.5\times k\times e^2$
The spring will stop extending when the 0.589J of KE have been stored (absorbed) so ${\displaystyle 0.589=0.5\times 21\times e^2}$
This gives e (extension) of 0.237m
This is the maximum extension and the spring will oscillate.

The equation for force
$$\begin{gathered} F=kx \\ =mg \\ k=\frac{mg}{x} \\ =\frac{0.150kg\times 9.8\frac{m}{s^2}}{0.050m} \\ =29.4\frac{N}{m} \end{gathered}$$
When the total energy is conserved then equation for velocity is
$$\begin{gathered} mgh=\frac{1}{2}m{v}^2 \\ {v}^2=2gh \\ v=\sqrt{2gh} \\ =\sqrt{2\times 9.8\times 0.3} \\ =2.42\frac{m}{s} \end{gathered}$$
Now by using law of conservation of momentum
$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
So the equation becomes
$$\begin{gathered} mv+M\left(0\right)=mv+Mv \\ 0.2\times 2.42+0=(0.15+0.2)v \end{gathered}$$